Characterization of orthogonality in an inner product space.

Question

How can I show the following assertion

In an inner product space, $x\perp y$ if and only if $\|x+\alpha y\|\ge \|x\|$ for all scalars $\alpha$.

Answer 1

$$ \langle x+\alpha y,x+\alpha y\rangle\geq\langle x,x\rangle $$ After rearranging, $$ \alpha\langle x,y\rangle+\bar \alpha\overline{\langle x,y\rangle}\geq-|\alpha|^2\langle y,y\rangle. $$ Choosing $\alpha=-\overline{\langle x,y\rangle}/n$, we get $$ -(1/n)2|\langle x,y\rangle|^2\geq-(1/n^2)|\langle x,y\rangle|^2\|y\|^2. $$ Multiply by $-n$ and take the limit $n\rightarrow \infty.$