Minimum of $\sum_{i=1}^n \frac{1}{\prod_{j\ne i} |x_j - x_i|}$ for $x_1,...,x_n \in \mathbb [-1, 1]$ Let's consider $x_1,\cdots,x_n \in \mathbb [-1, 1]$. I want to find minimum value of:
$$\frac{1}{|x_1-x_2|\cdots|x_1 - x_{n}|} + \frac{1}{|x_2-x_1||x_2-x_3|\cdots|x_2 - x_n|}+\cdots+\frac{1}{|x_n-x_1|\cdots|x_n-x_{n - 1}|}$$
i.e. in each denominator we have a product of $n-1$ differences, where we omit the difference with the same indices i.e. $x_j - x_j$.
My work so far
Without lose of generality we can say, that $x_1>x_2>...>x_n$, and therefore:
$$\frac{1}{(x_1-x_2)\cdots(x_1-x_n)} + \frac{1}{(x_1-x_2)(x_2-x_3)\cdots(x_2 - x_n)} + \frac{1}{(x_1 - x_n)\cdots(x_{n-1}-x_n)} =:K$$
What can be noticed, is that we obtain pretty bound, when applying AM-GM inequality:
$$K \ge n \frac{1}{\sqrt{(x_1-x_2)^2(x_1 - x_3)^2\cdots(x_n-x_{n-1})^2}} $$
And now we can use fact that, since $x_1,\cdots,x_n \in [-1, 1]$, then, for $i \neq j, x_i > x_j$:
$$|x_i - x_j| \le 2 \Rightarrow \frac{1}{(x_i-x_j)^2} \ge \frac 1 4 \Rightarrow \frac{1}{\sqrt{(x_i - x_j)^2}} \ge \frac 1 2$$
The number of terms under the root equals $\frac{(n - 1) \cdot n}{2}$, because in each difference we have $(n - 1)$ expressions, and we have $n$ sums. Also each square will be present only once. So finally, we obtain the bound:
$$K\ge n \cdot \frac{1}{2^{\frac{n(n-1)}{2}}}$$
And unfortunately, this bound is too weak to be a lower bound for each $n$. For example here The smallest value of sum of reversed absolute values we can find, that the minimum for this problem, for $n = 3$, equals $2$, and my bound suggests $\frac{3}{8}$. Do you know how my solution can be repaired to be accurate for all $n \in \mathbb N$?

Some thoughts:
Since the expression is symmetric, WLOG, assume that
$x_1 > x_2 > \cdots > x_n$.
Let
$$f_i = \prod_{j\ne i} |x_j - x_i|, \, i = 1, 2, \cdots, n.$$
Using AM-GM, we have
\begin{align*}
 \sum_{i=1}^n \frac{1}{f_i} &= 
 \frac{1}{f_1} + \sum_{i=2}^{n-1} \left(\frac{1}{2f_i} + \frac{1}{2f_i}\right) + \frac{1}{f_n}\\
 &\ge (2n - 2)\left(2^{2(n - 2)}f_1 f_n \prod_{i=2}^{n-1} f_i^2\right)^{-1/(2n - 2)}.
\end{align*}
Conjecture 1: We have
$$f_1 f_n \prod_{i=2}^{n-1} f_i^2 \le (n - 1)^{2n - 2}2^{-2n^2 + 6n - 2}$$
with equality if
\begin{align*}
 f_1 &= \frac{n - 1}{2^{n - 3}}, \\
 f_i &= \frac{n - 1}{2^{n - 2}}, \quad i = 2, 3, \cdots, n - 1;\\
 f_n &= \frac{n - 1}{2^{n - 3}}.
\end{align*}
If Conjecture 1 is true, the minimum of $\sum_{i=1}^n \frac{1}{f_i}$ is $2^{n - 2}$.

Remarks: I evaluated the minimum for $n=3, 4, \cdots, 9$ which is $2^{n - 2}$, and the minimizer $(x_1, x_2, \cdots, x_n) \in [-1, 1]^n$ (with $x_1 > x_2 > \cdots > x_n$) satisfies the system of equations
\begin{align*}
 f_1 &= \frac{n - 1}{2^{n - 3}}, \\
 f_i &= \frac{n - 1}{2^{n - 2}}, \quad i = 2, 3, \cdots, n - 1;\\
 f_n &= \frac{n - 1}{2^{n - 3}}.
\end{align*}
Also, clearly, $x_1 = 1$ and $x_n = -1$.
