How to find this minimum of the value 
Let $x_{i}$, where $i\in\{1,2,\cdots,n\}$ be distinct real  numbers. Find the minimum of the value of 
  $$\sum_{1\le i<j\le n}\left(\dfrac{1-x_{i}x_{j}}{x_{i}-x_{j}}\right)^2$$

It is  clear when $n=2$ minimum of the value is $0$, when $x_{1}x_{2}=1$. But for $n\ge 3$,I can't find it. 
Thanks.

We begin with a short proof for the case $n=3$ :
We make a substitution so we put :
$\frac{\lambda_1}{2}(\frac{1+(a-b)^2}{a-b})^2=(\frac{1-xy}{x-y})^2$
$\frac{\lambda_2}{2}(\frac{1+(b-c)^2}{c-b})^2=(\frac{1-zy}{z-y})^2$
$\frac{\lambda_3}{2}(\frac{1+(a-c)^2}{a-c})^2=(\frac{1-xz}{x-z})^2$
With $a,b,c$ and the condition $a\geq b \geq c$ real numbers and $\lambda_1,\lambda_2,\lambda_3$ real positive numbers with the condition $\lambda_1+\lambda_2+\lambda_3=1$
So we have to find the minimum value of :
$$\frac{\lambda_1}{2}(\frac{1+(a-b)^2}{a-b})^2+\frac{\lambda_2}{2}(\frac{1+(b-c)^2}{c-b})^2+\frac{\lambda_3}{2}(\frac{1+(a-c)^2}{a-c})^2$$
We could apply Jensen's inequality for $f(x)=\frac{1}{2}(\frac{1+x^2}{x})^2$ it becomes :
$$\frac{1}{2}[\lambda_1(\frac{1+(a-b)^2}{a-b})^2+\lambda_2(\frac{1+(b-c)^2}{c-b})^2+\lambda_3(\frac{1+(a-c)^2}{a-c})^2]\geq \frac{1}{2}[(\frac{1+(\lambda_1(a-b)+\lambda_2(b-c)+\lambda_3(a-c))^2}{((\lambda_1(a-b)+\lambda_2(b-c)+\lambda_3(a-c))})^2]$$
But $\lambda_1(a-b)+\lambda_2(b-c)+\lambda_3(a-c)>0$
So the minimum of the function $f(x)$ is 2 .
