= Solution
A <Lagrange top> is a <rigid body> that is symmetric about a principal axis, has a point on that axis fixed in space, and has its <center of mass> on the same axis while gravity acts uniformly. Here $I_1=I_2$ is the transverse <principal moment of inertia> about the fixed point, $I_3$ is the moment about the symmetry axis, $M$ is the total mass, and $l$ is the distance from the fixed point to the center of mass.
The <Euler angles for a symmetric top> use $\theta$ for inclination, $\phi$ for precession, and $\psi$ for spin about the body axis. Since $\psi$ is a <cyclic coordinate>, its <generalized momentum>
$$
\boxed{p_\psi=I_3(\dot\psi+\dot\phi\cos\theta)}
$$
is conserved. The coordinate $\phi$ is also cyclic, so
$$
\boxed{p_\phi=I_1\dot\phi\sin^2\theta+p_\psi\cos\theta}
$$
is a second integral. Finally, the Lagrangian has no explicit time dependence, and <conservation of energy from time-translation invariance> gives the independent integral
$$
\boxed{E=\frac12I_1(\dot\theta^2+\dot\phi^2\sin^2\theta)
+\frac{p_\psi^2}{2I_3}+Mgl\cos\theta.}
$$
For steady precession set $\theta$ constant and $\dot\phi=\Omega$ constant. The $\theta$ <Euler-Lagrange equation>, with $0<\theta<\pi/2$, reduces after division by $\sin\theta$ to
$$
I_1\Omega^2\cos\theta-p_\psi\Omega+Mgl=0.
$$
This quadratic has a real precession rate precisely when its <quadratic discriminant> is nonnegative. Hence <Steady precession of a Lagrange top> is possible if and only if
$$
\boxed{p_\psi^2\geq4MglI_1\cos\theta.}
$$
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