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.

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. $a = b = c = 2$
