Past exam of the options II course of the University of Oxford / 2004 / F / F2 / b

Positive definiteness gives a positive smallest eigenvalue $c$ with $x^{t}Ax\ge c|x{|}^2$. Hence the set $x^{t}Ax=y^{t}Ay=1$ is compact, and $h$ attains a maximum there. Its two constraint gradients are $(2Ax,0)$ and $(0,2Ay)$; they are linearly independent because $x,y$ are nonzero and $A$ is invertible.

Put $s=x^{t}Ay$. At any constrained extremum the multiplier equations areMultiplication by $x^t$ and $y^t$ gives $\lambda =\mu =s^2$. If $s=0$ the objective is zero. Otherwise $Ay=sAx$, so $y=sx$ and normalization gives $s^2=1$. Thus every maximum is at most one, and the value one is attained at $x=y$. This proves $\boxed{(x^{t}Ay{)}^2\le 1}$ on the normalized constraints.

For nonzero arbitrary vectors apply that result to $x/\sqrt{x^{t}Ax}$ and $y/\sqrt{y^{t}Ay}$. If either vector is zero the conclusion is immediate. ThereforeEquality for nonzero vectors holds precisely when they are linearly dependent. This is the Cauchy-Schwarz inequality for the inner product induced by $A$.