4

Part of 1969 Poland - Second Round

Problems(1)

1^m + 2^m + .. + n^m >= n\cdot ( \frac{n+1}{2})^m

Source: Polish MO second round 1969 p4

Prove that for any natural numbers min the inequality holds

1^m + 2^m + \ldots + n^m \geq n\cdot \left( \frac{n+1}{2}\right)^m

inequalitiesnumber theory