A condition equivalent to orthogonality

Question

Prove that in any inner product space: $x$ and $y$ are orthogonal if and only if $||x+\alpha y||\ge ||x||$ for every scalar $\alpha$.

Answer 1

$$||x+\alpha y||^2=\langle x+\alpha y,x+\alpha y \rangle= ||x||^2+|\alpha|^2||y||^2+\bar{\alpha} \langle x, y \rangle+\alpha \langle y, x \rangle.$$

Show that if $\langle x, y \rangle\neq 0$ there is $\alpha $ such that this expresion is smaller than $||x||$.

Clearly, if $\langle x, y \rangle= 0$ then $$||x+\alpha y||^2=||x||^2+|\alpha|^2||y||^2\geq ||x||^2$$

Answer 2

Let $\alpha = -(x,y)/(y,y)$, assuming $y \ne 0$. Then, by assumption, $$ \|x\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} $$ Using the Pythagorean Theorem: $$ \left\|\left(x-\frac{(x,y)}{(y,y)}y\right)+\frac{(x,y)}{(y,y)}y\right\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} \\ \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}+\left\|\frac{(x,y)}{(y,y)}y\right\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} $$ Hence $|(x,y)|=0$.