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

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 $\x,\y \in \mathbb{R}^n$.

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    $\begingroup$ Yes, it is a valid proof. As far as I know this is the only sane way to prove this! $\endgroup$ Mar 7 '20 at 0:08
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    $\begingroup$ Looks OK to me! $\endgroup$
    – Allawonder
    Mar 7 '20 at 0:26
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As far as I know that is the only proof of the reverse triangle inequality. At least, I've never seen a different one.

Compare with what Wikipedia writes.

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