Determine variables that fit this criterion... There is a unique triplet of positive integers $(a, b, c)$ such that $a ≤ b ≤ c$.
$$
\frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc}
$$
Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is:
Could we solve for $\frac{1}{a}$ in terms of the other variables? Then substitute that value in for each occurrence of $a$, to solve for $a$?
That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced.
Thanks.

Clearly $\displaystyle \frac{1}{a} < \frac{25}{84} < \frac{1}{3}$ so $a \geq 4$.
Then $b \geq 4$ and one of $a,b,c$ must be a multiple of $7$ so $c \geq 7$.
Hence $\displaystyle \frac{25}{84} \leq \frac{1}{a} + \frac{1}{4a} + \frac{1}{28a} = \frac{9}{7a}$ so $\displaystyle a \leq \frac{108}{25}<5$ so we must have $a = 4$.
Substitute $a=4$ into the original equation to obtain $\displaystyle \frac{c+1}{bc} = \frac{4}{21}.$ It follows that $4 \mid (c+1)$ and, since $c$ and $c+1$ are coprime, $c \mid 21$. Since we know that $c \geq 7$, we must have that $c = 7$ and so $b = 6$.
