Polar Norms and the Cauchy-Schwarz Inequality

Theorem: Suppose $x,y\in\mathbb{R}^n.$ Then $(x'y)^2\leq x'x.y'y,$ with equality if and only if $x$ and $y$ are proportional.

Proof: The result is trivially true if either $x$ or $y$ is zero. Thus we suppose both are non-zero. We have $h(\lambda)\mathop{=}\limits^{\Delta}(x-\lambda y)'(x-\lambda y)\geq 0$ for all $\lambda.$ Thus $$\min_\lambda h(\lambda)=x'x-\frac{(x'y)^2}{y'y}\geq 0,$$ which is the required result. QED