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Minimum Value of$\quad$ $x_1^2+y_1^2+x_2^2+y_2^2-2x_1x_2-2y_1y_2$ subject to condition $x_1,y_1$ and $x_2,y_2$ lies on curve $xy=1$. It is also given that $x_1\gt0$ and $x_2\lt0$

My Approach: $AM\geq GM$

$\frac{x_1^2+y_1^2+x_2^2+y_2^2-2x_1x_2-2y_1y_2}{6}\geq(4x_1^3.y_1^3.x_2^3.y_2^3)^\frac{1}{6}$

and I obtained minimum value as $6\sqrt[3]{2}$. But I think this is not correct minimum value as when minimum value will occur $AM=GM$ must satisfy and all number must be equal that is $x_1^2=x_2^2=-2x_1x_2=-2y_1y_2=y_1^2=y_2^2$

From first three relation I obtained that $x_1=-2x_2$ and $x_2=-2x_1$ which cannot be true except for zero.

Is my approach correct?

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  • $\begingroup$ yes those are non negative. you can check $\endgroup$
    – mathophile
    Commented May 8, 2021 at 12:02
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    $\begingroup$ Well, it is not correct. You haven't imposed the constraints $x_iy_i=1$ and therefore only located a minimum inside $x_1,y_1\geq0,x_2,y_2\leq0$. Impose the condition that those points are on the hyperbolas. A cheap way is to replace $y_i$ by $1/x_i$. The objective function becomes $x_1^2+1/x_1^2+x_2^2+1/x_2^2-2x_1x_2-2/(x_1x_2)=(x_1-x_2)^2+(1/x_1-1/x_2)^2=(x_1-x_2)^2[1+1/(x_1x_2)^2]$. Now apply AM-GM for the first factor and once again for what remains. $\endgroup$
    – plop
    Commented May 8, 2021 at 12:21
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    $\begingroup$ I know other method to solve this problem infact I've solved this using AM>=GM it self but with some modifications. I want to know Only mistake in my solution $\endgroup$
    – mathophile
    Commented May 8, 2021 at 16:43
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    $\begingroup$ AM-GM inequality is correct, but that doesn't mean it gives the minimum always. It gives a minimum when you can find allowable variables for the LHS which makes the LHS equal to the constant you get on the RHS. In this case there are no such points which make LHS = RHS, so the RHS value is just a lower bound, not the minimum. $\endgroup$
    – Macavity
    Commented May 8, 2021 at 17:15
  • $\begingroup$ @user710599 In my comment I told you the mistake, which is not having imposed the constraints $x_iy_i=1$. By not imposing them, you are solving an optimization problem with a larger set of feasible solutions. Therefore, the more relaxed problem has solutions that are not solutions of the original problem. In your case, this is the point $x_1=x_2=y_1=y_2=0$. All other aspects of the argument are fine, but they solve a different problem. $\endgroup$
    – plop
    Commented May 8, 2021 at 17:38

3 Answers 3

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Applying AM-GM, you have to consider if the equality can hold. If not, you only get a strictly lower bound rather than the minimum.

We may use AM-GM in this way: $$x_1^2 + y_1^2 + x_2^2 + y_2^2 + (- x_1x_2) + (- x_1x_2) + (- y_1 y_2) + (- y_1y_2) \ge 8\sqrt[8]{x_1^4 x_2^4 y_1^4 y_2^4} = 8$$ with equality if $x_1 = y_1 = 1, \ x_2 = y_2 = -1$.

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  • $\begingroup$ I know this method. Even I've solved using this method. I just wanted to Know that why $x_1=-2x_2$ and $x_2=-2x_1$ and $y_1=-2y_2$ and $y_2=-2y_1$ didn't work. How do i know that I've Solved Incorrectly. $\endgroup$
    – mathophile
    Commented May 9, 2021 at 3:06
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    $\begingroup$ @user710599 Because in your way, AM = GM does not hold given the conditions, so you know it is incorrect? $\endgroup$
    – River Li
    Commented May 9, 2021 at 3:13
  • $\begingroup$ i think when I'll multiply, I'll obtain the result $x_1*y_1=4y_2*x_2$ that is $1=4$ which will be incorrect $\endgroup$
    – mathophile
    Commented May 9, 2021 at 3:13
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    $\begingroup$ @user710599 You have said: $x_1 = -2 x_2$ and $x_2 = -2x_1$ give $x_1 = -2x_2 = -2(-2x_1)$ which results in $x_1 = 0$ which is impossible. $\endgroup$
    – River Li
    Commented May 9, 2021 at 3:16
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The objective function you wish to minimize is $$ x_1^2+x_2^2+y_1^2+y_2^2-2x_1x_2-2y_1y_2=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}^2, $$ which means that you wish to minimize the distance of two points on the curve $xy=1$ residing on the opposite branches $x,y>0$ and $x,y<0$. By intuition, we must have $x_1=y_1=1$ and $x_2=y_2=-1$ with a minimal distance of $2\sqrt 2.$

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    $\begingroup$ i know this method I've already solved using this method but my question is different. I don't need solution i need mistake in my approach $\endgroup$
    – mathophile
    Commented May 8, 2021 at 16:41
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$\min(x_1^2+x_2^2-2x_1x_2+y_1^2+y_2^2-2y_1y_2)=\min((x_1-x_2)^2+(y_1-y_2)^2)$,

We also have $y_i=\frac{1}{x_i}$ for $i=1,2$, subbing in the equation above we have

$\min\left(\left(x_1-x_2\right)^2\left(\frac{1}{x_1^2x_2^2}+1\right)\right)$. Then just use lagrange multiplies or Cauchy-Schwartz.

(P.S- from here on, it shouldn't actually be to hard to guess that the minimum occurs at $x_1=1,x_2=-1$ )

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    $\begingroup$ I know other method to solve this problem infact I've solved this using AM>=GM it self but with some modifications. I want to know Only mistake in my solution. $\endgroup$
    – mathophile
    Commented May 8, 2021 at 16:42

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