Show that $||x| - |y|| \le |x-y|$ for all $x, y, \in \mathbb{R}$

Question

Show that $||x| - |y|| \le |x-y|$ for all $x, y, \in \mathbb{R}$

Working:

So we know that $|x| = |(x-y) + y | \le |x-y|+|y|$

Thus, $|x|-|y| \le |x-y| \ \cdots (1)$ and similarly, $|y|-|x| = -(|x|-|y|) \le |x-y| \ \cdots (2)$.

We also know that $||x|-|y|| = \begin{cases} |x|-|y| \ \text{if} \ |x| - |y| \ge 0 \\ -(|x| - |y|) \ \text{if} \ |x| - |y| < 0 \end{cases}$

Query:

I'm not exactly sure how to finish off the proof. Can I say because $(1)$ and $(2)$ are both true, then $||x| - |y|| \le |x-y|$ as required?

Answer 1

Yes you can. If you are unsure, you can divide the proof in two cases: $|x|-|y|<0$ or $|x|-|y| \geq 0$. You can prove the inequality in both cases and these cases contain all possibilities, so you have proved the inequality in all cases.

Answer 2

Yes, because $\forall x, y \in \mathbb{R}$, $$|x|=|x-y+y|\leq |x-y|+|y|, $$ then $$|x|-|y|\leq |x-y| .$$ Analogously,

$$|y|=|y-x+x|\leq |y-x|+|x|=|-1(x-y)|+|x|=|x-y|+|x|\Rightarrow |y|-|x|\leq |x-y|, $$ i.e., $$-(|x|-|y|)\leq |x-y|. $$

Now, as for all $a\in \mathbb{R}$, $|a|=\max\{-a,a\}$, we have $$||x|-|y||=\max\{-(|x|-|y|),|x|-|y|\}\leq |x-y|, $$ in other words, $$||x|-|y||\leq |x-y|. $$