# Reverse triangle inequality question

This is a really silly question, but why do I need "$\max$" in the two spots below?

$|a+b+c|\geq |a|-|b+c|\geq |a|-(|b|+|c|)=|a|-|b|-|c|$

Taking that a little further, you could permute the $a,b,c$ to get

$|a+b+c| \geq \max(|a|-|b|-|c|,|b|-|a|-|c|,|c|-|a|-|b|)$

and using $|a|-|b|-|c| = 2|a| - (|a|+|b|+|c|)$, etc., you get

$|a+b+c| \geq 2\max(|a|,|b|,|c|) - (|a|+|b|+|c|)$

• I am not quite sure where you are going with this. I can tell you why it is true that there is a max there, but it seems that you are aiming at something bigger than this detail, but this bigger idea is not expressed in the question above. – user357980 Sep 18 '16 at 2:43
• I asked someone if $|a+b+c| \geq ||a|-|b|-|c||$ and he said no but you do have $|a+b+c|\geq 2\max(|a|,|b|,|c|)-(|a|+|b|+|c|)$. At the moment I've been doing questions that use Rouche's Theorem and I need to use the reverse triangle inequality – Faolan Sep 18 '16 at 2:58
• I think what my real question, perhaps, is say $|b|=\max(|a|,|b|,|c|)$. Is it still true that $|a+b+c| \geq 2|c|-(|a|+|b|+|c|)$? Also, why would I want to use the $\max$? – Faolan Sep 18 '16 at 3:01
• You don't need the max, but including the max gives a larger lower bound (hence a stronger mathematical statement). – Greg Martin Sep 18 '16 at 3:14
• Okay, so when using Rouche's Theorem, would using the max be better than to not use it? (when finding lower bound for |f| when showing |f-g|<|f|) – Faolan Sep 18 '16 at 3:20