I'm considering a quantum system with $N$ bosonic degrees of freedom labelled by pairs $(x^i,p^i)$, such that the Hilbert space is $\mathcal{H}=L^2(\mathbb{R}^N)$. Given an arbitrary quantum state $|\psi\rangle$, I can compute its 2-point function \begin{align} C^{ab}=\langle\psi|\xi^a\xi^b|\psi\rangle\,, \end{align} where $\xi^a$ labels $(x^1,\cdots,x^N,p^1,\cdots, p^N)$. There always exists a (possibly mixed) Gaussian state $\rho$, such that the $2$-point function of $\rho$ is given by $C^{ab}$, this means $\mathrm{tr}(\rho\xi^a\xi^b)=C^{ab}$.
It is a well-known fact that the entanglement entropy $S_A(|\psi\rangle)$ associated to a subsystem $A$ is bounded from above by the entropy $S(\rho_A)$ of $\rho_A$, where $\rho_A$ is the Gaussian state restricted to the subsystem. This means, we always have \begin{align} S(\rho_A)\geq S_A(|\psi\rangle) \end{align}
What I'm looking for: Is there a known relation that allows me to bound $S_A(|\psi\rangle)$ from below? I'm looking for something like \begin{align} S_A(|\psi\rangle)\geq S(\rho_A)-f(|\psi\rangle)\,, \end{align} where $f$ can be computed from the higher $n$-point functions of $|\psi\rangle$ or something similar.
Why I'm interested in this: I know a class of systems for which $S(\rho_A)$ increases linearly in time and from numerical computations, I find evidence that also the true entanglement entropy of an arbitrary initial state $S_A(|\psi\rangle)$ grows linearly. So far, I could only prove it for Gaussians, but such an inequality would allow me to generalize the result...
