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Is it true that for vectors $x$ and $y$ in $\mathbb R^n$ $|\Vert x\Vert -\Vert y\Vert| \ge \Vert x-y\Vert $?

Can I simply use the triangle inequality $\Vert x\Vert +\Vert y\Vert \ge \Vert x+y\Vert $ to prove it even though there is a minus sign between the norms of $x$ and $y$?

user273488
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  • The expression should be- $|\Vert x\Vert -\Vert y\Vert| \ge \Vert x-y\Vert$. No need of extra$|.|$. – Rajat Sep 23 '15 at 16:01
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    Let assume $n=1, \ x=2, \ y=-3$. Then, this inequality does not hold. I think the inequality sign will be opposite. – Rajat Sep 23 '15 at 16:05
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    "Can I simply use the triangle inequality..." Yes, you have to use the triangle inequality, but in a slightly more involved way. Note that $|x-y|\le |x| - |y|$ is false. – Giuseppe Negro Sep 23 '15 at 16:07

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No, simply consider two vectors $x \ne y$ that have the same norm.

However, the inequality $| \| x\| - \|y\|| \le \|x - y\|$ holds [it is often called reverse triangle inequality]. By the triangle inequality, the following inequalities hold: $$ \| x \| = \|x - y + y\| \le \| x - y \| + \|y\|$$ $$ \| y \| = \|y - x + x\| \le \| y - x \| + \|x\|$$

The statement now follows from observing that $|a| \le b$ is equivalent to "$a \le b$ and $-a \le b$".

Dominik
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