If $0\le x_1\le\dots\le x_n\le1$, then $\sum\limits_{i=1}^n\left(x_i-\frac{i}{n+1}\right)^2\le\sum\limits_{i=1}^n\left(\frac{i}{n+1}\right)^2$

Question

Given positive integer $n$, $0\leq x_1 \leq \dots \leq x_n \leq 1$. Prove that $$ \sum_{i=1}^n \left(x_i- \frac{i}{n+1} \right)^2 \leq \sum_{i=1}^n \left(\frac{i}{n+1}\right) ^2 $$

I try to use transformation $x_i=\sum_{k=1}^i y_k$, prove the case when n=2, with a few discussions. It seems hard for $n\geq 3$. Any help is appreciated. Thx.

Answer 1

Hint: you can use the Chebyshev sum inequality to prove that $$ \frac{1}{n}\sum_{i=1}^n x_i \cdot \frac{i}{n+1} \geq \left(\frac{1}{n}\sum_{i=1}^n x_i\right) \cdot \left(\frac{1}{n}\sum_{i=1}^n \frac{i}{n+1}\right) = \frac{1}{2n} \sum_{i=1}^n x_i. $$