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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.

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

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  • $\begingroup$ Thanks @Izralbu $\endgroup$ – user587389 May 15 at 7:41
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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.

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

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