We prove this form of the Cauchy-Schwarz inequality:

(x,y) ≤ ||x||⋅||y||where

||x||=sqrt((x,x)),and we use no properties of ||.|| as a norm. Exactly the same proof works for other formulations of the inequality, and something similar for the complex case. It also works for 2n numbers, so you can show it to your algebra teacher (and not have to learn derivatives!).

Consider the quadratic function f(t) = (x-ty,x-ty). It is clear that f(t) ≥ 0 (nonnegativity of the inner product). Using bilinearity (and symmetry), we also have that

f(t) = (x,x) - 2 (x,y) t + twhich is a quadratic function (of t). Therefore its discriminant must be nonpositive (otherwise it must have 2 roots, and take on negative values). So we must have 0 ≥ Δ = 4 (x,y)^{2}⋅(y,y) ,

^{2}- 4(x,x)*(y,y), hence (x,x)*(y,y) ≥ (x,y)

^{2}, as desired.