Others have asked in general about cases in which the action integral is not minimized, but my question is specific: Can we show that the conventional action integral is always maximized for a system at stable equilibrium?
My reasoning is as follows. The conventional action integral is
At static equilibrium (whether stable or unstable), we have that the kinetic energy is identically zero and the potential energy is constant in time. Thus it would seem that, for any static equilibrium configuration, the action reduces to
We know that for stable equilibrium configurations, the potential energy is minimized (the principle of minimum potential energy). Since $S=-U\Delta t$, and $\Delta t$ is positive, it would seem that $S$ and $U$ have opposite signs, so that when $U$ is minimized, $S$ is maximized.
Is my logic sound? Or am I oversimplifying?