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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 $|a-b| \leq |a|+|b|$. Thanks!

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Is this homework? –  mixedmath Jun 10 '11 at 7:52
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@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
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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! ${}{}{}{}$ –  mixedmath Jun 10 '11 at 7:54
    
Thanks everyone for the answers! Too bad I can accept only one... –  confused Jun 10 '11 at 8:22

2 Answers 2

up vote 2 down vote accepted

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

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Actually, I suppose that $|x| \geq |y|$. ;p –  mixedmath 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. –  mixedmath 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. –  mixedmath 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.

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

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