Giải thích các bước giải:
$\begin{cases}x^3+\dfrac{1}{3^3}+\dfrac{1}{3^3} \ge 3\sqrt[3]{x^3.\dfrac{1}{3^3}.\dfrac{1}{3^3}}=\dfrac{x}{3}\\y^3+\dfrac{1}{3^3}+\dfrac{1}{3^3} \ge 3\sqrt[3]{y^3.\dfrac{1}{3^3}.\dfrac{1}{3^3}}=\dfrac{y}{3}\\z^3+\dfrac{1}{3^3}+\dfrac{1}{3^3} \ge 3\sqrt[3]{z^3.\dfrac{1}{3^3}.\dfrac{1}{3^3}}=\dfrac{z}{3}\end{cases}$
$\rightarrow x^3+\dfrac{1}{3^3}+\dfrac{1}{3^3}+y^3+\dfrac{1}{3^3}+\dfrac{1}{3^3}+z^3+\dfrac{1}{3^3}+\dfrac{1}{3^3} \ge \dfrac{x}{3}+\dfrac{y}{3}+\dfrac{z}{3}$
$\rightarrow x^3+y^3+z^3+\dfrac{2}{9}\ge \dfrac{x+y+z}{3}=\dfrac{1}{3}$
$\rightarrow x^3+y^3+z^3\ge \dfrac{1}{9}$
