Prove that a inner product is zero

Question

Let $\langle, \rangle$ and $||,||$ an inner product and a norm over $\mathbb{R}^n$. Suppose for all $a\in \mathbb{R}$ the inequality $$ ||u||\le ||u+av|| $$ holds. How can I prove that $\langle u,v\rangle=0$? Any help would be appreciated.

Answer 1

Assume $\langle u,v \rangle \neq 0$. Square both sides of your inequality to get $$\|u\|^2 \leq \|u\|^2 + 2a\langle u,v \rangle + a^2\|v\|^2$$ or $$2a\langle u,v \rangle + a^2\|v\|^2 \geq 0$$ for all $a \in \mathbb{R}$. Choosing $a = -\frac{\langle u,v \rangle}{\|v\|^2}$ yields a contradiction.