if $c \ge 4$ then $1+\frac{1}{a} \leq 1+\frac{1}{b} \leq 1+\frac{1}{c} \leq \frac{5}{4}$ $\to \left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \leq \frac{125}{64}<2$ so $c=2$ or $c=3$ And...

$(5,4,3)$

Bosnia-Herzegovina JBMO TST 2015 P2 wrote: Find all triplets of positive integers $a$, $b$ and $c$ such that $a \geq b \geq c$ and $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=2$ Solution: WLOG assume, $a \geq b \geq c$ $\implies$ $\frac{1}{c} \geq \frac{1}{b} \geq \frac{1}{a}$ $$2=\left(1+\frac{1}{a} \right) \left(1+\frac{1}{b} \right) \left(1+\frac{1}{c} \right) \leq \left(1+\frac{1}{c} \right)^3 \implies \sqrt[3]{2} -1 \leq \frac{1}{c} \implies c=1,2,3$$For $c=1$, we have no solution! For $c=2$ $\implies$ $(a,b,c)$ $\equiv$ $(7,6,2),$ $(9,5,2),$ $(15,4,2)$ And for $c=3$ $\implies$ $(a,b,c)$ $\equiv$ $(8,3,3),$ $(5,4,3)$ hence the permutations of all these solutions are the complete solutions

How do we work after founding c=2,3???

sttsmet wrote: How do we work after founding c=2,3??? I think you could proceed similarly to find the bounds of a and b, e.g. if c=2 4/3=(1+1/a)(1+1/b)$\leq$(1+1/b)^2, meaning 2/sqrt3-1$\leq$1/b meaning b is less than or equal to 6, and now test values.

sttsmet wrote: How do we work after founding c=2,3??? For $c=2$ $(a-3)(b-3)=12$ For $c=3$ $(a-2)(b-2)=6$