$\ \forall x_1,x_2,...,x_n \in \mathbb{R} (x_i\not=x_j)$ in the range of $[-1,1]$ prove:$\sum_{i=1}^{n}\frac{1}{\Pi_{k\not=i}|x_k-x_i|}\ge2^{n-2}$

Question

$\ \forall$ $x_1,x_2,...,x_n$ $\in \mathbb{R}$ $(x_i\not=x_j)$ in the range of $[-1,1]$ prove : $$\sum_{i=1}^{n}\frac{1}{\Pi_{k\not=i}|x_k-x_i|}\ge2^{n-2}$$

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my attempt :

$$p(x) = \sum_{i=1}^{n}\left(p(x_i)\prod_{k\not=i}\frac{x-x_k}{x_i-x_k}\right)$$

by triangle inequality :

$$\left|p(x)\right| \leq \sum_{i=1}^{n}\left|p(x_i)\right|\prod_{k\not=i}\bigg|\frac{x-x_k}{x_i-x_k}\bigg|$$

now if $p(x) = \sum_{i=0}^{n-1}a_ix^i$ then :

$$\left|\frac{p(x)}{x^{n-1}}\right|\leq\sum_{k=1}^{n}\frac{\left|p(x_k)\right|}{\prod_{k\not=j}\left|x_k-x_j\right|}\Bigg|\prod_{k\not=j}(1-\frac{x_i}{x})\Bigg|$$

then if we $x\to\infty$ we get :

$$\left|a_{n-1}\right|\leq\sum_{k=1}^{n}\frac{\left|p(x_k)\right|}{\prod_{k\not=i}\left|x_k-x_i\right|}$$

Answer 1

One may follow a proof of Chebyshev's $\|\cdot\|_\infty$-minimality.

Suppose $j<k\implies x_j<x_k$ (w.l.o.g.). Denote our sum $S=\sum_{k=1}^n 1/\prod_{j\neq k}|x_k-x_j|$.

Assume $S<2^{n-2}$, and consider the polynomial $$P(x)=\sum_{k=1}^n(-1)^{n-k}\sum_{j\neq k}\frac{x-x_j}{x_k-x_j}.$$ Then $P(x_k)=(-1)^{n-k}$ for each $k$, and the leading coefficient of $P$ is exactly $S$.

Now, as in the proof, the degree of the polynomial $$F(x)=\frac{P(x)}{S}-\frac{T_{n-1}(x)}{2^{n-2}}$$ is less than $n-1$ (since the coefficient of $x^{n-1}$ vanishes), but $F$ must have at least $n-1$ zeros (since the sequence $k\mapsto F(x_k)$ changes sign; this follows from $|T_{n-1}(x_k)|\leqslant 1$ and the assumption $S<2^{n-2}$).

(Thus, $F$ must be identically zero, which contradicts the assumption.)