= Solution
The <generalized momentum>[generalized momenta] conjugate to the two <ignorable coordinates> are
$$
\boxed{
p_\phi
=I_1\dot\phi\sin^2\theta
+I_3(\dot\psi+\dot\phi\cos\theta)\cos\theta
}
$$
and
$$
\boxed{
p_\psi=I_3(\dot\psi+\dot\phi\cos\theta)
}.
$$
Since the Lagrangian contains neither $\phi$ nor $\psi$, both momenta are constant in time.
Eliminating the two angular velocities gives
$$
\dot\phi=\frac{p_\phi-p_\psi\cos\theta}
{I_1\sin^2\theta},
\qquad
\dot\psi+\dot\phi\cos\theta=\frac{p_\psi}{I_3}.
$$
Conservation of energy then yields the <heavy symmetric top reduction>
$$
\frac12I_1\dot\theta^2+V_{\rm eff}(\theta)=E,
$$
where
$$
\boxed{
V_{\rm eff}(\theta)
=\frac{(p_\phi-p_\psi\cos\theta)^2}
{2I_1\sin^2\theta}
+\frac{p_\psi^2}{2I_3}
+Mgl\cos\theta
}.
$$
As $\theta\to0$, the first term diverges unless
$$
\boxed{p_\phi=p_\psi}.
$$
Writing their common value as $p$, the small-angle expansion is
$$
V_{\rm eff}(\theta)
=\frac{p^2}{2I_3}+Mgl
+\left(\frac{p^2}{8I_1}-\frac{Mgl}{2}\right)\theta^2
+O(\theta^4).
$$
By the <effective potential stability criterion>, the vertical state is stable exactly when
$$
\boxed{p^2>4I_1Mgl}.
$$
At the vertical state $p=I_3(\dot\psi+\dot\phi)$, so this is the <gyroscopic stabilization of an inverted symmetric top> by sufficiently rapid spin.
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