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

Question

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.$\:\:\:\:$

Answer 1

$|x|-|y| \le |x-y|$ and $|y|-|x| \le |x-y|$. Since $||x|-|y||= \text{either } |x|-|y| \text{ or } |y|-|x|$, the desired result follows.
So, you don't need need to split up into various cases.

Answer 2

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.