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

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 $|(x,y)|=0$.
