TITLE: How to show that for any real numbers $x,y,z$ $|x|+|y|+|z|\le|x+y-z|+|y+z-x|+|z+x-y|?$
QUESTION [6 upvotes]: How to show that for any real numbers $x,y,z$$$|x|+|y|+|z|\le|x+y-z|+|y+z-x|+|z+x-y|?$$
I'm don't know how to split RHS.

REPLY [4 votes]: By Triangle Inequality we have: $$|x+y-z|+|y+z-x|\ge |(x+y-z)+(y+z-x)|=|2y|=2|y|\hspace{0.5cm} ...(1)$$
Similarly, we have:
$$|y+z-x|+|z+x-y|\ge 2|z|\hspace{0.5cm} ...(2)$$
$$|x+y-z|+|y+z-x|\ge 2|x|\hspace{0.5cm} ...(3)$$
Therefore, by adding $(1), (2)$ and $(3)$ we have: $$2(|x+y-z|+|y+z-x|+|z+x-y|)\ge 2(|x|+|y|+|z|)$$ $$\implies |x+y-z|+|y+z-x|+|z+x-y|\ge |x|+|y|+|z|$$ $$\implies |x|+|y|+|z|\le |x+y-z|+|y+z-x|+|z+x-y|.$$