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.
 A: 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 $c$.
A: 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. $a = b = c = 2$
A: 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 $a=b=c=2.$
A: Given the results you already claim, you only need to show that
$$abc - (a+b+c+2) > 0$$
For all triples integers two or greater, except for $a=b=c=2$, where it's satisfied with equality.
I think all you need to do is differentiate with respect to the variables. Take the $c$ derivative of the LHS:
$$ab - (a + b).$$
But you already claimed that this is positive in the regime of interest. Thus, the expression can only increase if we increase $c$ (implicitly, I'm extending the integers to real numbers). The same result applies to $a,b$. Since the inequality is satisfied with equality at $a=b=c=2$ and any other possibility would have to increase one of these variables without decreasing another, any other possibility would have to be greater and no longer satisfy the expression with equality.
A: I think your answer is correct. Assume that $a>2$, since $bc\geq b+c$, you get $$abc\geq ab+ac\geq a+b + a +c>a+b+c+2.$$
All $a,b,c$ must be less than or equal to $2$.
A: Let $a=x+2$, $b=y+2$ and $c=z+2$.
Thus, $x\geq0$, $y\geq0$, $z\geq0$ and $$(x+2)(y+2)(z+2)=x+y+z+8$$ or
$$xyz+2(xy+xz+yz)+3(x+y+z)=0,$$ which gives $x=y=z=0$ and $a=b=c=2$.
A: $bc > 4 \implies abc > 4a = a+a+a+a > a+b+c+2$ since wlog $a \geq b \geq c \geq 2$.
$bc = 4 \implies b,c = 2 \implies a=2 \implies abc=a+b+c+2 \implies bc=4$ using the contrapositive of the above statement and all is proved here.
