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.

$$
\begin{align}
\frac{25}{84} &= \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} \\
&= \frac{bc+c+1}{abc} = \frac{(b+1)c+1}{abc}
\end{align}
$$
Now $ 84 = 2\cdot 2 \cdot 3 \cdot 7$  so $b,c$. Let us try $(b+1)c= c = 24 = 2\cdot 2\cdot 2\cdot 3$ and $abc = 84$ (this is just a guess, it can also be a multiple each time). We are looking for $a,b,c$ that are in the factorization of $84$. So $b+1=8$ and $c =3$ would work with $b=7$ since $3$ and $7$ are in the factorization of $84$. So with $abc =84=2\cdot 2 \cdot 3 \cdot 7 = a\cdot 7\cdot 3$ it follows that $a = 4$.
So $\frac{25}{84} = \frac{1}{4} + \frac{1}{28} + \frac{1}{84}$.
Another approach would be trying to find an Eyptian fraction expansion, for example:
$$\frac{25}{84} = \frac{1}{4}+\frac{1}{21} = \frac{1}{4}+\frac{1}{22}+\frac{1}{462} = \frac{1}{4}+\frac{1}{24}+\frac{1}{168} $$
