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There is a following inequality

$$\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}{n}\geq\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}+2x_{1}x_{2}+...+2x_{1}x_{n}+2x_{2}x_{3}+...+2x_{2}x_{n}+...+2x_{n-1}x_{n}}{n^{2}}.$$

In my opinion, it is held for all $x_i>0$. It can be rewritten to the following form \begin{eqnarray*} \frac{n-1}{n}\frac{x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2}}{n} & \geq & \frac{2x_{1}x_{2}+...+2x_{1}x_{n}+2x_{2}x_{3}+...+2x_{2}x_{n}+...+2x_{n-1}x_{n}}{n^{2}},\\ x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} & \geq & \frac{2x_{1}x_{2}+...+2x_{1}x_{n}+2x_{2}x_{3}+...+2x_{2}x_{n}+...+2x_{n-1}x_{n}}{\left(n-1\right)}. \end{eqnarray*}

However, how to proof the last inequlity? I used \begin{eqnarray*} x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} & \geq & \frac{2}{n-1}\frac{n!}{\left(n-2\right)!2!}E(x_{i}x_{j}),\\ x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2} & \geq & nE(x_{i}x_{j}) \end{eqnarray*}

which may not be correct. However, there must be a more rigorous and simple proof.

tomb
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  • How do you know that the inequality holds even when you multiply the RHS by $\frac{n-1}{n}<1$? – Hayden Mar 29 '14 at 13:48

2 Answers2

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The inequality can be rewritten as: \begin{align*} \sum_{i=1}^n x_i^2\geq \frac{(\sum_{i=1}^nx_i)^2}{n} \end{align*}

Let us fix $\sum_{i=1}^n x_i = c$. Then, the problem: \begin{align*} \mbox{minimize} &\sum_{i=1}^n x_i^2 \\ \mbox{subject to} &\; \\ &\sum_{i=1}^n x_i = c,\\ & x_i \geq 0, \end{align*} has the solution $c^2/n$, when $x_1=x_2=...=x_n=c/n$. Now, it follows that \begin{align*} \sum_{i=1}^n x_i^2 \geq \frac{c^2}{n} = \frac{(\sum_ i x_i)^2}{n} \end{align*}

user137846
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Expand

$$ \sum_{i<j} (x_i - x_j)^2.$$

Calvin Lin
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