Find triples $(a,b,c)$ of positive integers such that... Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$
I found this on a local question paper, and I am unable to solve it.
Any help will be appreciated.
 A: We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $<3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.
Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2<\frac{3}{2}$ we must have $b<5$. Obviously we need $b>2$. We find $b=3$ gives the solution $(a,b,c)=(1,3,8)$ and $b=4$ gives the solution $(1,4,5)$.
Suppose $a=2$. Then we have $(1+\frac{1}{b})(1+\frac{1}{c})=2$. Since $(1+\frac{1}{3})^2<2$ we must have $b<3$. That gives the solution already noted of $(2,2,3)$.
A: 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.
Answer: $(a,b,c) \in \{(1,3,8),(1,4,5),(2,2,3)\}$
A: Let $a=b=2$ and $c=3$.  So you have $(3/2)(3/2)(4/3)=3$.
