**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](http://mathworld.wolfram.com/HlawkasInequality.html)
* [Absolute value inequality for complex numbers](https://mathoverflow.net/questions/167685/absolute-value-inequality-for-complex-numbers/167741#167741)



An inequality in product space $V$
It follows from the 1-dimensional case which you say is true: project everything to a randomly chosen line, apply 1d case and integrate.


