# Is this a valid proof for the reverse triangle inequality: Given $x,\,y\in\mathbb{R}^n$, $\|x-y\|\ge |\|x\| - \|y\||$.

$$\def\a{{\bf a}} \def\b{{\bf b}}$$ $$\def\x{{\bf x}}$$ $$\def\y{{\bf y}}$$

I've seen various methods of proving the reverse triangle inequality but I was wondering if this one is fine as well.

Prove the reverse triangle inequality: Given $$\x,\,\y\in\mathbb{R}^n$$, $$\|\x-\y\|\ge |\|\x\| - \|\y\||$$.

$$\textbf{Solution}$$ Let $$\x,\y \in \mathbb{R}^n$$. Now, $$\x=\x-\y+\y$$ and so $$||\x|| = ||\x-\y+\y|| \le ||\x-\y|| + ||\y||$$ by triangle inequality. Thus, $$||\x|| - ||\y|| \le ||\x-\y||$$ [*].

Now for $$\y$$. Again, $$\y = \y-\x+\x$$ so $$||\y|| = ||\y-\x+\x||$$ implies $$||\y|| \le ||\y-\x|| + ||\x||$$ by triangle inequality. So, $$||\y|| - ||\x||\le ||\x-\y||$$. Hence, $$-||\x-\y|| \le ||\x|| - ||\y||$$ [**].

By [*] and [**], $$-||\x-\y|| \le ||\x|| - ||\y|| < ||\x-\y||$$ implying $$\bigl ||\x|| - ||\y|| \bigr\rvert \le ||\x-\y||$$ for all $$\x,\y \in \mathbb{R}^n$$.

• Yes, it is a valid proof. As far as I know this is the only sane way to prove this! Mar 7 '20 at 0:08
• Looks OK to me! Mar 7 '20 at 0:26