Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$
Let $a$, $b$, $c$ be nonnegative real numbers.
Prove $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$.
I will prove the stronger inequality: $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$$ because $$(a^2+2)(b^2+2)=(a^2+1)(b^2+1)+a^2+b^2+3\ge (a+b)^2+\dfrac{1}{2}(a+b)^2+3 =\dfrac{3}{2}[(a+b)^2+2]$$ so $$(a^2+2)(b^2+2)(c^2+2)\ge \dfrac{3}{2}[(a+b)^2+2](c^2+2)\ge\dfrac{3}{2} [\sqrt{2}(a+b)+\sqrt{2}c]^2=3(a+b+c)^2$$ so $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$