Let $p, q$ be two distinct prime number. I'm trying to provide a non-trivial upper bound for the sum $$S(p, q) = \sum_{1 \leq x < p} \sum_{1 \leq y < q} \frac{1}{\|x / p\| \, \|y / q\| \, \|x/p + y/q\|},$$ where $\|t\|$ denotes the distance of $t \in \mathbb{R}$ from the nearest integer.
Precisely, I'm interested in having $S(p, q) = o((pq)^2)$ as $p, q$ go to infinity in some way.
I know that $\min(p, q) \to +\infty$ doesn't suffices, since $S(p, q) \geq (pq)^2 / (q - p)$ (considering $x = 1$ and $y = q - 1$), and we can take a sequence of primes $p_k < q_k$ such that $p_k \to +\infty$ and $q_k - p_k$ is bounded.
Maybe $S(p, q) = o((pq)^2)$ as $p \to +\infty$ and $q/p \to +\infty$ ?
The motivation comes from the fact that $$\sum_{\substack{1 \leq z < pq \\ (pq, z) = 1}} \frac1{|p^{-1}z \bmod q|\,|q^{-1}z \bmod p| \, z} = \frac{S(p, q)}{(pq)^2} ,$$ where $|p^{-1}z \bmod q| = \min\{|r| : r \in \mathbb{Z}, pr \equiv z \pmod q\}$, and similarly for $|q^{-1}z \bmod p|$. Therefore, $S(p, q) = o((pq)^2)$ means that, on average, $|p^{-1}z \bmod q|$, $|q^{-1}z \bmod p|$, and $z$ cannot be all small.
Thanks for any help