proof of cauchy schwarz for inner product 
could someone explain how to get from $\|x\|^2 - \beta \langle y,x\rangle - \bar{\beta}\langle x,y\rangle + |\beta|^2\|y\|^2$ to $\|x\|^2 - |\beta|^2\|y\|^2 $? I understand $$-\beta \langle y,x\rangle - \bar{\beta}\langle x,y\rangle = -2\mathcal{Re}(\overline{\beta}\langle x,y\rangle)$$ but I don't see how we conclude?
 A: The parameter $\beta$ is presumed to be given by $\beta=\frac{\langle x,y\rangle}{||y||^2}$. 
Noting that 
$$-\beta \langle y,x\rangle -\bar \beta \langle x,y\rangle =-2\text{Re}(\bar \beta \langle x,y \rangle)$$
we see that 
$$\begin{align}-\beta \langle y,x\rangle -\bar \beta \langle x,y\rangle&=-2\text{Re}(\overline{\beta} \langle x,y\rangle)\\\\
&=2\text{Re}\left(\frac{\overline{\langle x,y\rangle}}{||y||^2}\langle x,y\rangle\right)\\\\
&=-2\frac{|\langle x,y\rangle|^2}{||y||^2}
\end{align}$$
A: $$\|w\|^2=|x\|^2 - \beta <y,x> - \bar{\beta}\langle x,y\rangle + |\beta|^2\|y\|^2$$
$$=\|x\|^2-2\mathcal{Re}(\beta<x,y>)+ |\beta|^2\|y\|^2$$
Taking into account that $\beta:=\dfrac{\langle x,y\rangle}{\|y\|^2}$:
$$\|w\|^2=\|x\|^2-2\mathcal{Re}\dfrac{|\langle x,y\rangle|^2}{\|y\|^2}+ \dfrac{| \langle x,y\rangle |^2}{\|y\|^4}\|y\|^2$$
Dropping the now useless $\mathcal{Re}$, (and simplifying the last term by $\|y\|^2$) this expression, indeed, reduces to
$$\|w\|^2=\|x\|^2-\dfrac{|\langle x,y\rangle |^2}{\|y\|^2}$$
