Câu hỏi 120
Cho $x$, $y$, $z \gt 0$. Chứng minh rằng
$$
\frac{x^3}{z^3 + x^2 y} + \frac{y^3}{x^3 + y^2 z} + \frac{z^3}{y^3 + z^2 x} \ge \frac{3}{2}.
$$
$$
\frac{x^3}{z^3 + x^2 y} + \frac{y^3}{x^3 + y^2 z} + \frac{z^3}{y^3 + z^2 x} \ge \frac{3}{2}.
$$