TITLE: Stuck trying to prove an inequality
QUESTION [20 upvotes]: I have been trying to prove (the left half of) the following inequality:
$$ \underbrace{\sum_i \sum_j |x_i| \le \sum_i \sum_j |x_i + x_j|}_\textrm{?} \le 2 \sum_i \sum_j |x_i|$$
(All $x_i$s are arbitrary reals and sums are over $1, 2, \dots, n$)
The right half follows by a simple application of the triangle inequality, but the left half isn't as simple.
I tried induction, and I could reduce the above problem to proving the following statement:
$$ \sum_i |x_i + x_{n+1}| + \sum_{i=1}^{n+1} |x_i + x_{n+1}| \ge \sum_{i=1}^{n+1} |x_i| + n|x_{n+1}|$$
or more simply (replacing $x_{n+1}$ by $y$),
$$ 2\sum_i |x_i + y| \ge \sum_i |x_i| + (n-1)|y|$$
However, I don't know if the above "reduction" is any simpler than the original problem itself!
Can I continue the above line of thought, or is there an easier way to solve this problem? Any help is appreciated! :)

REPLY [13 votes]: Let $p$ be the number of $x_i\ge0$, $m$ be the number of $x_i<0$, and $n=p+m$.
$$
\small\begin{array}{}
&\sum_i\;\:\sum_j|x_i+x_j|\\
&=\sum_{x_i\ge0}\;\:\sum_{x_j\ge0}|x_i+x_j|
&+\sum_{x_i\ge0}\;\:\sum_{x_j<0}|x_i+x_j|
&+\sum_{x_i<0}\;\:\sum_{x_j\ge0}|x_i+x_j|
&+\sum_{x_i<0}\;\:\sum_{x_j<0}|x_i+x_j|\\
&\ge\sum_{x_i\ge0}\;\:\sum_{x_j\ge0}|x_i|+|x_j|
&+\left|\sum_{x_i\ge0}\;\:\sum_{x_j<0}|x_i|-|x_j|\right|
&+\left|\sum_{x_i<0}\;\:\sum_{x_j\ge0}|x_j|-|x_i|\right|
&+\sum_{x_i<0}\;\:\sum_{x_j<0}|x_i|+|x_j|\\
&=2p\sum_{x_i\ge0}|x_i|
&+2\left|m\sum_{x_i\ge0}|x_i|-p\sum_{x_i<0}|x_i|\right|
&&+2m\sum_{x_i<0}|x_i|
\end{array}
$$
Subtracting $\displaystyle\sum_i\;\:\sum_j|x_i|=n\sum_i\;|x_i|$ from the inequality above yields
$$
\small\begin{align}{}
&\sum_i\;\:\sum_j|x_i+x_j|-\sum_i\;\:\sum_j|x_i|\\
&\ge(p-m)\sum_{x_i\ge0}|x_i|
+\left|2m\sum_{x_i\ge0}|x_i|-2p\sum_{x_i<0}|x_i|\right|
+(m-p)\sum_{x_i<0}|x_i|\\
&=(p-m)\sum_{i}x_i
+\left|(m-p)\sum_{i}|x_i|+n\sum_ix_i\right|\\
&\ge0\tag{1}
\end{align}
$$
In the last inequality, if $p-m$ and $\sum\limits_ix_i$  have the same sign, we can ignore the quantity in absolute values.  If $p-m$ and $\sum\limits_ix_i$ have different signs, then the quantities in the absolute values have the same sign and $|m-p|\sum\limits_i|x_i|\ge(m-p)\sum\limits_ix_i$.
Therefore, $(1)$ says that
$$
\sum_i\;\:\sum_j|x_i|\le\sum_i\;\:\sum_j|x_i+x_j|
$$