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

Source: Polish MO second round 1969 p4

August 28, 2024
inequalitiesnumber theory

Problem Statement

Prove that for any natural numbers min the inequality holds 1m+2m+…+nm≥n⋅(n+12)m1^m + 2^m + \ldots + n^m \geq n\cdot \left( \frac{n+1}{2}\right)^m
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