A Power Mean Versus an Average

Problem

Let a>b>0a > b > 0 and n∈N∗n \in \mathbb{N}^*. Define

An=an+an−1b+⋯+abn−1+bnn+1,Bn=(a+b2)n.A_n = \frac{a^n + a^{n-1}b + \cdots + ab^{n-1} + b^n}{n+1}, \qquad B_n = \left(\frac{a+b}{2}\right)^n.

1. Prove A2>B2A_2 > B_2. 2. Compare AnA_n and BnB_n in general, with proof.

Answer

Solution

Difficulty8/10
TopicsSubstitution, algebra, Binomial Theorem, inequality

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