= Solution
Put $x=\omega/\omega_0$. Maximizing the output <wave amplitude> is equivalent to minimizing
$$
D(x)=|H|^{-2}=(1-x^2)^2+\frac{x^2}{Q^2}.
$$
Its derivative is $D'(x)=2x[2x^2-2+Q^{-2}]$. An interior minimum at positive <frequency> therefore requires $Q>1/\sqrt2$, and occurs at
$$
\boxed{\omega_{\rm res}=\omega_0\sqrt{1-\frac1{2Q^2}},\qquad Q>\frac1{\sqrt2}.}
$$
Indeed $D''(x_{\rm res})=8x_{\rm res}^2>0$, and the peak value is $|H|_{\max}=Q/\sqrt{1-1/(4Q^2)}$. For $0<Q\leq1/\sqrt2$, $D$ increases for $x>0$ and the largest output is the DC value $|H(0)|=1$. At equality the putative peak has reached zero <frequency>; there is no distinct nonzero-<frequency> <resonance>. This threshold concerns <capacitor> <voltage> peaking: the <series RLC circuit>'s <electric current> still has its maximum at $\omega_0$ for every positive $R$.
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