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If $V$ is a inner product space and $x,y\in V$, why is $\langle x,y \rangle = 0$ equivalent to say that for every $\alpha\in\Bbb R$, $|x|\le|x+\alpha y|$?

I understand that $\langle x,y \rangle = 0$ leads to $|x|\le|x+\alpha y|$ but not the opposite.

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Note that for $\langle x,y\rangle\ne 0$ ( and hence $|y|\ne 0$), $$ |x+\alpha y|^2-|x|^2=2\alpha \langle x,y\rangle +\alpha^2|y|^2=\left(|y|\alpha+\frac{\langle x,y\rangle}{|y|}\right)^2-\frac{\langle x,y\rangle^2}{|y|^2}$$ will be negative for suitable $\alpha$.

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  • $\begingroup$ Does this mean the equivalence relation in the OP question is false? $\endgroup$ – Learn_and_Share May 12 at 20:28
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$|x|\leq |x\pm \frac 1 n y|$ gives (after multiplication by $n^{2}$) $|y|^{2}\pm n \langle x, y \rangle \geq 0$ for every positive integer $n$. This implies $\langle x, y \rangle = 0$ (Otherwise you can make LHS tend to $-\infty$).

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