With $x_{1},x_{2},...,x_{n}\in[-1;1]$ and $x_{i}\neq x_{j}$. Prove $\sum_{i=1}^{n}\frac{1}{\prod_{j=1,j\neq i}^{n}|x_{i}-x_{j}|}\ge 2^{n-2}$
The solution must begin from lagrange interpolation but I do not know how to prove it.
I thinhk may be if we define $a_{i}\in \mathbb{R}$ then we will have a new polynomial but it seem not working.
Start with $T_{n-1}(x)$, the Chebyshev Polynomial of the First Kind.
From Lagrange Interpolation, we have $$T_{n-1}(x)=\sum_{i=1}^n T_{n-1}(x_i) \prod_{j \not= i} \frac{x-x_j}{x_i-x_j}$$
By Triangle Inequality, we have $$|T_{n-1}(x)| \le \sum_{i=1}^n |T_{n-1}(x_i)| \prod_{j \not= i} |\frac{x-x_j}{x_i-x_j}|$$
Dividing by $x^{n-1}$, we have $$|\frac{T_{n-1}(x)}{x^{n-1}}| \le \sum_{i=1}^n \frac{|T_{n-1}(x_i)|}{\prod_{j \not= i } |x_i-x_j|} \prod_{j \not= i} |1-\frac{x_j}{x}|$$
Now let $x \rightarrow \infty$. We now have (since the leading coefficient of $T_{n-1}(x)$ is $2^{n-2}$) $$2^{n-2} \le \sum_{i=1}^n \frac{|T_{n-1}(x_i)|}{\prod_{j \not= i } |x_i-x_j|} \le \sum_{i=1}^n \frac{1}{\prod_{j \not= i} |x_i-x_j|}$$ as desired. $\blacksquare$