Proving absolute value inequalities

Question

I need help proving the last two cases for the following inequality: $\bigl|\lvert x\rvert-\lvert y\rvert\bigr| \le \lvert x-y\rvert$.

Case 1: $x > 0$ and $y > 0$: the inequality simplifies to: $|x-y|\le |x-y|$ and we are done this case

Case 2: $x < 0$ and $y < 0$: the inequality simplifies to: $|-x + y| \le |x - y|$. Here we let $z = y-x$ and we see $|z| = |-z|$ and we are done this case

Could somebody help me out with the last two cases and provide a detailed explanation? I have trouble "splitting up" the cases.

Answer 1

Since both sides are positive, we can square them and still preserve the inequality:

$$(|x|-|y|)^2\leq (x-y)^2$$ $$x^2-2|x||y|+y^2\leq x^2-2xy+y^2$$ $$-2|x||y|\leq -2xy$$ $$xy\leq|x||y|$$

Answer 2

for all $x,y \in \mathbb{R}$ $|x-y|+|y| \ge|x-y+y|$(by triangular inequality) and done the proof. In fact, by symmetry, nomatter the case $x>y$ or $y>x$ would also yield the inequality so $|x-y| \ge ||x|-|y||$.