Câu hỏi 175
Cho $a, b, c \in [1, \infty)$. Chứng minh rằng:
$$
\frac{a\sqrt{b}}{a+b} + \frac{b\sqrt{c}}{b+c} + \frac{c\sqrt{a}}{c+a} + \frac{3}{2} \le a+b+c.
$$
$$
\frac{a\sqrt{b}}{a+b} + \frac{b\sqrt{c}}{b+c} + \frac{c\sqrt{a}}{c+a} + \frac{3}{2} \le a+b+c.
$$