Set $a_{n+1} = a_1$ and let $0 \le x \le y$. Since
$$f(y) - f(x) = (y - x) \sum_{i=1}^{n} \frac{a_{i+1} - a_i}{(a_{i+1} + x)(a_{i+1} + y)}$$
showing $f$ is decreasing amounts to showing
$$\sum_{i=1}^{n} \frac{a_{i+1}}{(a_{i+1} + x)(a_{i+1} + y)} \le \sum_{i=1}^{n} \frac{a_i}{(a_{i+1} + x)(a_{i+1} + y)}$$
Noticing that $a_i \le a_j$ if and only if $(a_i + x)^{-1}(a_i + y)^{-1} \ge (a_j + x)^{-1}(a_j + y)^{-1}$, the above inequality is a straightforward consequence of the rearrangement inequality for the $a_i$ and the $(a_i + x)^{-1}(a_i + y)^{-1}$.