The problem is as follows:
Find the unique ordered triple of positive integers $(a,b,c)$ with $a\leq b\leq c$ such that
$$\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{84}$$
The problem is easy to solve by first bounding $a$, then bounding $b$ and finally $c$. This is the solution I was given:
It's obvious that $a\geq 4$, otherwise $\frac{1}{ab}+\frac{1}{abc}$ would have to be negative, which we don't want. We also have
$$\frac{25}{84}=\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}\leq \frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=\frac{1+a+a^2}{a^3}$$
$$\frac{25}{84}\leq \frac{1+a+a^2}{a^3}$$
so $a\leq 4$ in order to satisfy the inequality, hence $a=4$. We can proceed in a similar manner to solve the rest of the problem.
Yes, this solution is simple and beautiful, but suppose I just want to solve this problem using a rigorous number-theoretic method (e.g. without inequalities). How would I go about doing this?
*EDIT* By the way, this problem was from part C of the 2013 COMC.
