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Prove $\|x-y\|\|x+y\|\le\|x\|^2+\|y\|^2$ for all $x,y\in\mathbb{R}^n$

I've been struggling with this for a while and haven't figured out a way to do it either geometrically or algebraically.

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Presumably what you mean is "for all $x$ and $y$ in ${\bf R}^n$". – Gerry Myerson Feb 26 '13 at 0:41
That is correct. – iuppiter Feb 26 '13 at 0:42
OK. Are you aware that $|x\cdot y|\le\|x\|\|y\|$? – Gerry Myerson Feb 26 '13 at 0:43
Yes. Shwartz inequality. – iuppiter Feb 26 '13 at 0:44
OK, now that you see how to do it, write it up and post it as an answer (this site encourages people to post answers to their own questions. Then, after some time has passed, you can accept your answer by clicking in the check mark next to it. – Gerry Myerson Feb 26 '13 at 2:03

\begin{align} \|x-y\|^2\|x+y\|^2 &=(x-y,x-y)(x+y,x+y)\\ &=(\|x\|^2-2(x, y) + \|y\|^2)(\|x\|^2+2(x,y) + \|y\|^2)\\ &=(\|x\|^2+\|y\|^2)^2-4(x, y)^2\\ &\le(\|x\|^2+\|y\|^2)^2 \end{align}

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