Exercise 2, chapter 3 of Barry Simon. A comprehensive course in analysis part 1. 

*For $x, y$ in $V$, an inner product space, and for $\lambda=re^{i\theta}\in\mathbb{C}$, use $|x+\lambda y|\geq 0$ for $\theta$ fixed to get a quadratic equation in $r$ whose roots must be
either equal or nonreal. Show that this implies $|Re[<re^{i\theta}y, x>]|\leq |x|^2|y|^2$ and so conclude the Schwarz inequality.


I have this:
$|x+\lambda y|^2\geq 0$
$<x+\lambda y,x+\lambda y>\geq 0$
$(e^{2i\theta}|y|^2)r^2+(e^{-i\theta}<x,y>+e^{i\theta}<y,x>)r+|x|^2\geq 0$
then discriminant $\Delta$ is $\Delta\leq 0$.
Therefore, 
$ (e^{-i\theta}<x,y>+e^{i\theta}<y,x>)^2-4(e^{2i\theta}|y|^2)|x|^2\leq 0$
$e^{-4i \theta}<x,y>^2+2<x,y><y,x>+<y,x>^2\leq 4|x|^2|y|^2$
And until here I arrived, I do not know how to continue to prove that  $|Re[<re^{i\theta}y, x>]|\leq |x|^2|y|^2$

A straightforward proof of the CS inequality is obtained from the orthogonal (right triangle) decomposition where $x$ is the hypotenuse:
$$
                 x = \left(x-\frac{\langle x,y\rangle}{\langle y,y\rangle}y\right)+\frac{\langle x,y\rangle}{\langle y,y\rangle}y.
$$
From this it follows that
$$
            \|x\| \ge \frac{|\langle x,y\rangle|}{\|y\|} \\
            |\langle x,y\rangle| \le \|x\|\|y\|
$$
with equality iff $x=\alpha y$ for some $\alpha$. The special case where $y=0$ is handled by reversing the roles of $x,y$. The case where $x=y=0$ is trivial.
