# Absolute value of the sum of two absolute values real analysis proof

This proof seems to be giving me much trouble. I know I must split it up into various cases, with no loss of generality probably fewer but after that I really have no clue. $||x|-|y|| \le |x-y|$

Thanks ahead everyone.$\:\:\:\:$

• Presumably it should be $||x|-|y||\leq |x-y|$. – carmichael561 Oct 20 '16 at 5:33
• yeah the inequality should be the other way. my bad – juv Oct 20 '16 at 5:33
• Well, you can be slick. wolog $|x| = |x - y + y| \le |x-y| +|y|$ so $|x| - |y| \le |x - y|$ and same for $|y| \le |y - x| +|x|$ so $|y| - |x| \le |x-y|$. But I always screw it up when I try it so I usually find it easier to break it in four cases and .... – fleablood Oct 20 '16 at 5:46
• This works since I already proved the triangle inequality. Once I get to that point just using the additive inverse on the absolute values work since |x| is real and so is |y|. Cool then i just need to split it up into 4 cases and thats it. – juv Oct 20 '16 at 6:01
• The result is true more generally, for any norm we have $|\|x\|-\|y\| | \le \|x-y\|$. It is a useful result that shows that a norm is Lipschitz continuous with rank one. – copper.hat Oct 20 '16 at 7:15

$|x|-|y| \le |x-y|$ and $|y|-|x| \le |x-y|$. Since $||x|-|y||= \text{either } |x|-|y| \text{ or } |y|-|x|$, the desired result follows.
Hint: Triangle inequality gives. $$|x| = |(x-y) + y| \leq |x-y| + |y| \implies |x| - |y| \leq |x-y|.$$ I will leave the rest to you.