I have seen other proofs of the inequality, but I haven’t seen the one I found. I would like to know if it’s correct, and what I could improve. Here’s the proof:
First we prove $(\lvert a\rvert - \lvert b\rvert)^2 \leq (a-b)^2 $, where $a, b \in \mathbb{R}$.
$$ \begin{align} (\lvert a\rvert - \lvert b\rvert)^2 &= \lvert a \rvert^2 - 2 \lvert a\rvert \lvert b\rvert + \lvert b\rvert^2 \\ &= a^2 - \lvert 2ab\rvert + b^2 \\ &\leq a^2 - 2ab + b^2 \\ &= (a-b)^2 \\ \Rightarrow\sqrt(\lvert a\rvert -\lvert b\rvert)^2 &= \lvert \lvert a \rvert - \lvert b\rvert\rvert \\ &\leq \sqrt(a-b)^2 \\ &= \lvert a - b\rvert \end{align}$$
Thus $\lvert \lvert a \rvert - \lvert b\rvert \rvert \leq \lvert a - b\rvert \ \forall a, b \in \mathbb{R}.$