<s>[INST] 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. 
 [/INST] Solve for $c$ from the constraint:
$$
   c = \frac{3- a b}{a+b}
$$
and substitute back into the inequality:
$$
 a + b + \frac{3-a b}{a+b} \geqslant 2 + \frac{a b(3-a b)}{a+b} \implies \frac{3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3}{a+b} \geqslant 0
$$
Multiply both sides by $(a+b)$:
$$
   3 + a^2 b^2 + a^2 + b^2 -2 a b -2a -2b+3 \geqslant 0
$$
The left-hand-side can be reduced to the shifted sum of squares:
$$
     (a b-2)^2 + (a+b-1)^2  - 2 \geqslant 0
$$
It is easy to see that the left-hand-side attains its minimum at $a=b=1$, where the inequality is saturated.
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