Cauchy-Schwarz inequality proof Say $V$ is a vector space with inner product $\langle x,y\rangle$ and norm $\|x\| = \langle x,x\rangle^{1/2}$. Then 
$$\langle x,y\rangle\leq \|x\| \|y\|.$$
Solution: say $x, y$ are not $\theta$, otherwise we have desired result easily. Letting $c = \frac{1}{\|x\|}$ and $d = \frac{1}{\|y\|}$. Then,
$$\|cx \pm dy\|^2 = \langle cx \pm dy, cx \pm dy\rangle$$
and this in turn equals: 
\begin{align*}
&\langle cx, cx\rangle\pm 2\langle cx,dy\rangle \pm \langle dy, dy\rangle \\
=& c^2 \langle x, x\rangle \pm 2cd \langle x,y\rangle \pm d^2 \langle y, y\rangle \\
=& 1 \pm 1 \pm \frac{2\langle x , y\rangle}{\|x\| \|y\|}.
\end{align*}
Here I'm stuck. Can someone help me see the light? Thanks.

I think you need to make sure that you are using a single value $u=cx\pm dy$ and compute $\left<u,u\right>$. This gets rid of most of the $\pm$s in the equation, because they are the same. Your argument is computing, amongst other things, $\left<cx+dy,cx-dy\right>$ which is unnecessary.
So $$0\leq \left<u,u\right> = 1 + 1 \pm \frac{2\left<x,y\right>}{||x||||y||}$$
So:
$$\mp \frac{2\left<x,y\right>}{||x||||y||}\leq 2$$
So:
$$\left|\frac{2\left<x,y\right>}{||x||||y||}\right|\leq 2$$
And you are basically done.
