= Solution
The <Hubbard model> combines hopping with a local repulsion,
$$
H=-t\sum_{\langle ij\rangle,\sigma}(c_{i\sigma}^\dagger c_{j\sigma}+\mathrm{h.c.})+U\sum_i n_{i\uparrow}n_{i\downarrow},\qquad U>0.
$$
Hopping favours a broad itinerant band; the interaction penalizes opposite-spin <double occupancy>. For <uniform Stoner decoupling of the Hubbard interaction>, let $s=n_\uparrow-n_\downarrow$ be the spin imbalance per site. The interaction is $U(n^2-s^2)/4$, so polarization lowers it by $Us^2/4$, at the expense of kinetic energy.
If $D(E_F)$ is the noninteracting <density of states> \b[per spin per site], small polarization costs $s^2/[4D(E_F)]$. Therefore
$$
\boxed{\Delta E=\frac{s^2}{4}[D(E_F)^{-1}-U]+O(s^4),\qquad UD(E_F)>1}
$$
is the mean-field ferromagnetic instability criterion. The exchange splitting is $\Delta=Us$, and the paramagnetic response is enhanced as $\chi=\chi_0/[1-UD(E_F)]$. A <density of states> including both spins changes the written factor in the criterion.
The model explains how itinerant bands can polarize without imposing pre-existing localized atomic spins. It is not a claim that every large-$U$ metal becomes ferromagnetic: near half filling the <Hubbard model> can instead form a <Mott insulator> with antiferromagnetic <superexchange> of scale $4t^2/U$. <Band filling>, <bandwidth> and fluctuations matter.<Quantum Gases by David Tong>[Section 3.6.6: the free-electron magnetic response]{ref} The interaction-energy and kinetic-cost expansion above is the stated mean-field derivation of the enhancement.<Electron correlations in narrow energy bands by John Hubbard>[Original local-repulsion model and the competition between atomic and itinerant limits]{ref}
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