Triangle Inequality with complex numbers: Prove that ||x|−|y||≤|x|-|y|.

Question

Prove that $ ||x| - |y|| \le |x| - |y| $ for all $ x,y \in \mathbb{C} $.

I fully understand the other inequality: $|x+y| \le |x|+|y| $ for all $ x,y \in \mathbb{C} $.

But I have no clue how to start this one.

Any help will be greatly appreciated.

Thanks in advance!

See also this question and the questions linked there. – Martin Sleziak Sep 09 '13 at 10:16 — Martin Sleziak, Sep 09 '13 at 10:16

score 5 · Answer 1 · edited Apr 13 '17 at 12:20

First off, the assertion $\vert \vert x \vert - \vert y \vert \vert \le \vert x \vert - \vert y \vert$ is false: to see this, just choose $\vert y \vert > \vert x \vert$. I think what you want is $ \vert \vert x \vert - \vert y \vert \vert \le \vert x - y \vert$, as was stated and correctly demonstrated by Citizen in his/her answer.

Having said these things, and to avoid re-writing what is essentially the same answer over and over, you might check out my answer to this question:

How can I derive this expression related to the triangle inequality?

Hope this helps! Cheers!

score 4 · Accepted Answer · answered Sep 09 '13 at 08:01

4

Notice $$|x| = |x + y - y| \leq |x - y| + |y| \implies |x| - |y| \leq |x-y| $$

By the same reasoning, you can do the same with $y$ instead of $x$, and obtain the desired result, the reversed triangle inequality.

answered Sep 09 '13 at 08:01

ILoveMath

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thanks! That was fast and really helpful. God bless! – PandaMan Sep 09 '13 at 08:02

Triangle Inequality with complex numbers: Prove that ||x|−|y||≤|x|-|y|.

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