# Proof of the Cauchy-Schwarz Inequality

For two vectors $$f$$ and $$g$$ in an inner product space, pick an arbitrary real scalar $$\lambda$$ and write:

which is true because it’s a length squared, so

The expression on the left is quadratic in $$\lambda$$, and since it’s always greater than or equal to zero, it has zero real roots (i.e. entirely above the x axis) or one real root (i.e. just touching the x axis).

It can’t have two real roots, because that would require that the expression be negative for some $$\lambda$$ (i.e. go underneath the x axis), and that can’t be the case since it’s a length squared.

Altogether, this means that the quadratic’s discriminant is less than or equal to zero:

Equality occurs when $$f$$ and $$g$$ are linearly dependent. This means $$f = \lambda g$$ for some scalar $$\lambda$$, and we can show this directly by writing:

See some similar/alternate proofs and applications of the Cauchy-Schwarz Inequality here: http://cnx.org/content/m10757/latest/