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 + t2⋅(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.