The Cauchy-Schwarz Inequalities (CSI) are
$$\langle a^{\dagger}a\,b^{\dagger}b\rangle^{2} \le \langle a^{\dagger 2}a^{2}\rangle\,\langle b^{\dagger 2}b^{2}\rangle$$
This is true (or "satisfied") for classical fields. In quantum physics, we are interested in violating such inequalities, in which case the $\le$ becomes a $>$ instead.
In terms of $g^{(2)}$ (satisfied): $$\left[g^{(2)}_{ab}\right]^2 \le g^{(2)}_{aa}\, g^{(2)}_{bb}$$
In terms of populations alone, the CSI reads: $$\langle n_a n_b\rangle^2 \le \langle n_a(n_a-1)\rangle\,\langle n_b(n_b-1)\rangle .$$
Clearly, this involves only the diagonal elements $p(n_a,n_b)=\langle n_a n_b|\rho|n_a n_b\rangle$ and up to its first two factorial moments (no phase, no quantum coherence, etc.)
Note that if one mode is antibunched while $\langle n_an_b\rangle\neq0$, then the rhs is zero and CSI are automatically violated. More generally: $$g^{(2)}_{ab}>\min(g^{(2)}_{aa},g^{(2)}_{bb})$$
Importantly, CS violation also depends on the choice of basis, e.g., $|1,1\rangle$ violates CSI but the state through a BS $(|20\rangle-|02\rangle)/\sqrt2$ does not.
The violation is immune to uncorrelated linear loss ($g^{(2)}$ are loss-independent) but is destroyed by uncorrelated background counts, which inflate $\langle n_a\rangle\langle n_b\rangle$ without feeding $\langle n_an_b\rangle$.
$p(n_a,n_b)$ is a bona fide classical probability distribution, so genuine Cauchy–Schwarz does apply to it, and it can never be violated: $$\langle n_a n_b\rangle^2 \le \langle n_a^2\rangle\,\langle n_b^2\rangle$$
Since $\langle n^2\rangle = \langle n(n-1)\rangle + \langle n\rangle$, one can bound the ratio $R$ of violation by: $$R \;\le\; \left(1+\frac{1}{\langle n_a\rangle\, g^{(2)}_{aa}}\right)\left(1+\frac{1}{\langle n_b\rangle\, g^{(2)}_{bb}}\right)$$ so that as both $\langle n_a\rangle g^{(2)}_{aa}$ and $\langle n_b\rangle g^{(2)}_{bb}$ get large, the violation ratio reduces to $R\to1$ and ultimately disappears: one needs antibunching and/or low occupancy.
If there is a single mode, then from Titulaer and Glauber[1], $g^{(2)}$ is one and there is no CSI: spatial CS is a witness of the mode structure, not of the state alone.
The violation is: $$\lvert \Psi(\mathbf{r}_1,\mathbf{r}_2)\rvert^2 \;>\; \lvert \Psi(\mathbf{r}_1,\mathbf{r}_1)\rvert\,\lvert \Psi(\mathbf{r}_2,\mathbf{r}_2)\rvert$$
The resolution of the camera sets a bound to the degree of violation: the larger the pixel, the less the violation (including the whole field reduces to a single mode with no violation whatsoever).
Pauli exclusion is a maximal CS violator by construction, so fermions always violate CSI.