Proving Young's Inequality for Inner Product Spaces I am trying to prove the following inequality, I have proven it for when my field is $\Bbb R$ but I am having trouble when my field $\Bbb F = \Bbb C$
$$|\langle u,v\rangle| \leq \frac{\lambda^2}{2} \lVert u\rVert^2 + \frac{1}{2\lambda^2} \lVert v\rVert^2$$

Using $2\mathrm{Re}(w)=w+\overline{w}$ and $\langle v,u\rangle=\overline{\langle u,v\rangle}$ we may FOIL out
$$ \begin{array}{ll} 
0 & \le \|u-v\|^2  \\
& =\|u\|^2-\langle u,v\rangle-\langle v,u\rangle+\|v\|^2 \\
& =\|u\|^2-2\mathrm{Re}\langle u,v\rangle+\|v\|^2
\end{array}$$
which gives $\mathrm{Re}\langle u,v\rangle\le \frac{1}{2}\big(\|u\|^2+\|v\|^2\big)$.
For real $\lambda\ne0$ we may replace $u\mapsto\lambda u$, $v\mapsto\lambda^{-1} v$ to get
$$ \mathrm{Re}\langle u,v\rangle \le \frac{\lambda^2}{2}\|u\|^2+\frac{1}{2\lambda^2}\|v\|^2. $$
Every complex number may be written in polar form as $\langle u,v\rangle=e^{i\theta}|\langle u,v\rangle|$.
If $\langle u,v\rangle$ is conjugate-linear in the first argument, replace $u\mapsto e^{i\theta}u$ to get
$$ |\langle u,v\rangle| \le \frac{\lambda^2}{2}\|u\|^2+\frac{1}{2\lambda^2}\|v\|^2. $$
I've heard Terence Tao call this trick "arbitrage."
