Let $ f: \mathbb{N} \to \mathbb{N} $ be a number-theoretic function satisfying $ f(xy) = f(x) + f(y) $ whenever $ \gcd(x,y) = 1 $.
How can I prove that $ \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p) \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\} \sim \frac{1}{2} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p), $ i.e., the ratio of these two sums, as $ n $ tends to $ \infty $, is equal to $ 1 $?
Notation: $ \lceil \cdot \rceil $ denotes the ceiling function, and $ \{ \cdot \} $ the fractional-part function.
If someone could just simplify the problem, not even necessarily prove it, I would greatly appreciate it.
Also, I don't know if the following relations might help in a proof/simplification: \begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \bigg\lceil \frac{n}{k} \bigg\rceil \left\{ \frac{n}{k} \right\} &= \frac{1}{2}, \\ \lim_{n \to \infty} \frac{1}{n} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\} &= \frac{1}{2}. \end{align}