Cauchy–Schwarz inequality in complex variables I have seen various proofs for Cauchy–Schwarz inequality but all of them discuss only of real numbers. Can someone please give the proof for it using complex numbers in simple steps?
 A: Assuming you have prove C-S for real numbers, here is a way to prove the complex case. Giving $x,y \in \mathbb{C}^n$, you want to prove that $$|\langle x,y \rangle| \leq \|x\|\|y\|$$ The problem is that $\langle x,y \rangle$ is not real. Consider $u =\frac{\overline{\langle x,y \rangle}}{|\langle x,y \rangle|}$ It is clear that $\langle ux,y \rangle$ is now real. Hence by real C_S, you get $$|\langle ux,y \rangle| \leq \|ux\|\|y\|.$$ But $|u|=1$, hence you get the desired inequality.
A: Let $\lambda \in \mathbb{C}$, and $x$ and $y$ elements of an inner product space.  Then, if $z=x-\lambda y$ we have
$$\begin{align}
||z||^2&=\langle x-\lambda y,x-\lambda y\rangle\\\\
&=||x||^2+|\lambda|^2||y||^2-2\text{Re}\left(\lambda \langle x, y\rangle\right) \tag 1
\end{align}$$
At the minimum value of $||z||^2$ with respect to $\lambda$ we must have both
$$\frac{d||z||^2}{d\text{Re}(\lambda)}=0 \tag 2$$
and
$$\frac{d||z||^2}{d\text{Im}(\lambda)}=0 \tag 3$$
From $(2)$ and $(3)$, we find that
$$\lambda=\frac{\overline{\langle x,y\rangle}}{||y||^2} \tag 4$$
Substituting $(4)$ into $(1)$ reveals that the minimum of $||z||^2$ with respect to $\lambda$ is 
$$\min_{\lambda \in \mathbb{C}}\left(||z||^2\right)=||x||^2-\frac{|\langle x,y\rangle|}{||y||^2} \tag 5$$
Inasmuch as $||z||$ is the norm of $z$, it must be non-negative for all values of $\lambda$, including the value of $\lambda$ that minimizes $z$.  Therefore, from $(5)$ we obtain the coveted Cauchy-Schwarz Inequality
$$\bbox[5px,border:2px solid #C0A000]{|\langle x,y\rangle|\le ||x||^2\,||y||^2}$$
with equality if and only if $z=x-\lambda y=0$, or equivalently $x=\lambda y$ for $\lambda \in \mathbb{C}$.
