Inequality $a+b+c \geqslant abc +2$ 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. 

For $x,y,z \geq 0 $, $ f(x,y,z) = x+y+z $ given that, $xy+yz+xz =3 = \phi(x, y, z)$
$$ \nabla f = \lambda \nabla \phi $$
$$ 1 = \lambda (y+z) \hspace {2 cm} (1) $$
$$ 1 = \lambda (x+z) \hspace {2 cm} (2) $$
$$ 1 = \lambda (x+y) \hspace {2 cm} (3) $$
$$ xy+yz+xz =3 \hspace{2 cm} (4)$$
Solveing $(1), (2), (3), \text{ and } (4)$ we get, $x=y=z=1$, so the minimum value of $f(x, y, z) = x+y+z$ under the constraint $xy+yz+xz = 3$ is $ 1 + 1 + 1 = 3$  
Again $g(x, y, z) = xyz + 2$ 
$$ \nabla g = \lambda \nabla \phi $$
$$ yz = \lambda (y+z) \hspace {2 cm} (5) $$
$$ xz = \lambda (x+z) \hspace {2 cm} (6) $$
$$ xy = \lambda (x+y) \hspace {2 cm} (7) $$
$$ xy+yz+xz =3 \hspace{2 cm} (4)$$
Solving these we get $x=y=z=1$, the maximum value of $g(x, y, z) = 1\cdot 1 \cdot 1 + 2 = 3 $
Since, $ \text{ min }( f) = \text{ max}(g) $, we have $a+b+c \geq abc+2$  
I hope there is a better method!!
