Câu hỏi 98
Cho các số thực dương $x,y,z$ thỏa mãn $x+y+z=1$. Chứng minh rằng
$$
\begin{aligned}
\frac{(1+xy+yz+zx)(1+3x^3+3y^3+3z^3)}{9(x+y)(y+z)(z+x)}
\\
&\ge\left(\frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\frac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}+\frac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2.
\end{aligned}
$$
$$
\begin{aligned}
\frac{(1+xy+yz+zx)(1+3x^3+3y^3+3z^3)}{9(x+y)(y+z)(z+x)}
\\
&\ge\left(\frac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\frac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}+\frac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2.
\end{aligned}
$$