$\|u\|\leq \|u+av\| \Longrightarrow \langle u,v\rangle=0$

Assuming that you mean that the inequality hold for all $a$, we have $$ \|u\|^2\le\|u+av\|^2\tag{1} $$ Let $a=t\langle u,v\rangle$ for $t\in\mathbb{R}$, then $(1)$ implies $$ \begin{align} \langle u,u\rangle &\le\langle u+av,u+av\rangle\\ &=\langle u,u\rangle+2\mathrm{Re}\left(a\,\overline{\langle u,v\rangle}\right)+|a|^2\langle v,v\rangle\\ &=\langle u,u\rangle+2t\left|\langle u,v\rangle\right|^2+t^2\left|\langle u,v\rangle\right|^2\langle v,v\rangle\\ 0&\le\left(2t+t^2\langle v,v\rangle\right)\left|\langle u,v\rangle\right|^2 \end{align} $$

Since the right hand side is $0$ when $t=0$, and the right side must always be greater than or equal to $0$, the derivative at $t=0$ must be $0$. That is, $$ \langle u,v\rangle=0 $$