Inequality in Inner product spaces I want to prove that for $u,v \in V$, $\langle u,v \rangle =0$ iff $\|u\| \le \|u+av\|$ for all scalars $a \in \mathbb{F}$. Here $\mathbb{F}$ is the field of $V$ which is either complex or real numbers. 
The ($\Rightarrow$) implication has been proved. What I need is ($\Leftarrow$). I tried expanding the norm $\|u+av\|$ but what I get is that  $\langle u,v \rangle \ge 0$. How do I prove it equal to 0?
Am I missing some small step? Please help, Thanks.
 A: Let $\langle u,v\rangle=re^{i\theta}$. Let $a=te^{i\theta}$, where $t\in \mathbb{R}$ will be varied. Then by assumption
$$
\|u\|^2\leq \|u+av\|^2=\langle u+av,u+av\rangle=\|u\|^2+a\langle v,u\rangle+\overline{a}\langle u,v\rangle +\vert a\vert^2\|v\|^2=\|u\|^2+2rt+t^2\|v\|^2.
$$
This implies that for all $t\in \mathbb{R}$,
$$
0\leq 2rt+t^2\|v\|^2,
$$
but if $r\neq 0$, this is a quadratic in $t$ with distinct roots, and hence the right side takes negative values, a contradiction. So $r=0$, which means $\langle u,v\rangle=0$, as desired.
A: Let $~u \ne 0~$.
$$ \|u+av\|^2=\langle u+av , u+av\rangle =\|u\|^2+a^2\|v\|^2+2Re(a\langle u , v\rangle)$$
$$\|u+av\|^2-\|u\|^2=a^2\|v\|^2+2Re(a\langle u , v\rangle)\tag1$$
If $~\langle u,v \rangle =0~$, then from $(1)$, $$\|u+av\|^2-\|u\|^2=a^2\|v\|^2 \ge 0$$
$$\implies \|u\| \le \|u+av\|\qquad \forall\quad a \in \mathbb{F}$$
Conversely suppose $$ \|u\| \le \|u+av\|\qquad \forall\quad a \in \mathbb{F}$$holds good. 
Let $~\langle u,v \rangle \ne 0~$ i.e., $~Re(\langle u , v\rangle)=k~(\ne 0)~$
So $(1) $ becomes $$\|u+av\|^2-\|u\|^2=a^2\|v\|^2+2ak\tag2$$
Then $~\|u+av\|^2-\|u\|^2 \lt 0~$, if we take $~a~$ as 
$(i)~~~$  if $~k > 0~$ then we pick $~a~$such that $-\frac{2k}{\|u\|}\lt a\lt 0$
$(ii)~~~$  if $~k < 0~$ then we pick $~a~$such that $0\lt a\lt -\frac{2k}{\|u\|}$
So both $~Re(\langle u , v\rangle)=0~$and $~Im(\langle u , v\rangle)=0~$, thus $~\langle u , v\rangle=0~$
