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

\begin{equation*} \frac{1}{2}(|y+z-x|+|x+z-y| + |y+x-z|) \le |x| + |y|+|z|+\frac{1}{2}|x+y+z| \end{equation*} (1)

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

**Note that:** *I have a proof of the inequality (1).*

>**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*}\frac{1}{2}(\|y+z-x\|+\|x+z-y\| + \|y+x-z\|) \le \|x\| + \|y\|+\|z\|+\frac{1}{2}\|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]: https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers/167741#167741