# Proving Schwarz inequality for complex numbers using calculus

I am working on this problem for real analysis and it's rather baffling. I've never really seen any of this kind of notation before which is part of it I think? But also it just seems utterly mysterious. Here's the problem:

For $j=1,2,...,n$, let $a_j,b_j$ be complex numbers. Define a real valued function $f(x,y)$ of two variables $x,y$ by $z=x+iy$ and $f(x,y)=\sum_j |a_j+zb_j|^2$. The idea is to make an optimal choice of $x,y$ and substitute it into $f(x,y)$.

a) Find $z=x+iy$ that solves $Lf=0$, where $L=\frac{1}{2}(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y})$, if it exists. Use notation $A=\sum_j a_j \bar{a_j}, B=\sum_j b_j \bar{b_j}$, $C=\sum_j a_j \bar{b_j}$.

b) Prove that $|\sum_j a_j \bar{b}_j|^2 \leq (\sum_j|a_j|^2)(\sum_j|b_j|^2)$

I've made some strides but I feel like I'm fumbling around in the dark. I'm pretty sure I got part (a)—I just included it for reference but I got $z=-\frac{C}{B}$. As per the suggestion, I plugged that into $f$ and got $f(z=-\frac{C}{B})=A-\frac{C \bar{C}}{B}$. So that's the minimum value of $f$ but I can't for the life of me understand how that helps me.

Can someone help me out?

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I think you mean $f\left(-\frac{C}{B}\right)$ not $f\left(z -\frac{C}{B}\right)$. –  Michael Albanese Sep 28 '12 at 8:32
yup. Well actually I meant $f(z=-\frac{C}{B})$, just cause I felt weird about saying $f(-\frac{C}{B})$ when $f$ is a function of $x,y$ –  crf Sep 28 '12 at 8:38
Fair enough. In order to prove part b, all you have left to do is argue that $f\left(-\frac{C}{B}\right) \geq 0$. –  Michael Albanese Sep 28 '12 at 8:42
Think quadratic formula - what is $-\frac{C}{B}?$ –  Taylor Sep 28 '12 at 8:45
@MichaelAlbanese !! I can do that. So I'm very close. But I don't see why that proves it. I guess it must be because somehow we can express $(\sum_j|a_j|^2)(\sum_j|b_j|^2)-|\sum_j a_j \bar{b}_j|^2$ in terms of $f$ but I don't see how.. –  crf Sep 28 '12 at 8:56
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You are already done. You know that $f(z)\geq0$ for all $z$. So in particular $A-|C|^2/B\geq0$, i.e. $|C|^2\leq A B$, which is what you wanted to prove.