TITLE: How to find this minimum of the value
QUESTION [15 upvotes]: 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.

REPLY [2 votes]: I have found the following nice proof for $n=3$.
I believe that it will help in the general case.
Indeed, let $x_1=a$, $x_2=b$ and $x_3=c$.
Hence, $$\sum_{cyc}\left(\frac{1-ab}{a-b}\cdot\frac{1-bc}{b-c}\right)=\frac{\sum\limits_{cyc}(1-ab)(1-bc)(c-a)}{\prod\limits_{cyc}(a-b)}=$$
$$=\frac{\sum\limits_{cyc}(a^2b-a^2c)}{\prod\limits_{cyc}(a-b)}=-1.$$
Id est,
$$\sum_{cyc}\frac{(1-ab)^2}{(a-b)^2}=\left(\sum_{cyc}\frac{1-ab}{a-b}\right)^2-2\sum_{cyc}\left(\frac{1-ab}{a-b}\cdot\frac{1-bc}{b-c}\right)=$$
$$=\left(\sum_{cyc}\frac{1-ab}{a-b}\right)^2+2\geq2.$$