= Solution
Let $m(r)$ be the background <enclosed mass>, so $g=Gm/r^2$ and $m'=4\pi r^2\rho$. The background <hydrostatic equilibrium> is $p'=-\rho g$. Use a Lagrangian <fluid displacement> $\xi=\xi_r$ of each spherical mass shell; $\Delta$ denotes its Lagrangian perturbation. Conservation of shell mass gives
$$
\frac{\Delta\rho}{\rho}=-\frac1{r^2}(r^2\xi)'=-(\xi'+2\xi/r),
$$
and the adiabatic <pressure> law gives the <Lagrangian pressure perturbation>
$$
\boxed{\Delta p=-\gamma p(\xi'+2\xi/r).}
$$
This retains perturbed self-gravity: a shell keeps its enclosed mass while its gravitational acceleration changes with its radius.
In mass coordinates, the exact radial shell equation is $\ddot r=-4\pi r^2\partial p/\partial m-Gm/r^2$. Its first-order perturbation is
$$
\ddot\xi=-4\pi r^2\partial_m\Delta p
-8\pi r\xi\partial_m p+2g\xi/r.
$$
Hydrostatic equilibrium gives $\partial_m p=-g/(4\pi r^2)$, and $\partial_m\Delta p=(\Delta p)'/(4\pi r^2\rho)$. Thus the perturbation of the <pressure>-force geometry contributes another $2g\xi/r$, in addition to the gravity term:
$$
\ddot\xi=-\frac1\rho(\Delta p)'+4g\xi/r
=\frac1\rho[\gamma p(\xi'+2\xi/r)]'+4g\xi/r.
$$
Put $\eta=\xi/r$, so $\xi'+2\xi/r=r\eta'+3\eta$. Expanding the derivative, using constant $\gamma$ and $p'=-\rho g$, yields the <radial stellar pulsation equation>:
$$
\boxed{\frac{\partial^2\xi_r}{\partial t^2}
=\frac1{\rho r^3}\frac{\partial}{\partial r}
\left[\gamma p r^4\frac{\partial}{\partial r}(\xi_r/r)\right]
-(3\gamma-4)\frac{g\xi_r}{r}.}
$$
The coefficient $4$ includes both geometric <pressure> and self-gravity effects; dropping the gravitational perturbation would not give this equation.
For a <normal mode> $\xi(r,t)=\xi(r)e^{-i\omega t}$, multiply by $\rho r^2\xi^*$ and integrate. Since $\xi^*/r=\eta^*$, integration by parts gives
$$
\omega^2\int_0^R\rho r^2|\xi|^2dr
=\int_0^R\left[\gamma p r^4|\eta'|^2
+(3\gamma-4)\rho gr|\xi|^2\right]dr
-\left[\gamma p r^4\eta^*\eta'\right]_0^R.
$$
Regularity at the center requires $\xi=O(r)$ with finite $\eta$, so the lower endpoint vanishes. For an isolated star take $p(R)=0$ and the free-surface condition $\Delta p=0$, with a regular finite-energy displacement; these eliminate the upper boundary term. Hence
$$
\boxed{\omega^2\int_0^R\rho r^2|\xi_r|^2dr
=\int_0^R\left[\gamma p r^4\left|\frac d{dr}(\xi_r/r)\right|^2
+(3\gamma-4)\rho gr|\xi_r|^2\right]dr.}
$$
These boundary conditions give the self-adjoint radial <Sturm-Liouville problem> behind the <weighted stellar pulsation Rayleigh quotient>.
Use the homologous trial <fluid displacement> $\xi_r=r$, or $\eta=1$. It is regular at the center, and $\Delta p=-3\gamma p$ vanishes at the free surface. The derivative term in the <Rayleigh quotient> is zero, leaving
$$
\omega_{\min}^2\leq\mathcal R[r]
=(3\gamma-4)\frac{\int_0^R\rho g r^3dr}{\int_0^R\rho r^4dr}.
$$
The ratio is a positive weighted average of $g/r$. Since the central <mass density> is its maximum,
$$
m(r)=4\pi\int_0^r\rho(s)s^2ds\leq\frac{4\pi}3\rho_c r^3,\qquad
\frac{g(r)}r=\frac{Gm(r)}{r^3}\leq\frac{4\pi}3G\rho_c.
$$
For $\gamma>4/3$ multiplication preserves this inequality, proving the <central-density upper bound for a radial stellar frequency>:
$$
\boxed{\omega_{\min}^2\leq(3\gamma-4)\frac{4\pi}3G\rho_c.}
$$
The full positive quadratic form also shows radial stability for constant $\gamma>4/3$ under these assumptions. A uniform-density equilibrium has constant $g/r$, making the homologous displacement an exact mode and saturating the bound.
If $\gamma<4/3$, the same trial value is strictly negative. Therefore \b[$\omega_{\min}^2<0$: the star has an exponentially growing radial instability], with growth rate $\sqrt{-\omega_{\min}^2}$. One must use this negative trial quotient, rather than carrying the central-density inequality across the sign change unchanged. At the limiting value $\gamma=4/3$, the homologous displacement has zero restoring force and is a neutral radial mode in this constant-exponent model.
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