Proving an inequality involving absolute values

Question

How can I prove the inequality

$\left|x\right|+\left|y\right|+\left|z\right|\le\left|x+y-z\right|+\left|y+z-x\right|+\left|z+x-y\right|$

for all $x, y, z$ being real number.

Can I prove this by using triangle inequality? Or do I have to use some other technique? Please help me solve this problem. Thanks in advance.

Answer 1

Note that

$$|2x|=|(x+y-z) + (z+x-y)| \le |x+y-z| + |z+x-y|$$ $$|2y|=|(x+y-z) + (y+z-x)| \le |x+y-z| + |y+z-x|$$ $$|2z|=|(y+z-x) + (z+x-y)| \le |y+z-x| + |z+x-y|$$

The result follows from adding up the inequalities.

Answer 2

Let $x\geq y\geq z$.

Thus, $$(x+y-z,x+z-y,y+z-x)\succ(x,y,z)$$ and since $f(x)=|x|$ is a convex function, our inequality it's just Karamata.