How to use triangle inequality to establish Reverse triangle inequality

I need to use $|a+b| \leq |a|+|b|$ to show that $||a|-|b|| \leq |a-b|$ .

I have tried to represent $||a|-|b||$ as $||a|+(-|b|)|$ , and then get $||a|+(-|b|)| \leq |a|+|-|b||$ , but that isn't leading me anywhere given $|a-b| \leq |a|+|b|$.

• Is this homework? – davidlowryduda Jun 10 '11 at 7:52
• @mixedmath: no, I would have marked it as such if that was the case. This is me solving Abbot's Understanding Analysis for fun over the summer. – confused Jun 10 '11 at 7:52
• I would still probably prefer hints over a complete answer though, given the nature of the problem... – confused Jun 10 '11 at 7:53
• No problem! ${}{}{}{}$ – davidlowryduda Jun 10 '11 at 7:54
• See also the related question math.stackexchange.com/q/193938/11994. – Marnix Klooster Jan 25 '14 at 21:04

HINT: supposing $x \geq y$, consider that $x = x - y + y$.

• Actually, I suppose that $|x| \geq |y|$. ;p – davidlowryduda Jun 10 '11 at 7:56
• hm, so by triangle inequality, I get $|a-b+b| \leq |a-b|+|b|$ which simplifies into $|a|-|b| \leq |a-b|$, while I need to prove that $||a|-|b|| \leq |a-b|$ -- am I missing something? – confused Jun 10 '11 at 8:13
• @confused: you're absolutely right. I let x > y, and you can generalize this (what if x < y? y = y - x + x, but that's the same case with different names, so it's a WLOG thing once you see it). Does that make sense? Alternatively, the way to take the maximum of $|x| - |y|$ and $|y| - |x|$ is to take its absolute value. – davidlowryduda Jun 10 '11 at 8:17
• so by showing both cases I can take the abs. value of the equation, since the absolute value is the max, and we have shown that even the max of the expression obeys $\leq |a-b|$? Sorry, I am still trying to develop the intuition for abs. values. – confused Jun 10 '11 at 18:45
• @confused: Yes, that's about right. – davidlowryduda Jun 10 '11 at 19:20

The answer is quite easy:

$|a-b|+|b|\geq |a|$

$|b-a|+|a|\geq |b|$

Then $|a-b| \geq \max\{|a|-|b|,|b|-|a|\}=||a|-|b||$.

This argument is quite standard and applies in proving the continuity of norms.

• never mind, got it. :) – confused Jun 10 '11 at 8:15