🧠:[<think>Alright, so I need to find the maximum and minimum values of the function y = |x² - x| + |x + 1| on the interval from -2 to 2. Hmm, okay. Let me start by understanding how to approach this. Since the function involves absolute values, it's piecewise-defined, which means the expression inside the absolute value can change its sign depending on the value of x. So, I need to figure out the points where the expressions inside the absolute values change their sign because those points will be critical points where the function's behavior changes. Then, I can break the interval [-2, 2] into subintervals based on these critical points and analyze each piece separately. After that, I can check the function's values at the critical points and endpoints to find the maximum and minimum. First, let's identify the critical points by setting the expressions inside each absolute value to zero.For |x² - x|: The expression inside is x² - x. Let's set that equal to zero:x² - x = 0x(x - 1) = 0So, x = 0 or x = 1. These are the points where the expression inside the first absolute value changes sign.For |x + 1|: The expression inside is x + 1. Setting that equal to zero:x + 1 = 0x = -1. So, this is the point where the second absolute value expression changes sign.Therefore, the critical points are at x = -1, 0, and 1. These divide the interval [-2, 2] into subintervals: [-2, -1), [-1, 0), [0, 1), [1, 2]. But since x = -1, 0, 1 are critical points, we should consider them as separate points in the analysis as well. So, we need to check the function on each of these intervals and at these critical points.Now, let's consider each interval and determine the expression of y without absolute values, then find the function's behavior on each interval.Starting with the first interval: [-2, -1)In this interval, let's check the sign of each expression inside the absolute values:For x² - x: Let's pick a test point, say x = -2 (the left endpoint). Compute x² - x: (-2)^2 - (-2) = 4 + 2 = 6, which is positive. So, in this interval, x² - x is positive, so |x² - x| = x² - x.For x + 1: At x = -2, x + 1 = -1, which is negative. So, |x + 1| = -(x + 1) = -x -1.Therefore, on [-2, -1), y = (x² - x) + (-x -1) = x² - x - x -1 = x² - 2x -1.Okay, so in this interval, y simplifies to x² - 2x -1. This is a quadratic function opening upwards. To find its extrema on this interval, we can check the derivative or check the endpoints and critical points. Since it's a quadratic, its vertex is at x = -b/(2a). Here, a = 1, b = -2, so vertex at x = 2/(2*1) = 1. But x = 1 is not in this interval (since we're in [-2, -1)), so the extrema here must occur at the endpoints. Therefore, we can just compute y at x = -2 and x = -1.Wait, but x = -1 is not included in this interval; it's the next interval. So, we can compute the limit as x approaches -1 from the left. But since we need to evaluate the function at critical points, including x = -1, we can just compute y at x = -1 when we get there.Second interval: [-1, 0)Here, check the expressions:For x² - x: Let's pick x = -0.5 (test point in this interval). Compute x² - x: (-0.5)^2 - (-0.5) = 0.25 + 0.5 = 0.75, positive. So, |x² - x| = x² - x.For x + 1: At x = -0.5, x + 1 = 0.5, positive. So, |x + 1| = x + 1.Thus, in [-1, 0), y = (x² - x) + (x + 1) = x² - x + x + 1 = x² + 1.So, here y simplifies to x² + 1. Again, quadratic opening upwards. The vertex is at x = 0, which is the right endpoint of this interval. To find extrema, check endpoints: x = -1 and x = 0 (but x=0 is not included here, so again, we'll check x approaching 0 from the left, but actual computation at x=0 will be in the next interval).Third interval: [0, 1)Check the expressions:For x² - x: Let's take x = 0.5. Compute 0.25 - 0.5 = -0.25, which is negative. So, |x² - x| = -(x² - x) = -x² + x.For x + 1: At x = 0.5, x + 1 = 1.5, positive. So, |x + 1| = x + 1.Thus, on [0, 1), y = (-x² + x) + (x + 1) = -x² + x + x + 1 = -x² + 2x + 1.This is a quadratic opening downward. Its vertex is at x = -b/(2a) = -2/(2*(-1)) = 1. But x = 1 is not in this interval [0,1); it's the next interval. Therefore, the maximum on this interval would occur at the vertex if it were included, but since it's not, the extrema will be at the endpoints. Compute y at x = 0 and approaching x = 1 from the left.Fourth interval: [1, 2]Check the expressions:For x² - x: Let's take x = 2. Compute 4 - 2 = 2, positive. So, |x² - x| = x² - x.For x + 1: At x = 2, x + 1 = 3, positive. So, |x + 1| = x + 1.Therefore, on [1, 2], y = (x² - x) + (x + 1) = x² - x + x + 1 = x² + 1.Same as the second interval. So, y is x² + 1 here. Quadratic opening upwards, so minimum at x = 1 (vertex here is x = 0, but x=0 isn't here. Wait, the function here is x² +1, which has its minimum at x=0, but our interval is [1,2], so the minimum here is at x=1, and maximum at x=2.Okay, so now we need to analyze each interval's function and compute the values at critical points and endpoints. Let me summarize the function in each interval:1. [-2, -1): y = x² - 2x -12. [-1, 0): y = x² + 13. [0, 1): y = -x² + 2x +14. [1, 2]: y = x² +1Additionally, we have the critical points at x = -1, 0, 1, which we need to evaluate separately.Let me tabulate the function at each critical point and endpoints:First, compute y at x = -2:From interval 1, y = (-2)^2 - 2*(-2) -1 = 4 +4 -1 = 7.At x = -1:We need to evaluate y here. Since x = -1 is a critical point, let's compute y:y = |(-1)^2 - (-1)| + |(-1) + 1| = |1 +1| + |0| = |2| + 0 = 2 + 0 = 2.At x = 0:From interval 3, approaching x=0 from the left would be y = 0² +1 =1. But at x=0, since it's a critical point:y = |0² -0| + |0 +1| = 0 +1 =1. So, same as the left limit.At x=1:From interval 4, approaching x=1 from the left: the expression in interval 3 is y = -1^2 +2*1 +1 = -1 +2 +1=2. At x=1:y = |1^2 -1| + |1 +1| = |0| +2 =0 +2=2.At x=2:From interval 4: y =2^2 +1=4+1=5.So now, let's check the endpoints and critical points:x=-2: y=7x=-1: y=2x=0: y=1x=1: y=2x=2: y=5But we also need to check for possible maxima and minima within the intervals, not just at the endpoints. For each interval, we need to see if there are any local extrema inside the interval.Starting with interval 1: [-2, -1), y =x² -2x -1. Since it's a quadratic opening upwards, its minimum is at x=1 (vertex), but x=1 is not in this interval. Therefore, on [-2, -1), the function is increasing? Wait, let's check the derivative.Wait, maybe instead of just relying on the parabola's vertex, since the interval is closed on the left and open on the right, but since the vertex is at x=1, which is outside the interval, the function is either increasing or decreasing throughout the interval. Let's take the derivative of y in interval 1: dy/dx = 2x -2. Setting derivative to zero: 2x -2=0 => x=1. But x=1 is not in [-2, -1). So, the function's derivative in this interval is 2x -2. Let's compute at x=-2: 2*(-2)-2= -4 -2 = -6. So, derivative is negative throughout the interval, meaning the function is decreasing on [-2, -1). Therefore, the maximum is at x=-2 (y=7) and the minimum as x approaches -1 from the left. But since x=-1 is evaluated separately (y=2), which is lower than y approaching from the left. Wait, let's compute the limit as x approaches -1 from the left in interval 1: lim x->-1- y = (-1)^2 -2*(-1) -1 =1 +2 -1=2. So, approaching from the left, y approaches 2, which matches the value at x=-1. So, in interval 1, the function decreases from 7 at x=-2 to 2 at x=-1.Next interval: [-1, 0), y =x² +1. This is a quadratic opening upwards with vertex at x=0. Since the interval is from -1 to 0, the function is decreasing from x=-1 to x=0. The derivative here is 2x, which is negative in this interval (since x is negative), so the function is decreasing. Therefore, maximum at x=-1 (y=(-1)^2 +1=2) and minimum as x approaches 0 from the left (y approaching 0 +1=1). The actual value at x=0 is 1, which is the minimum here.Third interval: [0,1), y=-x² +2x +1. This is a downward opening parabola. The derivative is dy/dx = -2x +2. Setting to zero: -2x +2=0 => x=1. But x=1 is not in [0,1), so the maximum would be at the critical point if it's within the interval, but since it's not, check endpoints. At x=0: y= -0 +0 +1=1. At x approaching 1 from the left: y= -1 +2 +1=2. The derivative at x=0 is dy/dx= -0 +2=2, which is positive, so the function is increasing throughout this interval. Therefore, the function increases from 1 at x=0 to 2 as x approaches 1. So the maximum in this interval is approaching 2, which matches the value at x=1.Fourth interval: [1,2], y=x² +1. This is upward opening parabola, derivative 2x, which is positive in [1,2], so the function is increasing. Therefore, minimum at x=1 (y=1 +1=2) and maximum at x=2 (y=4 +1=5).Putting all this together:The function's behavior is as follows:- From x=-2 to x=-1: decreasing from 7 to 2.- From x=-1 to x=0: decreasing from 2 to 1.- From x=0 to x=1: increasing from 1 to 2.- From x=1 to x=2: increasing from 2 to 5.So, looking for maxima and minima:The maximum values occur at x=-2 (7) and x=2 (5). But 7 is larger than 5, so the overall maximum is 7 at x=-2.The minimum values occur at x=0 (1) and at x=1 (2). Wait, but in the third interval, the function increases from 1 to 2, so the minimum there is 1 at x=0. Also, in the second interval, it decreases to 1 as approaching x=0. So, the overall minimum is 1 at x=0.Wait, but let me double-check. Are there any other points where the function could attain lower or higher values?We checked all critical points and endpoints. The critical points are x=-1,0,1. The endpoints are x=-2 and x=2. At each of these points, the function is 7,2,1,2,5 respectively. Additionally, within each interval, we checked if there are any local maxima or minima, but in each interval, the function was either strictly increasing or decreasing, so no internal extrema except at the critical points already considered.Therefore, the maximum value is 7 at x=-2, and the minimum value is 1 at x=0.But wait, let me verify this by testing some sample points in each interval to ensure there are no mistakes.First interval [-2,-1): Let's take x=-1.5. Compute y:|x² -x| + |x +1|. Compute x² -x: (-1.5)^2 - (-1.5) = 2.25 +1.5=3.75. |3.75|=3.75. x +1= -1.5 +1= -0.5, so | -0.5 | =0.5. Therefore, y=3.75 +0.5=4.25. According to our expression for interval 1, y=x² -2x -1. Let's compute that: (-1.5)^2 -2*(-1.5) -1=2.25 +3 -1=4.25. Correct. So, at x=-1.5, y=4.25. But according to our previous analysis, the function is decreasing from 7 to 2 here. 4.25 is between 2 and 7, which makes sense. So, no problem here.Second interval [-1,0): Take x=-0.5. Compute y: |(-0.5)^2 - (-0.5)| + |(-0.5)+1| = |0.25 +0.5| + |0.5| =0.75 +0.5=1.25. According to the expression x² +1: (-0.5)^2 +1=0.25 +1=1.25. Correct. And since the function is decreasing here, from 2 at x=-1 to 1 at x=0, so 1.25 is between 1 and 2. Good.Third interval [0,1): Take x=0.5. Compute y: |0.25 -0.5| + |0.5 +1| = | -0.25 | +1.5=0.25 +1.5=1.75. According to the expression -x² +2x +1: -0.25 +1 +1=1.75. Correct. The function increases from 1 to 2 here, so 1.75 is between 1 and 2. Good.Fourth interval [1,2]: Take x=1.5. Compute y: |(1.5)^2 -1.5| + |1.5 +1| = |2.25 -1.5| +2.5= |0.75| +2.5=0.75 +2.5=3.25. According to x² +1: (1.5)^2 +1=2.25 +1=3.25. Correct. And since the function is increasing from 2 to 5 here, 3.25 is between 2 and 5. Good.So, all test points check out with our expressions. Therefore, our analysis seems correct.Thus, the maximum value is 7 at x=-2, and the minimum value is 1 at x=0. However, let's check if there are any other critical points where derivative is zero but we might have missed.Wait, in each interval, we considered the expressions and their derivatives. For the first interval, derivative was 2x -2, which didn't have a critical point inside the interval. Second interval derivative was 2x, zero at x=0, which is the endpoint. Third interval derivative was -2x +2, zero at x=1, which is the endpoint. Fourth interval derivative was 2x, zero at x=0, which is outside. So, no critical points inside the open intervals where derivative is zero. Therefore, all extrema occur at endpoints or critical points already considered.Therefore, the conclusion is that the maximum y is 7 at x=-2, and the minimum y is 1 at x=0.Final AnswerThe maximum value of ( y ) is (boxed{7}) and the minimum value is (boxed{1}).</think>To find the maximum and minimum values of the function ( y = |x^2 - x| + |x + 1| ) on the interval ([-2, 2]), we first identify the critical points where the expressions inside the absolute values change sign. These critical points are ( x = -1 ), ( x = 0 ), and ( x = 1 ). The interval ([-2, 2]) is divided into subintervals: ([-2, -1)), ([-1, 0)), ([0, 1)), and ([1, 2]).We analyze each interval and simplify the function ( y ) without absolute values:1. Interval ([-2, -1)): - ( |x^2 - x| = x^2 - x ) (positive) - ( |x + 1| = -x - 1 ) (negative) - ( y = x^2 - 2x - 1 ) - The function is decreasing, so maximum at ( x = -2 ) (7) and approaches 2 as ( x to -1 ).2. Interval ([-1, 0)): - ( |x^2 - x| = x^2 - x ) (positive) - ( |x + 1| = x + 1 ) (positive) - ( y = x^2 + 1 ) - The function is decreasing, so maximum at ( x = -1 ) (2) and approaches 1 as ( x to 0 ).3. Interval ([0, 1)): - ( |x^2 - x| = -x^2 + x ) (negative) - ( |x + 1| = x + 1 ) (positive) - ( y = -x^2 + 2x + 1 ) - The function is increasing, so minimum at ( x = 0 ) (1) and approaches 2 as ( x to 1 ).4. Interval ([1, 2]): - ( |x^2 - x| = x^2 - x ) (positive) - ( |x + 1| = x + 1 ) (positive) - ( y = x^2 + 1 ) - The function is increasing, so minimum at ( x = 1 ) (2) and maximum at ( x = 2 ) (5).Evaluating the function at the critical points and endpoints:- ( x = -2 ): ( y = 7 )- ( x = -1 ): ( y = 2 )- ( x = 0 ): ( y = 1 )- ( x = 1 ): ( y = 2 )- ( x = 2 ): ( y = 5 )The maximum value is ( 7 ) at ( x = -2 ) and the minimum value is ( 1 ) at ( x = 0 ).[text{Maximum value: } boxed{7}][text{Minimum value: } boxed{1}],<think>Okay, so I need to find the maximum and minimum values of the function y = |x² - x| + |x + 1| on the interval [-2, 2]. Hmm, let me think. Since there are absolute value signs, the function might have different expressions depending on the intervals where the inside of the absolute values is positive or negative. I should probably break down the problem by considering different cases based on the critical points where the expressions inside the absolute values change their sign. First, let's identify the critical points for each absolute value term. Starting with |x² - x|. The expression inside is x² - x. To find when this is zero, set x² - x = 0. Factoring, x(x - 1) = 0, so x = 0 or x = 1. That means the expression x² - x changes sign at x = 0 and x = 1. So I need to check the intervals divided by these points: x < 0, 0 < x < 1, and x > 1. But since our domain is from -2 to 2, we'll have intervals [-2, 0], [0, 1], and [1, 2].Next, the other absolute value term is |x + 1|. The inside is x + 1, which is zero at x = -1. So this expression changes sign at x = -1. So we need to check intervals x < -1 and x > -1. Within our domain, that splits into [-2, -1] and [-1, 2].Therefore, combining both critical points, the overall intervals we need to consider are where both expressions inside the absolute values maintain a consistent sign. The critical points are x = -1, 0, and 1. So the intervals to check are:1. [-2, -1]2. [-1, 0]3. [0, 1]4. [1, 2]In each of these intervals, the expressions inside the absolute values will either be non-negative or non-positive, so we can rewrite the original function without absolute values by considering their signs. Then, we can find the function y in each interval and compute its maximum and minimum.Let's go through each interval one by one.Interval 1: [-2, -1]In this interval, x ≤ -1.First, evaluate x² - x. Let's check the sign here. Since x is negative, x² is positive, and -x is positive as well. So x² - x is positive + positive = positive. Therefore, |x² - x| = x² - x.Next, |x + 1|. Since x ≤ -1, x + 1 ≤ 0. So |x + 1| = -(x + 1) = -x - 1.Therefore, the function y in this interval is:y = (x² - x) + (-x - 1) = x² - x - x - 1 = x² - 2x - 1.So in [-2, -1], y = x² - 2x - 1. This is a quadratic function opening upwards. To find its extrema in the interval, we can check the vertex and endpoints.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). Here, a = 1, b = -2, so vertex at x = -(-2)/(2*1) = 2/2 = 1. But 1 is outside of this interval [-2, -1], so the maximum and minimum must occur at the endpoints.Calculate y at x = -2 and x = -1:At x = -2:y = (-2)^2 - 2*(-2) - 1 = 4 + 4 -1 = 7.At x = -1:y = (-1)^2 - 2*(-1) -1 = 1 + 2 -1 = 2.So in this interval, the maximum is 7 and the minimum is 2.Interval 2: [-1, 0]Now, x is between -1 and 0.First, x² - x again. x is negative here. x² is positive, -x is positive (since x is negative), so x² - x is still positive. So |x² - x| = x² - x.Next, |x + 1|. Since x ≥ -1 here, x + 1 ≥ 0. So |x + 1| = x + 1.Therefore, the function y is:y = (x² - x) + (x + 1) = x² - x + x + 1 = x² + 1.So in [-1, 0], y = x² + 1. This is a simple parabola opening upwards, shifted up by 1. The minimum occurs at the vertex, which is at x = 0 (since the x² term is zero there). But since the interval is from -1 to 0, the minimum is at x = 0, and maximum at x = -1.Calculating:At x = -1:y = (-1)^2 + 1 = 1 + 1 = 2.At x = 0:y = 0² + 1 = 0 + 1 = 1.So in this interval, the maximum is 2 and the minimum is 1.Interval 3: [0, 1]Now x is between 0 and 1.First, x² - x. Let's check the sign here. x is positive, but between 0 and 1. Let's pick x = 0.5: x² = 0.25, x = 0.5. So x² - x = -0.25, which is negative. So in [0, 1], x² - x is negative. Therefore, |x² - x| = -(x² - x) = -x² + x.Next, |x + 1|. Since x ≥ 0 here, x + 1 ≥ 1 > 0. So |x + 1| = x + 1.Therefore, the function y is:y = (-x² + x) + (x + 1) = -x² + x + x + 1 = -x² + 2x + 1.So in [0, 1], y = -x² + 2x + 1. This is a quadratic opening downward. Its vertex is at x = -b/(2a) = -2/(2*(-1)) = 1. So the vertex is at x = 1, which is the endpoint of this interval. Therefore, maximum at the vertex (x=1) and minimum at the other endpoint x=0.Calculating:At x = 0:y = -0 + 0 + 1 = 1.At x = 1:y = -1 + 2 + 1 = 2.Wait, but since the quadratic is opening downward, the maximum is at the vertex (x=1) which gives 2, and the minimum at x=0 which is 1. However, let's check if there's any other critical points. Since it's a quadratic, it's smooth, so only endpoints and vertex. But the vertex is at x=1, which is already considered. So yes, in this interval, max is 2, min is 1.Wait, but hold on. Let me check the value at x=0.5:y = -0.25 + 1 + 1 = 1.75. Which is between 1 and 2, so that's consistent. So the function increases from x=0 to x=1, reaching 2 at x=1. So yes, the minimum is at x=0 (1) and maximum at x=1 (2).Interval 4: [1, 2]Now x is between 1 and 2.First, x² - x. Let's check the sign. x ≥1, so x² - x = x(x - 1). Since x ≥1, x -1 ≥0, so x(x -1) ≥0. So x² - x ≥0 here. Therefore, |x² - x| = x² - x.Next, |x + 1|. x ≥1, so x +1 ≥2 >0. So |x + 1| = x +1.Therefore, the function y is:y = (x² - x) + (x + 1) = x² - x + x +1 = x² +1.So in [1, 2], y = x² +1. This is the same as in interval 2. The parabola opening upwards, minimum at x=1, maximum at x=2.Calculating:At x=1:y = 1 +1 = 2.At x=2:y = 4 +1 =5.So in this interval, the minimum is 2 and maximum is 5.Now, compiling the results from all intervals:- [-2, -1]: max 7, min 2- [-1, 0]: max 2, min 1- [0, 1]: max 2, min 1- [1, 2]: max 5, min 2So overall, the maximum value among all intervals is 7 (from the first interval) and the minimum value is 1 (from the second and third intervals).But wait, let me double-check. The minimum in [-1, 0] is 1 at x=0, and in [0,1] it's also 1 at x=0. So indeed the overall minimum is 1. The maximum in [-2, -1] is 7, which is higher than the other maxima (5, 2, 2). So yes, 7 is the overall maximum.However, just to be thorough, maybe I should check the points at the critical values x = -1, 0, 1 to make sure there's no discrepancies. For example, at x = -1, from the first interval, y=2, and in the second interval, also y=2. So consistent. At x=0, in the second interval y=1, in the third interval also y=1. At x=1, in the third interval y=2, in the fourth interval y=2. So all endpoints are consistent.Therefore, the conclusion is that the maximum value of y is 7 at x=-2, and the minimum value is 1 at x=0.Wait, but hold on. Let me confirm the maximum. At x=-2, y=7. Is that the only maximum? Let's check the value at x=2 in interval [1,2], which is 5. So 5 is less than 7. The maximum is indeed 7. The next highest is 5. So 7 is the maximum.Similarly, the minimum is 1 at x=0, and the next lowest is 2 in other intervals. So 1 is the minimum.But just to make sure, perhaps there are critical points inside the intervals where the derivative is zero? For example, in interval [-2, -1], y = x² - 2x -1. The derivative is 2x - 2. Setting to zero: 2x -2 =0 => x=1. But 1 is outside of this interval, so no critical points here.In [-1,0], y = x² +1. Derivative is 2x. Setting to zero: x=0. But x=0 is the endpoint, already considered.In [0,1], y = -x² +2x +1. Derivative is -2x +2. Setting to zero: -2x +2 =0 => x=1, which is the endpoint.In [1,2], y =x² +1. Derivative is 2x. Setting to zero: x=0, which is outside the interval.Therefore, there are no critical points inside the open intervals, only at endpoints. So the earlier conclusion holds.Therefore, the maximum value is 7 at x=-2 and the minimum value is 1 at x=0.Final AnswerThe maximum value of ( y ) is (boxed{7}) and the minimum value is (boxed{1}).</think>To find the maximum and minimum values of the function ( y = |x^2 - x| + |x + 1| ) on the interval ([-2, 2]), we need to consider the critical points where the expressions inside the absolute values change sign. These critical points are ( x = -1 ), ( x = 0 ), and ( x = 1 ). We break the interval ([-2, 2]) into subintervals and analyze the function in each interval.1. Interval ([-2, -1]): - ( |x^2 - x| = x^2 - x ) (since ( x^2 - x geq 0 )) - ( |x + 1| = -x - 1 ) (since ( x + 1 leq 0 )) - Function: ( y = x^2 - 2x - 1 ) - Values at endpoints: ( y(-2) = 7 ), ( y(-1) = 2 )2. Interval ([-1, 0]): - ( |x^2 - x| = x^2 - x ) (since ( x^2 - x geq 0 )) - ( |x + 1| = x + 1 ) (since ( x + 1 geq 0 )) - Function: ( y = x^2 + 1 ) - Values at endpoints: ( y(-1) = 2 ), ( y(0) = 1 )3. Interval ([0, 1]): - ( |x^2 - x| = -x^2 + x ) (since ( x^2 - x leq 0 )) - ( |x + 1| = x + 1 ) (since ( x + 1 geq 0 )) - Function: ( y = -x^2 + 2x + 1 ) - Values at endpoints: ( y(0) = 1 ), ( y(1) = 2 )4. Interval ([1, 2]): - ( |x^2 - x| = x^2 - x ) (since ( x^2 - x geq 0 )) - ( |x + 1| = x + 1 ) (since ( x + 1 geq 0 )) - Function: ( y = x^2 + 1 ) - Values at endpoints: ( y(1) = 2 ), ( y(2) = 5 )Summarizing the results:- Maximum values: 7 (at ( x = -2 )), 2 (at ( x = -1 ), ( x = 0 ), ( x = 1 )), 5 (at ( x = 2 ))- Minimum values: 2 (at ( x = -1 ), ( x = 1 ), ( x = 2 )), 1 (at ( x = 0 ))The maximum value of ( y ) is (boxed{7}) and the minimum value is (boxed{1}).]

🔑:(1) When 1 leq x leq 2, y=x^2-x+x+1=x^2+1, when x=1, the minimum value is 2, when x=2, the maximum value is 5; (2) When -2 leq x leq -1, y=x^2-2x-1=(x-1)^2-2, when x=-1, y reaches the minimum value of 2, when x=-2, y reaches the maximum value of 7; (3) When -1 leq x leq 0, y=x^2-x+x+1=x^2+1, when x=-1, y reaches the maximum value of 2, when x=0, y reaches the minimum value of 1; (4) When 0 leq x leq 1, y=x-x^2+x+1=-(x-1)^2+2, when x=1, y reaches the maximum value of 2, when x=0, y reaches the minimum value of 1; In summary: The maximum value of y is boxed{7}, and the minimum value is boxed{1}.

🧠:[<think>Okay, so I have this problem about an ellipse and a line intersecting it at two points A and B. The problem has two parts: first, finding the equation of the ellipse given some conditions about the distances from a point on the ellipse to the lower focus, and second, determining whether the area of triangle AOB is a constant value when certain vector conditions are met. Let me try to unpack this step by step.Starting with part (Ⅰ). The ellipse equation given is (frac{y^2}{a^2} + frac{x^2}{b^2} = 1) where (a > b > 0). So this is a standard vertical ellipse centered at the origin, right? The major axis is along the y-axis because the denominator under y² is larger. The foci of an ellipse are located at a distance of c from the center along the major axis, where (c^2 = a^2 - b^2). Since it's mentioned that the lower focus is considered, the coordinates of the lower focus should be (0, -c).Now, the problem states that the maximum and minimum distances from a point on the ellipse to the lower focus are (2 + sqrt{3}) and (2 - sqrt{3}), respectively. So, for any point on the ellipse, the distance to the lower focus (0, -c) ranges between these two values. I need to relate this to the ellipse parameters a and b.In an ellipse, the maximum distance from a focus to a point on the ellipse is (a + c), and the minimum distance is (a - c). Wait, is that correct? Let me recall: For an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant and equal to (2a). However, here we are talking about the distance to just one focus, not the sum. So, the maximum and minimum distances to a single focus would indeed be when the point is at the farthest and closest points along the major axis.Since the major axis is vertical, the farthest point from the lower focus would be the topmost point of the ellipse, which is (0, a). The distance from (0, a) to the lower focus (0, -c) is (a + c). Similarly, the closest point would be the bottommost point (0, -a), but wait, that's actually the focus itself. Wait, no. Wait, the ellipse is defined as (frac{y^2}{a^2} + frac{x^2}{b^2} =1), so the vertices along the major axis (y-axis) are at (0, ±a). The foci are at (0, ±c). So, the lower focus is (0, -c). Then, the closest point on the ellipse to the lower focus would be (0, -a), but the distance from (0, -c) to (0, -a) is (a - c), since (c < a). Similarly, the farthest point from the lower focus is (0, a), distance (a + c). Therefore, according to the problem, (a + c = 2 + sqrt{3}) and (a - c = 2 - sqrt{3}). So, solving these two equations:Adding them: ( (a + c) + (a - c) = (2 + sqrt{3}) + (2 - sqrt{3}) Rightarrow 2a = 4 Rightarrow a = 2 ).Subtracting them: ( (a + c) - (a - c) = (2 + sqrt{3}) - (2 - sqrt{3}) Rightarrow 2c = 2sqrt{3} Rightarrow c = sqrt{3} ).Now, since (c^2 = a^2 - b^2), we can solve for b:( c^2 = a^2 - b^2 Rightarrow (sqrt{3})^2 = 2^2 - b^2 Rightarrow 3 = 4 - b^2 Rightarrow b^2 = 1 Rightarrow b = 1 ).Therefore, the equation of the ellipse is (frac{y^2}{4} + x^2 = 1). So part (Ⅰ) is done. Let me check if that's correct.Wait, substituting back, a=2, c=√3, so the foci are at (0, ±√3). The maximum distance from the lower focus (0, -√3) to the topmost point (0, 2) is |2 - (-√3)| = 2 + √3. Similarly, the minimum distance to the lower focus is from (0, -2) to (0, -√3), which is |-2 - (-√3)| = | -2 + √3 | = 2 - √3, since √3 is approximately 1.732, so 2 - 1.732 ≈ 0.267, which is positive. So that checks out. Hence, the equation should be (frac{y^2}{4} + x^2 = 1). So part (Ⅰ) answer is that.Moving on to part (Ⅱ). We need to determine if the area of triangle AOB is a constant value, given that vectors (overrightarrow{m} = (a x_1, b y_1)) and (overrightarrow{n} = (a x_2, b y_2)) are perpendicular. So, (overrightarrow{m} perp overrightarrow{n}) implies their dot product is zero:( overrightarrow{m} cdot overrightarrow{n} = (a x_1)(a x_2) + (b y_1)(b y_2) = a^2 x_1 x_2 + b^2 y_1 y_2 = 0 ).So, we have the condition (a^2 x_1 x_2 + b^2 y_1 y_2 = 0). The problem states that line l intersects the ellipse at points A and B, so points A(x₁, y₁) and B(x₂, y₂) lie on the ellipse and on line l. The question is whether the area of triangle AOB is constant regardless of the line l, as long as the vectors m and n are perpendicular.First, let's recall that the area of triangle AOB can be calculated using the determinant formula:Area = (frac{1}{2} |x_1 y_2 - x_2 y_1|).We need to see if this expression is constant given the conditions. Since we know a and b from part (Ⅰ), which are a=2 and b=1, substituting these values might help.But before plugging in numbers, let me see if we can do this more generally. The vectors m and n are (a x₁, b y₁) and (a x₂, b y₂), so their perpendicularity gives:( a^2 x_1 x_2 + b^2 y_1 y_2 = 0 ).But since points A and B lie on the ellipse (frac{y^2}{a^2} + frac{x^2}{b^2} = 1), we have:For point A: (frac{y_1^2}{a^2} + frac{x_1^2}{b^2} = 1).Similarly for point B: (frac{y_2^2}{a^2} + frac{x_2^2}{b^2} = 1).Moreover, since A and B lie on a line l, there is a linear relationship between x and y for these points. Let me suppose that the line l has equation y = mx + c, but since it's a general line intersecting the ellipse, maybe we can parametrize points A and B in terms of parameters or use the condition of perpendicular vectors.Alternatively, since vectors m and n are perpendicular, their dot product is zero. So, given that:( a^2 x_1 x_2 + b^2 y_1 y_2 = 0 ).But from the ellipse equations, maybe we can express y₁² and y₂² in terms of x₁² and x₂².Wait, let's try to connect these. The area of triangle AOB is (1/2)|x₁ y₂ - x₂ y₁|. Let's denote this as (1/2)|det|, where det is the determinant.We need to see if det² = (x₁ y₂ - x₂ y₁)^2 is constant. Let's compute det²:det² = (x₁ y₂ - x₂ y₁)^2 = x₁² y₂² + x₂² y₁² - 2 x₁ x₂ y₁ y₂.Hmm. Maybe we can express this in terms of the ellipse equations.Given that for points A and B:( frac{y_1^2}{a^2} + frac{x_1^2}{b^2} = 1 Rightarrow y_1^2 = a^2 left(1 - frac{x_1^2}{b^2}right) ).Similarly, ( y_2^2 = a^2 left(1 - frac{x_2^2}{b^2}right) ).So, substituting into det²:det² = x₁² * [a²(1 - x₂² / b²)] + x₂² * [a²(1 - x₁² / b²)] - 2 x₁ x₂ y₁ y₂= a² x₁² - (a² / b²) x₁² x₂² + a² x₂² - (a² / b²) x₁² x₂² - 2 x₁ x₂ y₁ y₂= a²(x₁² + x₂²) - 2 (a² / b²) x₁² x₂² - 2 x₁ x₂ y₁ y₂.Hmm, not sure if that helps. Let's also note the condition from perpendicular vectors:a² x₁ x₂ + b² y₁ y₂ = 0 ⇒ y₁ y₂ = - (a² / b²) x₁ x₂.So, substituting y₁ y₂ into the det² expression:det² = a²(x₁² + x₂²) - 2 (a² / b²) x₁² x₂² - 2 x₁ x₂ (- (a² / b²) x₁ x₂ )= a²(x₁² + x₂²) - 2 (a² / b²) x₁² x₂² + 2 (a² / b²) x₁² x₂²= a²(x₁² + x₂²).So det² = a²(x₁² + x₂²). Therefore, the area is (1/2) sqrt(det²) = (1/2) a sqrt(x₁² + x₂²).Wait, but that seems to depend on x₁ and x₂. Wait, but we need to check if sqrt(x₁² + x₂²) is a constant. But how?Alternatively, maybe there's another relation. Since points A and B lie on the line l, which intersects the ellipse. Let me consider parametric equations for the line.Suppose the line l has equation y = kx + m. Since the line intersects the ellipse, substituting y into the ellipse equation:(frac{(kx + m)^2}{a^2} + frac{x^2}{b^2} = 1).This can be rewritten as:(left( frac{k^2}{a^2} + frac{1}{b^2} right) x^2 + frac{2 k m}{a^2} x + frac{m^2}{a^2} - 1 = 0).Let me denote this quadratic equation as A x² + B x + C = 0, where:A = (frac{k^2}{a^2} + frac{1}{b^2}),B = (frac{2 k m}{a^2}),C = (frac{m^2}{a^2} - 1).The roots x₁ and x₂ of this equation are the x-coordinates of points A and B. So, by Vieta's formula:x₁ + x₂ = -B/A,x₁ x₂ = C/A.Similarly, y₁ = k x₁ + m,y₂ = k x₂ + m.But how does this relate to the condition a² x₁ x₂ + b² y₁ y₂ = 0?Substituting y₁ y₂:y₁ y₂ = (k x₁ + m)(k x₂ + m) = k² x₁ x₂ + k m (x₁ + x₂) + m².Therefore, the condition becomes:a² x₁ x₂ + b² [k² x₁ x₂ + k m (x₁ + x₂) + m²] = 0.Let me substitute x₁ + x₂ = -B/A and x₁ x₂ = C/A.First, let's compute A, B, C in terms of k and m.Given that A = (k² / a² + 1 / b²),B = (2 k m) / a²,C = (m² / a² - 1).Therefore,x₁ + x₂ = -B/A = - (2 k m / a²) / (k² / a² + 1 / b²) = - (2 k m) / (k² + a² / b²).x₁ x₂ = C/A = (m² / a² - 1) / (k² / a² + 1 / b²) = [ (m² - a²) / a² ] / (k² / a² + 1 / b² ) = (m² - a²) / [ k² + a² / b² ].So, substituting into the condition:a² x₁ x₂ + b² [k² x₁ x₂ + k m (x₁ + x₂) + m²] = 0Let's compute each term:1. a² x₁ x₂ = a² * [ (m² - a²) / (k² + a² / b²) ]2. b² k² x₁ x₂ = b² k² * [ (m² - a²) / (k² + a² / b²) ]3. b² k m (x₁ + x₂) = b² k m * [ -2 k m / (k² + a² / b²) ) ]4. b² m² = b² m²So, combining all terms:a² x₁ x₂ + b² k² x₁ x₂ + b² k m (x₁ + x₂) + b² m² = 0Substituting the expressions:[ a² (m² - a²) + b² k² (m² - a²) - 2 b² k² m² + b² m² (k² + a² / b²) ] / (k² + a² / b²) = 0Simplify numerator:First term: a² (m² - a²)Second term: b² k² (m² - a²)Third term: -2 b² k² m²Fourth term: b² m² (k² + a² / b²) = b² k² m² + a² m²So combining all terms:a² m² - a⁴ + b² k² m² - a² b² k² - 2 b² k² m² + b² k² m² + a² m²Let's compute term by term:1. a² m²2. -a⁴3. b² k² m²4. -a² b² k²5. -2 b² k² m²6. +b² k² m²7. +a² m²So combining like terms:a² m² + a² m² = 2 a² m²-a⁴b² k² m² -2 b² k² m² + b² k² m² = (1 -2 +1) b² k² m² = 0- a² b² k²So total numerator:2 a² m² - a⁴ - a² b² k² = 0Therefore:2 a² m² - a⁴ - a² b² k² = 0Divide both sides by a²:2 m² - a² - b² k² = 0 ⇒ 2 m² = a² + b² k²Therefore, the condition for the line l is that 2 m² = a² + b² k².So, this is a relation between the slope k and the intercept m of the line l. Any line satisfying this condition will result in points A and B such that vectors m and n are perpendicular.Now, we need to check whether the area of triangle AOB is constant for all such lines l.Earlier, we found that the area is (1/2) |x₁ y₂ - x₂ y₁| = (1/2) sqrt(det²) = (1/2) a sqrt(x₁² + x₂²). Wait, but we need to verify this again.Wait, earlier computation:det² = (x₁ y₂ - x₂ y₁)^2 = a² (x₁² + x₂²). So, the area is (1/2) a sqrt(x₁² + x₂²).But how can we compute x₁² + x₂²?From Vieta's formula, x₁ + x₂ = -B/A, x₁ x₂ = C/A. Then,x₁² + x₂² = (x₁ + x₂)^2 - 2 x₁ x₂ = (B²/A²) - 2 (C/A) = (B² - 2 A C)/A².But let's compute that.First, A = (k² / a² + 1 / b²),B = (2 k m)/a²,C = (m² / a² -1).Therefore, B² = (4 k² m²)/a⁴,2 A C = 2 * (k² / a² + 1 / b²) * (m² / a² -1 )= 2 [ (k² m² / a⁴ - k² / a² + m² / (a² b²) - 1 / b² ) ]So,x₁² + x₂² = (B² - 2 A C)/A²= [ (4 k² m² / a⁴ ) - 2 (k² m² / a⁴ - k² / a² + m² / (a² b²) - 1 / b² ) ] / A²= [4 k² m² / a⁴ - 2 k² m² / a⁴ + 2 k² / a² - 2 m² / (a² b²) + 2 / b² ] / A²Simplify numerator:(4 k² m² / a⁴ - 2 k² m² / a⁴) = 2 k² m² / a⁴+ 2 k² / a²- 2 m² / (a² b²)+ 2 / b²So numerator:2 k² m² / a⁴ + 2 k² / a² - 2 m² / (a² b²) + 2 / b²Factor terms:2 k² / a² (m² / a² + 1 ) + 2 / b² (1 - m² / a² )But from the earlier condition 2 m² = a² + b² k² ⇒ m² = (a² + b² k²)/2.Let me substitute m² into the expression.First, replace m² with (a² + b² k²)/2:Numerator becomes:2 k² / a² * [ (a² + b² k²)/2 a² + 1 ] + 2 / b² [1 - (a² + b² k²)/(2 a² ) ]Simplify each part:First term inside the brackets:[ (a² + b² k²)/ (2 a²) + 1 ] = (a² + b² k² + 2 a² ) / (2 a²) ) = (3 a² + b² k²) / (2 a² )Second term inside the brackets:[ 1 - (a² + b² k²)/(2 a² ) ] = (2 a² - a² - b² k² ) / (2 a² ) = (a² - b² k² ) / (2 a² )So, substituting back:First part: 2 k² / a² * (3 a² + b² k²) / (2 a² ) = [2 k² / a² ] * [ (3 a² + b² k²) / (2 a² ) ] = k² (3 a² + b² k² ) / a⁴Second part: 2 / b² * (a² - b² k² ) / (2 a² ) ) = [2 / b² ] * [ (a² - b² k² ) / (2 a² ) ] = (a² - b² k² ) / (a² b² )Therefore, numerator becomes:k² (3 a² + b² k² ) / a⁴ + (a² - b² k² ) / (a² b² )Let me get a common denominator of a⁴ b²:First term: k² (3 a² + b² k² ) * b² / a⁴ b² = k² b² (3 a² + b² k² ) / a⁴ b²Second term: (a² - b² k² ) * a² / a⁴ b² = a² (a² - b² k² ) / a⁴ b² = (a² - b² k² ) / a² b²So total numerator:[ k² b² (3 a² + b² k² ) + a² - b² k² ] / a⁴ b²Expand the first term in the numerator:k² b² * 3 a² + k² b² * b² k² + a² - b² k²= 3 a² b² k² + b⁴ k⁴ + a² - b² k²Combine like terms:(3 a² b² k² - b² k² ) + b⁴ k⁴ + a²= b² k² (3 a² - 1) + b⁴ k⁴ + a²Hmm, this seems complicated. Maybe there's a better way.Alternatively, recall that from the condition 2 m² = a² + b² k². So maybe using this to replace m² in terms of a² and k².But even better, perhaps using specific values for a and b from part (Ⅰ). Since in part (Ⅰ), we found a=2 and b=1, so let's substitute these values.Given a=2, b=1, so the ellipse equation is (frac{y^2}{4} + x^2 =1). Then, the condition 2 m² = a² + b² k² becomes 2 m² = 4 + 1 * k² ⇒ m² = (4 + k²)/2.So, m = ±√[(4 + k²)/2]. So, the lines l are given by y = kx ± √[(4 + k²)/2].Now, the area of triangle AOB is (1/2)|x₁ y₂ - x₂ y₁|. Let's compute this.First, note that y₁ = k x₁ + m and y₂ = k x₂ + m. So,x₁ y₂ - x₂ y₁ = x₁ (k x₂ + m) - x₂ (k x₁ + m) = k x₁ x₂ + m x₁ - k x₁ x₂ - m x₂ = m(x₁ - x₂).Therefore, the area is (1/2)|m(x₁ - x₂)| = (1/2)|m| |x₁ - x₂|.Now, |x₁ - x₂| can be expressed in terms of the quadratic equation's roots. For quadratic equation A x² + B x + C =0, the difference of roots is sqrt[(x₁ + x₂)^2 -4 x₁ x₂] = sqrt[( -B/A )^2 -4 (C/A ) ] = sqrt[(B² -4 A C)/A²] = sqrt(B² -4 A C)/|A|.Therefore, |x₁ - x₂| = sqrt(B² -4 A C)/|A|.But in our case, with a=2, b=1, let's compute A, B, C:A = (k² /4 + 1/1 ) = k² /4 +1,B = (2 k m ) /4 = (k m)/2,C = (m² /4 -1 ).Given that m² = (4 +k²)/2, so substituting m² into C:C = [(4 +k²)/2 /4 -1 ] = [(4 +k²)/8 -1 ] = (4 +k² -8)/8 = (k² -4)/8.Therefore, discriminant B² -4 A C:B² -4 A C = (k² m² /4 ) -4 (k² /4 +1 ) ( (k² -4)/8 )First, compute B²:B² = (k m /2 )² = k² m² /4.But m² = (4 +k²)/2, so:B² = k² (4 +k² ) / 8.Then, compute 4 A C:4 A C =4 * (k² /4 +1 ) * (k² -4)/8 = (k² +4 ) * (k² -4 ) / 8 = (k⁴ -16 ) /8.Therefore, discriminant:B² -4 A C = [k² (4 +k² ) /8 ] - [ (k⁴ -16 ) /8 ] = [4 k² +k⁴ -k⁴ +16 ] /8 = (4 k² +16 ) /8 = (4(k² +4 )) /8 = (k² +4 ) /2.Therefore, |x₁ - x₂| = sqrt( (k² +4)/2 ) / |A|.But A = k² /4 +1 = (k² +4 )/4.So, |x₁ - x₂| = sqrt( (k² +4 ) /2 ) / [ (k² +4 )/4 ] = sqrt( (k² +4 ) /2 ) * 4 / (k² +4 )= 4 / sqrt(2(k² +4 )) = (4 / sqrt(2)) / sqrt(k² +4 ) = 2 sqrt(2) / sqrt(k² +4 ).Therefore, |x₁ - x₂| = 2 sqrt(2) / sqrt(k² +4 ).But recall the area is (1/2)|m| |x₁ -x₂|. Let's compute |m|. Since m² = (4 +k²)/2, |m| = sqrt( (4 +k²)/2 ).Therefore, area = (1/2) * sqrt( (4 +k²)/2 ) * 2 sqrt(2)/sqrt(k² +4 )Simplify:(1/2) * sqrt( (4 +k²)/2 ) * 2 sqrt(2)/sqrt(k² +4 ) = (1/2)*2 * sqrt( (4 +k²)/2 ) * sqrt(2)/sqrt(4 +k² )= sqrt( (4 +k²)/2 ) * sqrt(2)/sqrt(4 +k² )= sqrt( (4 +k²)/2 * 2 / (4 +k² ) )= sqrt(1) =1.Wow! So the area simplifies to 1. Therefore, the area of triangle AOB is constant, equal to 1, regardless of the value of k (i.e., regardless of the line l satisfying the given condition). Therefore, the answer to part (Ⅱ) is yes, the area is a constant value, and it's equal to 1.But let me check my steps again to ensure there's no error.First, we found that the area is (1/2)|m| |x₁ -x₂|.We computed |x₁ -x₂| as 2 sqrt(2)/sqrt(k² +4 ).Then |m| is sqrt( (k² +4)/2 ).Multiply them: sqrt( (k² +4)/2 ) * 2 sqrt(2)/sqrt(k² +4 )The sqrt( (k² +4)/2 ) and sqrt(k² +4 ) in the denominator:sqrt( (k² +4 ) /2 ) / sqrt(k² +4 ) = 1 / sqrt(2).Therefore, multiplying by 2 sqrt(2):1 / sqrt(2) * 2 sqrt(2 ) = 2 sqrt(2)/ sqrt(2) ) = 2. Then multiply by 1/2: 2 * 1/2 =1.Yes, that checks out.Therefore, regardless of k, the area is 1. Hence, it's a constant.Final AnswerⅠ. The equation of the ellipse is boxed{dfrac{y^2}{4} + x^2 = 1}.Ⅱ. Yes, the area of (triangle AOB) is a constant value, and the area is boxed{1}.</think>Given a line ( l ) intersects with the ellipse (frac{y^2}{a^2} + frac{x^2}{b^2} = 1) (where (a > b > 0)) at two points (A(x_1, y_1)) and (B(x_2, y_2)). The maximum and minimum distances from a point on the ellipse to the lower focus are (2 + sqrt{3}) and (2 - sqrt{3}), respectively. Vectors (overrightarrow{m} = (ax_1, by_1)) and (overrightarrow{n} = (ax_2, by_2)) are perpendicular. We need to find the equation of the ellipse and determine if the area of (triangle AOB) is a constant value. Part (Ⅰ)1. The maximum and minimum distances from a point on the ellipse to the lower focus are (2 + sqrt{3}) and (2 - sqrt{3}), respectively.2. For an ellipse, the maximum distance from a focus to a point on the ellipse is (a + c) and the minimum distance is (a - c).3. Solving (a + c = 2 + sqrt{3}) and (a - c = 2 - sqrt{3}): - Adding these equations: (2a = 4 Rightarrow a = 2). - Subtracting these equations: (2c = 2sqrt{3} Rightarrow c = sqrt{3}).4. Using (c^2 = a^2 - b^2): - ((sqrt{3})^2 = 2^2 - b^2 Rightarrow 3 = 4 - b^2 Rightarrow b^2 = 1 Rightarrow b = 1).5. The equation of the ellipse is (frac{y^2}{4} + x^2 = 1). Part (Ⅱ)1. Vectors (overrightarrow{m}) and (overrightarrow{n}) are perpendicular, implying (a^2 x_1 x_2 + b^2 y_1 y_2 = 0).2. The area of (triangle AOB) is (frac{1}{2} |x_1 y_2 - x_2 y_1|).3. For the line (y = kx + m) intersecting the ellipse, substituting (m^2 = frac{a^2 + b^2 k^2}{2}) (from the perpendicular condition) and simplifying the area expression: - The area simplifies to (frac{1}{2} |m| |x_1 - x_2|). - After detailed calculations, the area is found to be constant, equal to 1, regardless of the line's slope (k). Final AnswersⅠ. The equation of the ellipse is (boxed{dfrac{y^2}{4} + x^2 = 1}).Ⅱ. Yes, the area of (triangle AOB) is a constant value, and the area is (boxed{1}).,<think>Okay, so I need to solve this problem about an ellipse and some vectors. Let me start by reading the problem again carefully.The problem says: A line l intersects the ellipse y²/a² + x²/b² = 1 (with a > b > 0) at points A(x₁, y₁) and B(x₂, y₂). The maximum and minimum distances from a point on the ellipse to the lower focus are 2 + √3 and 2 - √3, respectively. Vectors m = (a x₁, b y₁) and n = (a x₂, b y₂) are perpendicular. Then part I asks for the equation of the ellipse, and part II asks if the area of triangle AOB is a constant value, with O being the origin.First, let's tackle part I. To find the equation of the ellipse, we need to find a and b. The given information is about the maximum and minimum distances from a point on the ellipse to the lower focus. Let's recall some properties of ellipses.An ellipse in standard form (y²/a² + x²/b² = 1 here, since the major axis is along the y-axis because a > b) has foci located at (0, ±c), where c² = a² - b². The lower focus is at (0, -c). The distance from any point (x, y) on the ellipse to this lower focus is sqrt((x - 0)² + (y + c)²). The problem states that the maximum and minimum distances are 2 + √3 and 2 - √3.So, the distance from a point (x, y) on the ellipse to the lower focus (0, -c) is sqrt(x² + (y + c)²). The maximum and minimum values of this distance occur at specific points on the ellipse. For an ellipse, the maximum distance from a focus is a + c, and the minimum distance is a - c. Wait, is that true?Wait, actually, for an ellipse with major axis along the y-axis, the closest and farthest points from the lower focus (0, -c) would be the vertices along the major axis. The top vertex is at (0, a), and the bottom vertex is at (0, -a). So the distance from the lower focus (0, -c) to the top vertex (0, a) is sqrt((0)^2 + (a + c)^2) = a + c. Similarly, the distance from the lower focus to the bottom vertex (0, -a) is sqrt(0 + (-a + c)^2) = | -a + c |. But since a > c (because in an ellipse, a² = b² + c², so a > c), this would be a - c.But wait, if the major axis is along the y-axis, then the distance from the lower focus (0, -c) to the top vertex (0, a) is a + c, and to the bottom vertex (0, -a) is | -a + (-c) |? Wait, no. Wait, the focus is at (0, -c). The bottom vertex is (0, -a). So the distance between (0, -c) and (0, -a) is | -a - (-c) | = | c - a | = a - c since a > c. Similarly, the top vertex is (0, a), so the distance is | a - (-c) | = a + c. So yes, the maximum distance is a + c, the minimum is a - c.Given that the maximum is 2 + √3 and the minimum is 2 - √3, so we can set up equations:a + c = 2 + √3a - c = 2 - √3If we solve these two equations, we can find a and c.Adding both equations: (a + c) + (a - c) = (2 + √3) + (2 - √3) ⇒ 2a = 4 ⇒ a = 2.Subtracting the second equation from the first: (a + c) - (a - c) = (2 + √3) - (2 - √3) ⇒ 2c = 2√3 ⇒ c = √3.Since c² = a² - b², we can solve for b:c² = (√3)² = 3a² = 4So, 3 = 4 - b² ⇒ b² = 1 ⇒ b = 1.Therefore, the equation of the ellipse is y²/4 + x²/1 = 1, which simplifies to y²/4 + x² = 1.Wait, but let me check again. The problem states the ellipse equation as y²/a² + x²/b² =1 with a > b. Since we found a=2, b=1, which satisfies a > b. So the equation is y²/4 + x² =1. So part I is done? That seems straightforward. Let me confirm.Yes, the maximum and minimum distances from the lower focus are a + c and a - c. Since the foci are along the major axis (y-axis here), the distances from the lower focus to the top and bottom vertices are indeed a + c and a - c, so solving those gives a=2 and c=√3, hence b² = a² - c² = 4 - 3 =1. Therefore, equation is y²/4 + x² =1. So part I answer is x² + y²/4 =1, but written as y²/4 +x²=1. Either way, same equation.Okay, moving to part II: Determine whether the area of triangle AOB is a constant value. AOB is the triangle formed by points A, B, and the origin O.Given that line l intersects the ellipse at points A and B, and vectors m = (a x1, b y1) and n = (a x2, b y2) are perpendicular. Since a=2 and b=1 from part I, vectors m=(2x1, 1*y1)=(2x1, y1) and n=(2x2, y2). Their dot product is (2x1)(2x2) + (y1)(y2) = 4x1x2 + y1y2 =0. So 4x1x2 + y1y2 =0.We need to check if the area of triangle AOB is a constant. The area can be calculated as 1/2 |x1 y2 - x2 y1|. So we need to see if |x1 y2 - x2 y1| is a constant for any such line l intersecting the ellipse at A and B with the condition 4x1x2 + y1y2 =0.Let me try to approach this. Let me recall that points A and B lie on the ellipse y²/4 +x²=1, so they satisfy y1²/4 +x1²=1 and y2²/4 +x2²=1. Also, they lie on a line l. So the line l has some equation, say y = mx + c, but since it's a chord of the ellipse, it can be parametrized or expressed in terms of some parameters.But perhaps there's a better approach. Since vectors m and n are perpendicular, which gives 4x1x2 + y1y2=0. Let's see if we can relate this to the area.The area of triangle AOB is 1/2 |x1 y2 - x2 y1|. Let's denote that determinant as D = x1 y2 - x2 y1. So the area is 1/2 |D|.We need to see if |D| is constant. Let's see if we can express D in terms of the given condition 4x1x2 + y1y2=0.From 4x1x2 + y1y2=0, we can write y1y2 = -4x1x2. Let's substitute this into D:D = x1 y2 - x2 y1. Let's square D:D² = (x1 y2 - x2 y1)² = x1² y2² - 2x1 x2 y1 y2 + x2² y1².We can try to express this in terms of the ellipse equations and the given condition.Since points A and B are on the ellipse, we have y1² = 4(1 - x1²) and y2² = 4(1 - x2²). Substitute these into D²:D² = x1² * 4(1 - x2²) - 2x1x2 y1y2 + x2² *4(1 -x1²)= 4x1²(1 -x2²) +4x2²(1 -x1²) -2x1x2 y1y2But we know from the given condition that y1y2 = -4x1x2. Substitute that:=4x1²(1 -x2²) +4x2²(1 -x1²) -2x1x2*(-4x1x2)=4x1² -4x1²x2² +4x2² -4x1²x2² +8x1²x2²Simplify terms:4x1² +4x2² -4x1²x2² -4x1²x2² +8x1²x2²=4x1² +4x2² + (-4 -4 +8)x1²x2²=4x1² +4x2² +0*x1²x2²=4x1² +4x2²Therefore, D²=4x1² +4x2², so D=±2√(x1² +x2²). Wait, but then the area is 1/2 |D|= √(x1² +x2²). Hmm, so unless x1² +x2² is a constant, the area would not be constant.So we need to check if x1² +x2² is constant for all such lines l where 4x1x2 + y1y2=0.Alternatively, maybe we can relate x1² +x2² using the ellipse equations.Since A and B are on the ellipse, x1² + y1²/4 =1 and x2² + y2²/4=1. So x1² =1 - y1²/4 and x2²=1 - y2²/4. Then x1² +x2²= 2 - (y1² + y2²)/4. So if we can express y1² + y2² in terms of other variables, maybe we can find a relation.Alternatively, let's consider that points A and B lie on a line, and the line intersects the ellipse. Let me parametrize the line. Suppose the line has the equation y = kx + m. Then intersecting with the ellipse y²/4 +x²=1.Substitute y =kx +m into the ellipse equation:(kx +m)^2 /4 +x² =1Multiply through by 4:(kx +m)^2 +4x² =4Expand:k²x² + 2kmx +m² +4x² =4Combine like terms:(k² +4)x² + 2kmx + (m² -4)=0This quadratic in x has solutions x1 and x2, so by Vieta's formula:x1 +x2 = -2km/(k² +4)x1x2= (m² -4)/(k² +4)Similarly, y1 =kx1 +m, y2=kx2 +mSo y1y2= (kx1 +m)(kx2 +m)=k²x1x2 +km(x1 +x2) +m²From Vieta, substitute x1 +x2 and x1x2:=k²*(m² -4)/(k² +4) + km*(-2km)/(k² +4) +m²Simplify:= [k²(m² -4) -2k²m² +m²(k² +4)] / (k² +4)Wait, let's compute numerator step by step:First term: k²(m² -4) =k²m² -4k²Second term: km*(-2km)= -2k²m²Third term: m²(k² +4)=k²m² +4m²So numerator:(k²m² -4k²) -2k²m² +k²m² +4m²Combine like terms:k²m² -4k² -2k²m² +k²m² +4m²The k²m² terms: (1 -2 +1)k²m²=0Then remaining: -4k² +4m²Therefore, numerator= -4k² +4m²=4(m² -k²)Hence, y1y2= [4(m² -k²)]/(k² +4)Similarly, from the given condition, 4x1x2 + y1y2=0.From Vieta, x1x2=(m² -4)/(k² +4). Therefore:4*(m² -4)/(k² +4) + [4(m² -k²)]/(k² +4)=0Multiply both sides by (k² +4):4(m² -4) +4(m² -k²)=0Divide by 4:(m² -4) + (m² -k²)=0Simplify:2m² -k² -4=0 ⇒ 2m² =k² +4So the lines l must satisfy 2m² =k² +4. Therefore, any line y=kx +m intersecting the ellipse must have this relationship between k and m for the vectors m and n to be perpendicular.Now, going back to the area. The area is 1/2 |x1 y2 -x2 y1|=1/2 |D|. As we found earlier, D²=4x1² +4x2². Let me check that again.Wait, previously I thought D²=4x1² +4x2², but let me verify.Earlier steps:D² = x1² y2² -2x1x2 y1y2 +x2² y1²We substituted y1²=4(1 -x1²) and y2²=4(1 -x2²), and y1y2=-4x1x2.So substituting:D² =x1² *4(1 -x2²) -2x1x2*(-4x1x2) +x2² *4(1 -x1²)=4x1²(1 -x2²) +8x1²x2² +4x2²(1 -x1²)=4x1² -4x1²x2² +8x1²x2² +4x2² -4x1²x2²Combine terms:4x1² +4x2² + (-4x1²x2² +8x1²x2² -4x1²x2²)Which is 4x1² +4x2² +0x1²x2²=4x1² +4x2²So yes, D²=4x1² +4x2², so D=±2√(x1² +x2²). Therefore, the area is 1/2 *2√(x1² +x2²)=√(x1² +x2²). So we need to check if √(x1² +x2²) is a constant. Which would mean x1² +x2² is a constant.Is x1² +x2² a constant for all such lines l? Let's see.From Vieta's formula on the quadratic equation (k² +4)x² +2kmx +m² -4=0:Sum of roots x1 +x2= -2km/(k² +4)Product x1x2=(m² -4)/(k² +4)We can compute x1² +x2²=(x1 +x2)^2 -2x1x2= [4k²m²/(k² +4)^2] -2*(m² -4)/(k² +4)= [4k²m² -2(m² -4)(k² +4)]/(k² +4)^2Let me compute the numerator:4k²m² -2(m² -4)(k² +4)Expand the second term:-2[m²(k² +4) -4(k² +4)] = -2k²m² -8m² +8k² +32Therefore, numerator:4k²m² -2k²m² -8m² +8k² +32= 2k²m² -8m² +8k² +32Factor terms:2k²m² -8m² +8k² +32= 2m²(k² -4) +8(k² +4)Hmm, let's see. From the condition 2m² =k² +4, so m²=(k² +4)/2. Let's substitute this into the numerator.First, substitute m²=(k² +4)/2:2*( (k² +4)/2 )*(k² -4) +8(k² +4)= (k² +4)(k² -4) +8(k² +4)= (k^4 -16) +8k² +32= k^4 +8k² +16Which is (k² +4)^2. Therefore, numerator=(k² +4)^2. Therefore,x1² +x2²=(k² +4)^2/(k² +4)^2=1So x1² +x2²=1. Therefore, D²=4*1=4, so D=±2. Therefore, the area is 1/2* |D|=1/2*2=1.Therefore, the area of triangle AOB is always 1, which is a constant.Wow, that's neat. So even though the line l varies, as long as the vectors m and n are perpendicular, the area remains constant. So the answer to part II is yes, it is a constant value, and the area is 1.Let me verify the key steps again:1. Established that maximum and minimum distances from lower focus give a=2, c=√3, hence b=1. Equation of ellipse is y²/4 +x²=1.2. For part II, found that due to the perpendicularity condition, the line l must satisfy 2m²=k² +4.3. Calculated x1² +x2² using Vieta's formulas and substituted m²=(k² +4)/2, which led to x1² +x2²=1.4. Hence, area=1, a constant.Therefore, both parts solved.Final AnswerⅠ. The equation of the ellipse is boxed{dfrac{y^2}{4} + x^2 = 1}.Ⅱ. Yes, the area of triangle AOB is a constant value, and the constant area is boxed{1}.</think>Given a line ( l ) intersects the ellipse (frac{y^2}{a^2} + frac{x^2}{b^2} = 1) (with ( a > b > 0 )) at points ( A(x_1, y_1) ) and ( B(x_2, y_2) ). The maximum and minimum distances from a point on the ellipse to the lower focus are ( 2 + sqrt{3} ) and ( 2 - sqrt{3} ), respectively. Vectors (overrightarrow{m} = (ax_1, by_1)) and (overrightarrow{n} = (ax_2, by_2)) are perpendicular. Part I: Finding the equation of the ellipse1. The maximum and minimum distances from a point on the ellipse to the lower focus are given as ( 2 + sqrt{3} ) and ( 2 - sqrt{3} ).2. These distances correspond to ( a + c ) and ( a - c ) respectively, where ( c ) is the distance from the center to the focus.3. Solving the equations: [ a + c = 2 + sqrt{3} quad text{and} quad a - c = 2 - sqrt{3} ] Adding these equations gives ( 2a = 4 ) hence ( a = 2 ). Subtracting them gives ( 2c = 2sqrt{3} ) hence ( c = sqrt{3} ).4. Using ( c^2 = a^2 - b^2 ), we find ( b ): [ (sqrt{3})^2 = 2^2 - b^2 implies 3 = 4 - b^2 implies b^2 = 1 implies b = 1 ]5. The equation of the ellipse is: [ boxed{frac{y^2}{4} + x^2 = 1} ] Part II: Determining if the area of (triangle AOB) is a constant value1. The area of (triangle AOB) is given by (frac{1}{2} |x_1 y_2 - x_2 y_1|).2. Given (overrightarrow{m} perp overrightarrow{n}), we have (4x_1x_2 + y_1y_2 = 0).3. Using the ellipse equations (y_1^2 = 4(1 - x_1^2)) and (y_2^2 = 4(1 - x_2^2)), and substituting (y_1y_2 = -4x_1x_2) into the area determinant:4. The determinant squared (D^2 = (x_1 y_2 - x_2 y_1)^2) simplifies to (4(x_1^2 + x_2^2)).5. Using Vieta's formulas for the line intersecting the ellipse, we find (x_1^2 + x_2^2 = 1) after substituting the condition (2m^2 = k^2 + 4).6. Therefore, the area is (frac{1}{2} times 2 = 1), a constant.The area of (triangle AOB) is a constant value, and the constant area is:[boxed{1}]]

🔑:(Ⅰ) According to the problem, we have begin{cases} a+c=2+ sqrt {3} a-c=2- sqrt {3}end{cases}, thus begin{cases} a=2 c= sqrt {3}end{cases}, and therefore b^{2}=a^{2}-c^{2}=1. Hence, the equation of the ellipse is frac {y^{2}}{4}+x^{2}=1.(Ⅱ) The area of triangle AOB is a constant value of 1. Since overrightarrow {m} perp overrightarrow {n}, we have a^{2}x_{1}x_{2}+b^{2}y_{1}y_{2}=0, thus 4x_{1}x_{2}+y_{1}y_{2}=0. ① If the slope of line l does not exist, let the equation of line l be x=p, then x_{1}=x_{2}=p, y_{1}=-y_{2}, Since 4x_{1}x_{2}+y_{1}y_{2}=0, we have 4x_{1}^{2}-y_{1}^{2}=0. Given frac {y_{1}^{2}}{4}+x_{1}^{2}=1, we get x_{1}=± frac { sqrt {2}}{2}, y_{1}=± sqrt {2}. Therefore, S_{triangle AOB}= frac {1}{2}|x_{1}||y_{1}|=1. ② If the slope of line l exists, let the equation of line l be y=kx+r, substituting into the ellipse equation, we get (4+k^{2})x^{2}+2krx+r^{2}-4=0. Thus, x_{1}+x_{2}=- frac {2kr}{4+k^{2}}, x_{1}x_{2}= frac {r^{2}-4}{4+k^{2}}. Since 4x_{1}x_{2}+y_{1}y_{2}=0, we have (4+k^{2})x_{1}x_{2}+kr(x_{1}+x_{2})+r^{2}=0. Therefore, 2r^{2}=4+k^{2}, thus r^{2} geq 2. Hence, Delta=16(k^{2}-r^{2}+4)>0. Let the distance from the origin O to the line l be d, then S_{triangle AOB}= frac {1}{2}d•|AB|= frac {1}{2}× frac {|r|}{ sqrt {k^{2}+1}}× sqrt {1+k^{2}}× sqrt {(x_{1}+x_{2})^{2}-4x_{1}x_{2}}= frac {2r^{2}}{k^{2}+4}=1. In conclusion, the area of triangle AOB is a constant value of boxed{1}.

🧠:[<think>Alright, so I need to prove this inequality: √(a² + 1/a²) - √2 ≥ a + 1/a - 2, where a > 0. Hmm, okay. Let me start by understanding what's going on here. Both sides of the inequality involve expressions with a and 1/a. Maybe there's some symmetry here. Let me try plugging in some values for a to see if it holds. First, let's try a = 1. Then the left-hand side (LHS) becomes √(1 + 1) - √2 = √2 - √2 = 0. The right-hand side (RHS) is 1 + 1 - 2 = 0. So, 0 ≥ 0, which is true. Okay, that's a good start. What if a is something else? Let's try a = 2. Then LHS is √(4 + 1/4) - √2 = √(17/4) - √2 ≈ √4.25 - 1.414 ≈ 2.0616 - 1.414 ≈ 0.6476. RHS is 2 + 1/2 - 2 = 0.5. So, 0.6476 ≥ 0.5, which holds. How about a = 1/2? Then a is 0.5, so 1/a is 2. LHS: √(0.25 + 4) - √2 = √4.25 - √2 ≈ 2.0616 - 1.414 ≈ 0.6476. RHS: 0.5 + 2 - 2 = 0.5. Again, 0.6476 ≥ 0.5. So that works. Maybe if a is approaching 1 from above or below, let's see. Let a approach 1: LHS approaches 0, RHS approaches 0. Maybe equality holds only at a = 1? Let's check a = 3. LHS: √(9 + 1/9) - √2 ≈ √(9.111) - 1.414 ≈ 3.018 - 1.414 ≈ 1.604. RHS: 3 + 1/3 - 2 ≈ 1.333. So 1.604 ≥ 1.333. True again. So far, all the test cases work. Now, how to prove this in general? Let's think. Maybe rearrange the inequality to bring all terms to one side. Let me write:√(a² + 1/a²) - √2 - a - 1/a + 2 ≥ 0.Let me denote x = a + 1/a. Since a > 0, by AM ≥ GM, a + 1/a ≥ 2. Equality when a = 1. So x ≥ 2. Maybe express the terms in terms of x. Let's see. Let's compute a² + 1/a². That's (a + 1/a)^2 - 2 = x² - 2. So the square root term becomes √(x² - 2). So the inequality can be rewritten as:√(x² - 2) - √2 - x + 2 ≥ 0.So now, the inequality is in terms of x, where x ≥ 2. Let me write f(x) = √(x² - 2) - √2 - x + 2. Need to prove that f(x) ≥ 0 for x ≥ 2.Let me check at x = 2: f(2) = √(4 - 2) - √2 - 2 + 2 = √2 - √2 - 2 + 2 = 0. So equality holds at x = 2, which corresponds to a = 1. Now, need to check if f(x) increases for x > 2. Maybe take derivative of f(x) with respect to x.f'(x) = (1/(2√(x² - 2)))(2x) - 0 - 1 + 0 = x / √(x² - 2) - 1.We need to check if f'(x) is positive for x > 2. If f'(x) > 0, then f(x) is increasing, so since f(2) = 0, f(x) ≥ 0 for x ≥ 2. Let's compute f'(x):x / √(x² - 2) - 1. Let's compare x / √(x² - 2) and 1. Let me see:x / √(x² - 2) = x / (√(x²(1 - 2/x²))) = x / (x√(1 - 2/x²)) = 1 / √(1 - 2/x²). Since x ≥ 2, 2/x² ≤ 2/4 = 0.5, so √(1 - 2/x²) ≤ √(1 - 0) = 1. Therefore, 1 / √(1 - 2/x²) ≥ 1. Hence, f'(x) = 1 / √(1 - 2/x²) - 1 ≥ 0. Therefore, f'(x) ≥ 0 for x ≥ 2. Therefore, f(x) is increasing on [2, ∞). Since f(2) = 0, then f(x) ≥ 0 for all x ≥ 2. Hence, the original inequality holds. Therefore, proved.Wait, let me verify this derivative step again. Let me compute f'(x):f(x) = sqrt(x² - 2) - sqrt(2) - x + 2.So derivative:d/dx sqrt(x² - 2) = (1/(2 sqrt(x² - 2))) * 2x = x / sqrt(x² - 2).Then derivative of -sqrt(2) is 0, derivative of -x is -1, derivative of +2 is 0. So f'(x) = x / sqrt(x² - 2) - 1. Yes, that's correct.Now, for x > 2, sqrt(x² - 2) < x, since x² - 2 < x². Therefore, x / sqrt(x² - 2) > x / x = 1. So f'(x) > 1 - 1 = 0. Therefore, f'(x) is positive. So f(x) is increasing for x > 2, starting at f(2) = 0, so f(x) ≥ 0. Therefore, the inequality holds. Therefore, the original inequality is true for all a > 0.Alternatively, maybe there's another approach. Let me think. Since both sides involve expressions of a and 1/a, perhaps substituting t = a + 1/a, but already tried that. Or maybe consider squaring both sides? But that might complicate because of the square roots. Let me see.Starting with the original inequality:sqrt(a² + 1/a²) - sqrt(2) ≥ a + 1/a - 2.Let me move the sqrt(2) to the RHS:sqrt(a² + 1/a²) ≥ a + 1/a - 2 + sqrt(2).But I don't know if that helps. Alternatively, move all terms to the LHS:sqrt(a² + 1/a²) - a - 1/a + 2 - sqrt(2) ≥ 0.But again, not sure. Alternatively, maybe let’s denote t = a + 1/a, so t ≥ 2. Then, a² + 1/a² = t² - 2. So the inequality becomes sqrt(t² - 2) - sqrt(2) ≥ t - 2. Which is similar to the previous substitution. So we can write sqrt(t² - 2) - t + 2 - sqrt(2) ≥ 0. Then, perhaps consider this as a function in t. But as before, taking derivatives shows that it's increasing.Alternatively, maybe square both sides? Let me check. If I have sqrt(t² - 2) ≥ t - 2 + sqrt(2). Let me square both sides. But need to be careful about the sign. Since t ≥ 2, t - 2 + sqrt(2) ≥ 0 + sqrt(2) > 0, so both sides are positive. Therefore, squaring is valid.Left side squared: t² - 2.Right side squared: (t - 2 + sqrt(2))² = (t - 2)^2 + 2(t - 2)sqrt(2) + 2.So need to check if t² - 2 ≥ (t - 2)^2 + 2(t - 2)sqrt(2) + 2.Expand the RHS:(t² -4t +4) + 2(t - 2)sqrt(2) + 2 = t² -4t +4 + 2(t - 2)sqrt(2) + 2 = t² -4t +6 + 2(t - 2)sqrt(2).So inequality becomes:t² - 2 ≥ t² -4t +6 + 2(t - 2)sqrt(2).Subtract t² from both sides:-2 ≥ -4t +6 + 2(t - 2)sqrt(2).Rearrange:-2 -6 +4t ≥ 2(t - 2)sqrt(2).So:4t -8 ≥ 2(t - 2)sqrt(2).Factor left side: 4(t - 2).Right side: 2(t - 2)sqrt(2).Divide both sides by 2(t - 2), which is positive since t ≥ 2 (and t -2 =0 only when t=2, but t=2 would need separate check). So dividing gives:2 ≥ sqrt(2).Which is true because sqrt(2) ≈1.414 <2. Therefore, the inequality after squaring reduces to 2 ≥ sqrt(2), which is true. Therefore, the original inequality holds.But wait, this seems like after squaring, we end up with a true statement, which would mean that the original inequality is true? Wait, but the steps are reversed. Because we started with sqrt(t² -2) ≥ t -2 + sqrt(2), squared both sides to get t² -2 ≥ [t -2 + sqrt(2)]², then simplified and found that 2 ≥ sqrt(2). Since the conclusion is true, does that mean that the original inequality holds? Wait, this seems like a correct approach. Let me recap.Assume that sqrt(t² -2) ≥ t -2 + sqrt(2). If we square both sides, we get t² -2 ≥ [t -2 + sqrt(2)]². Then, simplifying leads us to a true statement 2 ≥ sqrt(2). Therefore, the original inequality is valid. But is this reasoning correct? Because when we have an inequality A ≥ B, and we square both sides (assuming A and B are non-negative), then A² ≥ B² is a necessary condition for A ≥ B. But A² ≥ B² does not necessarily imply A ≥ B unless we know that A - B ≥0. Wait, maybe this is a bit circular.Alternatively, the steps show that if sqrt(t² -2) ≥ t -2 + sqrt(2), then it must be that 2 ≥ sqrt(2). Since 2 ≥ sqrt(2) is true, does that mean the original inequality holds? Wait, not exactly. The correct reasoning is that when we square both sides, we need to see if t² -2 ≥ [t -2 + sqrt(2)]². But after simplifying, this reduces to 4(t -2) ≥ 2(t -2)sqrt(2), which further reduces to 2 ≥ sqrt(2), which is always true. Therefore, the inequality after squaring holds, which implies that the original inequality holds. Wait, but squaring can sometimes lead to extraneous solutions. However, since all steps are reversible? Let me check.Starting from the original inequality, if we square both sides, but since both sides are positive (sqrt(a² +1/a²) ≥ sqrt(2), since a² +1/a² ≥ 2 by AM ≥ GM, equality at a=1. So LHS is at least sqrt(2). The RHS: a +1/a -2. Since a +1/a ≥2, then RHS is ≥0. So both sides are non-negative, so squaring is valid. Therefore, if the squared inequality holds, then the original inequality holds. Therefore, since we showed that after squaring, the inequality reduces to 2 ≥ sqrt(2), which is true, therefore the original inequality holds. Therefore, the proof is valid. Hmm, that's a different approach.Alternatively, maybe even another approach. Let me consider the function f(a) = sqrt(a² +1/a²) - sqrt(2) - (a +1/a -2). Need to show f(a) ≥0 for all a>0. Let me compute f(1) = 0, as we saw. Then, maybe check the derivative of f(a) with respect to a. Let's compute f'(a):f'(a) = derivative of sqrt(a² +1/a²) - derivative of sqrt(2) - derivative of a - derivative of 1/a + derivative of 2.Which is:[ (2a - 2/a³) / (2 sqrt(a² +1/a²)) ) ] - 0 -1 - (-1/a²) + 0.Simplify:[ (a - 1/a³) / sqrt(a² +1/a²) ) ] -1 + 1/a².Hmm, complicated. Let me see if this derivative is positive or negative. Maybe test at a=1:At a=1, f'(1) = [ (1 -1)/sqrt(2) ] -1 +1 = 0 -1 +1 =0. So critical point at a=1. Then, check for a>1, say a=2:f'(2) = [ (2 -1/8)/sqrt(4 + 1/4) ) ] -1 +1/4.Compute numerator: 2 -1/8 = 15/8 ≈1.875. Denominator: sqrt(17/4) ≈2.0616. So 15/8 / sqrt(17/4) ≈1.875 /2.0616≈0.908. Then subtract 1 and add 1/4: 0.908 -1 +0.25 ≈0.158. So positive. Therefore, f'(2) >0. Similarly, for a approaching infinity, the derivative approaches [ (a)/sqrt(a²) ] -1 +0 =1 -1 +0=0. Wait, but let's compute:As a approaches infinity, a -1/a³ ≈a. sqrt(a² +1/a²)≈a. Therefore, [a / a] -1 +1/a² ≈1 -1 +0=0. Hmm. So derivative approaches 0 from positive side?Wait, when a is large, 1/a² is negligible. So f'(a) ≈ [a / a] -1 +0 ≈0. So derivative approaches 0. But for a >1, the derivative is positive. Similarly, for a <1, let's check a=0.5:f'(0.5) = [ (0.5 - (1)/(0.5)^3 ) / sqrt(0.25 + 16) ) ] -1 + (1)/(0.25).Compute numerator: 0.5 - 1/(0.125) =0.5 -8= -7.5. Denominator: sqrt(16.25)=~4.031. So numerator/denominator≈ -7.5 /4.031≈-1.86. Then subtract 1 and add 4: -1.86 -1 +4≈1.14. So positive. Therefore, derivative is positive at a=0.5. Wait, but the numerator here was negative. Let me check:Wait, f'(a) for a=0.5:Numerator: a -1/a³=0.5 -1/(0.125)=0.5 -8= -7.5.Denominator: sqrt(a² +1/a²)=sqrt(0.25 +16)=sqrt(16.25)=~4.031.Then, [ -7.5 /4.031 ]≈-1.86. Then subtract 1 and add 1/a²=1/0.25=4. So total: -1.86 -1 +4≈1.14, which is positive. So f'(0.5)= positive. So even for a <1, derivative is positive. Wait, but when a approaches 0 from the right, what happens to f'(a)? Let's see:As a approaches 0+, a -1/a³≈ -1/a³, sqrt(a² +1/a²)≈1/a. So numerator≈ -1/a³, denominator≈1/a. So [ -1/a³ ] / [1/a ] = -1/a². Then subtract 1 and add 1/a²: -1/a² -1 +1/a²= -1. Therefore, as a approaches 0+, f'(a) approaches -1. So derivative is negative near 0. But in our previous test at a=0.5, it's positive. That suggests that the derivative might cross from negative to positive somewhere between 0 and1. Wait, conflicting with the previous result? Wait, perhaps miscalculations.Wait, let's recast f'(a) = [ (a - 1/a³) / sqrt(a² +1/a²) ) ] -1 +1/a².At a approaching 0+, the first term is [ -1/a³ / (1/a) ] = -1/a². Then -1/a² -1 +1/a² = -1. So derivative approaches -1. Hence, near zero, derivative is negative. But at a=0.5, derivative is positive. So there must be a critical point between 0 and1 where derivative crosses zero. Let's try a=0.6:Numerator:0.6 -1/0.6³≈0.6 -1/0.216≈0.6 -4.63≈-4.03. Denominator:sqrt(0.36 +1/0.36)=sqrt(0.36 +2.777)=sqrt(3.137)≈1.771. So first term: -4.03 /1.771≈-2.276. Then subtract 1 and add 1/a²=1/0.36≈2.777. Total: -2.276 -1 +2.777≈-0.5. Still negative.a=0.7:Numerator:0.7 -1/0.7³≈0.7 -1/0.343≈0.7 -2.915≈-2.215. Denominator:sqrt(0.49 +1/0.49)=sqrt(0.49 +2.041)=sqrt(2.531)≈1.591. First term≈-2.215 /1.591≈-1.393. Subtract 1 and add 1/a²=1/0.49≈2.041. Total≈-1.393 -1 +2.041≈-0.352. Still negative.a=0.8:Numerator:0.8 -1/0.8³≈0.8 -1/0.512≈0.8 -1.953≈-1.153. Denominator:sqrt(0.64 +1/0.64)=sqrt(0.64 +1.5625)=sqrt(2.2025)≈1.484. First term≈-1.153 /1.484≈-0.777. Subtract 1 and add 1/a²=1/0.64≈1.5625. Total≈-0.777 -1 +1.5625≈-0.214. Still negative.a=0.9:Numerator:0.9 -1/0.9³≈0.9 -1/0.729≈0.9 -1.372≈-0.472. Denominator:sqrt(0.81 +1/0.81)=sqrt(0.81 +1.2345)=sqrt(2.0445)≈1.43. First term≈-0.472 /1.43≈-0.33. Subtract 1 and add 1/a²≈1/0.81≈1.2345. Total≈-0.33 -1 +1.2345≈-0.095. Still negative.a=0.95:Numerator:0.95 -1/(0.95)^3≈0.95 -1/0.857≈0.95 -1.167≈-0.217. Denominator≈sqrt(0.9025 +1/0.9025)=sqrt(0.9025 +1.107)=sqrt(2.0095)≈1.417. First term≈-0.217 /1.417≈-0.153. Subtract 1 and add 1/(0.95)^2≈1/0.9025≈1.107. Total≈-0.153 -1 +1.107≈-0.046. Still negative.a=0.99:Numerator≈0.99 -1/(0.99)^3≈0.99 -1/0.970≈0.99 -1.031≈-0.041. Denominator≈sqrt(0.9801 +1/0.9801)≈sqrt(0.9801 +1.0203)=sqrt(2.0004)≈1.414. First term≈-0.041 /1.414≈-0.029. Subtract 1 and add 1/(0.99)^2≈1/0.9801≈1.0203. Total≈-0.029 -1 +1.0203≈-0.0087. Still negative.So approaching a=1 from below, derivative approaches f'(1)=0. So between a=0.99 and a=1, derivative goes from ≈-0.0087 to 0. So maybe derivative crosses zero at a=1. So for a <1, derivative is negative near a=1 but becomes positive as a moves away? Wait, no. Wait, our previous calculation at a=0.5 gave positive derivative. But when approaching a=1 from below, derivative is slightly negative. This is a contradiction. Wait, perhaps miscalculations.Wait, when a=0.5, the derivative was positive, but when approaching a=1 from below, derivative approaches zero from negative side. That suggests that for some a <1, the derivative is positive. Wait, but according to previous calculations, at a=0.5, f'(0.5)=approx1.14, which is positive. But when a approaches 1 from below, derivative approaches 0 from negative. Therefore, there must be a point between 0.5 and1 where derivative changes from positive to negative. That contradicts the earlier calculations. Wait, but let's check a=0.75:Numerator:0.75 -1/(0.75)^3=0.75 -1/0.421875≈0.75 -2.37≈-1.62. Denominator:sqrt(0.5625 +1/0.5625)=sqrt(0.5625 +1.777)=sqrt(2.339)≈1.529. First term≈-1.62 /1.529≈-1.06. Subtract 1 and add1/(0.75)^2≈1.777. Total≈-1.06 -1 +1.777≈-0.283. Negative. Hmm. So at a=0.5, the derivative is positive; at a=0.75, it's negative. So there must be a point between a=0.5 and a=0.75 where derivative crosses zero. Let me check a=0.6:Wait earlier at a=0.6, f'(0.6)≈-0.5. At a=0.5, f'(0.5)≈1.14. Wait, but how is that possible? How can the derivative go from positive at a=0.5 to negative at a=0.6? That suggests a maximum somewhere between 0.5 and 0.6. Wait, let me recalculate for a=0.5:a=0.5:Numerator:0.5 -1/(0.5)^3=0.5 -1/0.125=0.5 -8= -7.5. Denominator:sqrt(0.25 +1/0.25)=sqrt(0.25 +4)=sqrt(4.25)≈2.0616. First term: -7.5 /2.0616≈-3.638. Then subtract 1 and add 1/(0.5)^2=4. So total: -3.638 -1 +4≈-0.638. Wait, that contradicts my previous calculation where I thought f'(0.5)≈1.14. Wait, I must have made a mistake.Wait, hold on. Wait, the derivative is [ (a -1/a³)/sqrt(a² +1/a²) ] -1 +1/a². For a=0.5:[ (0.5 -1/(0.5)^3)/sqrt(0.25 +1/(0.5)^2) ] -1 +1/(0.5)^2.Compute step by step:0.5 -1/(0.5)^3 =0.5 -1/(0.125)=0.5 -8= -7.5.sqrt(0.25 +1/(0.5)^2)=sqrt(0.25 +4)=sqrt(4.25)≈2.0616.So first term: -7.5 /2.0616≈-3.638.Then, subtract 1: -3.638 -1≈-4.638.Add 1/(0.5)^2=4: -4.638 +4≈-0.638.So f'(0.5)=≈-0.638. So negative. Which contradicts my previous wrong calculation. So actually, f'(0.5) is negative. So maybe earlier conclusion was wrong. Let me recast.Wait, I must have miscalculated before. So actually, at a=0.5, f'(0.5)≈-0.638. Similarly, at a=0.6, f'(0.6)≈-0.5. At a=0.8, f'(0.8)≈-0.214. At a=0.95,≈-0.046. At a=1, f'(1)=0. Then for a>1, f'(a) is positive, as we saw at a=2, f'(2)≈0.158. So the derivative is negative for a <1 and positive for a >1. Therefore, the function f(a) has a minimum at a=1. Because the derivative goes from negative to positive at a=1. Hence, the function decreases on (0,1) and increases on (1, ∞). Therefore, the minimum value is at a=1, which is f(1)=0. Therefore, f(a) ≥0 for all a>0. Hence, the original inequality holds.Therefore, the proof is established by showing that the function f(a) has its minimum at a=1, where it is zero, and it's increasing for a>1 and decreasing for a<1, but since a<1 corresponds to x>2 through the substitution x =a +1/a (Wait, no. Wait, when a<1, 1/a>1, so x=a +1/a >2. Wait, but earlier substitution was x =a +1/a, which is always ≥2. So when a approaches 0, x approaches infinity. When a=1, x=2. So x is always ≥2. Wait, but in the derivative analysis with respect to a, we found that the derivative of f(a) is negative for a <1 and positive for a >1. Thus, the function f(a) decreases as a increases from 0 to1, reaching minimum at a=1, then increases as a increases beyond1. But since the minimum at a=1 is zero, f(a) is always ≥0. Therefore, the original inequality holds for all a>0. This completes the proof.So, the key steps are:1. Substitute t =a +1/a, which is ≥2.2. Express the inequality in terms of t, leading to showing sqrt(t² -2) -sqrt(2) ≥t -2.3. Square both sides (since both sides are positive) and simplify to a true statement 2 ≥ sqrt(2).4. Alternatively, analyze the function f(a) and show it has a minimum at a=1 where it is zero, hence f(a) ≥0 for all a>0.Either approach works. The derivative approach confirms the function's behavior, while substitution and squaring gives a direct algebraic proof.Final Answerboxed{sqrt{a^{2} + frac{1}{a^{2}}} - sqrt{2} geq a + frac{1}{a} - 2}</think>Given (a > 0), we need to prove that:[sqrt{a^{2} + frac{1}{a^{2}}} - sqrt{2} geq a + frac{1}{a} - 2.]Step-by-Step Proof:1. Substitution: Let (x = a + frac{1}{a}). By the AM-GM inequality, (x geq 2) with equality if and only if (a = 1).2. Express (a^2 + frac{1}{a^2}) in terms of (x): [ a^2 + frac{1}{a^2} = left(a + frac{1}{a}right)^2 - 2 = x^2 - 2. ]3. Rewrite the inequality in terms of (x): [ sqrt{x^2 - 2} - sqrt{2} geq x - 2. ]4. Define the function (f(x)): [ f(x) = sqrt{x^2 - 2} - sqrt{2} - x + 2. ] We need to show (f(x) geq 0) for (x geq 2).5. Check the value at (x = 2): [ f(2) = sqrt{4 - 2} - sqrt{2} - 2 + 2 = sqrt{2} - sqrt{2} = 0. ]6. Analyze the derivative (f'(x)): [ f'(x) = frac{x}{sqrt{x^2 - 2}} - 1. ] For (x > 2), (sqrt{x^2 - 2} < x), so (frac{x}{sqrt{x^2 - 2}} > 1), implying (f'(x) > 0). Thus, (f(x)) is increasing for (x geq 2).7. Conclusion from derivative analysis: Since (f(x)) is increasing for (x geq 2) and (f(2) = 0), it follows that (f(x) geq 0) for all (x geq 2).8. Alternative algebraic proof by squaring both sides: Starting from (sqrt{x^2 - 2} geq x - 2 + sqrt{2}), squaring both sides: [ x^2 - 2 geq (x - 2 + sqrt{2})^2. ] Simplifying leads to: [ 4(t - 2) geq 2(t - 2)sqrt{2} implies 2 geq sqrt{2}, ] which is true.Both approaches confirm that the inequality holds for all (a > 0).[boxed{sqrt{a^{2} + frac{1}{a^{2}}} - sqrt{2} geq a + frac{1}{a} - 2}],<think>Okay, so I need to prove this inequality: √(a² + 1/a²) - √2 ≥ a + 1/a - 2, given that a > 0. Hmm, let's see. First, maybe I can simplify the left side or the right side somehow. Let me write down both sides clearly.Left side: √(a² + 1/a²) - √2Right side: a + 1/a - 2I need to show that Left side ≥ Right side. Maybe I can rearrange the inequality to make it easier to handle. Let me move the √2 to the right side and the other terms to the left. Wait, no, actually, maybe I should consider bringing all terms to one side and see if the difference is non-negative.So, let's consider:√(a² + 1/a²) - √2 - (a + 1/a - 2) ≥ 0Simplify that:√(a² + 1/a²) - a - 1/a + (2 - √2) ≥ 0Hmm, not sure if that helps. Maybe another approach. Let me try squaring both sides, but before that, perhaps manipulate the inequality. Squaring can be tricky because I need to ensure both sides are non-negative. Let's check when a + 1/a - 2 is non-negative. Wait, a + 1/a - 2. Let's see, if a = 1, then a + 1/a = 2, so the right side becomes 0. For a ≠ 1, if a > 0, by AM ≥ GM, a + 1/a ≥ 2, so equality holds when a = 1. Therefore, a + 1/a - 2 ≥ 0 for all a > 0. Wait, that can't be, because if a is less than 1, then 1/a is greater than 1, but maybe their sum is still at least 2. Let me check with a = 2: 2 + 1/2 = 2.5, which is greater than 2. For a = 1/2: 1/2 + 2 = 2.5, same thing. So yes, a + 1/a ≥ 2 for all a > 0. Therefore, the right side, a + 1/a - 2, is always non-negative. Similarly, the left side: √(a² + 1/a²) - √2. Is this non-negative? Let's check when a = 1: √(1 + 1) - √2 = √2 - √2 = 0. For a ≠ 1, what is a² + 1/a²? Let's note that a² + 1/a² = (a + 1/a)^2 - 2. Since a + 1/a ≥ 2, then (a + 1/a)^2 ≥ 4, so a² + 1/a² ≥ 4 - 2 = 2. Therefore, √(a² + 1/a²) ≥ √2, so the left side is non-negative as well. Therefore, both sides of the original inequality are non-negative. So maybe squaring both sides is a valid operation here. Let's try that. Let me set:Left side: L = √(a² + 1/a²) - √2Right side: R = a + 1/a - 2We need to show that L ≥ R. Since both L and R are non-negative, then L ≥ R iff L² ≥ R². Let's compute L² and R².First, compute L²:L² = (√(a² + 1/a²) - √2)^2 = (a² + 1/a²) - 2√2√(a² + 1/a²) + 2Compute R²:R² = (a + 1/a - 2)^2 = (a + 1/a)^2 - 4(a + 1/a) + 4Expand that:(a² + 2 + 1/a²) - 4a - 4/a + 4 = a² + 1/a² + 2 + 4 - 4a - 4/aSimplify: a² + 1/a² + 6 - 4a - 4/aSo, L² - R² = [ (a² + 1/a²) - 2√2√(a² + 1/a²) + 2 ] - [a² + 1/a² + 6 - 4a - 4/a ]Simplify term by term:(a² + 1/a² cancels out)-2√2√(a² + 1/a²) + 2 - 6 + 4a + 4/aWhich simplifies to:-2√2√(a² + 1/a²) - 4 + 4a + 4/aSo, L² - R² = 4a + 4/a - 4 - 2√2√(a² + 1/a²)We need to show that L² - R² ≥ 0, which would imply L ≥ R. Therefore, we need to check if:4a + 4/a - 4 - 2√2√(a² + 1/a²) ≥ 0Divide both sides by 2:2a + 2/a - 2 - √2√(a² + 1/a²) ≥ 0Hmm, maybe this is a bit complicated. Let me see if there's another approach. Maybe instead of squaring both sides, we can manipulate the original inequality. Let's consider:√(a² + 1/a²) - √2 ≥ a + 1/a - 2Let me denote t = a + 1/a. Then, since a > 0 and a ≠ 1, t ≥ 2. When a = 1, t = 2.Also, note that a² + 1/a² = t² - 2. So, the left side becomes √(t² - 2) - √2, and the right side is t - 2.So the inequality becomes:√(t² - 2) - √2 ≥ t - 2Now, let's let t ≥ 2. So, the inequality is:√(t² - 2) - t ≥ √2 - 2Hmm, maybe let's rearrange the terms:√(t² - 2) - t + 2 ≥ √2But not sure. Alternatively, move all terms to the left:√(t² - 2) - √2 - (t - 2) ≥ 0So, similar to before. Let's consider this function f(t) = √(t² - 2) - √2 - (t - 2). We need to show that f(t) ≥ 0 for t ≥ 2.Let me check at t = 2: f(2) = √(4 - 2) - √2 - (2 - 2) = √2 - √2 - 0 = 0.So, equality holds at t = 2. Now, check the derivative of f(t) to see if it's increasing or decreasing for t > 2.Compute f'(t):f'(t) = (1/(2√(t² - 2)))(2t) - 0 - 1 = t / √(t² - 2) - 1We need to analyze t / √(t² - 2) - 1. Let's see:t / √(t² - 2) = √(t² / (t² - 2)) = √(1 + 2 / (t² - 2))Wait, that might not help. Let's compute t / √(t² - 2):For t > 2, t² - 2 > 0, so denominator is √(t² - 2). Therefore,t / √(t² - 2) = √(t² / (t² - 2)) = √(1 + 2/(t² - 2)).Therefore, f'(t) = √(1 + 2/(t² - 2)) - 1Since 2/(t² - 2) is positive for t > 2, so √(1 + positive) - 1 is positive. Therefore, f'(t) > 0 for t > 2.Therefore, f(t) is increasing for t > 2, and since f(2) = 0, then f(t) ≥ 0 for t ≥ 2.Hence, the original inequality holds.Wait, that seems too straightforward. Let me verify with an example. Let's take a = 2, so t = 2 + 1/2 = 2.5.Compute left side: √(4 + 1/4) - √2 = √(17/4) - √2 = (√17)/2 - √2 ≈ 4.123/2 - 1.414 ≈ 2.0615 - 1.414 ≈ 0.6475Right side: 2 + 1/2 - 2 = 0.5So 0.6475 ≥ 0.5, which holds.Another example, a = 1/2, which gives t = 0.5 + 2 = 2.5, same as above, so same result.Another example, a = 3, t = 3 + 1/3 ≈ 3.333. Then left side: √(9 + 1/9) - √2 ≈ √(9.111) - 1.414 ≈ 3.018 - 1.414 ≈ 1.604Right side: 3 + 1/3 - 2 ≈ 1.333. So 1.604 ≥ 1.333, holds.If a approaches infinity, what happens? Let a approach infinity. Then t = a + 1/a ≈ a. Then left side: √(a² + 1/a²) - √2 ≈ a - √2. Right side: a + 1/a - 2 ≈ a - 2. So compare a - √2 and a - 2. Since √2 ≈ 1.414 < 2, so a - √2 > a - 2, so the inequality holds as a approaches infinity.Similarly, when a approaches zero, which is similar to a approaching infinity (since a is replaced by 1/a). So same reasoning.Therefore, the derivative approach seems valid. Since f(t) is increasing for t ≥ 2, and f(2) = 0, then f(t) ≥ 0 for all t ≥ 2, which corresponds to a > 0. Therefore, the inequality holds.Alternatively, maybe there's another way using the AM-QM inequality or some algebraic manipulation.Let me try another approach. Let me set x = a + 1/a. Then, as before, x ≥ 2, and a² + 1/a² = x² - 2. The inequality becomes:√(x² - 2) - √2 ≥ x - 2Let me rearrange:√(x² - 2) ≥ x - 2 + √2But not sure. Alternatively, perhaps let me set y = x - 2. Since x ≥ 2, then y ≥ 0.So, substitute x = y + 2. Then:Left side: √((y + 2)^2 - 2) - √2 = √(y² + 4y + 4 - 2) - √2 = √(y² + 4y + 2) - √2Right side: (y + 2) - 2 = ySo, the inequality becomes:√(y² + 4y + 2) - √2 ≥ yWhich is equivalent to:√(y² + 4y + 2) ≥ y + √2Now, square both sides (since both sides are non-negative):y² + 4y + 2 ≥ y² + 2y√2 + 2Simplify:Left side: y² + 4y + 2Right side: y² + 2y√2 + 2Subtract right side from left side:( y² + 4y + 2 ) - ( y² + 2y√2 + 2 ) = 4y - 2y√2 = y(4 - 2√2)Since y ≥ 0 and 4 - 2√2 ≈ 4 - 2.828 ≈ 1.172 > 0, then 4y - 2y√2 ≥ 0 for all y ≥ 0.Therefore, the inequality holds, so the original inequality is true.Therefore, we have proven that for all a > 0, √(a² + 1/a²) - √2 ≥ a + 1/a - 2.I think this second approach is more straightforward. By substituting x = a + 1/a, then using substitution y = x - 2, which reduces the inequality to something that can be squared without complications. Then, after squaring, we find that the difference is non-negative because 4 - 2√2 is positive, and y is non-negative. Hence, the inequality holds.Alternatively, the first approach using derivatives shows that the function f(t) is increasing for t ≥ 2 and f(2) = 0, hence the inequality holds.Either way, the conclusion is that the given inequality is true for all a > 0.Final Answerboxed{sqrt {a^{2}+ frac {1}{a^{2}}} - sqrt {2} geq a+ frac {1}{a}-2}</think>Given (a > 0), we need to prove that:[sqrt{a^2 + frac{1}{a^2}} - sqrt{2} geq a + frac{1}{a} - 2.]First, we note that both sides of the inequality are non-negative. For the right side, (a + frac{1}{a} - 2), by the AM-GM inequality, (a + frac{1}{a} geq 2), so (a + frac{1}{a} - 2 geq 0). For the left side, (sqrt{a^2 + frac{1}{a^2}} - sqrt{2}), since (a^2 + frac{1}{a^2} geq 2) (also by AM-GM), the square root is at least (sqrt{2}), making the left side non-negative.We introduce (t = a + frac{1}{a}), which satisfies (t geq 2). Then, (a^2 + frac{1}{a^2} = t^2 - 2). The inequality becomes:[sqrt{t^2 - 2} - sqrt{2} geq t - 2.]Let (y = t - 2), so (y geq 0). Substituting (t = y + 2), the inequality becomes:[sqrt{(y + 2)^2 - 2} - sqrt{2} geq y.]Simplifying the left side:[sqrt{y^2 + 4y + 2} - sqrt{2} geq y.]We square both sides (both sides are non-negative):[sqrt{y^2 + 4y + 2} geq y + sqrt{2}.]Squaring both sides:[y^2 + 4y + 2 geq y^2 + 2ysqrt{2} + 2.]Simplifying:[4y geq 2ysqrt{2}.]Dividing both sides by (2y) (since (y geq 0)):[2 geq sqrt{2}.]This is true as (2 > sqrt{2}). Therefore, the original inequality holds for all (a > 0).Thus, the proof is complete and the inequality is:[boxed{sqrt {a^{2}+ frac {1}{a^{2}}} - sqrt {2} geq a+ frac {1}{a}-2}]]

🔑:To prove sqrt {a^{2}+ frac {1}{a^{2}}} - sqrt {2} geq a+ frac {1}{a}-2, it suffices to prove sqrt {a^{2}+ frac {1}{a^{2}}} + 2 geq a+ frac {1}{a} + sqrt {2}. Since a > 0, it is enough to prove that (sqrt {a^{2}+ frac {1}{a^{2}}} + 2)^2 geq (a+ frac {1}{a} + sqrt {2})^2, which reduces to proving that 2sqrt {a^{2}+ frac {1}{a^{2}}} geq sqrt {2}(a+ frac {1}{a}), and it is sufficient to prove that a^{2}+ frac {1}{a^{2}} geq 2, which is obviously true. Therefore, boxed{sqrt {a^{2}+ frac {1}{a^{2}}} - sqrt {2} geq a+ frac {1}{a}-2}.

🧠:[<think>Okay, so I need to solve this system of inequalities. Let me see. The problem is:[begin{cases}6 - 2x geq 0 2x + 4 > 0end{cases}]Alright, so there are two inequalities here. I need to solve each one separately and then find the intersection of their solutions because it's a system where both conditions have to be satisfied at the same time.Starting with the first inequality: (6 - 2x geq 0). Let me solve for x. Hmm, subtract 6 from both sides? Wait, no. Let me rearrange this step by step.First inequality:(6 - 2x geq 0)I want to isolate x. So, subtract 6 from both sides? Wait, actually, let's move the 2x to the right side and the 0 to the left. Wait, maybe it's easier to first add 2x to both sides. Let me think.Wait, another approach: Let's solve it like a linear equation. Let me subtract 6 from both sides, but since the variable term is negative, maybe I can handle it differently.Alternatively, subtract 6 from both sides? Wait, no. Let's see:(6 - 2x geq 0)If I add 2x to both sides, I get:(6 geq 2x)Which is the same as (2x leq 6). Then divide both sides by 2:(x leq 3).Okay, so the first inequality simplifies to x is less than or equal to 3. Got that.Now the second inequality: (2x + 4 > 0). Let's solve this one.Subtract 4 from both sides:(2x > -4)Then divide by 2:(x > -2)Alright, so the second inequality gives x is greater than -2.So now we have both solutions:First inequality: (x leq 3)Second inequality: (x > -2)So the solution set is all real numbers x such that x is greater than -2 and less than or equal to 3. In interval notation, that would be (-2, 3].But let me double-check my work. Let me verify each inequality again to make sure I didn't make a mistake.First inequality: (6 - 2x geq 0). Solving for x:6 - 2x ≥ 0→ -2x ≥ -6Hmm, when I divide both sides by a negative number, I have to reverse the inequality sign. Wait, I didn't do that in my first step. Let's check that again.Wait, so perhaps in my initial solution, I didn't have to divide by a negative number. Let me see. Let's re-solve the first inequality step by step.First inequality:(6 - 2x geq 0)Subtract 6 from both sides:(-2x geq -6)Now, divide both sides by -2, which reverses the inequality sign:(x leq 3)Yes, that's correct. So even though I divided by a negative number, I reversed the inequality, and got x ≤ 3. So that part's correct.Wait, but in my first approach, I added 2x to both sides instead. Let me see:Original inequality:(6 - 2x geq 0)Adding 2x to both sides gives:(6 geq 2x)Which is the same as:(2x leq 6)Then dividing by 2:(x leq 3)Same result. So both methods are correct. So no problem there.Second inequality: (2x + 4 > 0)Subtract 4:(2x > -4)Divide by 2:(x > -2)That's straightforward. The inequality direction doesn't change because we divided by a positive number. So that's correct.So combining both inequalities, x must be greater than -2 and less than or equal to 3. So the solution set is the interval from -2 to 3, excluding -2 (since it's a strict inequality) and including 3 (since it's a non-strict inequality). So in interval notation, that's (-2, 3].To visualize this on the number line, we'd have an open circle at -2 and a closed circle at 3, with a line connecting them. Any number between -2 and 3 is a solution.Let me check with a test value in each interval to be thorough.First, let's check the first inequality (6 - 2x ≥ 0):Test x = 0: 6 - 0 = 6 ≥ 0 ✔️Test x = 4: 6 - 8 = -2 which is not ≥ 0 ❌Test x = -3: 6 - (-6) = 12 ≥ 0 ✔️So x ≤ 3 works for the first inequality. So at x = 3, 6 - 2(3) = 0, which satisfies the inequality. At x = 4, it's -2, which doesn't. At x = -3, it's 12, which does. So first inequality is correct.Second inequality (2x + 4 > 0):Test x = 0: 0 + 4 = 4 > 0 ✔️Test x = -3: -6 + 4 = -2 > 0 ❌Test x = -2: 2*(-2) +4 = 0 > 0 ❌ (since it's not strictly greater)So x must be greater than -2. So x > -2 is correct.Therefore, the overlapping region is between -2 and 3, including 3 but not including -2.Thus, the solution set is (-2, 3].Final AnswerThe solution set is boxed{(-2, 3]}.</think>To solve the system of inequalities:[begin{cases}6 - 2x geq 0 2x + 4 > 0end{cases}]we start by solving each inequality separately.1. For the first inequality (6 - 2x geq 0): [ 6 - 2x geq 0 ] Subtract 6 from both sides: [ -2x geq -6 ] Divide by -2 and reverse the inequality sign: [ x leq 3 ]2. For the second inequality (2x + 4 > 0): [ 2x + 4 > 0 ] Subtract 4 from both sides: [ 2x > -4 ] Divide by 2: [ x > -2 ]Combining the solutions of both inequalities, we get:[-2 < x leq 3]In interval notation, the solution set is (boxed{(-2, 3]}).,<think>Okay, so I need to solve this system of inequalities. Let me see. The system is:[begin{cases}6 - 2x ge 0 2x + 4 > 0end{cases}]Alright, first step is to solve each inequality separately and then find the intersection of the solutions, right? Because both conditions have to be true at the same time. Let me start with the first inequality.First Inequality: 6 - 2x ≥ 0Hmm, let's solve for x. I need to isolate x. So, subtract 6 from both sides? Wait, no, maybe I should first move the term with x to the other side. Let me think.Starting with 6 - 2x ≥ 0. Let me add 2x to both sides to get rid of the negative x term. That gives:6 ≥ 2xNow, divide both sides by 2 to solve for x. So:6/2 ≥ xWhich simplifies to:3 ≥ xOr, written the other way:x ≤ 3Okay, so the first inequality tells me that x has to be less than or equal to 3. Got it.Second Inequality: 2x + 4 > 0Alright, moving on to the second inequality. Let's solve for x here as well. Subtract 4 from both sides first:2x + 4 - 4 > 0 - 4Which simplifies to:2x > -4Now, divide both sides by 2:x > -2So the second inequality tells me that x has to be greater than -2.Combining Both SolutionsNow, since both inequalities need to be satisfied simultaneously, the solution set is the intersection of x ≤ 3 and x > -2. So, x has to be greater than -2 and less than or equal to 3. In interval notation, that would be (-2, 3]. But let me double-check to make sure I didn't make any mistakes.Wait, let me verify each step again.For the first inequality:6 - 2x ≥ 0Subtract 6: -2x ≥ -6Then divide by -2. Oh, wait! Hold on, when you divide or multiply by a negative number, you have to reverse the inequality sign. Did I do that?Wait, in my original solution, I added 2x to both sides instead. Let me check that again.Original steps:6 - 2x ≥ 0Add 2x to both sides: 6 ≥ 2xThen divide by 2: 3 ≥ x, which is the same as x ≤ 3.Okay, that's correct. I avoided dividing by a negative number by rearranging the terms, so I didn't have to reverse the inequality sign. That's a good method. So x ≤ 3 is correct.Second inequality:2x + 4 > 0Subtract 4: 2x > -4Divide by 2: x > -2. That's straightforward. No issues here.So combining both, x must be greater than -2 and less than or equal to 3. So the solution set is all real numbers between -2 and 3, including 3 but not including -2. So in interval notation, (-2, 3]. But the problem says "the solution set is ____." Depending on the convention they want, sometimes written with inequalities. Let me see if they prefer inequality notation or interval notation. The problem just says "the solution set is ____," so both are possible, but in the Chinese system, maybe interval notation is preferred. Let me check again.But wait, in the original problem, they might just want the interval. But to be thorough, let me write both.In inequality form: -2 < x ≤ 3.In interval notation: (-2, 3].Since the problem is presented in Chinese, maybe the answer is expected in interval notation. But the instructions say to put the final answer within boxed{}, so probably in interval notation boxed.But let me make sure there's no mistake here. Let me test a value in each interval to confirm.For example, take x = 0, which should be in the solution set.First inequality: 6 - 2(0) = 6 ≥ 0 ✔️Second inequality: 2(0) + 4 = 4 > 0 ✔️Good.Take x = -3, which should be outside the solution set.First inequality: 6 - 2(-3) = 6 + 6 = 12 ≥ 0 ✔️Second inequality: 2(-3) + 4 = -6 + 4 = -2 ≯ 0 ❌So x = -3 is excluded, correct.Take x = 4, outside the upper bound.First inequality: 6 - 2(4) = 6 - 8 = -2 ≱ 0 ❌Second inequality: 2(4) + 4 = 8 + 4 = 12 > 0 ✔️But since the first inequality fails, x=4 is excluded.Take x = -2, the lower boundary.First inequality: 6 - 2(-2) = 6 + 4 = 10 ≥ 0 ✔️Second inequality: 2(-2) + 4 = -4 + 4 = 0 ≯ 0 ❌So x=-2 is excluded, which is why we use a parenthesis there.Take x = 3, the upper boundary.First inequality: 6 - 2(3) = 6 - 6 = 0 ≥ 0 ✔️Second inequality: 2(3) + 4 = 6 + 4 = 10 > 0 ✔️So x=3 is included, hence the bracket on 3.Therefore, the solution set is correctly (-2, 3].I think that's solid. Let me just check once more if I handled the inequalities correctly, especially the first one where dividing by a negative might have been a pitfall. But since I added 2x to both sides instead, I avoided that issue. So x ≤ 3 is correct.Yeah, I think that's right. So the answer is (-2, 3], which in boxed form would be boxed{(-2, 3]}.Final AnswerThe solution set is boxed{(-2, 3]}.</think>To solve the system of inequalities:[begin{cases}6 - 2x ge 0 2x + 4 > 0end{cases}]we start by solving each inequality separately.First Inequality: (6 - 2x ge 0)1. Add (2x) to both sides: (6 ge 2x)2. Divide both sides by 2: (3 ge x)3. This simplifies to: (x le 3)Second Inequality: (2x + 4 > 0)1. Subtract 4 from both sides: (2x > -4)2. Divide both sides by 2: (x > -2)Combining Both SolutionsThe solution set is the intersection of (x le 3) and (x > -2). In interval notation, this is ((-2, 3]).To verify, we check boundary values and intervals:- (x = 0) satisfies both inequalities.- (x = -3) fails the second inequality.- (x = 4) fails the first inequality.- (x = -2) fails the second inequality.- (x = 3) satisfies both inequalities.Thus, the solution set is (boxed{(-2, 3]}).]

🔑:To solve the given system of inequalities, we proceed as follows:1. First Inequality:Starting with the first inequality:6 - 2x geq 0We can solve for x by isolating it on one side:begin{align*}6 - 2x &geq 0 -2x &geq -6 x &leq 3end{align*}This manipulation involves dividing both sides by -2, and remembering that dividing or multiplying both sides of an inequality by a negative number reverses the inequality sign.2. Second Inequality:For the second inequality:2x + 4 > 0We solve for x in a similar manner:begin{align*}2x + 4 &> 0 2x &> -4 x &> -2end{align*}Here, we subtract 4 from both sides and then divide by 2 to isolate x.3. Combining the Solutions:Combining the solutions from both inequalities, we have:-2 < x leq 3This means x must be greater than -2 but less than or equal to 3. Therefore, the solution set of the original system of inequalities is:boxed{-2 < x leq 3}

🧠:[<think>Okay, so I have this equation to solve: (x - 13)² + (y - 14)² + (x - y)² = 1/3. Hmm, that looks like a combination of squared terms. I need to find all the real numbers x and y that satisfy this equation. Let me try to break it down step by step.First, maybe expanding all the terms would help. Let me start by expanding each squared part separately. Let's take (x - 13)² first. Expanding that, I get x² - 26x + 169. Then, (y - 14)² would be y² - 28y + 196. And the third term is (x - y)², which expands to x² - 2xy + y². Now, if I add all these together, the left-hand side becomes:(x² - 26x + 169) + (y² - 28y + 196) + (x² - 2xy + y²)Let me combine like terms. Let's collect the x² terms first. There's x² from the first term, another x² from the third term. So that's 2x². Then the y² terms: y² from the second term, and another y² from the third term, so that's 2y². Then the cross term: -2xy. Then the linear terms: -26x from the first term and -28y from the second term. And finally, the constants: 169 + 196.Let me compute 169 + 196. 169 + 196 is 365? Wait, 169 + 196. 169 + 100 is 269, plus 96 is 365. Yeah, so constants add up to 365. So putting it all together, the equation becomes:2x² + 2y² - 2xy - 26x - 28y + 365 = 1/3.Hmm, okay. Now, maybe I can move the 1/3 to the left side:2x² + 2y² - 2xy - 26x - 28y + 365 - 1/3 = 0.Calculating 365 - 1/3. Since 365 is 364 + 1, then 364 + 1 - 1/3 = 364 + 2/3. So 364 and 2/3. Which is also 1094/3. Because 364*3 = 1092, plus 2 is 1094. So 1094/3. So the equation becomes:2x² + 2y² - 2xy - 26x - 28y + 1094/3 = 0.Hmm, this looks a bit messy. Maybe I can divide both sides by 2 to simplify it? Let me try that.Dividing each term by 2:x² + y² - xy - 13x - 14y + 547/3 = 0.Still, dealing with fractions might complicate things. Alternatively, maybe I can multiply through by 3 to eliminate denominators. Let's see:Original equation after moving 1/3:2x² + 2y² - 2xy - 26x - 28y + 365 - 1/3 = 0.Multiply all terms by 3:3*2x² + 3*2y² + 3*(-2xy) + 3*(-26x) + 3*(-28y) + 3*365 - 1 = 0.Calculating each term:6x² + 6y² - 6xy - 78x - 84y + 1095 - 1 = 0.Which simplifies to:6x² + 6y² - 6xy - 78x - 84y + 1094 = 0.Hmm, that still seems complicated. Maybe this quadratic in x and y can be rewritten in a more manageable form. Let me see if completing the square would work here. Alternatively, since there's a term with xy, maybe we can diagonalize the quadratic form or use substitution.Alternatively, perhaps we can use substitution. Let me consider variables u and v such that u = x - a and v = y - b for some constants a and b. Maybe shifting coordinates to eliminate the linear terms. Let me try that.Let me rewrite the equation as:2x² + 2y² - 2xy - 26x - 28y + 1094/3 = 0.Wait, actually, let me step back. The original equation after expanding is:2x² + 2y² - 2xy - 26x - 28y + 365 = 1/3.But when I moved 1/3 to the left, it's 365 - 1/3 = 364 + 2/3, which is 1094/3, so 2x² + 2y² - 2xy - 26x - 28y + 1094/3 = 0. That's correct. So maybe multiplying by 3 gives 6x² + 6y² - 6xy - 78x - 84y + 1094 = 0. Hmm.Alternatively, maybe instead of expanding, there's a geometric interpretation. The original equation is (x -13)^2 + (y -14)^2 + (x - y)^2 = 1/3. That's the sum of three squared distances? Let me see:(x -13)^2 is the squared distance from (x, y) to (13, 14) in the plane? Wait, no, (x -13)^2 + (y -14)^2 would be the squared distance to (13,14), but here we have an extra term (x - y)^2, which is the squared difference between x and y. So it's not exactly a geometric distance in 2D space. Maybe we can interpret this in 3D? Not sure.Alternatively, perhaps think of variables x and y as coordinates, and this equation represents a quadratic curve. Since there are cross terms (the -2xy), it's a rotated conic. To analyze it, maybe we can rotate the coordinate system to eliminate the cross term. But that might be a bit involved.Alternatively, maybe we can use substitution. Let me set u = x - y. Then, the third term becomes u². Then, we have:(x -13)^2 + (y -14)^2 + u² = 1/3.But u = x - y, so maybe expressing x or y in terms of u and the other variable. Let's say u = x - y, so x = y + u. Substitute into the other terms:(y + u -13)^2 + (y -14)^2 + u² = 1/3.Expanding the first term: (y + u -13)^2 = (y -13 + u)^2 = (y -13)^2 + 2u(y -13) + u².So the equation becomes:(y -13)^2 + 2u(y -13) + u² + (y -14)^2 + u² = 1/3.Combine like terms:(y -13)^2 + (y -14)^2 + 2u(y -13) + u² + u².Which simplifies to:(y -13)^2 + (y -14)^2 + 2u(y -13) + 2u² = 1/3.But u = x - y, so maybe we can substitute back later, but this seems getting complicated. Maybe there's another substitution.Alternatively, let's think of this as a system of equations. If we consider that the equation is quadratic in both x and y, perhaps taking partial derivatives and setting them to zero to find minima? Wait, but we need to solve for x and y such that the equation holds. Not sure.Alternatively, maybe we can write this equation as a quadratic in x, treating y as a constant, and then solve for x using the quadratic formula, then impose that the discriminant is non-negative for real solutions, and then solve for y. That might work. Let's try that approach.Starting with the original equation:(x -13)^2 + (y -14)^2 + (x - y)^2 = 1/3.Let me expand this again, but maybe keep it as a quadratic in x:First, expand (x -13)^2: x² -26x +169.Then, (x - y)^2: x² -2xy + y².So combining these with (y -14)^2, which is y² -28y +196.Adding all together:x² -26x +169 + y² -28y +196 + x² -2xy + y² = 1/3.Combine like terms:2x² + 2y² -2xy -26x -28y + 365 = 1/3.So, moving 1/3 to left:2x² + 2y² -2xy -26x -28y + 365 - 1/3 = 0.Which is:2x² + 2y² -2xy -26x -28y + 1094/3 = 0.Now, treating this as a quadratic equation in x:2x² - (2y +26)x + (2y² -28y +1094/3) = 0.Let me write it as:2x² - (2y +26)x + (2y² -28y +1094/3) = 0.Divide through by 2 to simplify:x² - (y +13)x + (y² -14y +547/3) = 0.So, quadratic in x: x² - (y +13)x + (y² -14y +547/3) = 0.Using the quadratic formula, x = [ (y +13) ± sqrt( (y +13)^2 - 4*(1)*(y² -14y +547/3) ) ] / 2.Simplify the discriminant:Discriminant D = (y +13)^2 - 4*(y² -14y +547/3).Compute each part:First, expand (y +13)^2: y² +26y +169.Then, compute 4*(y² -14y +547/3): 4y² -56y + 4*(547/3) = 4y² -56y + 2188/3.So, D = (y² +26y +169) - (4y² -56y +2188/3) = y² +26y +169 -4y² +56y -2188/3.Combine like terms:-3y² +82y +169 -2188/3.Convert 169 to thirds: 169 = 507/3.So, D = -3y² +82y +507/3 -2188/3 = -3y² +82y -1681/3.Therefore, the discriminant is -3y² +82y -1681/3.For real solutions in x, the discriminant must be non-negative:-3y² +82y -1681/3 ≥ 0.Multiply both sides by -3 (which reverses the inequality):9y² -246y +1681 ≤ 0.So, we have a quadratic inequality: 9y² -246y +1681 ≤ 0.Let me check if this quadratic has real roots. Compute discriminant D':D' = (-246)^2 -4*9*1681.Compute 246^2: 246*246. Let's compute 200^2=40000, 40^2=1600, 6^2=36. But 246^2 = (200 + 46)^2 = 200^2 + 2*200*46 +46^2 = 40000 + 18400 + 2116 = 40000 +18400=58400, +2116=60516.Then, 4*9*1681 = 36*1681. Let's compute 36*1681:1681*36: 1681*30=50430, 1681*6=10086, total 50430+10086=60516.Therefore, D' = 60516 - 60516 = 0. So, discriminant is zero. So the quadratic 9y² -246y +1681 is a perfect square and has a double root.So, the quadratic can be written as (3y - a)^2 = 0. Let's find a:Expand (3y - a)^2 = 9y² -6a y +a². Comparing to 9y² -246y +1681:-6a = -246 ⇒ a = 41.Then, a² = 1681, which is 41²=1681. So yes, correct. Therefore, the quadratic is (3y -41)^2 ≤0.But since a square is always non-negative, the inequality (3y -41)^2 ≤0 holds only when 3y -41 =0 ⇒ y=41/3.Therefore, the discriminant is zero only when y=41/3, which is approximately 13.6667.Therefore, the only possible y value is y=41/3. Then, substituting back into the equation for x.Earlier, when we had the quadratic in x:x² - (y +13)x + (y² -14y +547/3) = 0.Given that discriminant D=0, there is exactly one solution for x. So:x = [ (y +13) ± sqrt(D) ] /2 = (y +13)/2.Since sqrt(D)=0. So x=(y +13)/2.Given that y=41/3, then x=(41/3 +13)/2. Convert 13 to thirds: 13=39/3. So x=(41/3 +39/3)/2=80/3 /2=40/3.Therefore, the only solution is x=40/3 and y=41/3.Wait, but let me verify this solution in the original equation to make sure.Compute each term:(x -13)^2: (40/3 -13)^2 = (40/3 -39/3)^2 = (1/3)^2 = 1/9.(y -14)^2: (41/3 -14)^2 = (41/3 -42/3)^2 = (-1/3)^2 =1/9.(x - y)^2: (40/3 -41/3)^2 = (-1/3)^2 =1/9.Adding them together:1/9 +1/9 +1/9=3/9=1/3. Which matches the right-hand side. So yes, that's correct.Therefore, the only solution is x=40/3 and y=41/3.But wait, the problem says "Find all values of x and y". So is this the only solution? Since we had the discriminant for the quadratic in x being non-negative only when y=41/3, and then x=40/3. So that's the only pair (x,y) that satisfies the equation.Therefore, the solution is x=40/3, y=41/3.But just to make sure I didn't make a mistake in the discriminant calculation. Let's recap:Original equation: (x -13)^2 + (y -14)^2 + (x - y)^2 =1/3.Expanded and simplified to quadratic in x: 2x² +2y² -2xy -26x -28y +1094/3=0.Then, treated as quadratic in x, discriminant D must be non-negative. Calculated D=-3y² +82y -1681/3, multiplied by -3 to get 9y² -246y +1681=0, which factors as (3y-41)^2=0, so y=41/3, then x=(y +13)/2=40/3.Verified the solution in original equation, and it works. Therefore, this is the only solution.Alternatively, another approach: since the equation is the sum of squares equal to 1/3, which is a positive number. Each square is non-negative, so each term must be less than or equal to 1/3. Therefore, each individual term must be less than or equal to 1/3. So:(x -13)^2 ≤1/3,(y -14)^2 ≤1/3,(x - y)^2 ≤1/3.Therefore, each of these inequalities must hold. So:|x -13| ≤√(1/3),|y -14| ≤√(1/3),|x - y| ≤√(1/3).So x is in [13 - √(1/3), 13 + √(1/3)],y is in [14 - √(1/3), 14 + √(1/3)],and x - y is in [-√(1/3), √(1/3)].Therefore, the possible x and y must lie within these intervals and satisfy all three conditions. But perhaps combining these inequalities, maybe we can find that the only point that satisfies all three equalities is when each term is exactly 1/3, hence each squared term is 1/3, so they must all be equal. But in the original equation, the sum is 1/3, so each term must be exactly 1/9. Wait, no: if three non-negative numbers add up to 1/3, it's possible that each is 1/9, but they could also be different. However, if we assume that the minimum case is when each term is equal, but given that the problem is to find all solutions, not just the minimal case.But in our previous calculation, we found only one solution. So perhaps there's only one solution where all three squared terms are 1/9, so their sum is 1/3.Indeed, in our solution, each term is (1/3)^2=1/9, so all three terms are 1/9, sum is 3*(1/9)=1/3.Therefore, that must be the only solution where all three squared terms reach their minimal possible values to sum up to 1/3. Since each term is a square, their minimum is zero, but since they have to add up to 1/3, there must be a balance.But according to our algebra, there's only one such point where all three terms are equal to 1/9. Therefore, the only solution is x=40/3, y=41/3.Hence, the answer is x=40/3 and y=41/3.Final AnswerThe values of (x) and (y) are (boxed{left(dfrac{40}{3}, dfrac{41}{3}right)}).</think>To solve the equation ((x - 13)^2 + (y - 14)^2 + (x - y)^2 = frac{1}{3}), we start by expanding and combining like terms:1. Expand each term: [ (x - 13)^2 = x^2 - 26x + 169 ] [ (y - 14)^2 = y^2 - 28y + 196 ] [ (x - y)^2 = x^2 - 2xy + y^2 ]2. Combine all terms: [ (x^2 - 26x + 169) + (y^2 - 28y + 196) + (x^2 - 2xy + y^2) = frac{1}{3} ]3. Simplify the left-hand side: [ 2x^2 + 2y^2 - 2xy - 26x - 28y + 365 = frac{1}{3} ]4. Move (frac{1}{3}) to the left-hand side: [ 2x^2 + 2y^2 - 2xy - 26x - 28y + 365 - frac{1}{3} = 0 ]5. Convert constants to a common denominator: [ 365 - frac{1}{3} = frac{1094}{3} ]6. The equation becomes: [ 2x^2 + 2y^2 - 2xy - 26x - 28y + frac{1094}{3} = 0 ]Next, we treat this as a quadratic equation in (x):1. Rewrite the equation: [ 2x^2 - (2y + 26)x + (2y^2 - 28y + frac{1094}{3}) = 0 ]2. Divide by 2: [ x^2 - (y + 13)x + (y^2 - 14y + frac{547}{3}) = 0 ]3. Use the quadratic formula to solve for (x): [ x = frac{(y + 13) pm sqrt{(y + 13)^2 - 4(y^2 - 14y + frac{547}{3})}}{2} ]4. Simplify the discriminant: [ (y + 13)^2 - 4(y^2 - 14y + frac{547}{3}) = -3y^2 + 82y - frac{1681}{3} ]5. For real solutions, the discriminant must be non-negative: [ 9y^2 - 246y + 1681 leq 0 ]6. Factor the quadratic: [ (3y - 41)^2 leq 0 ]This inequality holds only when (y = frac{41}{3}). Substituting (y = frac{41}{3}) back into the quadratic equation for (x):1. Solve for (x): [ x = frac{(y + 13)}{2} = frac{frac{41}{3} + 13}{2} = frac{40}{3} ]Verification shows that (x = frac{40}{3}) and (y = frac{41}{3}) satisfy the original equation. Therefore, the values of (x) and (y) are:[boxed{left(dfrac{40}{3}, dfrac{41}{3}right)}],<think>Okay, so I need to find all the values of x and y that satisfy this equation: (x - 13)^2 + (y - 14)^2 + (x - y)^2 = 1/3. Hmm, let's see. Alright, first, this looks like a problem involving the sum of squares equaling a constant. Since all the terms are squared, they must each be non-negative, right? So their sum can only be 1/3 if each term is individually less than or equal to 1/3. That might help. But maybe instead of thinking about each term separately, I can try expanding the equation and simplifying. Let me try that. Let's expand each of the squares step by step.First term: (x - 13)^2 = x^2 - 26x + 169.Second term: (y - 14)^2 = y^2 - 28y + 196.Third term: (x - y)^2 = x^2 - 2xy + y^2.So adding all three together:(x^2 - 26x + 169) + (y^2 - 28y + 196) + (x^2 - 2xy + y^2) = 1/3.Let me combine like terms. First, the x^2 terms: there are two in the first term and one in the third term, so total x^2 terms: 1 + 1 + 1? Wait, no. Wait, first term is (x -13)^2 which gives x^2. Second term is (y -14)^2 which gives y^2. Third term is (x - y)^2 which gives x^2 - 2xy + y^2. So actually:First term: x^2Second term: y^2Third term: x^2 + y^2So total x^2 terms: 1 (from first term) + 1 (from third term) = 2x^2.Total y^2 terms: 1 (from second term) + 1 (from third term) = 2y^2.Then the cross term: -2xy from the third term.Then linear terms: -26x from the first term, -28y from the second term.Constants: 169 + 196 from the first and second terms. Let's compute that: 169 + 196. 169 + 196: 169 + 200 is 369, minus 4 is 365. So 365.So putting it all together:2x^2 + 2y^2 - 2xy - 26x - 28y + 365 = 1/3.Hmm. Let me write that equation again:2x^2 + 2y^2 - 2xy - 26x - 28y + 365 = 1/3.First, I can subtract 1/3 from both sides to set the equation to zero:2x^2 + 2y^2 - 2xy - 26x - 28y + 365 - 1/3 = 0.Calculating 365 - 1/3: 365 is 364 + 1, so 364 + 1 - 1/3 = 364 + 2/3. So 364 and 2/3. As an improper fraction, that's 364 * 3 + 2 = 1092 + 2 = 1094, so 1094/3. Wait, 364*3 is 1092, so 364 + 2/3 is 1094/3? Wait, no: 364 is 364/1, so 364 - 1/3 = (364*3 - 1)/3 = (1092 - 1)/3 = 1091/3. Wait, original equation was 365 - 1/3. So 365 is 364 + 1, so 364 + 1 - 1/3 = 364 + (2/3). So 364 and 2/3, which is 364*3 + 2 = 1092 + 2 = 1094 over 3. So 1094/3. Wait, but 364 * 3 is 1092, then 1092 + 2 = 1094. So yes, 1094/3. So the equation becomes:2x^2 + 2y^2 - 2xy - 26x - 28y + 1094/3 = 0.Hmm, this seems a bit messy. Maybe I can multiply both sides by 3 to eliminate denominators. Let's try that:3*(2x^2 + 2y^2 - 2xy - 26x - 28y) + 1094 = 0.So:6x^2 + 6y^2 - 6xy - 78x - 84y + 1094 = 0.Hmm, not sure if that helps. Maybe it's better to think in terms of completing the square. Alternatively, perhaps making a substitution. Let me think.Alternatively, note that the original equation is a sum of three squared terms. Maybe this can be interpreted geometrically. For example, in coordinate geometry, maybe representing points where the sum of squared distances to certain points is a constant. But I need to recall that.Alternatively, perhaps we can set variables u = x - a and v = y - b for some a and b to eliminate the linear terms. Let me consider that.But first, let's look back at the original equation:(x -13)^2 + (y -14)^2 + (x - y)^2 = 1/3.Alternatively, maybe think of x and y as coordinates, and this equation represents points (x,y) such that the sum of the squares of their distances to (13,14), (x,y) themselves (since (x - y)^2 is like the square of the distance from x to y on the line), but that might not be straightforward. Wait, (x - y)^2 is similar to the square of the distance between (x, y) and (y, x), but maybe not helpful.Alternatively, perhaps consider substituting variables. Let me set u = x - 13 and v = y - 14. Then x = u + 13, y = v + 14.Then, substituting into the original equation:(u)^2 + (v)^2 + ( (u +13) - (v +14) )^2 = 1/3.Simplify the third term:(u +13 - v -14)^2 = (u - v -1)^2.So the equation becomes:u^2 + v^2 + (u - v -1)^2 = 1/3.Expanding the third term:(u - v -1)^2 = (u - v)^2 - 2(u - v) +1 = u^2 - 2uv + v^2 - 2u + 2v +1.So now substitute back into the equation:u^2 + v^2 + u^2 - 2uv + v^2 - 2u + 2v +1 = 1/3.Combine like terms:u^2 + u^2 = 2u^2.v^2 + v^2 = 2v^2.-2uv remains.-2u and +2v remain.+1 from the constant term.So total equation:2u^2 + 2v^2 - 2uv -2u + 2v +1 = 1/3.Subtract 1/3 from both sides:2u^2 + 2v^2 - 2uv -2u + 2v +1 -1/3 =0.Which simplifies to:2u^2 + 2v^2 - 2uv -2u + 2v +2/3 =0.Hmm. Maybe we can divide through by 2 to simplify:u^2 + v^2 - uv -u + v +1/3 =0.So the equation is:u^2 + v^2 - uv - u + v + 1/3 = 0.This seems a bit simpler. Maybe I can rearrange terms:u^2 - uv - u + v^2 + v + 1/3 =0.Hmm, perhaps group terms with u and terms with v:(u^2 - uv - u) + (v^2 + v) +1/3 =0.Alternatively, maybe factor the terms with u:u^2 - u(v +1) + (v^2 + v +1/3) =0.This looks like a quadratic in u. So maybe we can write it as:u^2 - u(v +1) + (v^2 + v +1/3) =0.If I consider this as a quadratic equation in u, then for real solutions to exist, the discriminant must be non-negative. Let's compute the discriminant:Discriminant D = [-(v +1)]^2 - 4*1*(v^2 + v +1/3).Compute D:D = (v +1)^2 -4(v^2 + v +1/3).Expand (v +1)^2: v^2 + 2v +1.So:D = v^2 + 2v +1 -4v^2 -4v -4/3.Combine like terms:v^2 -4v^2 = -3v^2.2v -4v = -2v.1 -4/3 = -1/3.So D = -3v^2 -2v -1/3.For the discriminant to be non-negative, we need:-3v^2 -2v -1/3 ≥0.Multiply both sides by -1 (which reverses the inequality):3v^2 +2v +1/3 ≤0.But 3v^2 +2v +1/3 is a quadratic in v. Let's check its discriminant:Discriminant D' = (2)^2 -4*3*(1/3) =4 -4 =0.So the quadratic 3v^2 +2v +1/3 has a double root. Let's find it:v = [-2 ± sqrt(0)]/(2*3) = -2/6 = -1/3.Therefore, 3v^2 +2v +1/3 =3(v +1/3)^2.Indeed, expanding 3(v +1/3)^2: 3(v^2 + (2/3)v +1/9) =3v^2 +2v +1/3.Thus, 3v^2 +2v +1/3 =3(v +1/3)^2 ≥0 for all real v.But we have 3v^2 +2v +1/3 ≤0. The only solution is when equality holds, which is when v = -1/3.Therefore, the discriminant D is non-negative only when v = -1/3, and then D =0. Thus, there's exactly one solution for u when v = -1/3.So let's substitute v = -1/3 into the quadratic equation for u:u^2 - u(v +1) + (v^2 + v +1/3) =0.Plugging in v = -1/3:u^2 - u(-1/3 +1) + ((-1/3)^2 + (-1/3) +1/3)=0.Simplify each term:First, compute (-1/3 +1): that's 2/3.Then, compute (-1/3)^2 =1/9. Then (-1/3) +1/3 =0. So the constant term is 1/9 +0=1/9.So the equation becomes:u^2 - (2/3)u +1/9 =0.This is a quadratic equation in u. Let's solve it:u = [ (2/3) ± sqrt( (2/3)^2 -4*1*(1/9) ) ]/(2*1).Compute discriminant:(4/9) - (4/9) =0.Thus, u = (2/3)/2 =1/3.So only solution is u=1/3.Therefore, the only solution in u and v is u=1/3, v=-1/3.Recall that u =x -13, v =y -14.So x = u +13 =1/3 +13=13 +1/3=40/3.Similarly, y =v +14= -1/3 +14=14 -1/3=41/3.Therefore, the only solution is x=40/3, y=41/3.Wait, but let's check this solution in the original equation to verify.Compute (x -13)^2: (40/3 -13)^2= (40/3 -39/3)=1/3. So (1/3)^2=1/9.Similarly, (y -14)^2= (41/3 -14)^2= (41/3 -42/3)= (-1/3)^2=1/9.Then (x - y)^2= (40/3 -41/3)^2= (-1/3)^2=1/9.Adding them up: 1/9 +1/9 +1/9=3/9=1/3. Which matches the right-hand side. So that works.Therefore, the only solution is x=40/3 and y=41/3.Wait, but the problem says "Find all values of x and y", so this is the only solution?But let me think again. When I made the substitution u =x -13 and v =y -14, then the equation became sum of squares of u, v, and (u -v -1). Then by expanding, I transformed the equation into a quadratic in u, and found that the discriminant must be zero, leading to v=-1/3, then u=1/3, leading to x=40/3, y=41/3.Alternatively, perhaps there's a different approach. For example, using calculus, setting partial derivatives to zero, but since this is a sum of squares, the minimum value is achieved at some point, and the question is when the sum equals 1/3. But maybe 1/3 is the minimal value, so there's only one point where it's achieved.Alternatively, consider that this is the equation of a circle (since it's quadratic and the coefficients are positive for x^2 and y^2 terms, but there's a cross term). But since there is only one solution, it's actually a single point, which makes sense because the sum of three squares equals a small positive number, which can only happen at one point if the three terms are minimized there.But to confirm, let's think of this geometrically. The first two terms, (x -13)^2 + (y -14)^2, represent the squared distance from (x,y) to (13,14). The third term, (x - y)^2, is the squared distance from (x,y) to the line y=x. So the equation is saying that the sum of the squared distance to the point (13,14) and the squared distance to the line y=x equals 1/3. But this might not directly help, unless there's a geometric interpretation for such a sum.Alternatively, perhaps we can parametrize variables. Let me set t =x - y. Then, (x - y)^2 =t^2. Then, express y as x - t, so substitute y =x -t into the original equation.Original equation:(x -13)^2 + (y -14)^2 + t^2 =1/3.Substitute y =x -t:(x -13)^2 + (x -t -14)^2 +t^2 =1/3.Let's simplify:First term: (x -13)^2.Second term: (x - (t +14))^2.Third term: t^2.So expanding the second term:(x - t -14)^2 =x^2 -2(t +14)x + (t +14)^2.First term is x^2 -26x +169.So adding all three terms:(x^2 -26x +169) + (x^2 -2(t +14)x + (t +14)^2) + t^2 =1/3.Combine like terms:x^2 +x^2 =2x^2.-26x -2(t +14)x= -26x -2tx -28x= (-26 -28)x -2tx= -54x -2tx.Constants:169 + (t +14)^2 +t^2.So the equation becomes:2x^2 - (54 +2t)x +169 + (t +14)^2 +t^2 =1/3.This seems complicated, but maybe we can see if this approach leads anywhere. However, perhaps this substitution complicates things further. Let me check.Alternatively, since we found a solution via substitution and completing the square, and verified it works, and given that the discriminant analysis shows that there's only one solution, perhaps that's the only solution.Alternatively, another method: think of the equation as a quadratic in x and y, and try to write it in matrix form, but that might be overcomplicating.Alternatively, notice that the equation is equivalent to:(x -13)^2 + (y -14)^2 + (x - y)^2 = 1/3.Let me denote A = (x -13), B = (y -14), C = (x - y). Then, the equation is A^2 + B^2 + C^2 =1/3.But note that C = (x - y) = (x -13) - (y -14) - (13 -14) = A - B +1.So C = A - B +1. Therefore, the equation becomes:A^2 + B^2 + (A - B +1)^2 =1/3.Expanding the third term:(A - B +1)^2 = A^2 -2AB + B^2 +2A -2B +1.So the equation is:A^2 + B^2 + A^2 -2AB + B^2 +2A -2B +1 =1/3.Combine like terms:2A^2 +2B^2 -2AB +2A -2B +1 =1/3.Which is the same equation we had earlier with u and v. Then, proceeding as before, we can write this as a quadratic in A or B. For example, treating it as a quadratic in A:2A^2 + (-2B +2)A +2B^2 -2B +1 -1/3 =0.But this seems similar to the previous approach. Then, again, solving for A would require the discriminant to be non-negative, leading to a quadratic in B, which would collapse to a single solution.So, regardless of the substitution method, we end up with a single solution. Therefore, the only solution is x=40/3 and y=41/3.Therefore, I think that's the answer. But just to confirm once more.Alternatively, let's consider the problem as minimizing the sum of the three squares. Since squares are non-negative, the minimal value would occur when each term is as small as possible. But the question states that the sum is 1/3, which is a very small number, suggesting that this is the minimal possible value, and thus the point (40/3,41/3) is the unique point where this minimal sum is achieved.Alternatively, think of the three terms as functions. Let's call f(x,y) = (x-13)^2 + (y-14)^2 + (x-y)^2. We can compute the gradient of f and set it to zero to find critical points.Compute partial derivatives:df/dx = 2(x -13) + 2(x - y).df/dy = 2(y -14) -2(x - y).Set them to zero:2(x -13) + 2(x - y) =0,2(y -14) -2(x - y) =0.Simplify both equations.First equation:2x -26 +2x -2y =0 => 4x -2y -26 =0.Second equation:2y -28 -2x +2y =0 => (-2x) +4y -28 =0.So the system is:4x -2y =26,-2x +4y =28.Let me write this as:Equation 1: 4x -2y =26.Equation 2: -2x +4y =28.We can solve this system. Let's multiply Equation 1 by 2 to eliminate y:Equation 1 *2: 8x -4y =52.Equation 2: -2x +4y =28.Now add them together:(8x -4y) + (-2x +4y) =52 +28.6x =80.Thus, x=80/6=40/3≈13.333...Then substitute x=40/3 into Equation 1:4*(40/3) -2y=26.Compute 4*(40/3)=160/3.160/3 -2y=26.So, -2y=26 -160/3=78/3 -160/3= -82/3.Thus, y= (-82/3)/(-2)=41/3≈13.666...Thus, the critical point is at (40/3,41/3). Since the function f(x,y) is a sum of squares, it is convex, so this critical point is the global minimum. Therefore, the minimal value of f(x,y) is f(40/3,41/3)=1/3 as computed earlier. Therefore, the equation f(x,y)=1/3 has exactly one solution: the point (40/3,41/3).Therefore, the answer is x=40/3 and y=41/3.Final AnswerThe values of (x) and (y) are (boxed{left(dfrac{40}{3}, dfrac{41}{3}right)}).</think>To find all values of (x) and (y) for which the equality ((x-13)^{2}+(y-14)^{2}+(x-y)^{2}=frac{1}{3}) holds, we start by expanding and simplifying the given equation.First, we expand each term:[(x - 13)^2 = x^2 - 26x + 169][(y - 14)^2 = y^2 - 28y + 196][(x - y)^2 = x^2 - 2xy + y^2]Adding these together, we get:[x^2 - 26x + 169 + y^2 - 28y + 196 + x^2 - 2xy + y^2 = frac{1}{3}]Combining like terms, we have:[2x^2 + 2y^2 - 2xy - 26x - 28y + 365 = frac{1}{3}]Subtracting (frac{1}{3}) from both sides:[2x^2 + 2y^2 - 2xy - 26x - 28y + frac{1094}{3} = 0]Next, we substitute (u = x - 13) and (v = y - 14), leading to (x = u + 13) and (y = v + 14). Substituting these into the equation, we get:[u^2 + v^2 + (u - v - 1)^2 = frac{1}{3}]Expanding and simplifying the third term:[(u - v - 1)^2 = u^2 - 2uv + v^2 - 2u + 2v + 1]Combining all terms:[u^2 + v^2 + u^2 - 2uv + v^2 - 2u + 2v + 1 = frac{1}{3}]Simplifying further:[2u^2 + 2v^2 - 2uv - 2u + 2v + 1 = frac{1}{3}]Subtracting (frac{1}{3}) from both sides:[2u^2 + 2v^2 - 2uv - 2u + 2v + frac{2}{3} = 0]Dividing by 2:[u^2 + v^2 - uv - u + v + frac{1}{3} = 0]We solve this quadratic equation in (u) by considering the discriminant. The discriminant must be non-negative for real solutions:[(v + 1)^2 - 4(v^2 + v + frac{1}{3}) geq 0]Simplifying, we find that the discriminant is zero only when (v = -frac{1}{3}). Substituting (v = -frac{1}{3}) back, we solve for (u) and find (u = frac{1}{3}).Converting back to (x) and (y):[x = u + 13 = frac{1}{3} + 13 = frac{40}{3}][y = v + 14 = -frac{1}{3} + 14 = frac{41}{3}]Verifying this solution in the original equation confirms it satisfies the equality. Thus, the values of (x) and (y) are (boxed{left(dfrac{40}{3}, dfrac{41}{3}right)}).]

🔑:Answer: x=13 frac{1}{3}, y=13 frac{2}{3}Solution: We use the inequality between the quadratic mean and the arithmetic mean:sqrt{frac{(x-13)^{2}+(14-y)^{2}+(x-y)^{2}}{3}} geqslant frac{|x-13|+|14-y|+|x-y|}{3} geqslant frac{x-13+14-y+y-x}{3}=frac{1}{3}That is, (x-13)^{2}+(14-y)^{2}+(x-y)^{2} geqslant frac{1}{3}. Equality is achieved only when x-13=14-y= y-x=frac{1}{3}. Then x=13 frac{1}{3}, y=13 frac{2}{3}.

🧠:[<think>Okay, so I need to find the negation of the proposition: "There exists a real number ( x_0 ) such that ( ln x_0 < x_0^2 - 1 )." The options are A, B, C, D. Let me recall how to negate statements in logic. The original statement is an existential quantifier: ( exists x_0 in mathbb{R} ) such that ( ln x_0 < x_0^2 - 1 ). The negation of an existential quantifier is a universal quantifier with the negated predicate. So, the negation should be: For all real numbers ( x ), ( ln x geq x^2 - 1 ). Let me check that.So, the negation of ( exists x P(x) ) is ( forall x neg P(x) ). So in this case, ( P(x) ) is ( ln x < x^2 - 1 ). Therefore, the negation would be ( forall x, ln x geq x^2 - 1 ). That would correspond to option B. Let me look at the options again.Option A: For any real number ( x ), ( ln x < x^2 - 1 ). Hmm, that's not the negation. That's actually a stronger version of the original statement. If the original was "there exists," then saying "for all" is a different statement, not the negation. So A is incorrect.Option B: For any real number ( x ), ( ln x geq x^2 - 1 ). That seems to fit with the negation I thought of. Since we changed "there exists" to "for all" and flipped the inequality. So that should be correct. Let me check the other options just to be sure.Option C: There does not exist a real number ( x_0 ) such that ( ln x_0 geq x_0^2 - 1 ). Wait, this is the negation of "there exists an ( x_0 ) where ( ln x_0 geq x_0^2 - 1 )." But our original statement was about ( ln x_0 < x_0^2 - 1 ). So the negation of the original is not directly about the existence of ( x_0 ) with the opposite inequality. Let me see. The original is ( exists x_0 ( ln x_0 < x_0^2 - 1 ) ). Its negation is ( forall x ( ln x geq x^2 -1 ) ). So Option C is saying there does not exist ( x_0 ) with ( ln x_0 geq x_0^2 -1 ), which is equivalent to saying for all ( x ), ( ln x < x^2 -1 ). Wait, but that would be the negation of "there exists an ( x ) with ( ln x geq x^2 -1 )", but our original negation is different. So C is not the correct negation here.Option D: There exists a real number ( x_0 ) such that ( ln x_0 geq x_0^2 - 1 ). This is the opposite of the original statement, but it's not the negation. The negation of "there exists" is not another "there exists," unless the predicate is contradictory. But in general, the negation of existence is universal. So D is incorrect because it's just an existence statement with the inequality flipped, but that doesn't cover all cases. If there exists an ( x_0 ) where the inequality is reversed, that doesn't necessarily negate the original existence; both could coexist. So D is not the negation.Therefore, the correct answer must be B. Let me confirm once again.Original: There exists an x where ln x < x² -1.Negation: For all x, ln x is not less than x² -1, which is equivalent to ln x ≥ x² -1. So yes, that's exactly option B. So I think B is the right answer.Final Answerboxed{B}</think>The original proposition is "There exists a real number ( x_0 ) such that ( ln x_0 < x_0^2 - 1 )." To find its negation, we need to negate both the existential quantifier and the predicate. The negation of an existential quantifier ((exists)) is a universal quantifier ((forall)), and the negation of the predicate ( ln x_0 < x_0^2 - 1 ) is ( ln x geq x^2 - 1 ). Therefore, the negation of the proposition is "For any real number ( x ), ( ln x geq x^2 - 1 )."Checking the options:- Option A states a universal quantifier with the original inequality, which is incorrect.- Option B states the universal quantifier with the negated inequality, which is correct.- Option C negates the existence of the opposite inequality, which is incorrect.- Option D states an existence of the opposite inequality, which is also incorrect.Thus, the correct answer is (boxed{B}).,<think>Okay, let me try to figure out the negation of this proposition. The original statement is: "There exists a real number ( x_0 ) such that ( ln x_0 < x_0^2 - 1 )." So, I need to find the logical negation of this statement. First, I remember that the negation of "there exists" is "for all." In other words, if the original statement says there is at least one element with a certain property, the negation would say that all elements do not have that property. So, symbolically, the original statement is ( exists x_0 in mathbb{R} , (ln x_0 < x_0^2 - 1) ). Therefore, the negation should be ( forall x in mathbb{R} , neg (ln x < x^2 - 1) ).Now, the next step is to negate the inequality ( ln x < x^2 - 1 ). To negate an inequality, I just flip the inequality sign. But wait, I need to recall how negation affects inequalities. The negation of ( a < b ) is ( a geq b ). Similarly, the negation of ( a > b ) is ( a leq b ). So, the negation of ( ln x < x^2 - 1 ) is ( ln x geq x^2 - 1 ).Putting this together, the negation of the original statement is: "For all real numbers ( x ), ( ln x geq x^2 - 1 )." Looking at the options given, that's option B. Let me verify again to make sure I didn't make a mistake.Original: There exists an ( x_0 ) such that ( ln x_0 < x_0^2 - 1 ).Negation: For all ( x ), it's not the case that ( ln x < x^2 - 1 ), which translates to ( ln x geq x^2 - 1 ). So yes, that's exactly option B.Wait, but let me check the other options just in case. Option A says "For any real number ( x ), ( ln x < x^2 - 1 )." That's not a negation; that's actually a stronger statement than the original. The original only claims existence, but option A claims it's true for all ( x ), which is different.Option C: "There does not exist a real number ( x_0 ) such that ( ln x_0 geq x_0^2 - 1 )." Hmm, this is the negation of existence of ( x_0 ) where ( ln x_0 geq x_0^2 - 1 ). But that's not directly related to the original statement. The original was about existence of ( x_0 ) with ( ln x_0 < x_0^2 -1 ). The negation of that isn't about the non-existence of ( x_0 ) with the opposite inequality. Let me think. If the original is ( exists x (P(x)) ), then the negation is ( forall x (neg P(x)) ). So, ( exists x (ln x < x^2 -1 ) ) becomes ( forall x (ln x geq x^2 -1 ) ), which is option B. Option C is ( neg exists x (ln x geq x^2 -1 ) ), which would be equivalent to ( forall x (ln x < x^2 -1 ) ), which is option A. So, option C is the negation of the existence of ( x ) with ( ln x geq x^2 -1 ), which is a different statement.Option D: "There exists a real number ( x_0 ) such that ( ln x_0 geq x_0^2 - 1 )." So that's just the existence of such an ( x_0 ), which is not the negation of the original statement. The original's negation requires that all ( x ) satisfy the opposite inequality, not just one. So option D is actually the opposite; it's asserting existence, which isn't the negation.Therefore, going back, the correct answer is B. Let me just verify once more. If the original says there is some x where ln x is less than x² -1, the negation must be that for all x, ln x is not less than x² -1, which is the same as ln x is greater or equal to x² -1. So yes, option B.Final Answerboxed{B}</think>The original proposition is: "There exists a real number ( x_0 ) such that ( ln x_0 < x_0^2 - 1 )." Symbolically, this is ( exists x_0 in mathbb{R} , (ln x_0 < x_0^2 - 1) ).To find the negation, we use the principle that the negation of an existential quantifier is a universal quantifier. Therefore, the negation becomes ( forall x in mathbb{R} , neg (ln x < x^2 - 1) ).Next, we negate the inequality ( ln x < x^2 - 1 ). The negation of ( a < b ) is ( a geq b ), so the inequality becomes ( ln x geq x^2 - 1 ).Thus, the negation of the original statement is: "For all real numbers ( x ), ( ln x geq x^2 - 1 )." This corresponds to option B.[boxed{B}]]

🔑:The negation of a particular proposition is a universal proposition. The negation of "There exists a real number x_0 such that ln x_0 < x_0^2 - 1" is "For any real number x, ln x geq x^2 - 1".Hence, the answer is boxed{text{B}}.Directly using the fact that the negation of a particular proposition is a universal proposition, we can write out the result. This question tests the negation of propositions and the relationship between particular and universal propositions, which is a basic knowledge inquiry.