<s>[INST] Prove that: $(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$ Let $a,b,c$ be real numbers such that $ab+bc+ca=3$. Prove that:$$(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$$
I have an imperfect solution:
Let $a+b+c=p;ab+bc+ca=q=3;abc=r$
The problem is:
$$r^2-2(q^2-2pr)+4(p^2-2q)+8-3(p^2-2q)\ge18$$
or:
$$r^2-4pr+p^2+2\ge0 $$
By $p^2\ge3q=9;pr\le\dfrac{q^2}{3}=3$, we have: $r^2-4pr+p^2+2\ge r^2-1$
So we need to prove $$r^2\ge1$$
I have no idea from here, pls help me!
 [/INST] Some Hints:
$(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$
$(a^2+2)(b^2+2)(c^2+2)\ge3(a^2+2+b^2+2+c^2+2)$
Also,
$(a+b+c)^2$= $a^2+b^2+c^2+6$  (as $ab+bc+ca=3$)
$(a+b+c)^2$= $a^2+2+b^2+2+c^2+2$
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