Câu hỏi 165
Cho các số nguyên $n \ge m \ge 1$. Chứng minh rằng
$$
\sum_{k=m}^{n} \left( \frac{1}{k^2} + \frac{1}{k^3} \right) \ge m \cdot \left( \sum_{k=m}^{n} \frac{1}{k^2} \right)^2 .
$$
$$
\sum_{k=m}^{n} \left( \frac{1}{k^2} + \frac{1}{k^3} \right) \ge m \cdot \left( \sum_{k=m}^{n} \frac{1}{k^2} \right)^2 .
$$