A Lower Bound for Ratio of Power Means

. Let n and m be natural numbers. Suppose { a i } n + m i =1 is an increasing, logarithmically convex, and positive sequence. Denote the power mean P n ( r ) for any given positive real number r by P n ( r ) = (cid:0) 1 n P ni =1 a ri (cid:1) 1 /r . Then P n ( r ) /P n + m ( r ) ≥ a n /a n + m . The lower bound is the best possible.


Introduction
It is well-known that the following inequality holds for r > 0 and n ∈ N. We call the left-hand side of this inequality Alzer's inequality [1], and the right-hand side Martins' inequality [8].
Let {a i } i∈N be a positive sequence.If a i+1 a i−1 ≥ a 2 i for i ≥ 2, we call {a i } i∈N a logarithmically convex sequence; if a i+1 a i−1 ≤ a 2 i for i ≥ 2, we call {a i } i∈N a logarithmically concave sequence.
In [2], Martins' inequality was generalized as follows: Let {a i } i∈N be an increasing, logarithmically concave, positive, and nonconstant sequence satisfying (a +1 /a ) ≥ (a /a −1 ) −1 for any positive integer > 1, then where r is a positive number, n, m ∈ N, and a i !denotes the sequence factorial n i=1 a i .The upper bound is best possible.
Let n, m ∈ N and {a i } n+m i=1 be an increasing, logarithmically concave, positive, and nonconstant sequence such that the sequence i ai+1 ai − 1 is increasing.
In [22], a general form of Alzer's inequality was obtained: Let {a i } ∞ i=1 be a strictly increasing positive sequence, and let m be a natural number.If {a i } ∞ i=1 is logarithmically concave and the sequence an+1 an n ∞ i=1 is increasing, then In this short note, utilizing the mathematical induction, we obtain the following Theorem 1.Let n and m be natural numbers.Suppose {a i } n+m i=1 is an increasing, logarithmically convex, and positive sequence.Denote the power mean P n (r) for any given positive real number r by Then the sequence Pi(r) ai n+m i=1 is decreasing for any given positive real number r, that is, The lower bound in (5) is the best possible.
Considering that the exponential functions a x α and a α x for given constants α ≥ 1 and a > 1 is logarithmically convex on [0, ∞), as a corollary of Theorem 1, we have Corollary 1.Let α ≥ 1 and a > 1 be two constants.For any given real number r, the following inequalities hold: where n and m are natural numbers, and k is a nonnegative integer.The lower bounds above are the best possible.

Proof of Theorem 1
The inequality ( 5) is equivalent to that is, This is also equivalent to Since inequality (10) reduces to It is easy to see that inequality (12) holds for n = 1.
Assume that inequality (12) holds for some n > 1.Using the principle of mathematical induction, considering equality (11) and the inductive hypothesis, it is easy to show that the induction for inequality (12) on n + 1 can be written as which can be rearranged as Since the sequence {a i } n+m i=1 is increasing, we have an+1 an+2 ≤ 1 and an+1 an+2 r ≤ 1.