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I'm sort of stuck here. At first I thought that this had to do with Cauchy-Schwartz, but realized that this is similar to the Triangle Inequality. Rudin 3ed has a proof of a similar statement: $$|z+w|\le|z|+|w|$$

I've gone through the proof several times, but I'm having some trouble understanding it.

Here's what I've done so far:

  1. Square both sides $$ ||x|^2-2|x||y|+|y|^2 |≤|(x-y)(x ̅-y ̅)|$$

  2. Expand RHS (I believe that I can drop the absolute value on both sides as I've done, since $|x|=\sqrt(x^2)$) $$ |x|^2-2|x||y|+|y|^2≤xx ̅-xy ̅-x ̅y+yy ̅ $$

I'm not sure where to go from here. Am I just going down the wrong hole? Thanks all in advance!

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    This is utterly trivial. You just need to prove $|x|-|y| \le |x-y|$ and $|y|-|x| \le |x-y|$ but these both are just the triangle inequality. – 2017-02-06
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    @mathworker21 Condescension is not necessary or in good taste. When new to the subject, this is not at all trivial. – 2017-02-06
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    There was no condescension. I find it helpful to learn what people find obvious. – 2017-02-06
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    I'm roughly two years out from my last math course. So, while I can appreciate that this might be a somewhat trivial proof, it's far from trivial for me :) Incidentally, I'm also trying to avoid simply memorizing proofs. I'd prefer to have a more thorough understanding. – 2017-02-06

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Hint: $(x - y) + y = x,$ so by the triangle inequality $\Vert x \Vert \le \Vert x - y\Vert + \Vert y \Vert, $ and reversing the roles of $y$ and $x$ we find that $\Vert y \Vert \le \Vert x - y \Vert + \Vert x \Vert.$

$\Vert x \Vert - \Vert y \Vert \le \Vert x - y\Vert,$ and $\Vert y \Vert - \Vert x \Vert \le \Vert x - y \Vert. $

Taking together this implies $\vert \Vert x \Vert - \Vert y \Vert \vert \le \Vert x - y \Vert$.

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    This seems almost too simple. Just to make sure I'm understanding: We're showing that the norm of each of x,y is less than the resulting RHS of the inequality. Intuitively, this makes sense as justification for the initial inequality. Is there really no need for a deeper proof? – 2017-02-06
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    @YevgeniyFeyman No, there isn't. See the other answers as well. – 2017-02-06
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Hint: $$|x| = |y + (x-y)| \leqslant |y| + |x - y|$$

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Note that $|z_1| = |(z_1 - z_2) + (z_2)|$. Now use the triangle inequality. to get the desired result.