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In a normed space $(E,\lVert \cdot\rVert)$ space we have the following inequality: $$\forall\, x,y\in E,\quad\|x\|^{2}-\|y\|^{2}\leq \lVert x-y\rVert\cdot \|x+y\|.$$ How can we prove it?

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We have $$2||x||=||2x||=||x+y+x-y||\leq||x+y||+||x-y|| $$ so $$4||x||^2\leq ||x+y||^2+2||x+y||\cdot||x-y||+||x-y||^2,$$ and $$|||x+y||-||x-y|||\leq 2||y||$$ so $$||x+y||^2-2||x+y||\cdot||x-y||+||x-y||^2\leq 4||y||^2.$$ We get \begin{align*}4(||x||^2-||y||^2)&\leq ||x+y||^2+2||x+y||\cdot||x-y||+||x-y||^2\\ &-(||x+y||^2-2||x+y||\cdot||x-y||+||x-y||^2)\\ &=4||x+y||\cdot||x-y||, \end{align*} hence $||x||^2-||y||^2\leq ||x+y||\cdot||x-y||$ for all $x,y$. (we don't need a Banach space, a normed space is enough)

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  • $\begingroup$ @Julian Thanks for editing! $\endgroup$ Dec 18 '11 at 17:12

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