**I found an inequality as following:** Let $x, y, z$ be three complex numbers then:

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

The inequality holds with equality if and only if $x+y+z=0$

>**My question:** I am looking for a proof of conjecture as following:

> Let $x, y, z$ in an inner product space $V$ then

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

> where the norm ||z|| denotes the norm induced by the inner product

**See also** 

* [Hlawka's inequality][1] 
* [Absolute value inequality for complex numbers][2]


[1]: http://mathworld.wolfram.com/HlawkasInequality.html
[2]: http://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers/167741#167741