= Solution
Take $\theta$ as the inclination of the symmetry axis to the upward vertical, $\phi$ as its azimuth, and $\psi$ as rotation of the body about that axis relative to the line of nodes. The <kinetic energy> and <potential energy> are
$$
T=\frac A2(\dot\theta^2+\dot\phi^2\sin^2\theta)+\frac C2(\dot\psi+\dot\phi\cos\theta)^2,\qquad V=mg\ell\cos\theta.
$$
The <cyclic coordinates> $\psi,\phi$ and time independence give three conserved quantities:
$$
L=C(\dot\psi+\dot\phi\cos\theta),\qquad
p_\phi=A\dot\phi\sin^2\theta+L\cos\theta,\qquad E=T+V.
$$
Their meanings are the component of <angular momentum> along the body axis, the component along the vertical, and the total <energy>.
For a vanishingly small perturbation of the upright spinning state, $L\simeq Cn$, $p_\phi\simeq L$, and $E\simeq L^2/(2C)+mg\ell$. Eliminating $\dot\phi$ gives the limiting <effective potential>
$$
U_{\rm eff}(\theta)=\frac{L^2}{2A}\tan^2(\theta/2)+\frac{L^2}{2C}+mg\ell\cos\theta.
$$
Its quadratic change near zero is $\{L^2/(8A)-mg\ell/2\}\theta^2$. Thus if \b[$C^2n^2>4mg\ell A$], zero is a strict local minimum and a sufficiently small perturbation remains near it.
In the unstable case, the limiting energy equation becomes
$$
\frac A2\dot\theta^2=(1-\cos\theta)\left[mg\ell-\frac{L^2}{2A(1+\cos\theta)}\right].
$$
At the nonzero turning point the bracket vanishes, yielding
$$
\boxed{\cos\theta_{\max}\simeq\frac{C^2n^2}{2mg\ell A}-1}.
$$
The approximation is the small-perturbation limit, not an assertion that arbitrary disturbances preserve the upright state's exact conserved quantities.
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