Câu hỏi 77
Chứng minh rằng nếu $0 \lt \frac{a}{b} \lt b \lt 2a$ thì
$$
\frac{2ab - a^2}{7ab - 3b^2 - 2a^2} + \frac{2ab - b^2}{7ab - 3a^2 - 2b^2} \geq 1 + \frac{1}{4}\left(\frac{a}{b} - \frac{b}{a}\right)^2
$$
$$
\frac{2ab - a^2}{7ab - 3b^2 - 2a^2} + \frac{2ab - b^2}{7ab - 3a^2 - 2b^2} \geq 1 + \frac{1}{4}\left(\frac{a}{b} - \frac{b}{a}\right)^2
$$