Câu hỏi 39
Cho $a, b, c, d \in [0, 1]$. Chứng minh rằng
$$
\frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd \le 3.
$$
$$
\frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+d} + \frac{d}{1+a} + abcd \le 3.
$$
Math Olympiad Vault · MathNet 2017-part-05 — 176 bài (EN–VI)