Proving an inequality involving absolute values

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.

• Yes, $\triangle$ inequality is enough. You need to show your efforts so far, though.. – Macavity May 14 at 23:19

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.

• Thanks @Izralbu – user587389 May 15 at 7:41

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.

• What is Karamata? – user587389 May 19 at 11:20
• Search this in the net. If would be troubles, so in evening I'll explain. I am very busy now. – Michael Rozenberg May 19 at 11:52