Why does the Cauchy-Schwarz inequality hold in any inner product space?

Wikipedia has the one I've usually seen.

Cauchy-Schwarz Inequality

Nothing especially tricky about it, just a really clever setup.

The proof on Wikipedia is valid, and is instructive because it uses ideas which are applicable elsewhere in linear algebra. The proof I've usually seen selects vectors u,v and defines

$$f(t) = \langle u + t v , u + t v \rangle$$

and notices that f is nonnegative for all t, by the positive definiteness of the inner product. Expanding out the inner product by bilinearity and combining like terms by symmetry, you get a quadratic function of t:

$$f(t) = \| v \|^2 t^2 + 2 \langle u,v\rangle t + \| u \|^2.$$

(In the complex case you have to do something slightly different.)

This is nonnegative, so its discriminant is nonpositive, in other words:

$$4 \langle u,v\rangle^2 - 4 \| u \|^2 \| v \|^2 \leq 0$$

from which the inequality follows.

An equivalent way of looking at this (for $v \neq 0$) is to find the minimum of $f$. This occurs at $t=-\langle u , v \rangle/\| v \|^2$. So:

$$\langle u , v \rangle^2/\| v \|^2 - 2 \langle u , v \rangle^2/\| v \|^2 + \| u \|^2 = \| u \|^2 - \langle u , v \rangle^2/\| v \|^2 \geq 0$$

A simple rearrangement yields the Cauchy-Schwarz inequality. This is somewhat like the proof on Wikipedia, because $u-\langle u,v \rangle/\| v \|^2 v$ is the projection of $u$ onto the orthogonal complement of the span of $v$.

This is an old question but here is an interesting proof that generalizes it a bit. This is adapted from Folland's Real Analysis. Consider some unit scalar $\alpha$ (that is $|\alpha| = 1$) and some $t \in (0,\infty)$. Then,

$$\begin{align*} \langle x -\alpha ty, x-\alpha ty\rangle &= \langle x, x-\alpha ty\rangle + \langle -\alpha ty, x-\alpha ty\rangle \\ &= \langle x,x\rangle - \left(\overline{t} \langle x, \alpha y\rangle +t \langle \alpha y,x\rangle\right) + \underbrace{\alpha\overline{\alpha}}_{=|\alpha|^2=1} t^2 \langle y,y\rangle \\ &= \langle x,x\rangle - \underbrace{\left(\overline{t \langle x, \alpha y\rangle} + t \langle x,\alpha y\rangle\right)}_{2t\operatorname{Re} \langle x,y\rangle} + t^2 \langle y,y\rangle \\ &= \|y\|^2 t^2 - 2\operatorname{Re} \langle x,\alpha y\rangle t + \|x\|^2. \end{align*}$$

Now, this is a quadratic in $t$, and we observe that it opens upward and is non-negative. Therefore, consider its discriminant, $b^2 -4ac \le 0$.

We have $b = 2\operatorname{Re} \langle x,\alpha y\rangle$. However, since $\alpha = \frac{z}{\|z\|} = \operatorname{sgn} z$ for some $z$, we have $\langle x, \alpha y\rangle = \langle \alpha y, x\rangle = |\langle x, y\rangle|$; hence $\operatorname{Re}\langle x,\alpha y\rangle = |\langle x, y\rangle|$.

Therefore, we obtain

$$4|\langle x,y\rangle|^2 \le 4\|x\|^2\|y\|^2 \implies |\langle x,y\rangle| \le \|x\|\|y\|.$$