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 + t^{2}⋅(y,y) ,

which 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} - 4(x,x)*(y,y), hence
(x,x)*(y,y) ≥ (x,y)

^{2}, as desired.