Prove that for any positive integers $x, m, n$:
$$\sum_{i=1}^n\min\left(\left\lfloor\frac{x}{i} \right\rfloor,m\right)=\sum_{i=1}^m\min\left(\left\lfloor\frac{x}{i}\right\rfloor,n\right)$$
Intuitively this kind of sounds like it would be correct, but I am not sure how to write the proof to put it into words. I was thinking of doing casework on when $\min(\lfloor \frac{x}{i}\rfloor,m)$ changes to $m$, but then I got stuck.
On a $yz$-plane (not reusing the letter $x$), consider the region
$$\begin{cases} 0 < y \le m\\ 0 < z \le n\\ yz \le x\\ \end{cases}$$
and count the number of integer lattice points inside that region. Counting points along the $y$ or $z$ directions give the two sides of the equation:
When adding the lattice points for each $y$, the region can also be written as:
$$\begin{cases} 0 < y \le m\\ 0 < z \le \min\left(\frac x y, n\right)\\ \end{cases}$$
So the sum is $\sum_{y = 1}^m \min\left(\left\lfloor\frac x y\right\rfloor, n\right)$.
Similarly, when adding the lattice points for each $z$, the region can be written as: $$\begin{cases} 0 < z \le n\\ 0 < y \le \min\left(\frac x z, m\right)\\ \end{cases}$$
So the sum is $\sum_{z = 1}^n \min\left(\left\lfloor\frac x z\right\rfloor, n\right)$.
The same sum represented in the two ways is equal.