TITLE: Inequality $a+b+c \geqslant abc +2$
QUESTION [15 upvotes]: Assuming $a,b,c \in (0, \infty)$, we need to prove that: 
$$a+b+c \geqslant a b c+2 \quad \text{if} \quad  ab+bc+ca=3$$
Can you give me an idea, please? This inequality seem to be known, but I didn't manage to solve it.

REPLY [8 votes]: It's easy to see that $(x+y+z)^2 \geq 3(xy+yz+zx)$.Denoting $S=a+b+c$ we cand deduce from this that $S\geq 3$.
Plugging in $x=ab$, $y=bc$, $z=ca$ we get $9\geq 3abc(a+b+c)$, so $abc \leq \frac{3}{a+b+c}$.So it's enough to prove that $a+b+c \geq \frac{3}{a+b+c}+2$.
The last inequality is equivalent to $S^2-2S-3 \geq 0$ which is true since $S\geq 3$.