= Solution
For a regular spherical star, <mass conservation> gives $m'=4\pi r^2\rho$. Multiplying the <hydrostatic equilibrium> equation by $r^2/\rho$ and differentiating therefore yields
$$
\boxed{\frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dP}{dr}\right)=-4\pi G r^2\rho.}
$$
At the center, finite density implies $m(r)=O(r^3)$, so a regular pressure satisfies $P(0)=P_c<\infty$ and $P'(0)=0$. At a free outer boundary in vacuum, $P(R)=0$; a surrounding medium would instead prescribe its external pressure.
Differentiate the auxiliary function, using both equations:
$$
F'=-\frac{Gm\rho}{r^2}+\frac{Gmm'}{4\pi r^4}-\frac{Gm^2}{2\pi r^5}=-\frac{Gm^2}{2\pi r^5}<0
$$
where enclosed mass is nonzero. Since $m^2/r^4\to0$ at the regular center, $F(0)=P_c$, whereas $F(R)=GM^2/(8\pi R^4)$. Integrating the strictly negative derivative gives
$$
\boxed{P_c>\frac{GM^2}{8\pi R^4}.}
$$
With nonzero external pressure, the corresponding bound is $P_c>P(R)+GM^2/(8\pi R^4)$.
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