Proof of the ‘second’ triangle inequality

Question

I am trying to prove the ‘second’ triangle inequality: $$\Big||x|-|y|\Big| \leq |x-y|$$

My attempt:

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Proof: \begin{align} |x-y|^2 &= (x-y)^2 \\ &= x^2 - 2xy + y^2 \\ &\geq |x|^2 - 2|x||y| + |y|^2 \\ &= \left(\Big||x|-|y|\Big|\right)^2. \end{align}

Therefore $\implies |x-y| \geq \Big||x|-|y|\Big|$.

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My questions are: Is this an acceptable proof?; and are there alternative proofs that are more efficient?

Answer 1

Don’t work too hard: $$|x| - |y| \geq -|x - y|$$ is true because $|x| + |y - x| \geq |y|$, and $$|x| - |y| \leq |x - y|$$ because $|y| + |x - y| \geq |x|$. These are true because of the triangle inequality, hence the answer.

Answer 2

It is true in $\mathbb{R}$, but for $\mathbb{R}^n$, consider $$|x| \le |x-y| +|y|.$$ Analogously, $$|y|-|x| \le |x-y|.$$ or \begin{align} |x-y|^2 &= \langle x-y,x-y\rangle \\ &= |x|^2 - 2\langle x, y \rangle + |y|^2 \\ &\geq |x|^2 - 2|x||y| + |y|^2 \\ &= \left(\Big||x|-|y|\Big|\right)^2. \end{align}