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

$\min(x_1^2+x_2^2-2x_1x_2+y_1^2+y_2^2-2y_1y_2)=\min((x_1-x_2)^2+(y_1-y_2)^2)$,
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 $x_1=1,x_2=-1$ )
