Inequality for absolute values How do you show either of the equivalent inequalities:
$$2(|a|+|b|+|c|)\leq |a+b+c|+|a+b-c|+|a-b+c|+|a-b-c|$$
or
$$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$
Hold for complex numbers or in $n$ dimensions ?
See 
https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers
where the same question is asked and the comments show that a proof must involve inner product properites of $\mathbb{C}$.
Further the mathoverflow comments of Bill Johnson give a proof in great generality using sophisticated Banach space techniques.
An elementary proof however would still be welcome. 
Before this question was edited, a number of correct proofs for real values have been given using a case analysis technique, which, it must be remarked, form the base case for the arguments in mathoverflow. 

Once the inequality is proved in ${\bf R}$,
it follows in any inner-product space by writing each $|z|$
as a multiple of the average of $|u \cdot z|$ over unit vectors $u$.
In ${\bf C}$ the formula is
$$
|z| = \frac14 \int_0^{2\pi} \bigl| {\rm Re}(e^{i\theta} z) \bigr| \, d\theta.
$$
Applying this to $z=a$, $b$, $c$, and $a \pm b \pm c$ reduces
the desired inequality to the one-dimensional case.
(Abridged from 
my answer
in mathoverflow.)
