Câu hỏi 104
Chứng minh rằng nếu $a, b, c \in (0, \infty)$ thì
$$
\frac{a^2b}{a^6 + b^2 + 4ac + 2} + \frac{b^2c}{b^6 + c^2 + 4ba + 2} + \frac{c^2a}{c^6 + a^2 + 4cb + 2} \le \frac{a+b+c}{8}.
$$
$$
\frac{a^2b}{a^6 + b^2 + 4ac + 2} + \frac{b^2c}{b^6 + c^2 + 4ba + 2} + \frac{c^2a}{c^6 + a^2 + 4cb + 2} \le \frac{a+b+c}{8}.
$$