# Proving absolute value inequalities

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.

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 Do you have to split up the cases? Actually, only two or a few more lines are sufficient for the proof if we do not split up! – sos440 Sep 11 '12 at 3:42 Thanks for the great answers! However, Could someone show how it would be done by cases anyway so I can get that experience? – CodeKingPlusPlus Sep 11 '12 at 11:33

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