= Solution
Write the mean interior density as
$$
\bar\rho(r)=\frac{3m_r}{4\pi r^3}.
$$
Since
$$
\frac{d\bar\rho}{dr}=\frac3r(\rho-\bar\rho),
$$
the assumed outward decrease of $\bar\rho$ implies $\rho\leq\bar\rho$. In mass coordinates, <hydrostatic equilibrium> is
$$
\frac{dP}{dm_r}=-\frac{Gm_r}{4\pi r^4}
=-\frac G3\left(\frac{4\pi}{3}\right)^{1/3}
\bar\rho^{4/3}m_r^{-1/3}.
$$
For every interior mass $m\leq m_r$, monotonicity gives
$$
\bar\rho(r)\leq\bar\rho(m)\leq\rho_c.
$$
Integrating from the centre and using $\int_0^{m_r}m^{-1/3}dm=3m_r^{2/3}/2$ proves
$$
\boxed{\frac G2\left(\frac{4\pi}{3}\right)^{1/3}
\bar\rho^{4/3}m_r^{2/3}
\leq P_c-P(r)\leq
\frac G2\left(\frac{4\pi}{3}\right)^{1/3}
\rho_c^{4/3}m_r^{2/3}}.
$$
Let $\mathcal R$ be the gas constant per mole. At the centre,
$$
\beta_cP_c=\frac{\mathcal R}{\mu}\rho_cT_c,
\qquad
(1-\beta_c)P_c=\frac{aT_c^4}{3}.
$$
Eliminating $T_c$ gives the <Eddington quartic relation> in its central form,
$$
\boxed{\frac{1-\beta_c}{\beta_c^4}
=\frac a3\left(\frac{\mu}{\mathcal R}\right)^4
\frac{P_c^3}{\rho_c^4}}.
$$
At the surface, set $P=0$ and $m_r=M$ in the upper pressure bound. Cubing it gives
$$
\frac{P_c^3}{\rho_c^4}\leq\frac{\pi}{6}G^3M^2.
$$
The function $(1-\beta)/\beta^4$ decreases strictly as $\beta$ increases on $0<\beta<1$. Define $\beta^*$ by equality in the resulting bound:
$$
\frac{1-\beta^*}{\beta^{*4}}
=\frac{\pi a}{18}
\left(\frac{\mu}{\mathcal R}\right)^4G^3M^2.
$$
Then $\beta_c\geq\beta^*$, or
$$
\boxed{1-\beta_c\leq1-\beta^*},
$$
and rearrangement gives exactly
$$
\boxed{M=\left(\frac6\pi\right)^{1/2}
\left[\left(\frac{\mathcal R}{\mu}\right)^4
\frac3a\,\frac{1-\beta^*}{\beta^{*4}}\right]^{1/2}
G^{-3/2}}.
$$
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