Minimum Value of $x_1^2+y_1^2+x_2^2+y_2^2-2x_1x_2-2y_1y_2$

Question

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?

Answer 1

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 $x_1 = y_1 = 1, \ x_2 = y_2 = -1$.

Answer 2

$\min(x_1^2+x_2^2-2x_1x_2+y_1^2+y_2^2-2y_1y_2)=\min((x_1-x_2)^2+(y_1-y_2)^2)$,

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 $x_1=1,x_2=-1$ )

Answer 3

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 $2\sqrt 2.$