= Solution
<Positive definiteness> gives a positive smallest <eigenvalue> $c$ with $x^tAx\geq c|x|^2$. Hence the set $x^tAx=y^tAy=1$ 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 $s=x^tAy$. At any constrained extremum the multiplier equations are
$$
sAy=\lambda Ax,\qquad sAx=\mu Ay.
$$
Multiplication 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 $\boxed{(x^tAy)^2\leq1}$ on the normalized constraints.
For nonzero arbitrary vectors apply that result to $x/\sqrt{x^tAx}$ and $y/\sqrt{y^tAy}$. If either vector is zero the conclusion is immediate. Therefore
$$
\boxed{(x^tAy)^2\leq(x^tAx)(y^tAy).}
$$
Equality for nonzero vectors holds precisely when they are <linearly dependent>. This is the <Cauchy-Schwarz inequality> for the <inner product> induced by $A$.
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