Question 23
Let $a$, $b$ and $c$ be positive real numbers such that $abc = 1$. Prove that
$$
\begin{aligned}
& \frac{1}{\sqrt{a^3 + 2b^3 + 6}} + \frac{1}{\sqrt{b^3 + 2c^3 + 6}} + \frac{1}{\sqrt{c^3 + 2a^3 + 6}} \\
\le & \frac{a^2}{\sqrt{a^4 + 4b + 4c}} + \frac{b^2}{\sqrt{b^4 + 4c + 4a}} + \frac{c^2}{\sqrt{c^4 + 4a + 4b}}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{\sqrt{a^3 + 2b^3 + 6}} + \frac{1}{\sqrt{b^3 + 2c^3 + 6}} + \frac{1}{\sqrt{c^3 + 2a^3 + 6}} \\
\le & \frac{a^2}{\sqrt{a^4 + 4b + 4c}} + \frac{b^2}{\sqrt{b^4 + 4c + 4a}} + \frac{c^2}{\sqrt{c^4 + 4a + 4b}}
\end{aligned}
$$