If $\|u+\lambda v\| \ge \|u\| \forall \lambda$ then $u \perp v$ Let $H$ be a Hilbert space and $u,v \in H$. I need to prove:

If $\|u+\lambda v\| \ge \|u\|$ $\forall \lambda \in K = \mathbb{C},\mathbb{R}$ then $u \perp v$.

I need this to close a triple equivalence and I'm not sure if this way is evident or I should follow another way. 
 A: \begin{align}
||u+\lambda v||^2 &\ge ||u||^2 \quad \forall \, \lambda \in K \\
2Re\langle u,\lambda v\rangle + \lambda^2 ||v||^2 &\ge 0 \\
Re\langle u,\lambda v\rangle &\ge -\lambda^2 ||v||^2 /2 \tag{*} \label1
\end{align}


*

*Choose $\lambda > 0$.  Divide both sides by $\lambda$ to get $Re\langle u, v \rangle \ge -\lambda  ||v||^2$.

*Choose $\lambda' = -\lambda < 0$.  Divide both sides by $\lambda'$ to get $Re\langle u, v \rangle \le -\lambda' ||v||^2 = \lambda ||v||^2$.

*Now, we have $|Re\langle u, v \rangle| \le \lambda ||v||^2$.  Since the choice of $\lambda>0$ is arbitrary, we conclude that $Re\langle u, v \rangle = 0$.


Similarly, by making suitable choice of $\lambda$, we can show that $Im\langle u, v \rangle = 0$.  Hence $\langle u, v \rangle = 0$.

 Recall: $Re(iz) = Re(i(a+ib)) = Re(-b+ia) = -Im(z)$, so for $\lambda \in \Bbb{R}$, $Im\langle u, \lambda v \rangle = Re\langle u, i\lambda v \rangle$ Use $i\lambda$ instead of $\lambda$ in \eqref{1} and repeat the above arguments to get $Im\langle u, v \rangle \ge 0$; use $-i\lambda$ instead of $\lambda$ in \eqref{1} and repeat the above arguments to get $Im\langle u, v \rangle \le 0$.  As a result, $Im\langle u, v \rangle = 0$.

