Could I use thhe triangle inequality to perform the following:

$\displaystyle \frac{|a-b|}{2}\leq \frac{|a|}{2} - \frac{|b|}{2}$?

I cant seem to find a proof of why you can do this is if you can...

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- Apr 24th 2010, 08:57 AMsfspitfire23triangle inequality
Could I use thhe triangle inequality to perform the following:

$\displaystyle \frac{|a-b|}{2}\leq \frac{|a|}{2} - \frac{|b|}{2}$?

I cant seem to find a proof of why you can do this is if you can... - Apr 24th 2010, 09:16 AMnimon
I don't think so. The triangle inequality implies that

$\displaystyle |a| = |(a-b)+b| \leq |a-b|+|b|$

so that

$\displaystyle

|a|-|b| \leq |a-b| \Rightarrow \frac{|a|}{2}-\frac{|b|}{2} \leq \frac{|a-b|}{2},

$

which is the opposite of what you wanted!

In fact you can show that

$\displaystyle |a|-|b| \leq |a-b|,|a+b| \leq |a|+|b|,$

which is quite a nice way to remember them all. - Apr 24th 2010, 09:58 AMsfspitfire23
Ah, I must be able to do this then by the triangle inequality:

$\displaystyle \frac{|a-b|}{2}\leq \frac{|a|}{2} + \frac{|b|}{2}$

yes? - Apr 25th 2010, 01:41 AMnimon
Yep! To see why:

$\displaystyle |a-b| = |a+ (-b)| \leq |a|+|-b| = |a|+|b|. $

For practice, try also to show that

$\displaystyle |a|-|b| \leq |a+b|,$

the four inequalities will come in handy for the rest of your life!