Let $a,b,c$ positive integers such that $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3$. Find those triples. Full question: Let $a$,$b$,$c$ be three positive integers such that $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3$. Find those triples.
This is actually a national competition question in Vietnam (Violympic), which I have attended (and did poorly, but had a lot of fun). 
I have understood almost every questions asked that day, but this one really makes my head pop, because I haven't learn much about integer solution equation and how to solve hard cases (like that one)
I have solved that a,b,c can't be all the same, because:
$$a = b = c$$
$$\implies(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})=(1+\frac{1}{a})^3=3$$
$$⟺1+ \frac{1}{a}=\sqrt[3]{3}$$
$$⟺\frac{1}{a}=\sqrt[3]{3}-1$$
$$⟺a=b=c= \frac{1}{(\sqrt[3]{3}-1)}$$
And using wolframalpha.com, I found out that $(a,b,c) \in \{(1,3,8),(1,4,5),(2,2,3)\}$, but stuck at how to solve it.
Thank you in advance for checking out. I really appreciate your effort.

Suppose $a\geq 3$, then
$$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\leq\left(1+\frac{1}{3}\right)^3=\frac{64}{27}<3$$
(note that $a\leq b\leq c$), a contradiction. Hence $a=1,2$.  
If $a=1$, it comes to solve $(1+1/b)(1+1/c)=3/2$. The same trick shows $b<5$. Now one may simply list all possible values of $b$.
The case $a=2$ can be solved similarly.
