Journal of Inequalities and Applications
Volume 2010 (2010), Article ID 905679, 6 pages
doi:10.1155/2010/905679
Research Article

Optimal Power Mean Bounds for the Weighted Geometric Mean of Classical Means

Bo-Yong Long1,2 and Yu-Ming Chu3

1College of Mathematics and Econometrics, Hunan University, Changsha 410082, China
2School of Mathematical Sciences, Anhui University, Hefei 230039, China
3Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Received 13 November 2009; Accepted 25 February 2010

Academic Editor: Andrea Laforgia

Copyright © 2010 Bo-Yong Long and Yu-Ming Chu. 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

For 𝑝 ∈ ℝ , the power mean of order 𝑝 of two positive numbers 𝑎 and 𝑏 is defined by 𝑀 𝑝 ( 𝑎 , 𝑏 ) = ( ( 𝑎 𝑝 + 𝑏 𝑝 ) / 2 ) 1 / 𝑝 , for 𝑝 ≠ 0 , and 𝑀 𝑝 √ ( 𝑎 , 𝑏 ) = 𝑎 𝑏 , for 𝑝 = 0 . In this paper, we answer the question: what are the greatest value 𝑝 and the least value 𝑞 such that the double inequality 𝑀 𝑝 ( 𝑎 , 𝑏 ) ≤ 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) ≤ 𝑀 𝑞 ( 𝑎 , 𝑏 ) holds for all 𝑎 , 𝑏 > 0 and 𝛼 , 𝛽 > 0 with 𝛼 + 𝛽 < 1 ? Here 𝐴 ( 𝑎 , 𝑏 ) = ( 𝑎 + 𝑏 ) / 2 , √ 𝐺 ( 𝑎 , 𝑏 ) = 𝑎 𝑏 , and 𝐻 ( 𝑎 , 𝑏 ) = 2 𝑎 𝑏 / ( 𝑎 + 𝑏 ) denote the classical arithmetic, geometric, and harmonic means, respectively.

1. Introduction

For 𝑝 ∈ ℝ , the power mean of order 𝑝 of two positive numbers 𝑎 and 𝑏 is defined by

𝑀 𝑝 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑎 ( 𝑎 , 𝑏 ) = 𝑝 + 𝑏 𝑝 2 1 / 𝑝 √ , 𝑝 ≠ 0 , 𝑎 𝑏 , 𝑝 = 0 . ( 1 . 1 )

Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities for 𝑀 𝑝 ( 𝑎 , 𝑏 ) can be found in literatures [1–12]. It is well known that 𝑀 𝑝 ( 𝑎 , 𝑏 ) is continuous and increasing with respect to 𝑝 ∈ ℝ for fixed 𝑎 and 𝑏 . Let √ 𝐴 ( 𝑎 , 𝑏 ) = ( 𝑎 + 𝑏 ) / 2 , 𝐺 ( 𝑎 , 𝑏 ) = 𝑎 𝑏 , and 𝐻 ( 𝑎 , 𝑏 ) = 2 𝑎 𝑏 / ( 𝑎 + 𝑏 ) be the classical arithmetic, geometric, and harmonic means of two positive numbers 𝑎 and 𝑏 , respectively. Then

m i n { 𝑎 , 𝑏 } ≤ 𝐻 ( 𝑎 , 𝑏 ) = 𝑀 − 1 ( 𝑎 , 𝑏 ) ≤ 𝐺 ( 𝑎 , 𝑏 ) = 𝑀 0 ( 𝑎 , 𝑏 ) ≤ 𝐴 ( 𝑎 , 𝑏 ) = 𝑀 1 ( 𝑎 , 𝑏 ) ≤ m a x { 𝑎 , 𝑏 } . ( 1 . 2 )

In [13], Alzer and Janous established the following sharp double inequality (see also [14, page 350]):

𝑀 l o g 2 / l o g 3 2 ( 𝑎 , 𝑏 ) ≤ 3 1 𝐴 ( 𝑎 , 𝑏 ) + 3 𝐺 ( 𝑎 , 𝑏 ) ≤ 𝑀 2 / 3 ( 𝑎 , 𝑏 ) ( 1 . 3 ) for all 𝑎 , 𝑏 > 0 .

In [15], Mao proved

𝑀 1 / 3 1 ( 𝑎 , 𝑏 ) ≤ 3 2 𝐴 ( 𝑎 , 𝑏 ) + 3 𝐺 ( 𝑎 , 𝑏 ) ≤ 𝑀 1 / 2 ( 𝑎 , 𝑏 ) ( 1 . 4 ) for all 𝑎 , 𝑏 > 0 , and 𝑀 1 / 3 ( 𝑎 , 𝑏 ) is the best possible lower power mean bound for the sum ( 1 / 3 ) 𝐴 ( 𝑎 , 𝑏 ) + ( 2 / 3 ) 𝐺 ( 𝑎 , 𝑏 ) .

The following sharp bounds for ( 2 / 3 ) 𝐺 + ( 1 / 3 ) 𝐻 and ( 1 / 3 ) 𝐺 + ( 2 / 3 ) 𝐻 in terms of power mean are proved in [16]:

𝑀 − 1 / 3 2 ( 𝑎 , 𝑏 ) ≤ 3 1 𝐺 ( 𝑎 , 𝑏 ) + 3 𝐻 ( 𝑎 , 𝑏 ) ≤ 𝑀 0 𝑀 ( 𝑎 , 𝑏 ) , − 2 / 3 1 ( 𝑎 , 𝑏 ) ≤ 3 2 𝐺 ( 𝑎 , 𝑏 ) + 3 𝐻 ( 𝑎 , 𝑏 ) ≤ 𝑀 0 ( 𝑎 , 𝑏 ) ( 1 . 5 ) for all 𝑎 , 𝑏 > 0 .

The purpose of this paper is to answer the question: what are the greatest value 𝑝 and the least value 𝑞 such that the double inequality

𝑀 𝑝 ( 𝑎 , 𝑏 ) ≤ 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) ≤ 𝑀 𝑞 ( 𝑎 , 𝑏 ) ( 1 . 6 ) holds for all 𝑎 , 𝑏 > 0 and 𝛼 , 𝛽 > 0 with 𝛼 + 𝛽 < 1 ?

2. Main Result

In order to establish our main results we need the following lemma.

Lemma 2.1. If ⋃ 𝜆 ∈ ( − 1 , 0 ) ( 0 , 1 ) , 𝑡 ≥ 1 and 𝑓 ( 𝑡 ) = ( 1 / 𝜆 ) l o g ( ( 𝑡 𝜆 + 1 ) / 2 ) − 𝜆 l o g ( ( 𝑡 + 1 ) / 2 ) − ( ( 1 − 𝜆 ) / 2 ) l o g 𝑡 , then ( 1 ) 𝑓 ( 𝑡 ) > 0 for 𝜆 ∈ ( 0 , 1 ) and 𝑡 > 1 ;( 2 ) 𝑓 ( 𝑡 ) < 0 for 𝜆 ∈ ( − 1 , 0 ) and 𝑡 > 1 .

Proof. Simple computations lead to 𝑓 𝑓 ( 1 ) = 0 , ( 2 . 1 ) ( 𝑡 ) = 𝑔 ( 𝑡 ) 𝑡 𝑡 ( 𝑡 + 1 ) 𝜆 , + 1 ( 2 . 2 ) where 𝑔 ( 𝑡 ) = ( ( 1 − 𝜆 ) / 2 ) 𝑡 𝜆 + 1 + ( ( 1 + 𝜆 ) / 2 ) 𝑡 𝜆 − ( ( 1 + 𝜆 ) / 2 ) 𝑡 − ( ( 1 − 𝜆 ) / 2 ) : 𝑔 ( 1 ) = 0 , ( 2 . 3 ) 𝑔 ′ ( 𝑡 ) = ( 1 − 𝜆 ) ( 1 + 𝜆 ) 2 𝑡 𝜆 + 𝜆 ( 1 + 𝜆 ) 2 𝑡 𝜆 − 1 − 1 + 𝜆 2 𝑔 , ( 2 . 4 ) 𝑔 ( 1 ) = 0 , ( 2 . 5 ) ( 𝑡 ) = 𝜆 ( 1 − 𝜆 ) ( 1 + 𝜆 ) 2 ( 𝑡 − 1 ) 𝑡 𝜆 − 2 . ( 2 . 6 ) ( 1 )If 𝜆 ∈ ( 0 , 1 ) and 𝑡 > 1 , then (2.6) implies 𝑔 ( 𝑡 ) > 0 . ( 2 . 7 ) Therefore, Lemma 2.1( 1 ) follows from (2.1)–(2.3) and (2.5) together with (2.7).( 2 )If 𝜆 ∈ ( − 1 , 0 ) and 𝑡 > 1 , then (2.6) yields 𝑔 ( 𝑡 ) < 0 . ( 2 . 8 ) Therefore, Lemma 2.1(2) follows from (2.1)–(2.3) and (2.5) together with (2.8).

Theorem 2.2. For all 𝑎 , 𝑏 > 0 and 𝛼 , 𝛽 > 0 with 𝛼 + 𝛽 < 1 , one has ( 1 ) 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) = 𝑀 0 ( 𝑎 , 𝑏 ) = 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) for 2 𝛼 + 𝛽 = 1 ;(2) 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) ≥ 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) ≥ 𝑀 0 ( 𝑎 , 𝑏 ) for 2 𝛼 + 𝛽 > 1 , and 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) ≤ 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) ≤ 𝑀 0 ( 𝑎 , 𝑏 ) for 2 𝛼 + 𝛽 < 1 , each equality occurs if and only if 𝑎 = 𝑏 , and 𝑀 0 ( 𝑎 , 𝑏 ) and 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) are the best possible power mean bounds for the product 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) .

Proof. (1) If 2 𝛼 + 𝛽 = 1 , then simple computations lead to 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) = 𝑎 + 𝑏 2 2 𝛼 + 𝛽 − 1 ( 𝑎 𝑏 ) 1 − ( 𝛼 + ( 𝛽 / 2 ) ) = √ 𝑎 𝑏 = 𝑀 0 ( 𝑎 , 𝑏 ) = 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) . ( 2 . 9 )
( 2 ) If 2 𝛼 + 𝛽 ≠ 1 and 𝑎 = 𝑏 , then we clearly see that
𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) = 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) = 𝑀 0 ( 𝑎 , 𝑏 ) = 𝑎 . ( 2 . 1 0 )
If 2 𝛼 + 𝛽 ≠ 1 and 𝑎 ≠ 𝑏 , without loss of generality, we assume that 𝑎 > 𝑏 . Let 𝑡 = ( 𝑎 / 𝑏 ) > 1 and 𝜆 = 2 𝛼 + 𝛽 − 1 , then ⋃ 𝜆 ∈ ( − 1 , 0 ) ( 0 , 1 ) , and simple computations lead to
l o g 𝑀 2 𝛼 + 𝛽 − 1 𝐴 ( 𝑎 , 𝑏 ) − l o g 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 = 1 ( 𝑎 , 𝑏 ) 𝑡 2 𝛼 + 𝛽 − 1 l o g 2 𝛼 + 𝛽 − 1 + 1 2 − ( 2 𝛼 + 𝛽 − 1 ) l o g 1 + 𝑡 2 − 𝛽 1 − 𝛼 − 2 = 1 l o g 𝑡 𝜆 𝑡 l o g 𝜆 + 1 2 − 𝜆 l o g 𝑡 + 1 2 − 1 − 𝜆 2 𝐴 l o g 𝑡 , ( 2 . 1 1 ) 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) 𝑀 0 = ⎛ ⎜ ⎜ ⎝ √ ( 𝑎 , 𝑏 ) √ 𝑡 + 1 / 𝑡 2 ⎞ ⎟ ⎟ ⎠ 2 𝛼 + 𝛽 − 1 . ( 2 . 1 2 )
Therefore, 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) > 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) > 𝑀 0 ( 𝑎 , 𝑏 ) for 2 𝛼 + 𝛽 > 1 follows from (2.11) and Lemma 2.1(1) together with (2.12), and 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) < 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) < 𝑀 0 ( 𝑎 , 𝑏 ) for 2 𝛼 + 𝛽 < 1 follows from (2.11) and Lemma 2.1(2) together with (2.12).

Next, we prove that 𝑀 0 ( 𝑎 , 𝑏 ) and 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) are the best possible power mean bounds for the product 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) .

Firstly, we prove that 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) is the best possible upper power mean bound for the product 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) if 2 𝛼 + 𝛽 > 1 .

For any 𝜖 ∈ ( 0 , 2 𝛼 + 𝛽 − 1 ) and 𝑥 > 0 , one has

𝑀 2 𝛼 + 𝛽 − 1 − 𝜖 ( 1 , 1 + 𝑥 ) 2 𝛼 + 𝛽 − 1 − 𝜖 − 𝐴 𝛼 ( 1 , 1 + 𝑥 ) 𝐺 𝛽 ( 1 , 1 + 𝑥 ) 𝐻 1 − 𝛼 − 𝛽 ( 1 , 1 + 𝑥 ) 2 𝛼 + 𝛽 − 1 − 𝜖 = ( 1 + 𝑥 ) 2 𝛼 + 𝛽 − 1 − 𝜖 + 1 2 − 𝑥 1 + 2 ( 2 𝛼 + 𝛽 − 1 ) ( 2 𝛼 + 𝛽 − 1 − 𝜖 ) ( 1 + 𝑥 ) ( 1 − 𝛼 − ( 𝛽 / 2 ) ) ( 2 𝛼 + 𝛽 − 1 − 𝜖 ) . ( 2 . 1 3 )

Let 𝑥 → 0 , then the Taylor expansion leads to

( 1 + 𝑥 ) 2 𝛼 + 𝛽 − 1 − 𝜖 + 1 2 − 𝑥 1 + 2 ( 2 𝛼 + 𝛽 − 1 ) ( 2 𝛼 + 𝛽 − 1 − 𝜖 ) ( 1 + 𝑥 ) ( 1 − 𝛼 − ( 𝛽 / 2 ) ) ( 2 𝛼 + 𝛽 − 1 − 𝜖 ) 1 = − 8 𝜖 ( 2 𝛼 + 𝛽 − 1 − 𝜖 ) 𝑥 2 𝑥 + 𝑜 2 . ( 2 . 1 4 )

Equations (2.13) and (2.14) imply that if 2 𝛼 + 𝛽 > 1 , then for any 𝜖 ∈ ( 0 , 2 𝛼 + 𝛽 − 1 ) there exists 𝛿 1 = 𝛿 1 ( 𝜖 , 𝛼 , 𝛽 ) > 0 , such that 𝑀 2 𝛼 + 𝛽 − 1 − 𝜖 ( 1 , 1 + 𝑥 ) < 𝐴 𝛼 ( 1 , 1 + 𝑥 ) 𝐺 𝛽 ( 1 , 1 + 𝑥 ) 𝐻 1 − 𝛼 − 𝛽 ( 1 , 1 + 𝑥 ) for 𝑥 ∈ ( 0 , 𝛿 1 ) .

Secondly, we prove that 𝑀 0 ( 𝑎 , 𝑏 ) is the best possible lower power mean bound for the product 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) if 2 𝛼 + 𝛽 > 1 .

For any 𝜖 > 0 and 𝑡 > 1 , one has

𝐴 𝛼 ( 𝑡 , 1 ) 𝐺 𝛽 ( 𝑡 , 1 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑡 , 1 ) 𝑀 𝜖 = ( 𝑡 , 1 ) 1 + 𝑡 − 1 / 2 2 𝛼 + 𝛽 − 1 ( ( 1 + 𝑡 − 𝜖 ) / 2 ) 1 / 𝜖 𝑡 𝛼 + ( 𝛽 / 2 ) − 1 . ( 2 . 1 5 )

From (2.15) and 𝛼 + ( 𝛽 / 2 ) < 1 , we clearly see that

l i m 𝑡 → + ∞ 𝐴 𝛼 ( 𝑡 , 1 ) 𝐺 𝛽 ( 𝑡 , 1 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑡 , 1 ) 𝑀 𝜖 ( 𝑡 , 1 ) = 0 . ( 2 . 1 6 )

Equation (2.16) implies that if 2 𝛼 + 𝛽 > 1 , then for any 𝜖 ∈ ( 0 , 2 𝛼 + 𝛽 − 1 ) there exists 𝑇 1 = 𝑇 1 ( 𝜖 , 𝛼 , 𝛽 ) > 1 , such that 𝐴 𝛼 ( 𝑡 , 1 ) 𝐺 𝛽 ( 𝑡 , 1 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑡 , 1 ) < 𝑀 𝜖 ( 𝑡 , 1 ) for 𝑡 ∈ ( 𝑇 1 , + ∞ ) .

Thirdly, we prove that 𝑀 2 𝛼 + 𝛽 − 1 ( 𝑎 , 𝑏 ) is the best possible lower power mean bound for the product 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) if 2 𝛼 + 𝛽 < 1 .

For any 𝜖 ∈ ( 0 , 1 − 2 𝛼 − 𝛽 ) and 𝑥 > 0 , one has

𝑀 2 𝛼 + 𝛽 − 1 + 𝜖 ( 1 , 1 + 𝑥 ) 1 − 2 𝛼 − 𝛽 − 𝜖 − 𝐴 𝛼 ( 1 , 1 + 𝑥 ) 𝐺 𝛽 ( 1 , 1 + 𝑥 ) 𝐻 1 − 𝛼 − 𝛽 ( 1 , 1 + 𝑥 ) 1 − 2 𝛼 − 𝛽 − 𝜖 = 𝑔 ( 𝑥 ) 1 + ( 1 + 𝑥 ) 1 − 2 𝛼 − 𝛽 − 𝜖 ( 1 + ( 𝑥 / 2 ) ) ( 1 − 2 𝛼 − 𝛽 ) ( 1 − 2 𝛼 − 𝛽 − 𝜖 ) , ( 2 . 1 7 ) where 𝑔 ( 𝑥 ) = 2 ( 1 + 𝑥 ) 1 − 2 𝛼 − 𝛽 − 𝜖 ( 1 + ( 𝑥 / 2 ) ) ( 1 − 2 𝛼 − 𝛽 ) ( 1 − 2 𝛼 − 𝛽 − 𝜖 ) − ( 1 + 𝑥 ) ( 1 − 𝛼 − ( 𝛽 / 2 ) ) ( 1 − 2 𝛼 − 𝛽 − 𝜖 ) [ 1 + ( 1 + 𝑥 ) 1 − 2 𝛼 − 𝛽 − 𝜖 ] .

Let 𝑥 → 0 , then the Taylor expansion leads to

1 𝑔 ( 𝑥 ) = 4 𝜖 ( 1 − 2 𝛼 − 𝛽 − 𝜖 ) 𝑥 2 𝑥 + 𝑜 2 . ( 2 . 1 8 )

Equations (2.17) and (2.18) imply that if 2 𝛼 + 𝛽 < 1 , then for any 𝜖 ∈ ( 0 , 1 − 2 𝛼 − 𝛽 ) there exists 0 < 𝛿 2 = 𝛿 2 ( 𝜖 , 𝛼 , 𝛽 ) < 1 , such that 𝑀 2 𝛼 + 𝛽 − 1 + 𝜖 ( 1 , 1 + 𝑥 ) > 𝐴 𝛼 ( 1 , 1 + 𝑥 ) 𝐺 𝛽 ( 1 , 1 + 𝑥 ) 𝐻 1 − 𝛼 − 𝛽 ( 1 , 1 + 𝑥 ) for 𝑥 ∈ ( 0 , 𝛿 2 ) .

Finally, we prove that 𝑀 0 ( 𝑎 , 𝑏 ) is the best possible upper power mean bound for the product 𝐴 𝛼 ( 𝑎 , 𝑏 ) 𝐺 𝛽 ( 𝑎 , 𝑏 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑎 , 𝑏 ) if 2 𝛼 + 𝛽 < 1 .

For any 𝜖 > 0 and 𝑡 > 1 , one has

𝐴 𝛼 ( 𝑡 , 1 ) 𝐺 𝛽 ( 𝑡 , 1 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑡 , 1 ) 𝑀 − 𝜖 = ( 𝑡 , 1 ) 1 + 𝑡 − 1 / 2 2 𝛼 + 𝛽 − 1 ( ( 1 + 𝑡 − 𝜖 ) / 2 ) − 1 / 𝜖 𝑡 𝛼 + ( 𝛽 / 2 ) . ( 2 . 1 9 ) From (2.19) and 𝛼 + ( 𝛽 / 2 ) > 0 we clearly see that

l i m 𝑡 → + ∞ 𝐴 𝛼 ( 𝑡 , 1 ) 𝐺 𝛽 ( 𝑡 , 1 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑡 , 1 ) 𝑀 − 𝜖 ( 𝑡 , 1 ) = + ∞ . ( 2 . 2 0 )

Equation (2.20) implies that if 2 𝛼 + 𝛽 < 1 , then for any 𝜖 > 0 there exists 𝑇 2 = 𝑇 2 ( 𝜖 , 𝛼 , 𝛽 ) > 1 , such that 𝐴 𝛼 ( 𝑡 , 1 ) 𝐺 𝛽 ( 𝑡 , 1 ) 𝐻 1 − 𝛼 − 𝛽 ( 𝑡 , 1 ) > 𝑀 − 𝜖 ( 𝑡 , 1 ) for 𝑡 ∈ ( 𝑇 2 , + ∞ ) .

Acknowledgments

This work is partly supported by the National Natural Science Foundation of China (Grant no. 60850005) and the Natural Science Foundation of Zhejiang Province (Grant no. D7080080, Y607128).

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