While I was driving in my car today, I thought of a proof of the Cauchy-Schwartz theorem. I’m sure that it is completely unoriginal, but it has the advantages of both being longer and requiring more background than the usual proof (which you can find on the Wikipedia page).

Let (,) be an inner product. From the definition, we know that for two vectors *x, y* and two scalars *a, b* that

(ax+by, ax+by) = a^{2}(x,x) + 2ab (x,y) + b^{2}(y,y) ≥ 0

This is a positive-definite quadratic form in *a, b*, which means that its associated matrix has positive determinant:

(x,x) (y,y) – (x,y) (x,y) ≥ 0,

which is the result.

The real advantage of the proof, I suppose, is that if you already have the linear algebra background there’s no trick involved. It also means that using the same determinant argument there are analogues of the inequality that involve *n* vectors instead of two.