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. 

I will consider a different form of:
$$|x+y|+|x+z|+|y+z|\leq |x|+|y|+|z|+|x+y+z|$$
Note that $$\begin{align}|x+y+z| &\leq |x+y| + |z| \leq |x| + |y| + |z|\\&\leq |x+z| + |y| \leq |x| + |y| + |z|\\&\leq |y+z| + |x|  \leq |x| + |y| + |z|\end{align}$$
Summing the inequalities:
$$3|x+y+z| \leq |x+y|+|x+z|+|y+z| + |x| + |y| + |z|  \leq 3|x| + 3|y| + 3|z|$$
We will first use: $$3|x+y+z| \leq 3|x| + 3|y| + 3|z| \qquad \qquad\ \ \ \  \ \ \ \ \ (1)$$
And then $$|x+y|+|x+z|+|y+z| + |x| + |y| + |z|  \leq 3|x| + 3|y| + 3|z| \qquad\ \ \ \  \ \ \ \ \ (2)$$
Now I will prove that:
$$3|x+y|+3|x+z|+3|y+z|\leq 3|x|+3|y|+3|z|+3|x+y+z|$$
Using the first inequality $(1)$ we find that
$$3|x+y|+3|x+z|+3|y+z|\leq 6|x|+6|y|+6|z|$$
$$|x+y|+|x+z|+|y+z|\leq 2|x|+2|y|+2|z|$$
Add $|x| + |y| + |z|$ to both sides to find
$$|x+y|+|x+z|+|y+z| + |x| + |y| + |z|  \leq 3|x| + 3|y| + 3|z|$$
Which is true from above $(2)$.
