= Solution
At a circular orbit, $V'=0$. Differentiating the Schwarzschild <effective potential> at fixed $h$ gives
$$
V'=\frac{r_s}{2r^2}-\frac{h^2}{r^3}+\frac{3r_sh^2}{2r^4},\qquad\boxed{h^2=\frac{r^2r_s}{2r-3r_s}.}
$$
A timelike circular orbit requires $r>3r_s/2$. Evaluating the second derivative after this substitution gives
$$
V''=\frac{r_s(r-3r_s)}{r^3(2r-3r_s)}.
$$
A stable nondegenerate minimum therefore requires \b[$r>3r_s$]. The boundary $r=3r_s$ is the marginal <innermost stable circular orbit>, not a strict stable minimum.
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