= Solution
The components in <Euler equations for a torque-free rigid body> are evaluated in the body-fixed principal-axis frame, with $L_i=I_i\omega_i$. In an inertial space frame the <angular momentum> vector is constant; its components in the rotating body frame need not be.
Assume $\Omega\ne0$ and distinct positive <principal moments of inertia>. Write $\omega=(\Omega+\delta_1,\delta_2,\delta_3)$. To first order,
$$
\dot\delta_1=0,\qquad
\dot\delta_2=\frac{I_3-I_1}{I_2}\Omega\delta_3,
\qquad\dot\delta_3=\frac{I_1-I_2}{I_3}\Omega\delta_2,
$$
so
$$
\ddot\delta_2=\Omega^2\frac{(I_3-I_1)(I_1-I_2)}{I_2I_3}\delta_2.
$$
The coefficient is positive when $I_1$ lies between the other moments, giving an exponentially growing mode and instability. When $I_1$ is an extreme moment, it is negative and the transverse perturbations oscillate.
To establish actual nonlinear <Lyapunov stability> in the latter case, use the conserved <kinetic energy> $E=\frac12\sum I_j\omega_j^2$ and squared <angular momentum> $L^2=\sum I_j^2\omega_j^2$. Their combination is
$$
L^2-2I_1E=I_2(I_2-I_1)\omega_2^2+I_3(I_3-I_1)\omega_3^2.
$$
For an extreme $I_1$ its two coefficients have the same sign. Conservation controls both transverse components by their initially small values, while conservation of $E$ controls $\omega_1^2$ near $\Omega^2$. <Continuity> preserves its initial sign for sufficiently small perturbations. Thus \b[nonzero steady rotation is stable exactly about the largest or smallest principal moment], the <intermediate axis theorem>. It is not asymptotically stable, since <energy> is conserved. If $\Omega=0$, <energy> makes the zero angular-velocity state stable for every ordering; the nonzero-rotation assumption is essential to the stated equivalence.
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