Integer solution of $abc=a+b+c+2$.

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 $a=b=c=2.$

If $a>2$, then $ab>a+b$, so following your chain of inequalities, the final inequality is strict. Thus, if they’re equal than $a=2$ and by symmetry so are $b$ and $c$.

WLOG we assume $c \geq b \geq a \geq 2$

Then $3c \geq a + b + c$

We have,

$(ab-3) c \geq 2$

$abc \geq 2 + 3c \geq 2 + a + b + c$

For the equality case,

we are given, $abc = a + b + c +2 \leq 2 + 3 c$

$ \implies (ab-3) c \leq 2$

$ab - 3 \leq \dfrac{2}{c} \leq 1 \ ( \text {as } c \geq 2)$

That leads to $ab = 4, c = 2$.

i.e. $a = b = c = 2$