Proof with inner product of vectors

Question

Admit that $V$ is a linear space with inner product. Show that given $x,y \in V$ we have $<x,y>=0$ if and only if for every scalar $\alpha \in \mathbb{K}$ we have $|x|\le |x+ \alpha y |$

My attempt:

My first though was to use the triangular inequality so I have

$|x+ \alpha y | \le |x| + |\alpha||y|$

Then I get to

$|x|\le |x+ \alpha y | \le |x| + |\alpha||y|$

So:

$|x|\le |x| + |\alpha||y|$

But I don't if this is the way to keep going with the proof or how do I go from here... Can someone please help me, giving me 1 or 2 hints? Thanks!

Answer 1

The given inequality is equivalent to $$0\leq2\alpha\langle x,y\rangle+\alpha^2|y|^2.$$ If $\langle x,y\rangle=0$, the inequality holds trivially. Otherwise, wlog $y\neq0$. Now choose $$\alpha=-\frac{\langle x,y\rangle}{|y|}.$$

Answer 2

We have

\begin{align}\vert x\vert\le \vert x+\alpha y\vert&\iff \vert x\vert^2\le\langle x+\alpha y,x+\alpha y\rangle=\vert x\vert^2+2\alpha \langle x,y\rangle+\alpha^2\vert y\vert^2\\&\iff P(\alpha):=\alpha^2\vert y\vert^2+2\alpha\langle x,y\rangle\ge0,\;\forall \alpha\end{align} Since the polynomial $P$ is positive then $\Delta'=\langle x,y\rangle^2\le 0\iff \langle x,y\rangle=0$