= Solution
At fixed total <density>, the unpolarized and fully polarized candidate energies are
$$
\frac{E_u}{V}=A2^{-2/3}n^{5/3}-\frac{\gamma n^2}{4},\qquad
\frac{E_p}{V}=An^{5/3}-\frac{\gamma n^2}{2}.
$$
Polarization increases the <kinetic energy> because one <spin> species must occupy a larger <Fermi sphere>, but gains negative exchange <energy>. The fully polarized candidate wins this endpoint comparison if
$$
A n^{5/3}(1-2^{-2/3})-\frac{\gamma n^2}{4}<0.
$$
Using the defined $r_s$ gives
$$
\boxed{r_s>4\,2^{-1/3}(1-2^{-2/3})=2^{5/3}-2=1.174802\ldots.}
$$
Thus the specified two-state comparison indeed predicts a polarized state below the unpolarized state's <energy> for $r_s>1.175$.
The <endpoint and continuous polarization thresholds of a contact electron gas> are different. To check the actual onset within the same uniform Hartree-Fock model, allow $P=(n_\uparrow-n_\downarrow)/n$ to vary continuously. In units $2A(n/2)^{5/3}$,
$$
f(P)=\frac{(1+P)^{5/3}+(1-P)^{5/3}}2-\frac{r_s}{2}(1+P^2),
$$
$$
f(P)-f(0)=\left(\frac59-\frac{r_s}{2}\right)P^2+\frac5{243}P^4+O(P^6).
$$
The unpolarized state loses local stability already at \b[$r_s=10/9\simeq1.111$]. Full polarization becomes the minimizing endpoint only when $f'(1)\le0$, or $r_s\ge(5/6)2^{2/3}\simeq1.323$. Between those values the optimum is partly polarized. Consequently $1.175$ is the requested endpoint crossing, not the exact onset of spontaneous polarization after all <spin> populations are allowed.
\Image[/past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2015/paper-1-polarization-energy.png]
{title=Contact-interaction spin energies showing partial polarization between the local onset and full saturation}
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