Absolute value proof in Spivak

Question

Prove (1) using the fact that $|x|-|y| \le |x-y|$

(1) $|(|x|-|y|)| \le |x-y|$

Attempt: as I know $|a| \le b \iff -b \le a \le b$, then proving (2) would allow me to prove (1)

(2): $-|x-y| \le |x| - |y| \le |x-y|$.

From testing different cases of $x$ and $y$, it seems that (2) is true. Furthermore, as $-|a| \le a \le |a|$, it seems that (2) is true.

However, could someone show me a proof why we can assume (2) is true? I can't figure out how to write a rigorous one.

citation: Spivak Calculus 3rd edition, chapter 1 question 12 vi

Answer 1

$2$ is true because of the triangular inequality :

$$|x|=|x-y+y|\le |x-y|+|y|\implies$$

$$|x|-|y|\le |x-y|$$ and by symetry, $$|y|-|x|\le |x-y|$$

Remark $$A\le B \wedge -A\le B \implies |A|\le |B|$$

Answer 2

Hint: It is much simpler to prove if you observe that $|x-y|=|y-x|$, as there results that we also have $$|y|-|x| \le |x-y|,$$ whence $\;|x-y|\ge\max\bigl(|x|-|y|, |y|-|x|\bigr)$.