For any given $x\in\mathbb{R}^n$, let $\nabla{x}\in\mathbb{R}^{{n}\choose{2}}$ be the vector that its ${\{i,j\}}$th entry is $|x_i-x_j|$. I think the following claim is true. 

>**Claim.** If $f,g\in\mathbb{R}^n$ are vectors with zero mean (i.e. $\sum_{i}{f_i}=\sum_{i}{g_i}=0$) and the angle between them is at most $\pi/2$, then $dist(\nabla{f},\nabla{g}) \ge dist(f,g)$. 

If anybody has any idea about how to approach to prove this, please share it with me.

Thanks