Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $abc\geq ab+2c\geq ab+c+c\geq a+b+c+c\geq a+b+c+2$.
The main problem I face is to justify that $$abc=a+b+c+2\implies a=b=c=2.$$
My idea is to assume that $a>2$ and try to get a contradiction.

Another way,
The condition gives:
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1.$$
Thus, $$1=\sum_{cyc}\frac{1}{1+a}\leq\sum_{cyc}\frac{1}{1+2}=1,$$ which gives $a=b=c=2.$
