Question 87
Let $a,b,c$ be positive real numbers. Prove that
$$\begin{aligned}
&\frac{8a^2+2ab}{(b+\sqrt{6ac}+3c)^2}
+\frac{2b^2+3bc}{(3c+\sqrt{2ab}+2a)^2}\\
&\qquad+\frac{18c^2+6ac}{(2a+\sqrt{3bc}+b)^2}\ge1.
\end{aligned}$$
$$\begin{aligned}
&\frac{8a^2+2ab}{(b+\sqrt{6ac}+3c)^2}
+\frac{2b^2+3bc}{(3c+\sqrt{2ab}+2a)^2}\\
&\qquad+\frac{18c^2+6ac}{(2a+\sqrt{3bc}+b)^2}\ge1.
\end{aligned}$$