Table of Contents Author Guidelines Submit a Manuscript
Abstract and Applied Analysis
Volume 2010, Article ID 128934, 19 pages
http://dx.doi.org/10.1155/2010/128934
Research Article

A New Method to Prove and Find Analytic Inequalities

1Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
2Department of Mathematics, Inner Mongolia University for the Nationalities, Tongliao 028000, China

Received 19 October 2009; Revised 26 January 2010; Accepted 2 February 2010

Academic Editor: John Rassias

Copyright © 2010 Xiao-Ming Zhang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We present a new method to study analytic inequalities. As for its applications, we prove the well-known Hölder inequality and establish several new analytic inequalities.

1. Monotonicity Theorem

Throughout the paper denotes the set of real numbers and + denotes the set of strictly positive real numbers. Let 𝑛 2 , 𝑛 , and 𝐱 = ( 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 ) 𝑛 ; the arithmetic mean 𝐴 ( 𝐱 ) and the power mean 𝑀 𝑟 ( 𝐱 ) of order 𝑟 with respect to the positive real numbers 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 are defined by 𝐴 ( 𝐱 ) = ( 1 / 𝑛 ) 𝑛 𝑖 = 1 𝑥 𝑖 , 𝑀 𝑟 ( 𝐱 ) = ( ( 1 / 𝑛 ) 𝑛 𝑖 = 1 𝑥 𝑟 𝑖 ) 1 / 𝑟 for 𝑟 0 , and 𝑀 0 ( 𝐱 ) = ( 𝑛 𝑖 = 1 𝑥 𝑖 ) 1 / 𝑛 , respectively.

In [1], Pachpatte gave many basic methods and tools for researchers working in inequalities. In this section, we present a monotonicity theorem which can be used as powerful tool to prove and find analytic inequalities.

Lemma 1.1. Suppose that 𝑚 < 𝑀 , 𝐷 = { ( 𝑥 1 , 𝑥 2 ) 𝑚 𝑥 2 𝑥 1 𝑀 } . If 𝑓 𝐷 has continuous partial derivatives, then 𝜕 𝑓 / 𝜕 𝑥 1 ( ) 𝜕 𝑓 / 𝜕 𝑥 2 holds in 𝐷 if and only if 𝑓 ( 𝑎 , 𝑏 ) ( ) 𝑓 ( 𝑎 𝑙 , 𝑏 + 𝑙 ) holds for all ( 𝑎 , 𝑏 ) 𝐷 and 𝑙 > 0 with 𝑏 < 𝑏 + 𝑙 𝑎 𝑙 < 𝑎 .

Proof. We only prove the case of 𝜕 𝑓 / 𝜕 𝑥 1 𝜕 𝑓 / 𝜕 𝑥 2 .
Necessity. For all ( 𝑥 1 , 𝑥 2 ) 𝐷 and 𝑙 + with 𝑚 𝑥 2 < 𝑥 2 + 𝑙 𝑥 1 𝑙 < 𝑥 1 𝑀 , by the assumption we have 𝑓 ( 𝑥 1 𝑙 , 𝑥 2 + 𝑙 ) 𝑓 ( 𝑥 1 , 𝑥 2 ) 0 . Then from the Langrange's mean value theorem we know that there exists 𝜉 𝑙 ( 0 , 𝑙 ) such that 𝑙 𝑥 𝜕 𝑓 1 𝜉 𝑙 , 𝑥 2 + 𝜉 𝑙 𝜕 𝑥 1 + 𝑥 𝜕 𝑓 1 𝜉 𝑙 , 𝑥 2 + 𝜉 𝑙 𝜕 𝑥 2 𝑥 0 , 𝜕 𝑓 1 𝜉 𝑙 , 𝑥 2 + 𝜉 𝑙 𝜕 𝑥 1 + 𝑥 𝜕 𝑓 1 𝜉 𝑙 , 𝑥 2 + 𝜉 𝑙 𝜕 𝑥 2 0 . ( 1 . 1 ) Letting 𝑙 0 + , we get 𝑥 𝜕 𝑓 1 , 𝑥 2 𝜕 𝑥 1 𝑥 𝜕 𝑓 1 , 𝑥 2 𝜕 𝑥 2 . ( 1 . 2 ) According to the continuity of partial derivatives, we know that 𝑥 𝜕 𝑓 1 , 𝑥 1 𝜕 𝑥 1 𝑥 𝜕 𝑓 1 , 𝑥 1 𝜕 𝑥 2 ( 1 . 3 ) holds also.
Sufficiency. For all ( 𝑎 , 𝑏 ) 𝐷 and 𝑙 > 0 with 𝑏 < 𝑏 + 𝑙 𝑎 𝑙 < 𝑎 , from the assumption and the Langrange's mean value theorem we know that there exists 𝜉 𝑙 ( 0 , 𝑙 ) such that 𝑓 ( 𝑎 , 𝑏 ) 𝑓 ( 𝑎 𝑙 , 𝑏 + 𝑙 ) = ( 𝑓 ( 𝑎 𝑙 , 𝑏 + 𝑙 ) 𝑓 ( 𝑎 , 𝑏 ) ) = 𝑙 𝜕 𝑓 𝑎 𝜉 𝑙 , 𝑏 + 𝜉 𝑙 𝜕 𝑥 1 + 𝜕 𝑓 𝑎 𝜉 𝑙 , 𝑏 + 𝜉 𝑙 𝜕 𝑥 2 = 𝑙 𝜕 𝑓 𝑎 𝜉 𝑙 , 𝑏 + 𝜉 𝑙 𝜕 𝑥 1 𝜕 𝑓 𝑎 𝜉 𝑙 , 𝑏 + 𝜉 𝑙 𝜕 𝑥 2 0 . ( 1 . 4 )
Therefore the proof of Lemma 1.1 is completed.

Theorem 1.2. Suppose that 𝐷 𝑛 is a symmetric convex set with nonempty interior, 𝑓 𝐷 has continuous partial derivatives, and 𝐷 𝑖 = 𝐱 𝐷 𝑥 𝑖 = m a x 1 𝑗 𝑛 𝑥 𝑗 𝐱 𝐷 𝑥 1 = 𝑥 2 = = 𝑥 𝑛 , 𝐷 𝑖 = 𝐱 𝐷 𝑥 𝑖 = m i n 1 𝑗 𝑛 𝑥 𝑗 𝑥 𝐷 𝑥 1 = 𝑥 2 = = 𝑥 𝑛 , ( 1 . 5 ) 𝑖 = 1 , 2 , , 𝑛 . If for all 𝑖 , 𝑗 = 1 , 2 , , 𝑛 with 𝑖 𝑗 , 𝜕 𝑓 𝜕 𝑥 𝑖 > ( < ) 𝜕 𝑓 𝜕 𝑥 𝑗 ( 1 . 6 ) holds in 𝐷 𝑖 𝐷 𝑗 , then 𝑓 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ( ) 𝑓 ( 𝐴 ( 𝐚 ) , 𝐴 ( 𝐚 ) , , 𝐴 ( 𝐚 ) ) ( 1 . 7 ) for all 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝐷 , with equality if only if 𝑎 1 = 𝑎 2 = = 𝑎 𝑛 .

Proof. If 𝑛 = 2 , then Theorem 1.2 follows from Lemma 1.1 and 𝑙 = | 𝑎 1 𝑎 2 | / 2 . We assume that 𝑛 3 in the next discussion. Without loss of generality, we only prove the case of 𝜕 𝑓 / 𝜕 𝑥 𝑖 > 𝜕 𝑓 / 𝜕 𝑥 𝑗 with 𝑖 𝑗 .
If 𝑎 1 = 𝑎 2 = = 𝑎 𝑛 , then inequality (1.7) is clearly true. If m a x 1 𝑗 𝑛 { 𝑎 𝑗 } m i n 1 𝑗 𝑛 { 𝑎 𝑗 } , then without loss of generality we assume that 𝑎 1 𝑎 2 𝑎 𝑛 1 𝑎 𝑛 .
(1)   If 𝑎 1 > m a x 2 𝑗 𝑛 { 𝑎 𝑗 } and 𝑎 𝑛 < m i n 1 𝑗 𝑛 1 { 𝑎 𝑗 } , then ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝐷 1 𝐷 𝑛 . From Lemma 1.1 and the conditions in Theorem 1.2 we know that there exist 𝑎 1 ( 1 ) and 𝑎 𝑛 ( 1 ) such that 𝑙 = 𝑎 1 𝑎 1 ( 1 ) = 𝑎 𝑛 ( 1 ) 𝑎 𝑛 > 0 , 𝑎 1 ( 1 ) = 𝑎 2 or 𝑎 𝑛 ( 1 ) = 𝑎 𝑛 1 , and 𝑓 𝑎 1 , 𝑎 2 , 𝑎 3 , , 𝑎 𝑛 𝑎 𝑓 1 ( 1 ) , 𝑎 2 , 𝑎 3 , , 𝑎 𝑛 ( 1 ) . ( 1 . 8 ) For the sake of convenience, we denote 𝑎 𝑖 ( 1 ) = 𝑎 𝑖 , 2 𝑖 𝑛 1 . Consequently, 𝑓 𝑎 1 , 𝑎 2 , 𝑎 3 , , 𝑎 𝑛 𝑎 𝑓 1 ( 1 ) , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑎 𝑛 ( 1 ) . ( 1 . 9 ) If 𝑎 1 ( 1 ) = 𝑎 2 ( 1 ) = = 𝑎 𝑛 ( 1 ) , then Theorem 1.2 holds. Otherwise, for the case of 𝑎 1 ( 1 ) = 𝑎 2 ( 1 ) > 𝑎 𝑛 ( 1 ) , ( 𝑎 1 ( 1 ) , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑎 𝑛 ( 1 ) ) 𝐷 1 𝐷 𝑛 and 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 1 | | | | 𝐱 = ( 𝑎 1 ( 1 ) , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑎 𝑛 ( 1 ) ) > 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 𝑛 | | | | 𝐱 = ( 𝑎 1 ( 1 ) , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑎 𝑛 ( 1 ) ) . ( 1 . 1 0 ) From the continuity of partial derivatives we know that there exists 𝜀 > 0 such that 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 1 | | | | 𝐱 = ( 𝑠 , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑡 ) > 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 𝑛 | | | | | 𝐱 = ( 𝑠 , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑡 ) , ( 1 . 1 1 ) where 𝑠 [ 𝑎 1 ( 1 ) 𝜀 , 𝑎 1 ( 1 ) ] and 𝑡 [ 𝑎 𝑛 ( 1 ) , 𝑎 𝑛 ( 1 ) + 𝜀 ] . Denote 𝑎 1 ( 2 ) = 𝑎 1 ( 1 ) 𝜀 , 𝑎 𝑛 ( 2 ) = 𝑎 𝑛 ( 1 ) + 𝜀 , 𝑎 𝑖 ( 2 ) = 𝑎 𝑖 ( 1 ) ( 2 𝑖 𝑛 1 ) . By Lemma 1.1, we get 𝑓 𝑎 1 ( 1 ) , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑎 𝑛 ( 1 ) 𝑎 𝑓 1 ( 2 ) , 𝑎 2 ( 2 ) , 𝑎 3 ( 2 ) , , 𝑎 𝑛 ( 2 ) , ( 1 . 1 2 ) and 𝑎 2 ( 2 ) = m a x 1 𝑖 𝑛 { 𝑎 𝑖 ( 2 ) } . For the case of 𝑎 1 ( 1 ) > 𝑎 ( 1 ) 𝑛 1 = 𝑎 𝑛 ( 1 ) , after a similar argument, we get inequality (1.12) with 𝑎 ( 2 ) 𝑛 1 = m i n 1 𝑖 𝑛 { 𝑎 𝑖 ( 2 ) } .
Repeating the above steps, we get { 𝑎 1 ( 𝑖 ) , 𝑎 2 ( 𝑖 ) , , 𝑎 𝑛 ( 𝑖 ) } ( 𝑖 = 1 , 2 , ) such that 𝑛 𝑗 = 1 𝑎 𝑗 ( 𝑖 ) is a constant and { 𝑎 𝑗 ( 𝑖 ) } ( 𝑖 = 1 , 2 , ) are monotone increasing (decreasing) sequences if 𝑎 𝑗 ( ) 𝐴 ( 𝐚 ) , 𝑗 = 1 , 2 , 3 , , 𝑛 , and 𝑓 𝑎 1 ( 1 ) , 𝑎 2 ( 1 ) , 𝑎 3 ( 1 ) , , 𝑎 𝑛 ( 1 ) 𝑎 𝑓 1 ( 𝑖 ) , 𝑎 2 ( 𝑖 ) , 𝑎 3 ( 𝑖 ) , , 𝑎 𝑛 ( 𝑖 ) . ( 1 . 1 3 ) If there exists 𝑖 such that 𝑎 1 ( 𝑖 ) = 𝑎 2 ( 𝑖 ) = = 𝑎 𝑛 ( 𝑖 ) , then the proof of Theorem 1.2 is completed. Otherwise, we denote 𝛼 = i n f 𝑖 { m a x { 𝑎 1 ( 𝑖 ) , 𝑎 2 ( 𝑖 ) , , 𝑎 𝑛 ( 𝑖 ) } } ; without loss of generality, we assume that 𝑎 m a x ( 𝑖 𝑗 ) 1 , 𝑎 ( 𝑖 𝑗 ) 2 , , 𝑎 ( 𝑖 𝑗 ) 𝑛 = 𝑎 ( 𝑖 𝑗 ) 1 𝛼 , l i m 𝑗 + 𝑎 ( 𝑖 𝑗 ) 1 , 𝑎 ( 𝑖 𝑗 ) 2 , , 𝑎 ( 𝑖 𝑗 ) 𝑛 = 𝛼 , 𝑏 2 , 𝑏 3 , , 𝑏 𝑛 , ( 1 . 1 4 ) where { 𝑖 𝑗 } + 𝑗 = 1 is a subsequence of . Then from the continuity of function 𝑓 , we get 𝑓 𝑎 1 , 𝑎 2 , 𝑎 3 , , 𝑎 𝑛 𝑓 𝛼 , 𝑏 2 , 𝑏 3 , , 𝑏 𝑛 . ( 1 . 1 5 ) If 𝛼 m i n { 𝑏 2 , 𝑏 3 , , 𝑏 𝑛 } , then we repeat the above arguments and get a contradiction to the definition of 𝛼 . Hence 𝛼 = 𝑏 2 = 𝑏 3 = = 𝑏 𝑛 . From 𝛼 + 𝑛 𝑖 = 2 𝑏 𝑖 = 𝑛 𝑖 = 1 𝑎 𝑖 we get 𝛼 = 𝑏 2 = 𝑏 3 = = 𝑏 𝑛 = 𝐴 ( 𝐚 ) ; the proof of Theorem 1.2 is completed.
( 2 ) The proof for the case of 𝑎 1 = m a x 2 j 𝑛 { 𝑎 𝑗 } or 𝑎 𝑛 = m i n 1 𝑗 𝑛 1 { 𝑎 𝑗 } is implied in the proof of (1).

In particular, according to Theorem 1.2 the following corollary holds.

Corollary 1.3. Suppose that 𝐷 𝑛 is a symmetric convex set with nonempty interior, 𝑓 𝐷 is a symmetric function with continuous partial derivatives, and 𝐷 1 = 𝐱 𝐷 𝑥 1 = m a x 1 𝑗 𝑛 𝑥 𝑗 𝐱 𝐷 𝑥 1 = 𝑥 2 = = 𝑥 𝑛 , 𝐷 2 = 𝐱 𝐷 𝑥 2 = m i n 1 𝑗 𝑛 𝑥 𝑗 𝐱 𝐷 𝑥 1 = 𝑥 2 = = 𝑥 𝑛 , 𝐷 = 𝐷 1 𝐷 2 . ( 1 . 1 6 ) If 𝜕 𝑓 / 𝜕 𝑥 1 > ( < ) 𝜕 𝑓 / 𝜕 𝑥 2 holds in 𝐷 , then 𝑓 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ( ) 𝑓 ( 𝐴 ( 𝐚 ) , 𝐴 ( 𝐚 ) , , 𝐴 ( 𝐚 ) ) ( 1 . 1 7 ) for all 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝐷 , and equality holds if and only if 𝑎 1 = 𝑎 2 = = 𝑎 𝑛 .

2. Comparing with Schur's Condition

The Schur convexity was introduced by I. Schur [2] in 1923; the following Definitions 2.1 and 2.2 can be found in [2, 3].

Definition 2.1. For 𝐮 = ( 𝑢 1 , 𝑢 2 , 𝑢 𝑛 ) , 𝐯 = ( 𝑣 1 , 𝑣 2 , 𝑣 𝑛 ) 𝑛 , without loss of generality one assumes that 𝑢 1 𝑢 2 𝑢 𝑛 and 𝑣 1 𝑣 2 𝑣 𝑛 . Then 𝐮 is said to be majorized by 𝐯 (in symbols 𝐮 𝐯 ) if 𝑘 𝑖 = 1 𝑢 𝑖 𝑘 𝑖 = 1 𝑣 𝑖 for 𝑘 = 1 , 2 , , 𝑛 1 and 𝑛 𝑖 = 1 𝑢 𝑖 = 𝑛 𝑖 = 1 𝑣 𝑖 .

Definition 2.2. Suppose that Ω 𝑛 . A real-valued function 𝜑 Ω is said to be Schur convex (Schur concave) if 𝐮 𝐯 implies that 𝜑 ( 𝐮 ) ( ) 𝜑 ( 𝐯 ) .
Recall that the following so-called Schur's condition is very useful for deter m i n ing whether or not a given function is Schur convex or concave.

Theorem 2.3 (see [2, page 57]). Suppose that Ω 𝑛 is a symmetric convex set with nonempty interior i n t Ω . If 𝜑 Ω is continuous on Ω and differentiable in i n t Ω , then 𝜑 is Schur convex (Schur concave) on Ω if and only if it is symmetric and 𝑢 1 𝑢 2 𝜕 𝜑 𝜕 𝑢 1 𝜕 𝜑 𝜕 𝑢 2 ( ) 0 ( 2 . 1 ) holds for any 𝑢 = ( 𝑢 1 , 𝑢 2 , , 𝑢 𝑛 ) i n t Ω .

It is well known that a convex function is not necessarily a Schur convex function, and a Schur convex function need notbe convex in the ordinary sense either. However, under the assumption of ordinary convexity, 𝑓 is Schur convex if and only if it is symmetric [4].

Although the Schur convexity is an important tool in researching analytic inequalities, but the restriction of symmetry cannot be used in dealing with nonsymmetric functions. Obviously, Theorem 1.2 is the generalization and development of Theorem 2.3; the following results in Sections 35 show that a large number of inequalities can be proved, improved, and found by Theorem 1.2.

3. A Proof for the Hölder Inequality

Using Theorem 1.2 and Corollary 1.3, we can prove some well-known inequalities, for example, power mean inequality, Hölder inequality, and Minkowski inequality. In this section, we only prove the Hölder inequality.

Proposition 3.1 (Hölder inequality). Suppose that 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 , 𝑦 1 , 𝑦 2 , , 𝑦 𝑛 𝑛 + . ( 3 . 1 ) If 𝑝 , 𝑞 > 1 and 1 / 𝑝 + 1 / 𝑞 = 1 , then 𝑛 𝑘 = 1 𝑥 𝑝 𝑘 1 / 𝑝 𝑛 𝑘 = 1 𝑦 𝑞 𝑘 1 / 𝑞 𝑛 𝑘 = 1 𝑥 𝑘 𝑦 𝑘 . ( 3 . 2 )

Proof. Let ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + and 𝑓 𝐛 𝑛 𝑘 = 1 0 𝑥 0 2 0 0 𝑑 𝑎 𝑘 1 / 𝑝 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑘 1 / 𝑞 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑘 1 / 𝑞 , 𝑏 𝑛 + . ( 3 . 3 ) Then 𝜕 𝑓 𝜕 𝑏 𝑖 = 1 𝑞 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑝 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑘 1 / 𝑞 1 𝑎 𝑖 1 𝑞 𝑎 𝑖 𝑏 𝑖 1 / 𝑞 1 , 𝜕 𝑓 𝜕 𝑏 𝑖 𝜕 𝑓 𝜕 𝑏 𝑗 = 1 𝑞 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 𝑘 = 1 𝑏 𝑘 𝑎 𝑘 1 / 𝑝 𝑎 𝑖 𝑎 𝑗 1 𝑞 𝑎 𝑖 𝑏 𝑖 1 / 𝑝 𝑎 𝑗 𝑏 𝑗 1 / 𝑝 . ( 3 . 4 ) Let 𝐛 𝐷 𝑖 𝐷 𝑗 (see (1.5)).
(1) If 𝑎 𝑖 𝑎 𝑗 , then 𝜕 𝑓 𝜕 𝑏 𝑖 𝜕 𝑓 𝜕 𝑏 𝑗 1 𝑞 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑖 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑝 𝑎 𝑖 𝑎 𝑗 1 𝑞 𝑎 𝑖 𝑏 𝑖 1 / 𝑝 𝑎 𝑗 𝑏 𝑗 1 / 𝑝 = 1 𝑞 𝑎 𝑗 𝑏 𝑗 1 / 𝑝 𝑏 𝑖 1 / 𝑝 > 0 . ( 3 . 5 )
( 2 ) If 𝑎 𝑖 𝑎 𝑗 , then 𝜕 𝑓 𝜕 𝑏 𝑖 𝜕 𝑓 𝜕 𝑏 𝑗 1 𝑞 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑗 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑝 𝑎 𝑖 𝑎 𝑗 1 𝑞 𝑎 𝑖 𝑏 𝑖 1 / 𝑝 𝑎 𝑗 𝑏 𝑗 1 / 𝑝 = 1 𝑞 𝑎 𝑖 𝑏 𝑗 1 / 𝑝 𝑏 𝑖 1 / 𝑝 > 0 . ( 3 . 6 ) From Theorem 1.2 we get 𝑓 ( 𝐛 ) 𝑓 ( 𝐴 ( 𝐛 ) , 𝐴 ( 𝐛 ) , , 𝐴 ( 𝐛 ) ) , ( 3 . 7 ) that is, 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑝 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑘 1 / 𝑞 𝑛 𝑘 = 1 𝑎 𝑘 𝑏 𝑘 1 / 𝑞 0 . ( 3 . 8 ) Therefore, the Hölder inequality follows from (3.8) with 𝑎 𝑘 = 𝑥 𝑝 𝑘 and 𝑏 𝑘 = 𝑦 𝑞 𝑘 / 𝑥 𝑝 𝑘 .

4. Improvement of the Sierpiński Inequality

In the section, we give some improvements of the well-known Sierpiński inequality:

𝑀 1 ( 𝐚 ) ( 𝑛 1 ) / 𝑛 [ ] 𝐴 ( 𝐚 ) 1 / 𝑛 𝑀 0 𝑀 ( 𝐚 ) 1 ( 𝐚 ) 1 / 𝑛 [ ] 𝐴 ( 𝐚 ) ( 𝑛 1 ) / 𝑛 . ( 4 . 1 )

Theorem 4.1. Suppose that 𝑛 3 , 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + , 𝛽 > 0 > 𝛼 . If 𝜆 = 2 𝛼 / 𝑛 ( 𝛽 𝛼 ) for 𝛽 + 𝛼 > 0 and 𝜆 = 1 / 𝑛 for 𝛽 + 𝛼 0 , then 𝑀 𝛼 ( 𝐚 ) 1 𝜆 𝑀 𝛽 ( 𝐚 ) 𝜆 𝑀 0 ( 𝐚 ) . ( 4 . 2 )

Proof. Let 𝑓 ( 𝐱 ) = ( 1 / 𝑛 𝛽 ) l n ( 𝑛 𝑖 = 1 𝑥 𝑖 ) ( ( 1 𝜆 ) / 𝛼 ) l n ( ( 1 / 𝑛 ) 𝑛 𝑖 = 1 𝑥 𝑖 𝛼 / 𝛽 ) , 𝐱 𝑛 + . Then 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 𝑗 = 1 𝑛 𝛽 𝑥 𝑗 1 𝜆 𝛽 𝑥 𝑗 𝛼 / 𝛽 1 𝑛 𝑖 = 1 𝑥 𝑖 𝛼 / 𝛽 , 𝑗 = 1 , 2 , 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 1 𝜕 𝑓 ( 𝐱 ) 𝜕 𝑥 2 = 𝑥 2 𝑥 1 𝑛 𝛽 𝑥 1 𝑥 2 1 𝜆 𝛽 𝑥 1 𝛼 / 𝛽 1 𝑥 2 𝛼 / 𝛽 1 𝑛 𝑖 = 1 𝑥 𝑖 𝛼 / 𝛽 . ( 4 . 3 )
Case 1. 𝛼 + 𝛽 > 0 . Let 𝑔 ( 𝑡 ) = 𝛽 + 𝛼 𝑡 𝛽 𝛼 𝛽 𝛼 𝑡 𝛽 + 𝑡 𝛼 𝛽 + 𝛼 𝛽 𝛼 , 𝑡 ( 1 , + ) . ( 4 . 4 ) Then 𝑡 𝛼 + 1 𝑔 ( 𝑡 ) = ( 𝛽 + 𝛼 ) 𝑡 𝛽 𝛽 𝑡 𝛽 + 𝛼 𝑡 𝛼 , 𝛼 + 1 𝑔 ( 𝑡 ) = ( 𝛽 + 𝛼 ) 𝛽 𝑡 𝛽 + 𝛼 1 ( 𝑡 𝛼 1 ) > 0 . ( 4 . 5 ) Therefore, 𝑡 𝛼 + 1 𝑔 ( 𝑡 ) is monotone increasing in ( 1 , + ) . From l i m 𝑡 1 + 𝑡 𝛼 + 1 𝑔 ( 𝑡 ) = l i m 𝑡 1 + ( 𝛽 + 𝛼 ) 𝑡 𝛽 𝛽 𝑡 𝛽 + 𝛼 𝛼 = 0 , ( 4 . 6 ) we know that 𝑡 𝛼 + 1 𝑔 ( 𝑡 ) > 0 , 𝑔 ( 𝑡 ) > 0 . Then l i m 𝑡 1 + 𝑔 ( 𝑡 ) = 0 leads to 𝑔 ( 𝑡 ) > 0 and 𝛽 + 𝛼 𝑡 𝛽 𝛼 𝛽 𝛼 𝑡 𝛽 + 𝑡 𝛼 𝛽 + 𝛼 𝛽 𝛼 > 0 , 𝛽 + 𝛼 𝑡 𝛽 𝛼 𝛽 1 + 2 𝛼 𝑡 𝛽 𝛼 𝛼 𝑡 𝛼 + 𝛽 + 1 > 0 , 𝛽 + 𝛼 𝑡 𝛽 𝛼 𝛽 𝑛 1 + 2 𝛼 𝑡 𝛽 𝛼 𝛼 𝑡 𝛼 + 𝛽 + ( 𝑛 1 ) > 0 , ( 1 𝑛 𝜆 ) 𝑡 𝛽 ( 𝑛 1 𝑛 𝜆 ) 𝑡 𝛼 𝑡 𝛼 + 𝛽 + ( 𝑛 1 ) > 0 , ( 4 . 7 ) ( 1 𝜆 ) 1 𝑡 𝛼 𝛽 𝑡 𝛼 + > 𝑡 ( 𝑛 1 ) 𝛽 1 𝑛 𝑡 𝛽 . ( 4 . 8 ) We assume that 𝐱 𝐷 (see (1.16)). Let 𝑡 = ( 𝑥 1 / 𝑥 2 ) 1 / 𝛽 . Then inequality (4.8) becomes 𝑥 ( 1 𝜆 ) 2 𝛼 / 𝛽 1 𝑥 1 𝛼 / 𝛽 1 𝑥 1 𝛼 / 𝛽 + ( 𝑛 1 ) 𝑥 2 𝛼 / 𝛽 > 𝑥 1 𝑥 2 𝑛 𝑥 1 𝑥 2 , 1 𝜆 𝛽 𝑥 2 𝛼 / 𝛽 1 𝑥 1 𝛼 / 𝛽 1 𝑛 𝑖 = 1 𝑥 𝑖 𝛼 / 𝛽 > 𝑥 1 𝑥 2 𝑛 𝛽 𝑥 1 𝑥 2 . ( 4 . 9 ) Combining inequalities (4.3) and (4.9) yields that 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 1 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 2 > 0 . Using Corollary 1.3 we have 𝑓 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 1 𝑓 ( 𝐴 ( 𝐱 ) , 𝐴 ( 𝐱 ) , , 𝐴 ( 𝐱 ) ) , 𝑛 𝛽 l n 𝑛 𝑖 = 1 𝑥 𝑖 1 𝜆 𝛼 1 l n 𝑛 𝑛 𝑖 = 1 𝑥 𝑖 𝛼 / 𝛽 𝜆 𝛽 1 l n 𝑛 𝑛 𝑖 = 1 𝑥 𝑖 . ( 4 . 1 0 ) Letting 𝑎 𝑖 = 𝑥 𝑖 1 / 𝛽 , 𝑖 = 1 , 2 , , 𝑛 , we get 𝑀 𝛼 ( 𝐚 ) 1 𝜆 𝑀 𝛽 ( 𝐚 ) 𝜆 𝑀 0 ( 𝐚 ) . ( 4 . 1 1 ) Case 2. 𝛼 + 𝛽 < 0 . Let 𝑡 > 1 . Then from 𝛼 < 0 and 𝛼 + 𝛽 < 0 , one has ( 𝑛 1 ) > ( 𝑛 2 ) 𝑡 𝛼 + 𝑡 𝛼 + 𝛽 . ( 4 . 1 2 ) Hence inequality (4.8) holds. The rest is similar to above, so we omit it.
The proof of Theorem 4.2 is similar to the proof of Theorem 4.1, and so we omit it.

Theorem 4.2. Suppose that 𝑛 3 , 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + , 𝛽 > 0 > 𝛼 . If 𝜃 = ( 𝑛 1 ) / 𝑛 for 𝛽 + 𝛼 > 0 and 𝜃 = 1 2 𝛽 / 𝑛 ( 𝛽 𝛼 ) for 𝛽 + 𝛼 0 , then 𝑀 0 𝑀 ( 𝐚 ) 𝛼 ( 𝐚 ) 1 𝜃 𝑀 𝛽 ( 𝐚 ) 𝜃 . ( 4 . 1 3 )

Theorem 4.3. Suppose that 𝑛 3 , 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + . If 𝑟 = l n 𝑛 / ( 𝑛 1 ) [ l n 𝑛 l n ( 𝑛 1 ) ] , then 𝑟 < 1 and 𝑀 1 / 𝑟 ( 𝐚 ) ( 𝑛 1 ) / 𝑛 [ ] 𝐴 ( 𝐚 ) 1 / 𝑛 𝑀 0 𝑀 ( 𝐚 ) 𝑟 ( 𝐚 ) 1 / 𝑛 [ ] 𝐴 ( 𝐚 ) ( 𝑛 1 ) / 𝑛 . ( 4 . 1 4 )

Proof. Let 𝑛 3 and 𝑓 𝐱 ( 0 , + ) 𝑛 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 𝑛 𝑛 𝑘 = 1 𝑥 𝑘 𝑛 1 / 𝑟 ( 𝑛 1 ) 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) . ( 4 . 1 5 ) Then 𝑟 = l n 𝑛 l n ( 1 + 1 / ( 𝑛 1 ) ) 𝑛 1 < l n 𝑛 l n 𝑒 = l n 𝑛 < 1 , 𝜕 𝑓 𝜕 𝑥 1 = 𝑥 1 ( 1 𝑟 ) / 𝑟 𝑟 𝑛 𝑛 𝑘 = 1 𝑥 𝑘 𝑛 1 / 𝑟 ( 𝑛 1 ) + 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 𝑟 𝑛 2 ( 𝑛 1 ) 𝑛 𝑘 = 1 𝑥 𝑘 𝑛 ( 1 / 𝑟 ( 𝑛 1 ) ) 1 1 𝑟 ( 𝑛 1 ) 𝑥 1 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) . ( 4 . 1 6 ) Therefore, we get 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 = 1 𝑥 𝑟 𝑛 1 ( 1 𝑟 ) / 𝑟 𝑥 2 ( 1 𝑟 ) / 𝑟 𝑛 𝑘 = 1 𝑥 𝑘 𝑛 1 / 𝑟 ( 𝑛 1 ) 1 𝑟 ( 𝑛 1 ) 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) 1 𝑥 1 1 𝑥 2 , 𝑛 𝑘 = 1 𝑥 𝑘 1 / ( 𝑟 ( 𝑛 1 ) ) 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 = 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑟 𝑛 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑛 𝑛 𝑘 = 1 𝑥 𝑘 𝑛 𝑘 = 1 𝑥 𝑘 1 / ( 𝑟 ) ( 𝑛 1 ) + 𝑥 1 𝑥 2 𝑟 ( 𝑛 1 ) 𝑥 1 𝑥 2 = 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑟 𝑛 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑛 𝑛 𝑘 = 1 𝑛 𝑖 = 1 , 𝑘 𝑥 𝑖 1 1 / ( 𝑟 ) ( 𝑛 1 ) + 𝑥 1 𝑥 2 𝑟 ( 𝑛 1 ) 𝑥 1 𝑥 2 . ( 4 . 1 7 ) We assume that 𝐱 𝐷 (see (1.16)). Then we have 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑟 𝑛 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑛 𝑥 2 ( 𝑛 1 ) + ( 𝑛 1 ) 𝑥 1 1 𝑥 2 ( 𝑛 2 ) 1 / ( 𝑟 ) ( 𝑛 1 ) + 𝑥 1 𝑥 2 𝑟 ( 𝑛 1 ) 𝑥 1 𝑥 2 . ( 4 . 1 8 ) Letting 𝑥 1 / 𝑥 2 = 𝑡 > 1 , from 𝑛 1 + 1 / 𝑟 ( 𝑛 1 ) = 𝑛 1 ( l n 𝑛 l n ( 𝑛 1 ) ) / l n 𝑛 = 𝑛 1 , we get 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 1 𝑥 𝑛 1 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑟 𝑥 1 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 2 ( 1 𝑟 ) / ( 𝑟 ) 𝑥 1 𝑥 2 𝑛 1 𝑥 1 + ( 𝑛 1 ) 𝑥 2 ( l n 𝑛 l n ( 𝑛 1 ) ) / l n 𝑛 + 𝑥 1 𝑥 2 𝑟 ( 𝑛 1 ) 𝑥 1 𝑥 2 = 1 𝑟 ( 𝑛 1 ) 𝑥 2 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 1 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 𝑡 𝑡 + 𝑛 1 ( l n 𝑛 l n ( 𝑛 1 ) ) / l n 𝑛 𝑡 1 𝑡 = 1 𝑟 ( 𝑛 1 ) 𝑥 2 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 1 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 1 + 𝑛 1 𝑡 ( l n 𝑛 l n ( 𝑛 1 ) ) / l n 𝑛 𝑡 1 𝑡 . ( 4 . 1 9 )
According to Bernoulli's inequality ( 1 + 𝑥 ) 𝛼 < 1 + 𝛼 𝑥 with 𝑥 1 , 𝑥 0 , and 0 < 𝛼 < 1 , one has 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 > 1 𝑟 ( 𝑛 1 ) 𝑥 2 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 1 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 1 1 + ( l n 𝑛 l n ( 𝑛 1 ) ) / l n 𝑛 ( 𝑛 1 ) / 𝑡 𝑡 1 𝑡 = 1 𝑟 ( 𝑛 1 ) 𝑥 2 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 1 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 𝑡 1 / ( 𝑟 ) / 𝑟 𝑡 1 𝑡 = 1 𝑟 ( 𝑛 1 ) 𝑥 2 ( 1 + 1 / 𝑟 ) 𝑡 1 / ( 𝑟 ) 1 / 𝑟 𝑡 ( 1 + 𝑟 ) / ( 𝑟 ) 1 𝑡 ( 1 𝑟 ) / ( 𝑟 ) 𝑡 1 / ( 𝑟 ) . / 𝑟 ( 4 . 2 0 ) For 0 < 𝑠 < 1 and 𝑡 > 1 , it is not difficult to verify that ( 1 𝑠 ) 𝑡 𝑠 + 𝑠 𝑡 𝑠 1 1 > 0 . Letting 𝑠 = 1 / 𝑟 , we have 1 1 + 𝑟 𝑡 1 / ( 𝑟 ) 1 𝑟 𝑡 ( 1 + 𝑟 ) / ( 𝑟 ) 1 > 0 , 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 > 0 . ( 4 . 2 1 ) Using Corollary 1.3, we know that 𝑓 𝑥 1 , 𝑥 2 , , 𝑥 𝑛 𝑓 ( 𝐴 ( 𝐱 ) , 𝐴 ( 𝐱 ) , , 𝐴 ( 𝐱 ) ) , 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 𝑛 𝑛 𝑘 = 1 𝑥 𝑘 𝑛 1 / 𝑟 ( 𝑛 1 ) 𝑛 𝑘 = 1 𝑥 𝑘 1 / 𝑟 ( 𝑛 1 ) . ( 4 . 2 2 ) Letting 𝑎 𝑖 = 𝑥 𝑖 1 / 𝑟 ( 𝑖 = 1 , 2 , , 𝑛 ) , we get 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 𝑛 𝑘 = 1 𝑎 𝑟 𝑘 𝑛 1 / 𝑟 ( 𝑛 1 ) 𝑛 𝑘 = 1 𝑎 𝑘 1 / ( 𝑛 1 ) , [ ] ( 4 . 2 3 ) 𝐴 ( 𝐚 ) ( 𝑛 1 ) / 𝑛 𝑀 𝑟 ( 𝐚 ) 1 / 𝑛 𝑀 0 ( 𝐚 ) . ( 4 . 2 4 )
From (4.23), we get 𝑛 𝑘 = 1 𝑎 𝑘 𝑟 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 ( 𝑛 1 ) 𝑟 𝑛 𝑘 = 1 𝑎 𝑟 𝑘 𝑛 . ( 4 . 2 5 ) Letting 𝑎 𝑖 𝑎 𝑖 1 / 𝑟 ( 𝑖 = 1 , 2 , , 𝑛 ) , we have 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑟 𝑛 ( 𝑛 1 ) 𝑟 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 , 𝑀 0 𝑀 ( 𝐚 ) 1 / 𝑟 ( 𝐚 ) ( 𝑛 1 ) / 𝑛 [ ] 𝐴 ( 𝐚 ) 1 / 𝑛 . ( 4 . 2 6 ) Inequality (4.14) is proved.

5. Five New Inequalities

Let 𝑛 3 and 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + . Then

𝑘 𝑛 ( 𝐚 ) = 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 𝑛 𝑘 ) 1 / ( ( 5 . 1 )

References [5, 6] is the third symmetric mean of 𝐚 .

Theorem 5.1. If 2 𝑘 𝑛 1 , 𝑝 = ( 𝑘 1 ) / ( 𝑛 1 ) , then 𝑘 𝑛 [ ] ( 𝐚 ) 𝐴 ( 𝐚 ) 𝑝 𝑀 0 ( 𝐚 ) 1 𝑝 ( 5 . 2 ) with the best possible constant 𝑝 = ( 𝑘 1 ) / ( 𝑛 1 ) .

Proof. Let 𝑎 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + and 𝑓 ( 𝐚 ) = 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 . ( 5 . 3 ) Then 𝜕 𝑓 𝜕 𝑎 1 = ( 𝑛 𝑘 ) ( 𝑛 𝑘 ) 𝑛 ( 𝑛 1 ) 𝑎 1 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 + 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 2 𝑖 1 < 𝑖 𝑘 1 𝑛 1 𝑎 1 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 , ( 5 . 4 ) 𝜕 𝑓 𝜕 𝑎 1 𝜕 𝑓 𝜕 𝑎 2 = ( 𝑛 𝑘 ) ( 𝑛 𝑘 ) 𝑛 ( 𝑛 1 ) 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 1 𝑎 1 1 𝑎 2 + 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 3 𝑖 1 < 𝑖 𝑘 1 𝑛 1 𝑎 1 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 1 𝑎 2 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 = 𝑎 1 𝑎 2 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 1 𝑘 𝑘 𝑗 = 1 𝑎 𝑖 𝑗 ( 𝑛 𝑘 ) ( 𝑛 𝑘 ) 𝑛 ( 𝑛 1 ) 𝑎 1 𝑎 2 3 𝑖 1 < 𝑖 𝑘 1 𝑛 1 𝑎 1 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 𝑎 2 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 . ( 5 . 5 ) If 𝐚 𝐷 (see (1.16)), then 𝑎 1 + ( 𝑘 1 ) 𝑎 2 > 𝑎 1 , ( 𝑛 𝑘 ) ( 𝑛 𝑘 ) 𝑛 ( 𝑛 1 ) 𝑎 1 𝑎 2 > 𝑛 2 𝑘 1 𝑘 𝑎 2 𝑎 1 + ( 𝑘 1 ) 𝑎 2 , ( 𝑛 𝑘 ) ( 𝑛 𝑘 ) 𝑛 ( 𝑛 1 ) 𝑎 1 𝑎 2 > 3 𝑖 1 < 𝑖 𝑘 1 𝑛 1 𝑎 1 + ( 𝑘 1 ) 𝑎 2 𝑘 𝑎 2 , ( 𝑛 𝑘 ) ( 𝑛 𝑘 ) 𝑛 ( 𝑛 1 ) 𝑎 1 𝑎 2 > 3 𝑖 1 < 𝑖 𝑘 1 𝑛 1 𝑎 1 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 𝑎 2 + 𝑘 1 𝑖 = 1 𝑎 𝑖 𝑗 . ( 5 . 6 )
Combining inequalities (5.5) and (5.6) yields that 𝜕 𝑓 / 𝜕 𝑎 1 𝜕 𝑓 / 𝜕 𝑎 2 > 0 . Then from Corollary 1.3 we have 𝑓 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 𝑓 ( 𝐴 ( 𝐚 ) , 𝐴 ( 𝐚 ) , , 𝐴 ( 𝐚 ) ) ( 5 . 7 ) for all 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + , which implies that 𝑛 𝑖 = 1 𝑎 𝑖 𝑛 𝑘 ( 𝑛 𝑘 ) ( ) / 𝑛 ( 𝑛 1 ) 1 𝑖 1 < < 𝑖 𝑘 𝑛 𝑘 𝑘 1 𝑗 = 1 𝑎 𝑖 𝑗 [ ] 𝐴 ( 𝐚 ) 𝑛 𝑘 ( 𝑘 1 ) ( ) / ( 𝑛 1 ) . ( 5 . 8 ) Therefore, inequality (5.2) is proved.

Taking 𝑎 1 = 𝑎 2 = = 𝑎 𝑛 1 = 1 and 𝑎 𝑛 = 𝑥 in inequality (5.2), we get

𝑥 + 𝑘 1 𝑘 ( 𝑛 𝑘 ) 𝑛 1 𝑘 1 ) / ( 𝑥 + 𝑛 1 𝑛 𝑝 𝑛 𝑥 1 𝑝 , 𝑝 ( 𝑘 / 𝑛 ) l n ( ( 𝑥 + 𝑘 1 ) / 𝑘 ) ( 1 / 𝑛 ) l n 𝑥 𝑛 l n ( 𝑥 + 𝑛 1 ) l n 𝑛 𝑥 . ( 5 . 9 ) Letting 𝑥 + , we get

𝑝 l i m 𝑥 + 𝑘 / 𝑛 1 / ( 𝑥 + 𝑘 1 ) 1 / 𝑛 𝑥 1 / ( 𝑥 + 𝑛 1 ) 1 / 𝑛 𝑥 = l i m 𝑥 + 𝑘 𝑥 / ( 𝑥 + 𝑘 1 ) 1 = 𝑛 𝑥 / ( 𝑥 + 𝑛 1 ) 1 𝑘 1 . 𝑛 1 ( 5 . 1 0 ) So 𝑝 = ( 𝑘 1 ) / ( 𝑛 1 ) is the best possible constant.

For 𝑛 2 , 𝐚 = ( 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 ) 𝑛 + , Alzer [7] established the following inequality:

𝑛 1 𝑛 1 𝐴 ( 𝐚 ) + 𝑛 𝑀 1 ( 𝐚 ) 𝑀 0 ( 𝐚 ) . ( 5 . 1 1 )

Theorems 5.2 and 5.3 are the improvements of Alzer's inequality.

Theorem 5.2. If 𝑝 = 𝑛 2 / ( 𝑛 2 + 4 𝑛 4 ) , then 𝑝 𝐴 ( 𝐚 ) + ( 1 𝑝 ) 𝑀 1 ( 𝐚 ) 𝑀 0 ( 𝐚 ) . ( 5 . 1 2 )

Proof. Firstly, let 𝑝 > 𝑛 2 / ( 𝑛 2 + 4 𝑛 4 ) , and 𝑓 ( 𝐱 ) = 𝑝 / 𝑛 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 + ( 1 𝑝 ) 𝑛 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 1 𝑥 , 𝑥 = 1 , 𝑥 2 , , 𝑥 𝑛 𝑛 . ( 5 . 1 3 ) Then 𝜕 𝑓 𝜕 𝑥 1 = 𝑝 𝑛 𝑒 𝑥 1 𝑛 + ( 1 𝑝 ) 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 2 𝑒 𝑥 1 , 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 = 𝑝 𝑛 ( 𝑒 𝑥 1 𝑒 𝑥 2 ) 𝑛 ( 1 𝑝 ) 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 2 ( 𝑒 𝑥 2 𝑒 𝑥 1 ) . ( 5 . 1 4 ) If 𝑥 1 = m a x 1 𝑖 𝑛 { 𝑥 𝑖 } > 𝑥 2 = m i n 1 𝑖 𝑛 { 𝑥 𝑖 } , 𝑡 = 𝑒 𝑥 1 𝑥 2 > 1 , then 𝜕 𝑓 𝜕 𝑥 1 𝜕 𝑓 𝜕 𝑥 2 𝑝 𝑛 ( 𝑒 𝑥 1 𝑒 𝑥 2 𝑛 ) ( 1 𝑝 ) ( ( 𝑛 1 ) 𝑒 𝑥 1 + 𝑒 𝑥 2 ) 2 ( 𝑒 𝑥 2 𝑒 𝑥 1 ) = ( 𝑒 𝑥 1 𝑒 𝑥 2 ) 𝑛 ( ( 𝑛 1 ) 𝑒 𝑥 2 + 𝑒 𝑥 1 ) 2 𝑝 ( ( 𝑛 1 ) 𝑒 𝑥 2 + 𝑒 𝑥 1 ) 2 ( 1 𝑝 ) 𝑛 2 𝑒 𝑥 1 𝑒 𝑥 2 = 𝑒 3 𝑥 2 ( 𝑡 1 ) 𝑛 ( ( 𝑛 1 ) 𝑒 𝑥 2 + 𝑒 𝑥 1 ) 2 𝑝 ( 𝑛 1 + 𝑡 ) 2 𝑛 2 𝑡 + 𝑝 𝑛 2 𝑡 > 𝑒 3 𝑥 2 ( 𝑡 1 ) 𝑛 ( ( 𝑛 1 ) 𝑒 𝑥 2 + 𝑒 𝑥 1 ) 2 𝑛 2 𝑛 2 + 4 𝑛 4 ( 𝑛 1 + 𝑡 ) 2 𝑛 2 𝑛 𝑡 + 2 𝑛 2 𝑛 + 4 𝑛 4 2 𝑡 = 𝑛 𝑒 3 𝑥 2 ( 𝑡 1 ) ( 𝑡 𝑛 + 1 ) 2 𝑛 2 + 4 𝑛 4 ( ( 𝑛 1 ) 𝑒 𝑥 2 + 𝑒 𝑥 1 ) 2 0 . ( 5 . 1 5 ) Then from Corollary 1.3, we get 𝑝 𝑓 ( 𝐱 ) 𝑓 ( 𝐴 ( 𝐱 ) , 𝐴 ( 𝐱 ) , , 𝐴 ( 𝐱 ) ) , 𝑛 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 𝑛 + ( 1 𝑝 ) 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 𝑒 𝐴 ( 𝐱 ) = 𝑛 𝑛 𝑖 = 1 𝑒 𝑥 𝑖 . ( 5 . 1 6 ) Let 𝑒 𝑥 𝑖 = 𝑎 𝑖 in above inequality. Then we know that inequality (5.12) holds. From continuity we know that inequality (5.12) holds also for 𝑝 = 𝑛 2 / ( 𝑛 2 + 4 𝑛 4 ) .

Theorem 5.3. If 𝑝 = ( 1 𝑛 5 𝑛 2 6 𝑛 + 1 ) / ( 2 𝑛 ) , then 𝑛 1 𝑛 1 𝐴 ( 𝐚 ) + 𝑛 𝑀 𝑝 ( 𝐚 ) 𝑀 0 ( 𝐚 ) . ( 5 . 1 7 )

Proof. Let 𝑓 ( 𝐚 ) = 𝑛 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑝 ( 𝑛 1 ) 𝑛 2 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑝 , 𝐚 𝑛 + . ( 5 . 1 8 ) Then 𝜕 𝑓 𝜕 𝑎 1 = 1 𝑛 𝑝 𝑎 1 𝑛 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑝 𝑛 1 𝑛 2 𝑝 𝑎 1 1 / 𝑝 1 , 𝜕 𝑓 𝜕 𝑎 1 𝜕 𝑓 𝜕 𝑎 2 𝑎 = 1 𝑎 2 𝑛 𝑝 𝑎 1 𝑎 2 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑛 𝑝 𝑛 1 𝑛 2 𝑝 𝑎 1 1 / 𝑝 1 𝑎 2 1 / 𝑝 1 . ( 5 . 1 9 ) If 𝑎 1 = m a x 1 𝑖 𝑛 { 𝑎 𝑖 } > 𝑎 2 = m i n 1 𝑖 𝑛 { 𝑎 𝑖 } > 0 and 𝑎 1 / 𝑎 2 = 𝑡 > 1 , then from 𝑝 < 0 and ( 𝑎 1 𝑎 2 ) / 𝑛 𝑝 𝑎 1 𝑎 2 > 0 we get 𝜕 𝑓 𝜕 𝑎 1 𝜕 𝑓 𝜕 𝑎 2 𝑎 1 𝑎 2 𝑛 𝑝 𝑎 1 𝑎 2 𝑎 1 1 / ( 𝑛 𝑝 ) 𝑎 2 ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑛 1 𝑛 2 𝑝 𝑎 1 1 / 𝑝 1 𝑎 2 1 / 𝑝 1 = 𝑎 1 1 / 𝑝 1 𝑛 2 𝑝 𝑛 𝑡 1 𝑡 𝑡 1 ( 𝑛 1 ) / ( 𝑛 𝑝 ) ( 𝑛 1 ) 1 𝑡 1 1 / 𝑝 . ( 5 . 2 0 )
Let 𝑔 ( 𝑡 ) = 𝑛 𝑡 1 ( 𝑛 1 ) / ( 𝑛 𝑝 ) + 𝑛 𝑡 ( 𝑛 1 ) / ( 𝑛 𝑝 ) + ( 𝑛 1 ) 𝑡 1 1 / 𝑝 ( 𝑛 1 ) , 𝑡 > 1 . Then 𝑔 ( 𝑡 ) = 𝑛 + 𝑛 1 𝑝 𝑡 ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑛 1 𝑝 𝑡 1 ( 𝑛 1 ) / ( 𝑛 𝑝 ) 1 + ( 𝑛 1 ) 1 𝑝 𝑡 1 / 𝑝 , 𝑡 1 + ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑔 ( 𝑡 ) = 𝑛 + 𝑛 1 𝑝 𝑡 𝑛 1 𝑝 1 + ( 𝑛 1 ) 1 𝑝 𝑡 1 1 / ( 𝑛 𝑝 ) , 𝑡 1 + ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑔 ( 𝑡 ) = 𝑛 + 𝑛 1 𝑝 1 + ( 𝑛 1 ) 1 𝑝 1 1 𝑡 𝑛 𝑝 1 / ( 𝑛 𝑝 ) > 𝑛 + 𝑛 1 𝑝 1 + ( 𝑛 1 ) 1 𝑝 1 1 1 𝑛 𝑝 = 𝑝 2 𝑝 2 + 1 1 𝑛 1 𝑝 1 + 𝑛 = 0 . ( 5 . 2 1 ) Thus 𝑡 1 + ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑔 ( 𝑡 ) is a monotone increasing function. This monotonicity and l i m 𝑡 1 + 𝑡 1 + ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑔 ( 𝑡 ) = l i m 𝑡 1 + 𝑛 + 𝑛 1 𝑝 𝑡 𝑛 1 𝑝 1 + ( 𝑛 1 ) 1 𝑝 𝑡 1 ( 1 / 𝑛 𝑝 ) = 1 𝑛 1 𝑝 0 ( 5 . 2 2 ) lead to 𝑡 1 + ( 𝑛 1 ) / ( 𝑛 𝑝 ) 𝑔 ( 𝑡 ) > 0 . Therefore 𝑔 ( 𝑡 ) > 0 and 𝑔 ( 𝑡 ) is a monotone increasing function. From l i m 𝑡 1 + 𝑔 ( 𝑡 ) = 0 and the monotonicity of 𝑔 ( 𝑡 ) we know that 𝑔 ( 𝑡 ) > 0 . By (5.20), we know that 𝜕 𝑓 / 𝜕 𝑎 1 𝜕 𝑓 / 𝜕 𝑎 2 < 0 . According to Corollary 1.3 we get 𝑓 ( 𝐚 ) 𝑓 ( 𝐴 ( 𝐚 ) , 𝐴 ( 𝐚 ) , , 𝐴 ( 𝐚 ) ) , 𝑛 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑝 𝑛 1 𝑛 2 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑝 1 𝑛 𝐴 1 / 𝑝 ( 𝐚 ) . ( 5 . 2 3 ) Finally, let 𝑎 𝑖 𝑎 𝑝 𝑖 ( 𝑖 = 1 , 2 , , 𝑛 ) in the above inequality. Then we know that Theorem 5.3 holds.
If 𝑛 2 and 0 < 𝑎 1 𝑎 2 𝑎 𝑛 , then the following inequalities can be found in [810]: 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 2 𝑛 2 𝑎 𝑛 𝐴 ( 𝐚 ) 𝑀 0 ( 𝐚 ) 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 2 𝑛 2 𝑎 1 , 𝑎 3 1 2 𝑛 2 𝑎 4 𝑛 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 𝑀 0 ( 𝐚 ) 𝑀 1 ( 𝑎 𝐚 ) 3 𝑛 2 𝑛 2 𝑎 4 1 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 . ( 5 . 2 4 ) Theorems 5.4 and 5.5 are the improvements of inequalities (5.24).

Theorem 5.4. If 𝑛 2 and 0 < 𝑚 𝑎 1 , 𝑎 2 , , 𝑎 𝑛 𝑀 , then 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 2 𝑛 2 𝑀 ( 𝑛 1 ) / 𝑛 𝐴 1 / 𝑛 ( 𝐚 ) 𝐴 ( 𝐚 ) 𝑀 0 ( 𝐚 ) 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 2 𝑛 2 𝑚 ( 𝑛 1 ) / 𝑛 𝐴 1 / 𝑛 . ( 𝐚 ) ( 5 . 2 5 )

Proof. Let [ ] 𝑓 𝐚 𝑚 , 𝑀 𝑛 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑛 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 𝑖 < 𝑗 𝑛 𝑎 𝑖 𝑎 𝑗 2 2 𝑛 2 𝑀 ( 𝑛 1 ) / 𝑛 . ( 5 . 2 6 ) Then 𝜕 𝑓 𝜕 𝑎 1 = 1 𝑛 2 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑛 1 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 + 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑛 1 𝑛 1 𝑛 𝑎 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 2 𝑖 𝑛 𝑎 1 𝑎 𝑖 𝑛 2 𝑀 ( 𝑛 1 ) / 𝑛 , 𝜕 𝑓 𝜕 𝑎 1 𝜕 𝑓 𝜕 𝑎 2 = 𝑎 1 𝑎 2 𝑛 𝑎 1 𝑎 2 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 / 𝑛 𝑎 1 𝑎 2 𝑛 𝑀 ( 𝑛 1 ) / 𝑛 = 𝑎 1 𝑎 2 𝑛 𝑎 1 𝑎 2 𝑀 ( 𝑛 1 ) / 𝑛 𝑀 ( 𝑛 1 ) / 𝑛 𝑛 𝑛 𝑘 = 1 𝑎 𝑘 1 𝑛 𝑛 𝑘 = 1 𝑎