Prove that if$\ ||x|| \le ||x +\alpha y||$ then $x$ is perpendicular to $y.$ Prove that if$\ ||x|| \le ||x +\alpha y||$ (for every$\ \alpha \in\mathbb{ R}$  ) then $x$ is perpendicular to $y$ ($x$ and$ y$ are not$\ 0$)
We need to prove that $\ xy=0$, so I tried:
$\ ||x|| \le ||x +\alpha y||$
$\ ||x||^2 \le ||x +\alpha y||^2$
$\ x^2 \le (x +\alpha y)^2$
$\ x^2 \le x^2 + 2\alpha xy +\alpha^2y^2$
$\ 0 \le 2\alpha xy +\alpha^2y^2$
$\ 0 \le 2x +\alpha y$
But, this didn't got me anywhere and I'm confused on how to proceed
 A: Hint Write $x=x_1+x_2$ with $x_1$ parallel to $y$ and $x_2$ perpendicular to $y$. Then $x_1=ay$ for some scalar $a$.
Then
$$ ||x||^2 \le ||x -a y||^2 \\
||x_1+x_2||^2 \le ||x_2||^2 \\
||x_1\|^2+\|x_2||^2 \le ||x_2||^2$$
with the last line following from the fact that $x_1 \perp x_2$.
This implies that $x_1=0$ and hence $x=x_2$. Since $x_2 \perp y$ you are done.
A: $$
||x||^2=\langle x,x\rangle
$$
Using that and your previous work I find
$$
0\leq 2\alpha \langle x,y\rangle +\alpha^2||y||^2
$$
Let
$$
f(\alpha)=2\alpha\langle x,y\rangle +\alpha^2||y||^2
$$
using methods of calculus I find that $f$ is extremized when
$$
\alpha=-\frac{\langle x,y\rangle}{||y||^2}
$$
If I substitute this in 
$$
f(-\frac{\langle x,y\rangle}{||y||^2})=-\frac{\langle x,y\rangle^2}{||y||^2}
$$ 
so the inequality at this value of $\alpha$ reads
$$
0\leq -\frac{\langle x,y\rangle^2}{||y||^2}
$$
But $\langle x,y\rangle^2/||y||^2$ is a nonnegative number and if the inequality is to be satisfied we must conclude that $\langle x, y\rangle = 0$.
