For any given $x \in \mathbb{R}^n$, let $\nabla{x} \in \mathbb{R}^{n \choose 2}$ be the vector whose $\{i,j\}$-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., $$\sum_{i=1}^n f_i = \sum_{i=1}^n g_i = 0$$ and the angle between them is at most $\frac{\pi}{2}$, then $$\operatorname{dist} (\nabla{f},\nabla{g}) \ge \operatorname{dist}(f,g).$$ 

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