Prove that for any real number $x$ and for any positive integer $n, \lfloor nx \rfloor \ge \sum_{i=1}^n \frac{\lfloor x i\rfloor}{i}$.
I found a solution to this problem, but why is f constant and equal to $0$ for subintervals of $[0,1)$ that don't contain numbers of the form $p/q, 2\leq q\leq n$ and $1\leq p\leq q-1, \gcd(p,q) = 1$? I know that the rational numbers satisfying this requirement must have a denominator of at least $n+1$ when written in reduced form, but for a number like $x = \frac{n}{n+1}$ and $n=2$, we get $f(x) = 1- \frac{1}2 \neq 0$.
If the solution is incorrect, then is there another solution?
