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Why exactly can you take the absolute value of one side of this inequality and assume it is still true?

Why is $||a|-|b|| \ge |a|-|b|$, tried a lot (like comparing to quadratics), but couldn't find the answer. Can anyone help me in the right direction?

Regards, Kevin

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marked as duplicate by Qiaochu Yuan Sep 22 '11 at 19:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2  
I think you want $||a|-|b||\leq|a-b|$. –  bobobinks Sep 22 '11 at 15:23
    
Thijs, Yes, you are correct. This question is not about reverse triangle inequality. I misread. –  Srivatsan Sep 22 '11 at 15:29
    
@Sri: The title is misleading, so I don't blame you :) –  TMM Sep 22 '11 at 15:36
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4 Answers

up vote 1 down vote accepted

Apply the common triangle inequality $$ \|x\| + \|y\| \geq \|x+y\| $$ to the variables $x=(|a|-|b|)$ and $y=|b|$

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no triangle inequality needed, its much simpler, much more basic! –  Tomas Sep 22 '11 at 15:55
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HINT: Writing $x = |a| - |b|$, the expression reduces to $|x| \geq x$...

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yes, this is much more elegant way of what I tried to say! +1! –  Tomas Sep 22 '11 at 15:54
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Thanks. Stupid me! It was indeed $||a|-|b||\le|a-b|$

I did it this way, is it OK?

$||a|-|b||^2=(|a|-|b|)^2=|a|^2+|b|^2-2|a||b|$

$=a^2+b^2-2|ab|$

$\le a^2+b^2-2ab$

$=(a-b)^2=|a-b|^2$

So:

$||a|-|b||^2 \le |a-b|^2$

And therefor:

$||a|-|b|| \le |a-b|$

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1  
Yes, that works. –  Arturo Magidin Sep 22 '11 at 16:45
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You want to prove $|a| - |b| \le ||a| - |b||$.

It is very simple! No triangle equality complications! Just simply split it into two cases:

  1. $|a| - |b| \le 0$, then the theorem is obvious because clearly $0 \le ||a| - |b||$.
  2. $|a| - |b| > 0$, then $||a| - |b|| = |a| - |b|$.

QED.

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