I asked this question on stackexchange, but despite much effort on my part have been unsuccesful in finding a solution.


Does the inequality 
$$2(|a|+|b|+|c|) \leq |a+b+c|+|a+b-c|+|a+c-b|+|b+c-a|$$
hold for all complex numbers $a,b,c$ ?
For real values a case analysis will verify the inequality. 
What is desired is a proof using the triangle inequality or a counterexample.
Thanks in advance. 



Absolute value inequality for complex numbers
It seems that your inequality is just an incarnation of [Hlawka's inequality](http://mathworld.wolfram.com/HlawkasInequality.html) 
which says that for any vectors $x, y, z$ in an inner product space $V$ we have


\begin{equation\*}
 \|x+y\| + \|y+z\|+\|z+x\| \le \|x\|+\|y\| + \|z\| + \|x+y+z\|.
\end{equation\*}


Using $x=a+b-c$, $y=a+c-b$, and $z=b+c-a$ we obtain the inequality in the OP.




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**Additional remarks:**


To add some more context, please see the [paper linked here](http://link.springer.com/article/10.1007/s00010-012-0178-2), which provides quite a nice summary of work related to Hlawka's inequality, which apparently stems back to a 1942 paper of Hornich (also cited by *Zurab* below). The paper linked to above explores the interesting generalization:
\begin{equation\*}
 f(x+y) + f(y+z) + f(z+x) \le f(x+y+z) + f(x)+f(y)+f(z),
\end{equation\*}
where $x,y,z$ may come from an Abelian group, or a linear space, or the real line---each with its own set of conditions on the mapping $f$. The functional form of Hlawka's inequality is credited to a 1978 paper of Witsenhausen.


