Last modified 2017-09-25 00:25:11 CDT

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:


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