= Solution
Both the magnitude $J$ of <angular momentum> and rotational <kinetic energy> are conserved after the impulse. At fixed $J$, eliminate $J_3^2=J^2-J_1^2-J_2^2$ to write
$$
T=\frac{J^2}{2I_3}
+\frac{J_1^2}{2}\left(\frac1{I_1}-\frac1{I_3}\right)
+\frac{J_2^2}{2}\left(\frac1{I_2}-\frac1{I_3}\right).
$$
The <fixed-angular-momentum energy test for principal-axis rotation> makes the stability geometric. The largest moment gives a constrained energy minimum, the smallest a maximum, and the intermediate moment a saddle. Conservation of energy keeps a nearby motion near either isolated extremum, but a saddle permits departure in one direction while compensating in another. Thus both extremes are stable without dissipation; it would be incorrect to demand an energy minimum alone in this conservative problem.
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