To prove:
$$(a^2+2)(b^2+2)(c^2+2)\ge9(ab+bc+ca)$$ where $a,b,c$ are positive real numbers. When does equality hold?
I thought equality would hold when $a = b = c$, but that doesn't seem to fit the statement.
To prove the inequality, I tried substituting $a = \tan A, b=\tan B$ and $c=\tan C$ in the hope of using some trigonometric manipulation. Also, my observational skills suggest that the LHS is the obvious cube of the GM of $3$ numbers; but that didn't help either.
The above inequality is same as proving $(a^2+b^2+c^2+6)^3≥81(ab+bc+ca)$
I know that $a^2+b^2+c^2\ge ab+bc+ca$, but I'm not able to take it from here.
Is this approach fine, or is there a shorter method for the problem? I doubt is $AM\ge GM$ is the most elegant way of arriving at the required statement. Please help.
By C-S $$\left(\frac{(a+b)^2}{2}+1\right)(2+c^2)\geq(a+b+c)^2$$ and since $$(a^2+2)(b^2+2)\geq3\left(\frac{(a+b)^2}{2}+1\right)$$ it's $$(a-b)^2+2(ab-1)^2\geq0,$$ we obtain: $$(a^2+2)(b^2+2)(c^2+2)\geq3(a+b+c)^2\geq9(ab+ac+bc).$$