a	b	c
2	4	15
2	5	9
2	6	7
3	3	8
3	4	5

We can argue using inequalities to find all the possible cases. Let

${\displaystyle E=(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})}$

Notice that if $a$ is too small then $E$ will be too big and if a is too big then $E$ will be too small so, knowing that a is a whole number, we only have a few possibilities. We use similar reasoning to find $b$ and then calculate $c$ in each case.

If $a=1$ then $E>2$ so we know $a\ne 1$. If $a$ is 4 or more then $E\le (5/4)(5/4)(5/4)<2$ so we know that $a$ is equal to $2$ or $3$. By similar arguments we can now find all possible solutions.

In the first case, $a=2$ and we have to find positive integers $b$ and $c$ to satisfy

${\displaystyle F=(1+\frac{1}{b})(1+\frac{1}{c})=\frac{4}{3}}$

If $b$ is equal to 1, 2 or 3 then $F>4/3$ so we know b is at least 4. If b is 7 or more then $F\le (8/7)(8/7)=(64/49)=(4/3)(48/49)<4/3$ so we know that b must be 4, 5, or 6.

Taking b = 4 we have

${\displaystyle 1+\frac{1}{c}=\frac{4}{3}.\frac{4}{5}=(1+\frac{1}{15})}$

So $c=15$. Similarly if $b=5$ then $c=9$ and if $b=6$ then $c=7$.

In the second case $a=3$ and we have to find positive integers $b$ and $c$ to satisfy

${\displaystyle G=(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}}$

Again we find all the possible cases from considering inequalities. If $b$ is equal to 1 or 2 then $G>3/2$ and if $b$ is 5 or more then $G\le (6/5)(6/5)=(3/2)(24/25)<3/2$ so we know that $b=3\mathrm{or}4$ and we can calculate the corresponding values $c=8$ and $c=5$.

It is not hard work here to 'exhaust' all the possible cases and we have now found all the solutions.