Prove that for any $x,y \in \Bbb R$, we have $||x|-|y||\leq |x-y|$.

Question

Original question: Prove that for any $x,y \in \Bbb R,$ we have $$||x|- |y||\leq |x-y|.$$ I am confused about how to prove this. Could I simply use an example with real numbers $x$ and $y$, or do I have to prove it formally since it says for any $x$ and $y$? I have no idea how to prove this without a specific example!

Answer 1

Particular examples never prove anything - they only serve as counter examples for false assertions. Use the triangular inequality twice. Use that $$|x| = |x-y+y| \leq |x-y|+|y|,$$so that $|x|-|y| \leq |x-y|$. Repeat this reasoning to conclude that $|y|-|x| \leq |x-y|$.

But what does $-(|x|-|y|)\leq |x-y|$ and $|x|-|y|\leq |x-y|$ imply?

Answer 2

Since both sides are positives we can square: $$ ||x|- |y||\leq|x-y| \Leftrightarrow (|x|-|y|)^2\leq (x-y)^2 \Leftrightarrow x^2-2|xy|+y^2\leq x^2-2xy+y^2 \Leftrightarrow \\ \Leftrightarrow -2|xy|\leq -2xy \Leftrightarrow |xy|\geq xy $$ which is obviously true for $x,y\in\mathbb{R}$.