Here is an attempt for a special case. Let me write your problem as the following:
$$
\begin{align}
\max_\mathbf{x}&\quad \sum_{n} \log\left(1+\frac{x_n}{f_n(x)}\right)\\
\text{subject to}&\quad\mathbf{A}\mathbf{x}\leq \mathbf{p}.
\end{align}.
$$
Assume the following: (i) $x> 0$, (ii)the coefficients of $f_n(x)$ are all positive $\forall ~n$, and (iii) all elements of $A$ and $p$ are positive.

Firstly, using AM-GM inequality:
$$
1+\frac{x_n}{f_n(x)} \geq 2\sqrt{\frac{x_n}{f_n(x)}}.
$$
This leads to the relaxed problem (barring constants added/multiplied):
$$
\begin{align}
\max_\mathbf{x}&\quad \sum_{n} \log\left(\frac{x_n}{f_n(x)}\right)\\
\text{subject to}&\quad\mathbf{A}\mathbf{x}\leq \mathbf{p}, ~\mathbf{x}>0.
\end{align}.
$$
Now, introduce variables $t_n> 0, \forall n$, such that:
$$
\frac{x_n}{f_n(x)} \geq \frac{1}{t_n} \Rightarrow x^{-1}_nt_n^{-1}f_n(x) \leq 1.
$$
And then the relaxed problem is equivalent to (removing $\log$ as its a monotonic):
$$
\begin{align}
\min_\mathbf{x}&\quad \prod_{n} t_n\\
\text{subject to}&\quad\mathbf{A}\mathbf{x}\leq \mathbf{p},~\mathbf{x}> 0\\
&~~~~~\mathbf{t}>0,~x^{-1}_nt_n^{-1}f_n(x) \leq 1~ \forall n.
\end{align}.
$$
Note that the cost function and constraints are all *posynomials*, with the assumptions made. Thus, this is in the form of a Geometric Program (see https://en.wikipedia.org/wiki/Geometric_programming), which can be converted to a convex program. In fact, softwares like CVXPY will readily solve a problem in this form.

As far as relaxation gap is concerned, that still needs some thought. Anyway, hope this helps.