How to show that for any real numbers $x,y,z$ $|x|+|y|+|z|\le|x+y-z|+|y+z-x|+|z+x-y|?$

Write $x=a+b$, $y=b+c$, $z=c+a$. Then the following equivalent inequality is clear.

$$|a+b|+|b+c|+|c+a|\le 2|a|+2|b|+2|c|$$

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: $$2(|x+y-z|+|y+z-x|+|z+x-y|)\ge 2(|x|+|y|+|z|)$$ $$\implies |x+y-z|+|y+z-x|+|z+x-y|\ge |x|+|y|+|z|$$ $$\implies |x|+|y|+|z|\le |x+y-z|+|y+z-x|+|z+x-y|.$$