TITLE: How prove this inequality: $\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|$?
QUESTION [6 upvotes]: Let $x_{1},x_{2},\cdots,x_{n}$ be real numbers. Show that

$$\sum_{i,j=1}^{n}|x_{i}+x_{j}|\ge n\sum_{i=1}^{n}|x_{i}|.$$

I think this problem may be solved using nice methods, but I can't find them yet; I know this may be of use:
$$|a|+|b|\ge |a+b|.$$
But I can't make it work. Thank you everyone.

REPLY [9 votes]: The following solution has been presented by one of the participants of Romanian team selection test. Essentially, the problem boils down to the following completing the square. Note that $|x_i|+|x_j|-|x_i+x_j|=0$ when $x_ix_j\ge 0$ and $=2\min{(|x_i|,|x_j|)}$ when $x_ix_j< 0.$
With this in hands, we can estimate
\begin{align}
&\phantom{=}\sum_{1\leq i,j\leq n}(|x_i|+|x_j|-|x_i+x_j|)
\le \sum_{x_ix_j< 0}2\min\{|x_i|,|x_j|\}\\
&=4\sum_{x_i>0>x_j}\min\{x_i,-x_j\}
\leq 4\sum_{x_i>0>x_j}\sqrt{-x_ix_j}\\
&\le 4\sum_{x_i>0}\sqrt{x_i}\sum_{0>x_j}\sqrt{-x_j}
\leq \left (\sum_{x_i>0}\sqrt{x_i}+\sum_{0>x_j}\sqrt{-x_j}\right )^2\\
&= \left (\sum_{1\leq k\leq n}\sqrt{|x_k|}\right )^2
\leq n\sum_{1\leq k\leq n}|x_k|.
\end{align}
And the result follows.