# Minimum of $\sum_{i=1}^n \frac{1}{\prod_{j\ne i} |x_j – x_i|}$ for $x_1,…,x_n \in \mathbb [-1, 1]$

inequalitymaxima-minimaoptimizationreal-analysis

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 $$x_n = -1$$.