Prove that for all positive real $a,b,c$, we have $$(a^2+2)(b^2+2)(c^2+2) \geq 9(ab+bc+ca).$$
Because of the term $a^2+2$, this motiveates me to substitute $a=\sqrt{2}\tan A, b=\sqrt{2}\tan B, c=\sqrt{2}\tan C$ and the fact that $1+\tan^2x=\sec^2x$, then the desired inequality becomes $$2^2\sec^2 A\sec^2 B \sec^2 C\geq 3^2(\tan A\tan C + \tan B\tan C + \tan C \tan A).$$
But then I was stuck and couldn't move further, please helps.
Ok,if you use trigonometry to solve it
as you $$\Longleftrightarrow \cos{A}\cos{B}\cos{C}(\cos{A}\sin{B}\sin{C}+\sin{A}\cos{B}\sin{C}+\sin{A}\sin{B}\cos{C})\le\dfrac{4}{9}$$ $$\cos{A}\cos{B}\cos{C}(\cos{A}\cos{B}\cos{C}-\cos{(A+B+C)})\le\dfrac{4}{9}$$ By Jenson inequality we have $$\cos{A}\cos{B}\cos{C}\le\cos^3{\dfrac{A+B+C}{3}}$$ so let $$\cos{\dfrac{A+B+C}{3}}=x,\Longrightarrow \cos{(A+B+C)}=4x^3-3x$$ $$\Longleftrightarrow x^4(1-x^2)\le \dfrac{4}{27},0<x<1$$ Use AM-GM inequality we have $$x^4(1-x^2)=\dfrac{1}{2}x^2\cdot x^2\cdot (2-2x^2)\le\dfrac{4}{27}$$
PS:you can prove this stronger $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$
