Câu hỏi 48
Chứng minh rằng nếu $a, b, c \in [\frac{1}{\sqrt{6}}, 6]$ thì
$$
\frac{4}{a+3b} + \frac{4}{b+3c} + \frac{4}{c+3a} \ge \frac{3}{a+2b} + \frac{3}{b+2c} + \frac{3}{c+2a}.
$$
$$
\frac{4}{a+3b} + \frac{4}{b+3c} + \frac{4}{c+3a} \ge \frac{3}{a+2b} + \frac{3}{b+2c} + \frac{3}{c+2a}.
$$