TITLE: A natural proof of the Cauchy-Schwarz inequality
QUESTION [27 upvotes]: Most of the proofs of the Cauchy-Schwarz inequality on a pre-Hilbert space use a fact that if a quadratic polynomial with real coefficients takes positive values everywhere on the real line, then its discriminant is negative(e.g. Conway: A course in functional analysis).
I think this is somewhat tricky.
Moreover I often forget its proof when the pre-Hilbert space is defined over the field of complex numbers.
Is there a more natural proof (hence it's easy to remember) which is based on a completely different idea?

REPLY [6 votes]: Here's my favorite proof, mainly because it's nicely symmetric, easy to remember and not impossible to come up with (the main trick is that $2=1+1$):
We want to prove
$$|\langle x,y\rangle|\le\|x\|\|y\|$$
This is linear in $x$ or $y$ and obviously holds for $x=0$ or $y=0$. Therefore without loss of generality we can suppose that $\|x\|=\|y\|=1$:
$$|\langle x,y\rangle|\le\|x\|\|y\|\Longleftrightarrow|\langle x,y\rangle|\le1\Longleftrightarrow2|\langle x,y\rangle|\le2\Longleftrightarrow2|\langle x,y\rangle| \le \|x\|^2+\|y\|^2$$
Let $u\in\mathbb C$ be a complex unit (i.e. $|u|=1$), then
$$\begin{align}0\le\|x-uy\|^2&=\langle x-uy,x-uy\rangle=\langle x,x\rangle-u\langle x,y\rangle-\overline{u\langle x,y\rangle}+\langle y,y\rangle\\&=\|x\|^2+\|y\|^2-2\,\text{Re}(u\langle x,y\rangle)\end{align}$$
Now just set $u$ so that $\text{Re}(u\langle x,y\rangle)=|\langle x,y\rangle|$ and you are done.
Of course in the real case you can just expand $0\le\|x\pm y\|^2$.