# 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! – Sangchul Lee 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

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

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

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As an experiment, I tried to prove this using $$(0) \;\;\; |p| = p \max -p$$ together with the properties of $\;\max\;$ (and $\;\min\;$). The result is not too pretty, but the upside is that I could do it almost mechanically, and without any case distinctions.

Let's first start with the following subexpression: \begin{align} & |x| - |y| \\ = & \;\;\;\;\;\text{"definition $(0)$, twice"} \\ & (x \max -x) - (y \max -y) \\ = & \;\;\;\;\;\text{"$\;-\;$ distributes over $\;\max\;$ to the left"} \\ & (x - (y \max -y)) \;\max\; (-x - (y \max -y)) \\ = & \;\;\;\;\;\text{"a DeMorgan-like rule"} \\ & (x + (-y \min y)) \;\max\; (-x + (-y \min y)) \\ = & \;\;\;\;\;\text{"$\;+\;$ distributes over $\;\min\;$"} \\ & ((x - y) \min (x + y)) \;\max\; ((-x - y) \min (-x + y)) \\ \end{align} so we've proven $$(1) \;\;\; |x| - |y| \;=\; ((x - y) \min (x + y)) \;\max\; ((-x - y) \min (-x + y))$$ Now we use this to rewrite the left hand side of our original equation: \begin{align} & \big||x| - |y|\big| \\ = & \;\;\;\;\;\text{"definition $(0)$"} \\ & (|x| - |y|) \;\max\; (|y| - |x|) \\ = & \;\;\;\;\;\text{"property $(1)$, twice"} \\ & ((x - y) \min (x + y)) \;\max\; ((-x - y) \min (-x + y)) \;\max\; \\ & ((y - x) \min (y + x)) \;\max\; ((-y - x) \min (-y + x)) \\ = & \;\;\;\;\;\text{"exchange the second and third terms; rewrite some sums"} \\ & ((x - y) \min (x + y)) \;\max\; ((y - x) \min (x + y)) \;\max\; \\ & ((-x - y) \min (y - x)) \;\max\; ((-x - y) \min (x - y)) \\ = & \;\;\;\;\;\text{"extract common parts: $\;\dots \min (x+y)\;$, $\;(-x -y) \min \dots\;$"} \\ & (((x - y) \;\max\; (y - x)) \min (x + y)) \;\max\; \\ & ((-x - y) \min ((y - x) \;\max\; (x - y))) \\ = & \;\;\;\;\;\text{"extract common part: $\;((x - y) \;\max\; (y - x)) \min \dots\;$"} \\ & ((x - y) \max (y - x)) \;\min\; ((x + y) \max (-x - y)) \\ = & \;\;\;\;\;\text{"definition $(0)$, twice"} \\ & |x - y| \;\min\; |x + y| \\ \leq & \;\;\;\;\;\text{"$\;p \min q \leq p\;$"} \\ & |x - y| \\ \end{align} which completes this proof.

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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 $|x-y| \ge ||x|-|y||$.

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