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Notice that in order for the fractions to be integers, we must have that $$a|c+1 \qquad b|a+1 \qquad c|b+1.$$These in turn imply that $$a \leq c+1 \qquad b\leq a+1 \qquad c\leq b+1.$$Temporarily assume that $a<c,b<a,$ and $c<b.$ Adding these three inequalities and simplifying yields $0<0$ which is absurd, so we must have a combination of the nonzero amount of the following three cases: Case 1: $a=c$ or $a=c+1$ Note that if any two variables are equal, then the fraction involving the two variables will not be an integer unless the variable in the denominator is equivalent to $1.$ Substituting the former condition in this case yields that $\frac{a+1}{b},\frac{b+1}{a},\frac{a+1}{a}$ must be integers. As expected, the third fraction forces $a=1=c,$ and the first fraction forces $b=1,2.$ So $(a,b,c)=\boxed{(1,1,1),(1,2,1)}$ are valid solutions. For the latter, we obtain that the third fraction is always an integer ($1$) and that $\frac{c+2}{b},\frac{b+1}{c}$ must be integers. This implies that $c+2\equiv 0 \pmod{b}$ and $b+1\equiv 0 \pmod{c}.$ Need help finishing Case 2: $b=a$ or $b=a+1$ $b=a$ implies that $a,b$ must be $1$ and that $c=1,2$ so we have the new solution $\boxed{(1,1,2)}.$ Need help finishing Case 3: $c=b$ or $c=b+1$ $b=c$ implies that $c,b$ must be $1$ and that $a=1,2$ so we have the new solution $\boxed{(2,1,1)}.$ Need help finishing No solutions exist that are not a part of the conditions imposed in the constructed cases, so we have found all possible triples.

WLOG let $a\leq b\leq c.$ We observe that $\dfrac{(a+1)(b+1)(c+1)}{abc}$ is also an integer (which is at least $2$), so $\dfrac{(a+1)}{a}\leq\dfrac{4}{3}$ (if $\dfrac{(a+1)}{a}>\dfrac{5}{4}$ then $\dfrac{(a+1)(b+1)(c+1)}{abc}$ is at most $\left(\dfrac{5}{4}\right)^3<2,$ a contradiction.) So we must have $a=1,2$ or $3.$ $a=1$ gives $b=1$ or $b=2,$ since $b\mid a+1.$ If $b=1,$ $c\mid b+1$ gives $c=1$ or $c=2.$ If $b=2,$ then $c\mid b+1$ gives $c=3$ ($c\neq 1$ since $b\leq c$) and all of these are valid solutions. $a=2$ along with $b\mid a+1$ gives $b=3,$ since $a\leq b.$ And $b\leq c$ along with $c\mid b+1$ gives $c=4$ but then $\frac{c+1}{a}$ is not an integer. $a=3$ along with $b\mid a+1$ gives $b=4,$ since $a\leq b.$ And $b\leq c$ along with $c\mid b+1$ gives $c=5,$ which is a valid solution. So the solutions are $(1,1,1),(1,1,2),(1,2,3),(3,4,5)$ and their permutations.