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.

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There is a small error in your calculation. The second $\pm$ sign in your $\langle cx, cx\rangle\pm 2\langle cx,dy\rangle \pm \langle dy, dy\rangle$ (and the following two lines) should be a plus sign. –  user1551 Jan 9 '13 at 19:35

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.

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