= Solution
Let $a_\xi=\boldsymbol\xi\cdot\nabla\Psi$, $d_\xi=\nabla\cdot\boldsymbol\xi$ and $\rho_\Psi=d\rho/d\Psi$. Since $\nabla p=-\rho\nabla\Psi$, the <pressure> and <mass density> perturbations are $\delta p_\xi=\rho a_\xi-\gamma p d_\xi$ and $\delta\rho_\xi=-\rho d_\xi-\rho_\Psi a_\xi$. Integrating the <pressure>-<gradient> term by parts in the <mass density>-weighted <inner product> gives
$$
\langle\eta,\mathcal F\xi\rangle=\int\left[-\gamma p d_\eta^*d_\xi+\rho(a_\eta^*d_\xi+d_\eta^*a_\xi)+\rho_\Psi a_\eta^*a_\xi-\rho\kappa^2\eta_R^*\xi_R\right]d\tau.
$$
The boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma p d$. Completing the <pressure> square gives
$$
\langle\eta,\mathcal F\xi\rangle=-\int\left[\frac{\delta p_\eta^*\delta p_\xi}{\gamma p}+\rho\mathcal N^2a_\eta^*a_\xi+\rho\kappa^2\eta_R^*\xi_R\right]d\tau,
$$
where
$$
\mathcal N^2=-\frac{\rho}{\gamma p}-\frac{\rho_\Psi}{\rho}=-\frac1\rho\frac{dp}{d\Psi}\left(\frac1\gamma\frac{d\ln p}{d\Psi}-\frac{d\ln\rho}{d\Psi}\right).
$$
This <effective-potential stratification coefficient> is real. All coefficients of the bilinear form are real, so $\langle\eta,\mathcal F\xi\rangle=\langle\mathcal F\eta,\xi\rangle$: the operator is symmetric on the stated boundary domain, giving the usual <self-adjoint> realization of the stellar normal-mode problem.
Set $\eta=\xi$ and use $\mathcal F\xi=-\omega^2\xi$. The <Cowling energy principle for a rotating barotropic star> is
$$
\boxed{\omega^2\int\rho|\xi|^2d\tau=Q[\xi]=\int\left[\frac{|\delta p|^2}{\gamma p}+\rho\mathcal N^2|\xi\cdot\nabla\Psi|^2+\rho\kappa^2|\xi_R|^2\right]d\tau.}
$$
The <Rayleigh quotient> $Q[\xi]/\int\rho|\xi|^2$ tests stability. If $Q\ge0$ for every admissible displacement, no mode has $\omega^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the <Rayleigh-Ritz variational principle> puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega=i\sigma$, $\sigma>0$. Equality allows neutral modes, rather than establishing strictly positive <frequencies>. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the <pressure> and other positive terms.
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