Show that if $\|u\| \leq \|u+av\|$ for all $a \in F$, then $\langle u,v\rangle=0$, without using derivatives Before stating the question, I'd like to address the fact that there are other solutions to this problem. Robjohn's answer uses derivatives in their argument to show that the vectors are orthagonal, and 6005's solution appears to assume the inner product is real if I'm understanding it correctly.
I assume that $u$ and $v$ are vectors in an inner product space $V$ over the field $F$, a subfield of $\mathbb{C}$.
I know that assuming the inequality is true, I can show that for all $a\in F$, $0\leq\text{Re}(\bar{a}\langle u,v\rangle )+|a|^2\|v\|^2$. From here, I can proceed by contradiction and assume $\langle u,v\rangle\not= 0$.
Then, this means that either $\text{Re}(\langle u,v\rangle )\not=0$ or $\text{Im}(\langle u,v\rangle )\not=0$. If the real part of the inner product is not zero, then depending on whether it is positive or negative, I can pick a real $a$ so that $\text{Re}(\bar{a}\langle u,v\rangle )+|a|^2\|v\|^2<0$ and arrive at a contradiction. Thus, the real part must be $0$, but how could I show that the imaginary part of the inner product must be zero as well?
Or, am I approaching this the wrong way?
 A: 
Then, this means that either Re(⟨u,v⟩)≠0 or Im(⟨u,v⟩)≠0. If the real part of the inner product is not zero, then depending on whether it is positive or negative, I can pick a real $a$ so that $\text{Re}(\bar{a}\langle u,v\rangle )+|a|^2\|v\|^2<0$ and arrive at a contradiction.

You can't separate into these two cases since the value of $\text{Re}(\bar{a}\langle u,v\rangle )$ depends both on the real and imaginary parts of $\left<u,v\right>$. You should choose $a$ so that $\bar{a}\langle u,v\rangle$ is rotated onto the negative real  axis. Then choose the magnitude of $a$ to arrive at the contradiction.
So you should get something like  $-r|\left<u,v\right>| + r^2\|v\|^2 \geq 0$ where $r = |a|$. This will be false for small enough $r$.
Edit: Changed "positive real axis" to "negative real axis".
A: As you have calculated, the inequality $\|u\|^2 \le \|u + av\|^2$ constrains the real part of $\mathrm{Re}\left(\overline{a}\langle u, v\rangle\right)$, because
$$
\|u\|^2 \le \|u + av\|^2 = \langle u+av,u+av\rangle = \|u\|^2 +2\,\mathrm{Re}\left(\overline{a}\langle u, v\rangle\right) +|a|^2\|v\|^2
$$
gives us
$$
0 \le 2\,\mathrm{Re}\left(\overline{a}\langle u, v\rangle\right) +|a|^2\|v\|^2\tag1.
$$
We can use the freedom in choosing $a$ to compute both the real and imaginary parts of $\langle u, v\rangle$. We accomplish this by choosing real $a$ in the first case and imaginary $a$ in the second.
Specifically, let $\frac{\langle u, v\rangle}{\|v\|} = x + iy$ for $x,y\in \mathbb{R}$ and consider two cases $a=-\frac{x}{\|v\|}$ and $a=-\frac{iy}{\|v\|}$. Substituting into $(1)$, we get
$$
0 \le -2\,\mathrm{Re}(x(x+iy)) + x^2 = -2x^2 + x^2 = -x^2 \\
0 \le 2\,\mathrm{Re}(iy(x+iy)) + y^2 = -2y^2 + y^2 = -y^2 \\
$$
which means that $x^2=y^2=0$ and hence $\langle u, v\rangle=0$.
