If there exists $x_i>n$, this means that there is an element $j$ of the set $\{1,2,...,n\}$ not equal to any of the $x_i$. Now consider $x_i^2-\frac{2n+1}{3}x_i>j^2-\frac{2n+1}{3}j$. This implies that $x_1^2+x_2^2+\cdots + x_n^2-\frac{2n+1}3( x_1+x_2+\cdots+x_n)$ is minimized when $x_i$ is a permutation of $(1,2,..n)$. The conclusion follows then.

This problem is posted in section Solved problems and put in Old and New ineqs

Nttu could you post the topic of this ineq?

Let $x_1<x_2<\cdots<x_n$ without loss of generality. Set $x_i = y_i +i$. Check that since $x_{i+1} = y_{i+1}+i+1 \ge x_i+1 = y_i+i+1$, we have $0\le y_1\le \cdots \le y_n$. Equipped with this, we next observe \[ \sum_{1\le i\le n}x_i^2 = \sum_{1\le i\le n}y_i^2 + 2\sum_{1\le i\le n}iy_i + \frac{n(n+1)(2n+1)}{6}. \]Likewise \[ \frac{2n+1}{3}\sum_{1\le i\le n}x_i = \frac{n(n+1)(2n+1)}{6} + \frac{2n+1}{3} \sum_{1\le i\le n}y_i. \]Hence, it boils down proving \[ \sum_{1\le i\le n}y_i^2+2\sum_{1\le i\le n}iy_i \ge \frac{2n+1}{3}\sum_{1\le i\le n}y_i. \]Now, using Chebyshev's sum inequality, we have \[ \sum_{1\le i\le n}i y_i \ge \frac13 \left(\sum_{1\le i\le n}i \right)\left(\sum_{1\le i\le n}y_i\right). \]Hence \[ \sum_{1\le i\le n}y_i^2+2\sum_{1\le i\le n}iy_i \ge\sum_{1\le i\le n}y_i^2+ \frac{n(n+1)}{6} \sum_{1\le i\le n}y_i. \]With this, it boils down checking \[ \sum_{1\le i\le n}y_i^2 + \frac{(n-1)(n-2)}{6}\sum_{1\le i\le n}y_i \ge 0, \]which is obvious. The equality holds only when (a) either $n=1,2$ or (b) $n\ge 3$ and $y_i=0$, that is $(x_1,\dots,x_n)$ is a permutation of $(1,2,\dots,n)$.