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.

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This is not a full solution, but an upper bound to the minimum for given $n$ is given by:
$$M_n = \text{Minimum}(n) \leq \left\{ \begin{array}{ll} \frac{1}{2} n(n-2) & \text{for n even} \\ \frac{1}{2} (n-1)^2 & \text{for n odd} \end{array} \right.$$
I will first show that for a particular set of numbers $\{x_i\}$ these values would be an under limit to the sum and then give an example of such a set. I also will assume that all values are strictly positive $x_i > 0$.

As a short-hand notation let for $x \neq y$ the function $f(x,y)$ be given by
$$f(x,y) = \left( \frac{1 – x y}{x-y} \right)^2$$
Then it follows that for $x \neq \frac{1}{y}$
$$f(x,\frac{1}{y}) = \left( \frac{1 – \frac{x}{y}}{x-\frac{1}{y}} \right)^2 = \left( \frac{y -x}{x y-1} \right)^2 = \frac{1}{f(x,y)}$$
and hence we also have $f(x,y)=f(\frac{1}{x},\frac{1}{y})$ and $f(x,\frac{1}{y})=f(\frac{1}{x},y)$. Furthermore we have both $f(x,\frac{1}{x}) = 0$ and $f(x,1)=1$ for $x \neq 1$ as well as the inequality $f(x,y)+f(x,\frac{1}{y}) \geq 2$.

To combine the even an odd cases of $n$, I will set $n = 2 k + \sigma$, with $k \geq 1$ and $\sigma \in \{0,1\}$.

Now consider a set $\{x_i\}$ that satisfies
$$\begin{array}{ll} 1 <x_1 < x_2 < \dots < x_k \\ x_{k+i} = \frac{1}{x_i} & \text{for 1 \leq i \leq k}\\ x_n = 1 & \text{if n is odd} \end{array}$$
Then we find for the sum
$$\sum_{1\leq i<j\leq n} \left( \frac{1 – x_i x_j}{x_i-x_j} \right)^2 = \sum_{1\leq i<j\leq k} \left[ f(x_i,x_j) + f(x_i,x_{k+j}) + f(x_{k+i},x_j) + f(x_{k+i},x_{k+j})\right] \\ + \sum_{1\leq i \leq k} f(x_i,x_{k+i}) + \sigma \sum_{1 \leq i \leq n} f(x_i,1) \\ = 2 \sum_{1\leq i<j\leq k} \left[ f(x_i,x_j) + f(x_i,\frac{1}{x_j}) \right] + \sum_{1\leq i \leq k} f(x_i,\frac{1}{x_i}) + \sigma n \\ \geq 4 \binom{k}{2} + \sigma n = 2 k (k-1+\sigma)$$
which are the limits mentioned above.

As an example for such a set consider $x_i=1 + \delta^i$ for $1 \leq i \leq k$. Then all elements are distinct, in particular for $0<\delta \ll 1$.

The only contribution to the sum that we have to consider is:
$$f(x_i,x_j) + f(x_i,\frac{1}{x_j}) = 2 + {\cal O}(\delta^2) > 2$$ and hence we can get arbitrary close to the limit obtained.

Equality only holds when these terms do not exist, which is in the case $n=2$ and $n=3$, whereas for $n>4$ this type of set can not reach the lower limit.

It might be possible that other sets of $\{x_i\}$ provide lower values for the sum, but a numerical minimisation of the sum for various $n$ did not give me any.

We can also give a lower bound for $M_n$. If we consider the same
$$S_n(\{x_i\}) = \sum_{1 \leq i<j \leq n} \left( \frac{1 – x_i x_j}{x_i-x_j} \right)^2 \geq M_n$$
we can also write this as a some over partitions of the $n$ values, in particular partitions of $n-1$ elements.
$$S_n(\{x_i\}) = \frac{1}{n-2} \sum_{k=1}^n S_{n-1}(\{x_i\}_{i\neq k})$$
The factor $\frac{1}{n-2}$ is from over counting and follows directly from realising that each pair $(x_i,x_j)$ appears $n-2$ times in the various sums.
From this it immediately follows that
$$M_n \geq \frac{n}{n-2} M_{n-1}$$
Since we already know $M_2=0$ and from Michaels result that $M_3=2$, we can by repeatedly applying this lower bound obtain
$$M_n \geq \frac{n(n-1)}{6} M_3 = \frac{n(n-1)}{3} ~~~~ \forall n \geq 4$$
This gives $M_4 \geq 4$ and with the constructed solution above we find
$$S_4(\{x_i\}) – 4 < \delta ~~~ \forall \delta >0$$
In all other cases $n\geq 5$ the upper and lower bound are separated.

Not a full proof, but perhaps somebody else can use this result to continue.

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 .

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,
$$\sum_{cyc}\frac{(1-ab)^2}{(a-b)^2}=\left(\sum_{cyc}\frac{1-ab}{a-b}\right)^2-2\sum_{cyc}\left(\frac{1-ab}{a-b}\cdot\frac{1-bc}{b-c}\right)=$$
$$=\left(\sum_{cyc}\frac{1-ab}{a-b}\right)^2+2\geq2.$$