# Almost the reverse triangle inequality

I'm wondering if one can get such an inequality (for $n$ a positive integer) : $$\exists C>0\ \forall(x,y,u,v)\in \left(\mathbb{R}^n\right)^4,\quad \left| \left|x-y \right|^2-\left|u-v \right|^2 \right|\le C\left( \left|x-u \right|^2+\left|y-v \right|^2 \right)$$

I've tried using the fact that $|x-y|^2-|u-v|^2 = \left( |x-y|+|u-v|\right)\left( |x-y|-|u-v|\right)$ but I've got nothing...

Thanks a lot.

Suppose $x,u,v,y$ are colinear and displayed in order, and suppose $|u-v|=a$ together with $|x-u|=|y-v|=b$ are both alternative. Thus $|x-y|=a+2b$. Now your inequality is $$4ab+4b^2=(a+2b)^2-a^2\leq 2Cb^2$$ which is obviously invalid for all $a,b$.