# $\langle x,y \rangle = 0 \iff \forall \alpha\in\Bbb R, |x|\le|x+\alpha y|$

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.

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 $$\alpha$$.
$$|x|\leq |x\pm \frac 1 n y|$$ 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 $$-\infty$$).