Question 98
Let $x$, $y$, $z$ be positive real numbers satisfying $x + y + z = 1$. Prove that
$$
\begin{aligned}
\frac{(1 + xy + yz + zx)(1 + 3x^3 + 3y^3 + 3z^3)}{9(x + y)(y + z)(z + x)} \\
&\ge \left( \frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}} + \frac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}} + \frac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}} \right)^2
\end{aligned}
$$
$$
\begin{aligned}
\frac{(1 + xy + yz + zx)(1 + 3x^3 + 3y^3 + 3z^3)}{9(x + y)(y + z)(z + x)} \\
&\ge \left( \frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}} + \frac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}} + \frac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}} \right)^2
\end{aligned}
$$