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...
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.