Show that in a Hilbert space, $||x+\alpha y|| \ge ||x|| \,\forall\,\alpha\in\mathbb{C}\implies (x,y)=0$ We have a Hilbert space, equipped with the norm $||\cdot||=\sqrt{(\cdot,\cdot)}$,
and we're given the following inequality holds for all $\alpha\in\mathbb{C}: ||x+\alpha y||\ge ||x||$
How would you go about showing this inequality leads to $(x,y)=0$
Squaring both sides I get to
$(x,x)+\overline{\alpha}(x,y)+\alpha\overline{(x,y)}+|\alpha|^2(y,y)\ge (x,x)\implies \overline{\alpha}(x,y)+\alpha\overline{(x,y)}+|\alpha|^2(y,y)\ge 0$
I can write this in terms of real or imaginary parts depending on choice of $\alpha$ but I'm perplexed how to make this something meaningful. Any help appreciated.
 A: Consider $f : \Bbb{R} \to \Bbb{R}$ given by
$$f(\alpha) = \|x+\alpha y\|^2 = \|x\|^2 + 2\operatorname{Re} (\overline{\alpha} \langle x,y\rangle) + \|y\|^2 = \|x\|^2 + 2\alpha \operatorname{Re} \langle x,y\rangle + \|y\|^2.$$
$f$ is an affine function so it is clearly differentiable with
$$f'(\alpha) = 2\operatorname{Re} \langle x,y\rangle.$$
By your condition we have that $f$ has a minimum at $\alpha =0$ so in particular
$$0=f'(0) = 2\operatorname{Re} \langle x,y\rangle \implies \operatorname{Re} \langle x,y\rangle = 0.$$
Similarly, consider $g : \Bbb{R} \to \Bbb{R}$ given by
$$g(\alpha) = \|x+i\alpha y\|^2 = \|x\|^2 + 2\operatorname{Re} (\overline{i\alpha} \langle x,y\rangle) + \|y\|^2 = \|x\|^2 + 2\alpha \operatorname{Im} \langle x,y\rangle + \|y\|^2.$$
so as above we conclude $\operatorname{Im} \langle x,y\rangle = 0$.
A: It suffices to assume that $y\neq0$. By assumption,
$$\|x\|^2\leq \|x\|^2 + 2\operatorname{Re}\big(\overline{\alpha}(x|y)\big)+|\alpha|^2\|y\|^2$$
it follows that
$$0\leq2\operatorname{Re}\big(\overline{\alpha}(x|y)\big)+|\alpha|^2\|y\|^2$$
for $\alpha=-t\frac{(x|y)}{\|y\|^2}$, with $t\in\mathbb{R}$ we obtain
$$
0\leq (-2t+t^2) \frac{|(x|y)|^2}{\|y\|^2}
$$
fot $t=1$,   $0\leq -\frac{|(x|y)|^2}{\|y\|^2}$; for $t=3$, $0\leq 3\frac{|(x|y)|^2}{\|y\|^2}$. Fromm all this, it follows that $(x|y)=0$.
A: In Hilbert space, given $x,y$, with $(x,y) \ne 0$, project $y$ onto $x$
to see that there exist:
scalar $\tau$ and vector $z$ with
$$
y = \tau x + z, \qquad z \perp x,\qquad \tau \ne 0 .
$$
Since $\tau \ne 0$, choose $\alpha = 1/\tau$ to get
$$
\|x+\alpha y\|^2 = \|(1+\alpha\tau)x + \alpha z\|^2 =
\|2x + \alpha z\|^2 =
\|2x\|^2+\|\alpha z\|^2 \ge 4\|x\|^2 > \|x\|^2.
$$
A: Denote $\langle x,y \rangle = r_0e^{i\lambda}$.
By your computation we are given that the function $f: \alpha \rightarrow 2Re(\overline{\alpha}\langle x,y \rangle) + |\alpha|^2||y||^2$ is nonnegative for all $\alpha$. Setting  $\alpha = re^{i\theta}$ and fixing $\theta$, our function is $f: (r,\theta) \rightarrow 2rr_0\cos(\lambda - \theta) + r^2||y||^2$. Unless $r_0 = 0$, we could choose $\theta$ so that $\cos(\lambda - \theta)$ is negative and $r$ small to get $f(\alpha) < 0$, which is a contradiction.
Thus $r_0 = 0$, so $\langle x,y \rangle = 0$
