Journal of Inequalities and Applications
Volume 2010 (2010), Article ID 806825, 10 pages
doi:10.1155/2010/806825
Research Article

Optimal Inequalities for Generalized Logarithmic, Arithmetic, and Geometric 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 20 October 2009; Revised 16 December 2009; Accepted 1 February 2010

Academic Editor: Alexander I. Domoshnitsky

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 generalized logarithmic mean 𝐿 𝑝 ( 𝑎 , 𝑏 ) , arithmetic mean 𝐴 ( 𝑎 , 𝑏 ) , and geometric mean 𝐺 ( 𝑎 , 𝑏 ) of two positive numbers 𝑎 and 𝑏 are defined by 𝐿 𝑝 ( 𝑎 , 𝑏 ) = 𝑎 , for 𝑎 = 𝑏 , 𝐿 𝑝 ( 𝑎 , 𝑏 ) = [ ( 𝑏 𝑝 + 1 − 𝑎 𝑝 + 1 ) / ( ( 𝑝 + 1 ) ( 𝑏 − 𝑎 ) ) ] 1 / 𝑝 , for 𝑝 ≠ 0 , 𝑝 ≠ − 1 , and 𝑎 ≠ 𝑏 , 𝐿 𝑝 ( 𝑎 , 𝑏 ) = ( 1 / 𝑒 ) ( 𝑏 𝑏 / 𝑎 𝑎 ) 1 / ( 𝑏 − 𝑎 ) , for 𝑝 = 0 , and 𝑎 ≠ 𝑏 , 𝐿 𝑝 ( 𝑎 , 𝑏 ) = ( 𝑏 − 𝑎 ) / ( l o g 𝑏 − l o g 𝑎 ) , for 𝑝 = − 1 , and 𝑎 ≠ 𝑏 , 𝐴 ( 𝑎 , 𝑏 ) = ( 𝑎 + 𝑏 ) / 2 , and √ 𝐺 ( 𝑎 , 𝑏 ) = 𝑎 𝑏 , respectively. In this paper, we find the greatest value 𝑝 (or least value 𝑞 , resp.) such that the inequality 𝐿 𝑝 ( 𝑎 , 𝑏 ) < 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) (or 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) < 𝐿 𝑞 ( 𝑎 , 𝑏 ) , resp.) holds for 𝛼 ∈ ( 0 , 1 / 2 ) (or 𝛼 ∈ ( 1 / 2 , 1 ) , resp.) and all 𝑎 , 𝑏 > 0 with 𝑎 ≠ 𝑏 .

1. Introduction

For 𝑝 ∈ ℝ , the generalized logarithmic mean 𝐿 𝑝 ( 𝑎 , 𝑏 ) and power mean 𝑀 𝑝 ( 𝑎 , 𝑏 ) with parameter 𝑝 of two positive numbers 𝑎 and 𝑏 are defined by

𝐿 𝑝 ( ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ 𝑏 𝑎 , 𝑏 ) = 𝑎 , 𝑎 = 𝑏 , 𝑝 + 1 − 𝑎 𝑝 + 1 ( 𝑝 + 1 ) ( 𝑏 − 𝑎 ) 1 / 𝑝 1 , 𝑝 ≠ 0 , 𝑝 ≠ − 1 , 𝑎 ≠ 𝑏 , 𝑒 𝑏 𝑏 𝑎 𝑎 1 / ( 𝑏 − 𝑎 ) , 𝑝 = 0 , 𝑎 ≠ 𝑏 , 𝑏 − 𝑎 l o g 𝑏 − l o g 𝑎 , 𝑝 = − 1 , 𝑎 ≠ 𝑏 , ( 1 . 1 ) and

𝑀 𝑝 ⎧ ⎪ ⎨ ⎪ ⎩ 𝑎 ( 𝑎 , 𝑏 ) = 𝑝 + 𝑏 𝑝 2 1 / 𝑝 √ , 𝑝 ≠ 0 , 𝑎 𝑏 , 𝑝 = 0 , ( 1 . 2 ) respectively. It is well known that both means are continuous and increasing with respect to 𝑝 ∈ ℝ for fixed 𝑎 and 𝑏 . Recently, both means have been the subject of intensive research. In particular, many remarkable inequalities involving 𝐿 𝑝 ( 𝑎 , 𝑏 ) and 𝑀 𝑝 ( 𝑎 , 𝑏 ) can be found in the literature [1–9]. Let

𝐴 ( 𝑎 , 𝑏 ) = 𝑎 + 𝑏 2 ⎧ ⎪ ⎨ ⎪ ⎩ 1 , 𝐼 ( 𝑎 , 𝑏 ) = 𝑒 𝑏 𝑏 𝑎 𝑎 1 / ( 𝑏 − 𝑎 ) , 𝑏 ≠ 𝑎 , 𝑎 , 𝑏 = 𝑎 , 𝐿 ( 𝑎 , 𝑏 ) = 𝑏 − 𝑎 l o g 𝑏 − l o g 𝑎 , 𝑏 ≠ 𝑎 , 𝑎 , 𝑏 = 𝑎 , ( 1 . 3 ) √ 𝐺 ( 𝑎 , 𝑏 ) = 𝑎 𝑏 , and 𝐻 ( 𝑎 , 𝑏 ) = 2 𝑎 𝑏 / ( 𝑎 + 𝑏 ) be the arithmetic, identric, logarithmic, 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 ) = 𝐿 − 2 ( 𝑎 , 𝑏 ) < 𝐿 ( 𝑎 , 𝑏 ) = 𝐿 − 1 ( 𝑎 , 𝑏 ) < 𝐼 ( 𝑎 , 𝑏 ) = 𝐿 0 ( 𝑎 , 𝑏 ) < 𝐴 ( 𝑎 , 𝑏 ) = 𝑀 1 ( 𝑎 , 𝑏 ) = 𝐿 1 ( 𝑎 , 𝑏 ) < m a x { 𝑎 , 𝑏 } ( 1 . 4 ) for all 𝑎 ≠ 𝑏 .

In [10], Carlson proved that

1 𝐿 ( 𝑎 , 𝑏 ) < 3 2 𝐴 ( 𝑎 , 𝑏 ) + 3 𝐺 ( 𝑎 , 𝑏 ) ( 1 . 5 ) for all 𝑎 , 𝑏 > 0 with 𝑎 ≠ 𝑏 .

The following inequality is due to Sándor [11, 12]:

2 𝐼 ( 𝑎 , 𝑏 ) > 3 1 𝐴 ( 𝑎 , 𝑏 ) + 3 𝐺 ( 𝑎 , 𝑏 ) . ( 1 . 6 )

In [13], Lin established the following results: (1) 𝑝 ≥ 1 / 3 implies that 𝐿 ( 𝑎 , 𝑏 ) < 𝑀 𝑝 ( 𝑎 , 𝑏 ) for all 𝑎 , 𝑏 > 0 with 𝑎 ≠ 𝑏 ; (2) 𝑝 ≤ 0 implies that 𝐿 ( 𝑎 , 𝑏 ) > 𝑀 𝑝 ( 𝑎 , 𝑏 ) for all 𝑎 , 𝑏 > 0 with 𝑎 ≠ 𝑏 ; (3) 𝑝 < 1 / 3 implies that there exist 𝑎 , 𝑏 > 0 such that 𝐿 ( 𝑎 , 𝑏 ) > 𝑀 𝑝 ( 𝑎 , 𝑏 ) ; (4) 𝑝 > 0 implies that there exist 𝑎 , 𝑏 > 0 such that 𝐿 ( 𝑎 , 𝑏 ) < 𝑀 𝑝 ( 𝑎 , 𝑏 ) . Hence the question was answered: what are the least value 𝑞 and the greatest value 𝑝 such that the inequality 𝑀 𝑝 ( 𝑎 , 𝑏 ) < 𝐿 ( 𝑎 , 𝑏 ) < 𝑀 𝑞 ( 𝑎 , 𝑏 ) holds for all 𝑎 , 𝑏 > 0 with 𝑎 ≠ 𝑏 .

Pittenger [14] established that

𝑀 𝑝 1 ( 𝑎 , 𝑏 ) ≤ 𝐿 𝑝 ( 𝑎 , 𝑏 ) ≤ 𝑀 𝑝 2 ( 𝑎 , 𝑏 ) ( 1 . 7 ) for all 𝑎 , 𝑏 > 0 , where

𝑝 1 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ m i n 𝑝 + 2 3 , 𝑝 l o g 2 2 l o g ( 𝑝 + 1 ) , 𝑝 > − 1 , 𝑝 ≠ 0 , 3 , 𝑝 = 0 , m i n 𝑝 + 2 3 𝑝 , 0 , 𝑝 ≤ − 1 , 2 = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ m a x 𝑝 + 2 3 , 𝑝 l o g 2 l o g ( 𝑝 + 1 ) , 𝑝 > − 1 , 𝑝 ≠ 0 , l o g 2 , 𝑝 = 0 , m a x 𝑝 + 2 3 , 0 , 𝑝 ≤ − 1 . ( 1 . 8 )

Here, 𝑝 1 and 𝑝 2 are sharp and inequality (1.7) becomes equality if and only if 𝑎 = 𝑏 or 𝑝 = 1 , − 2 or − 1 / 2 . The case 𝑝 = − 1 reduces to Lin's results [13]. Other generalizations of Lin's results were given by Imoru [15].

Recently, some monotonicity results of the ratio between generalized logarithmic means were established in [16–18].

The aim of this paper is to prove the following Theorem 1.1.

Theorem 1.1. Let 𝛼 ∈ ( 0 , 1 ) and 𝑎 , 𝑏 > 0 with 𝑎 ≠ 𝑏 , then (1) 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) = 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 𝛼 = 1 / 2 ;(2) 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) < 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 0 < 𝛼 < 1 / 2 , and 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) > 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 1 / 2 < 𝛼 < 1 , moreover, in each case, the bound 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) for the sum 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) is optimal.

2. Proof of Theorem 1.1

In order to prove our Theorem 1.1 we need a lemma, which we present in this section.

Lemma 2.1. For 𝛼 ∈ ( 0 , 1 ) and ℎ ( 𝑡 ) = ( 6 𝛼 − 1 ) 𝑡 6 𝛼 − 4 − ( 6 𝛼 − 1 ) 𝑡 6 𝛼 − 5 − ( 6 𝛼 − 5 ) 𝑡 6 𝛼 − 6 + ( 6 𝛼 − 5 ) 𝑡 6 𝛼 − 7 one has (1)If 𝛼 ∈ [ 1 / 6 , 1 ) , then ℎ ( 𝑡 ) > 0 for 𝑡 > 1 ;(2)If 𝛼 ∈ ( 0 , 1 / 6 ) , then ℎ ( 𝑡 ) < 0 for √ 𝑡 > ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) , ℎ ( 𝑡 ) > 0 for √ 1 < 𝑡 < ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) , and ℎ ( 𝑡 ) = 0 for √ 𝑡 = ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) .

Proof. (1) If 𝛼 = 1 / 6 , then we clearly see that ℎ ( 𝑡 ) = 4 𝑡 − 6 ( 𝑡 − 1 ) > 0 ( 2 . 1 ) for 𝑡 > 1 .
If 𝛼 ∈ ( 1 / 6 , 1 ) , then 𝑡 ℎ ( 𝑡 ) = ( 6 𝛼 − 1 ) ( 𝑡 − 1 ) 2 4 − 1 + 𝑡 6 𝛼 − 1 6 𝛼 − 7 > 0 ( 2 . 2 ) for 𝑡 > 1 .
Therefore, Lemma 2.1( 1 ) follows from (2.1) and (2.2).
(2) If 𝛼 ∈ ( 0 , 1 / 6 ) , then ℎ ( 𝑡 ) = ( 6 𝛼 − 1 ) ( 𝑡 − 1 ) 𝑡 + 5 − 6 𝛼 1 − 6 𝛼 𝑡 − 5 − 6 𝛼 𝑡 1 − 6 𝛼 6 𝛼 − 7 . ( 2 . 3 )
Therefore, Lemma 2.1(2) follows from (2.3).

Proof of Theorem 1.1. Proof. ( 1 ) If 𝛼 = 1 / 2 , then (1.1) leads to 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) = 𝐿 − 1 / 2 ( 𝑎 , 𝑏 ) = 𝑎 + 𝑏 4 + √ 𝑎 𝑏 2 = 1 2 1 𝐴 ( 𝑎 , 𝑏 ) + 2 𝐺 ( 𝑎 , 𝑏 ) = 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) . ( 2 . 4 )
( 2 ) We divide the proof into two cases.Case 1. 𝛼 = 1 / 3 or 𝛼 = 2 / 3 . From inequalities (1.5) and (1.6) we clearly see that 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) < 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) ( 2 . 5 ) for 𝛼 = 1 / 3 , and 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) > 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) ( 2 . 6 ) for 𝛼 = 2 / 3 .Case 2. 𝛼 ∈ ( 0 , 1 ) ⧵ { 1 / 3 , 1 / 2 , 2 / 3 } . Without loss of generality, we assume that 𝑎 > 𝑏 . Let √ 𝑡 = 𝑎 / 𝑏 > 1 , then (1.1) leads to l o g 𝐿 3 𝛼 − 2 [ ] = 1 ( 𝑎 , 𝑏 ) − l o g 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) 𝑡 3 𝛼 − 2 l o g 6 𝛼 − 2 − 1 𝑡 ( 3 𝛼 − 1 ) 2 𝛼 − 1 − l o g 2 1 + 𝑡 2 . + ( 1 − 𝛼 ) 𝑡 ( 2 . 7 )
Let 𝑓 ( 𝑡 ) = ( 1 / ( 3 𝛼 − 2 ) ) l o g [ ( 𝑡 6 𝛼 − 2 − 1 ) / ( ( 3 𝛼 − 1 ) ( 𝑡 2 − 1 ) ) ] − l o g [ ( 𝛼 / 2 ) ( 1 + 𝑡 2 ) + ( 1 − 𝛼 ) 𝑡 ] , then simple computations yield l i m 𝑡 → 1 𝑓 𝑓 ( 𝑡 ) = 0 , ( 2 . 8 ) ( 𝑡 ) = 𝑔 ( 𝑡 ) 𝑡 ( 3 𝛼 − 2 ) 6 𝛼 − 2 𝑡 − 1 2 − 1 ( 𝛼 / 2 ) 1 + 𝑡 2 , + ( 1 − 𝛼 ) 𝑡 ( 2 . 9 ) where 𝑔 ( 𝑡 ) = ( 1 − 𝛼 ) ( 3 𝛼 − 2 ) 𝑡 6 𝛼 + 3 𝛼 ( 𝛼 − 1 ) 𝑡 6 𝛼 − 1 − 3 𝛼 ( 1 − 𝛼 ) 𝑡 6 𝛼 − 2 − 𝛼 ( 3 𝛼 − 1 ) 𝑡 6 𝛼 − 3 + 𝛼 ( 3 𝛼 − 1 ) 𝑡 3 + 3 𝛼 ( 1 − 𝛼 ) 𝑡 2 + 3 𝛼 ( 1 − 𝛼 ) 𝑡 − ( 1 − 𝛼 ) ( 3 𝛼 − 2 ) . ( 2 . 1 0 ) Note that 𝑔 𝑔 ( 1 ) = 0 , ( 2 . 1 1 ) ( 𝑡 ) = 6 𝛼 ( 1 − 𝛼 ) ( 3 𝛼 − 2 ) 𝑡 6 𝛼 − 1 + 3 𝛼 ( 𝛼 − 1 ) ( 6 𝛼 − 1 ) 𝑡 6 𝛼 − 2 − 6 𝛼 ( 1 − 𝛼 ) ( 3 𝛼 − 1 ) 𝑡 6 𝛼 − 3 − 3 𝛼 ( 3 𝛼 − 1 ) ( 2 𝛼 − 1 ) 𝑡 6 𝛼 − 4 + 3 𝛼 ( 3 𝛼 − 1 ) 𝑡 2 𝑔 + 6 𝛼 ( 1 − 𝛼 ) 𝑡 + 3 𝛼 ( 1 − 𝛼 ) , ( 2 . 1 2 ) 𝑔 ( 1 ) = 0 , ( 2 . 1 3 ) ( 𝑡 ) = 6 𝛼 ( 1 − 𝛼 ) ( 3 𝛼 − 2 ) ( 6 𝛼 − 1 ) 𝑡 6 𝛼 − 2 + 6 𝛼 ( 𝛼 − 1 ) ( 6 𝛼 − 1 ) × ( 3 𝛼 − 1 ) 𝑡 6 𝛼 − 3 − 1 8 𝛼 ( 1 − 𝛼 ) ( 3 𝛼 − 1 ) ( 2 𝛼 − 1 ) 𝑡 6 𝛼 − 4 − 6 𝛼 ( 3 𝛼 − 1 ) ( 2 𝛼 − 1 ) ( 3 𝛼 − 2 ) 𝑡 6 𝛼 − 5 𝑔 + 6 𝛼 ( 3 𝛼 − 1 ) 𝑡 + 6 𝛼 ( 1 − 𝛼 ) , ( 2 . 1 4 ) 𝑔 ( 1 ) = 0 , ( 2 . 1 5 ) ( 𝑡 ) = 1 2 𝛼 ( 1 − 𝛼 ) ( 3 𝛼 − 2 ) ( 6 𝛼 − 1 ) ( 3 𝛼 − 1 ) 𝑡 6 𝛼 − 3 + 1 8 𝛼 ( 𝛼 − 1 ) ( 6 𝛼 − 1 ) ( 3 𝛼 − 1 ) ( 2 𝛼 − 1 ) 𝑡 6 𝛼 − 4 − 3 6 𝛼 ( 1 − 𝛼 ) ( 3 𝛼 − 1 ) ( 2 𝛼 − 1 ) ( 3 𝛼 − 2 ) 𝑡 6 𝛼 − 5 − 6 𝛼 ( 3 𝛼 − 1 ) ( 2 𝛼 − 1 ) ( 3 𝛼 − 2 ) ( 6 𝛼 − 5 ) 𝑡 6 𝛼 − 6 𝑔 + 6 𝛼 ( 3 𝛼 − 1 ) , ( 2 . 1 6 ) 𝑔 ( 1 ) = 0 , ( 2 . 1 7 ) ( 4 ) ( 𝑡 ) = 3 6 𝛼 ( 3 𝛼 − 1 ) ( 3 𝛼 − 2 ) ( 2 𝛼 − 1 ) ( 1 − 𝛼 ) ℎ ( 𝑡 ) , ( 2 . 1 8 ) where ℎ ( 𝑡 ) is defined as in Lemma 2.1.
We divide the proof into five subcases.
Subcase 2 A. 𝛼 ∈ ( 0 , 1 / 6 ) . From (2.18) and Lemma 2.1( 2 ) we clearly see that 𝑔 ( 4 ) ( 𝑡 ) < 0 for √ 𝑡 ∈ ( 1 , ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) ) and 𝑔 ( 4 ) ( 𝑡 ) > 0 for √ 𝑡 ∈ ( ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) , + ∞ ) , then we know that 𝑔 ( 𝑡 ) is strictly decreasing in √ ( 1 , ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) ) and strictly increasing in ( √ ( 5 − 6 𝛼 ) / ( 1 − 6 𝛼 ) , + ∞ ) . Now from the monotonicity of 𝑔 ( 𝑡 ) and (2.17) together with the fact that l i m 𝑡 → + ∞ 𝑔 ( 𝑡 ) = 6 𝛼 ( 3 𝛼 − 1 ) < 0 we clearly see that 𝑔 ( 𝑡 ) < 0 for 𝑡 > 1 , then from (2.7)–(2.15) and ( 3 𝛼 − 2 ) ( 𝑡 6 𝛼 − 2 − 1 ) > 0 for 𝑡 > 1 we get 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) < 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 𝛼 ∈ ( 0 , 1 / 6 ) .Subcase 2 B. 𝛼 ∈ [ 1 / 6 , 1 / 3 ) . Then (2.18) and Lemma 2.1(1) lead to 𝑔 ( 4 ) ( 𝑡 ) < 0 ( 2 . 1 9 ) for 𝑡 > 1 .
From (2.7)–(2.17) and (2.19) together with the fact that ( 3 𝛼 − 2 ) ( 𝑡 6 𝛼 − 2 − 1 ) > 0 for 𝑡 > 1 we know that 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) < 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 𝛼 ∈ [ 1 / 6 , 1 / 3 ) .Subcase 2 C. 𝛼 ∈ ( 1 / 3 , 1 / 2 ) . Then (2.18) and Lemma 2.1(1) imply that 𝑔 ( 4 ) ( 𝑡 ) > 0 ( 2 . 2 0 ) for 𝑡 > 1 .
From (2.7)–(2.17), (2.20) and ( 3 𝛼 − 2 ) ( 𝑡 6 𝛼 − 2 − 1 ) < 0 for 𝑡 > 1 we know that 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) < 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 𝛼 ∈ ( 1 / 3 , 1 / 2 ) .Subcase 2 D. 𝛼 ∈ ( 1 / 2 , 2 / 3 ) . Then (2.19) again yields, and 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) > 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 𝛼 ∈ ( 1 / 2 , 2 / 3 ) follows from (2.7)–(2.17) and (2.19) together with ( 3 𝛼 − 2 ) ( 𝑡 6 𝛼 − 2 − 1 ) < 0 .Subcase 2 E. 𝛼 ∈ ( 2 / 3 , 1 ) . Then (2.20) is also true, and 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) > 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) for 𝛼 ∈ ( 2 / 3 , 1 ) follows from (2.7)–(2.17), (2.20) and the fact that ( 3 𝛼 − 2 ) ( 𝑡 6 𝛼 − 2 − 1 ) > 0 .
Next, we prove that the bound 𝐿 3 𝛼 − 2 ( 𝑎 , 𝑏 ) for the sum 𝛼 𝐴 ( 𝑎 , 𝑏 ) + ( 1 − 𝛼 ) 𝐺 ( 𝑎 , 𝑏 ) is optimal in each case. The proof is divided into six cases.
Case 1. 𝛼 = 1 / 3 . For any 𝜖 ∈ ( 0 , 1 ) and 𝑥 ∈ ( 0 , 1 ) , then (1.1) leads to 𝐿 3 𝛼 − 2 + 𝜖 ( 1 , 1 + 𝑥 ) 1 − 𝜖 − [ ] 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) 1 − 𝜖 = 𝐿 𝜖 − 1 ( 1 , 1 + 𝑥 ) 1 − 𝜖 − 1 3 2 𝐴 ( 1 , 1 + 𝑥 ) + 3 𝐺 ( 1 , 1 + 𝑥 ) 1 − 𝜖 = 𝜖 𝑥 ( 1 + 𝑥 ) 𝜖 − 1 − 1 3 + 𝑥 6 + 2 3 ( 1 + 𝑥 ) 1 / 2 1 − 𝜖 = 𝑓 1 ( 𝑥 ) ( 1 + 𝑥 ) 𝜖 , − 1 ( 2 . 2 1 ) where 𝑓 1 ( 𝑥 ) = 𝜖 𝑥 − [ ( 1 + 𝑥 ) 𝜖 − 1 ] [ 1 / 3 + 𝑥 / 6 + ( 2 / 3 ) ( 1 + 𝑥 ) 1 / 2 ] 1 − 𝜖 .
Let 𝑥 → 0 ; making use of Taylor expansion, one has 𝑓 1 1 ( 𝑥 ) = 𝜖 2 4 2 ( 1 − 𝜖 ) 𝑥 3 𝑥 + 𝑜 3 . ( 2 . 2 2 )
Equations (2.21) and (2.22) imply that for any 𝜖 ∈ ( 0 , 1 ) , there exists 0 < 𝛿 1 = 𝛿 1 ( 𝜖 ) < 1 , such that 𝐿 3 𝛼 − 2 + 𝜖 ( 1 , 1 + 𝑥 ) > 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) for any 𝑥 ∈ ( 0 , 𝛿 1 ) and 𝛼 = 1 / 3 .Case 2. 𝛼 = 2 / 3 . For any 𝜖 ∈ ( 0 , 1 ) and 𝑥 ∈ ( 0 , 1 ) , from (1.1) we have [ ] 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) 𝜖 − 𝐿 3 𝛼 − 2 − 𝜖 ( 1 , 1 + 𝑥 ) 𝜖 = 2 3 1 𝐴 ( 1 , 1 + 𝑥 ) + 3 𝐺 ( 1 , 1 + 𝑥 ) 𝜖 − 𝐿 − 𝜖 ( 1 , 1 + 𝑥 ) 𝜖 = 2 3 + 𝑥 3 + 1 3 ( 1 + 𝑥 ) 1 / 2 𝜖 − ( 1 − 𝜖 ) 𝑥 ( 1 + 𝑥 ) 1 − 𝜖 = 𝑓 − 1 2 ( 𝑥 ) ( 1 + 𝑥 ) 1 − 𝜖 , − 1 ( 2 . 2 3 ) where 𝑓 2 ( 𝑥 ) = [ ( 1 + 𝑥 ) 1 − 𝜖 − 1 ] [ 2 / 3 + 𝑥 / 3 + ( 1 / 3 ) ( 1 + 𝑥 ) 1 / 2 ] 𝜖 − ( 1 − 𝜖 ) 𝑥 .
Let 𝑥 → 0 ; making use of Taylor expansion, one has 𝑓 2 1 ( 𝑥 ) = 𝜖 2 4 2 ( 1 − 𝜖 ) 𝑥 3 𝑥 + 𝑜 3 . ( 2 . 2 4 )
Equations (2.23) and (2.24) imply that for any 𝜖 ∈ ( 0 , 1 ) , there exists 0 < 𝛿 2 = 𝛿 2 ( 𝜖 ) < 1 , such that 𝐿 3 𝛼 − 2 − 𝜖 ( 1 , 1 + 𝑥 ) < 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) for 𝑥 ∈ ( 0 , 𝛿 2 ) and 𝛼 = 2 / 3 .Case 3. 𝛼 ∈ ( 0 , 1 / 3 ) . For 𝜖 ∈ ( 0 , 1 − 3 𝛼 ) and 𝑥 ∈ ( 0 , 1 ) , we get 𝐿 3 𝛼 − 2 + 𝜖 ( 1 , 1 + 𝑥 ) 2 − 3 𝛼 − 𝜖 − [ ] 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) 2 − 3 𝛼 − 𝜖 = ( 1 − 3 𝛼 − 𝜖 ) 𝑥 ( 1 + 𝑥 ) 1 − 3 𝛼 − 𝜖 ( 1 + 𝑥 ) 1 − 3 𝛼 − 𝜖 − 𝛼 − 1 𝛼 + 2 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 2 − 3 𝛼 − 𝜖 = 𝑓 3 ( 𝑥 ) ( 1 + 𝑥 ) 1 − 3 𝛼 − 𝜖 , − 1 ( 2 . 2 5 ) where 𝑓 3 ( 𝑥 ) = ( 1 − 3 𝛼 − 𝜖 ) 𝑥 ( 1 + 𝑥 ) 1 − 3 𝛼 − 𝜖 − [ ( 1 + 𝑥 ) 1 − 3 𝛼 − 𝜖 − 1 ] [ 𝛼 + ( 𝛼 / 2 ) 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 ] 2 − 3 𝛼 − 𝜖 .
Let 𝑥 → 0 ; making use of Taylor expansion, one has 𝑓 3 1 ( 𝑥 ) = 2 4 𝜖 ( 1 − 3 𝛼 − 𝜖 ) ( 2 − 3 𝛼 − 𝜖 ) 𝑥 3 𝑥 + 𝑜 3 . ( 2 . 2 6 )
Equations (2.25) and (2.26) imply that for any 𝛼 ∈ ( 0 , 1 / 3 ) and any 𝜖 ∈ ( 0 , 1 − 3 𝛼 ) , there exists 0 < 𝛿 3 = 𝛿 3 ( 𝜖 , 𝛼 ) < 1 , such that 𝐿 3 𝛼 − 2 + 𝜖 ( 1 , 1 + 𝑥 ) > 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) for 𝑥 ∈ ( 0 , 𝛿 3 ) .Case 4. 𝛼 ∈ ( 1 / 3 , 1 / 2 ) . For any 𝜖 ∈ ( 0 , 2 − 3 𝛼 ) and 𝑥 ∈ ( 0 , 1 ) , we get 𝐿 3 𝛼 − 2 + 𝜖 ( 1 , 1 + 𝑥 ) 2 − 3 𝛼 − 𝜖 − [ ] 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) 2 − 3 𝛼 − 𝜖 = ( 3 𝛼 − 1 + 𝜖 ) 𝑥 ( 1 + 𝑥 ) 3 𝛼 + 𝜖 − 1 − 𝛼 − 1 𝛼 + 2 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 2 − 3 𝛼 − 𝜖 = 𝑓 4 ( 𝑥 ) ( 1 + 𝑥 ) 3 𝛼 − 1 + 𝜖 , − 1 ( 2 . 2 7 ) where 𝑓 4 ( 𝑥 ) = ( 3 𝛼 − 1 + 𝜖 ) 𝑥 − [ ( 1 + 𝑥 ) 3 𝛼 − 1 + 𝜖 − 1 ] [ 𝛼 + ( 𝛼 / 2 ) 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 ] 2 − 3 𝛼 − 𝜖 .
Let 𝑥 → 0 ; using Taylor expansion we have 𝑓 4 1 ( 𝑥 ) = 2 4 𝜖 ( 3 𝛼 − 1 + 𝜖 ) ( 2 − 3 𝛼 − 𝜖 ) 𝑥 3 𝑥 + 𝑜 3 . ( 2 . 2 8 )
Equations (2.27) and (2.28) show that for any 𝛼 ∈ ( 1 / 3 , 1 / 2 ) and any 𝜖 ∈ ( 0 , 2 − 3 𝛼 ) , there exists 0 < 𝛿 4 = 𝛿 4 ( 𝜖 , 𝛼 ) < 1 , such that 𝐿 3 𝛼 − 2 + 𝜖 ( 1 , 1 + 𝑥 ) > 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) for 𝑥 ∈ ( 0 , 𝛿 4 ) .Case 5. 𝛼 ∈ ( 1 / 2 , 2 / 3 ) . For any 𝜖 ∈ ( 0 , 3 𝛼 − 1 ) and 𝑥 ∈ ( 0 , 1 ) , we have [ ] 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) 2 − 3 𝛼 + 𝜖 − 𝐿 3 𝛼 − 2 − 𝜖 ( 1 , 1 + 𝑥 ) 2 − 3 𝛼 + 𝜖 = 𝛼 𝛼 + 2 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 2 − 3 𝛼 + 𝜖 − ( 3 𝛼 − 1 − 𝜖 ) 𝑥 ( 1 + 𝑥 ) 3 𝛼 − 𝜖 − 1 = 𝑓 − 1 5 ( 𝑥 ) ( 1 + 𝑥 ) 3 𝛼 − 1 − 𝜖 , − 1 ( 2 . 2 9 ) where 𝑓 5 ( 𝑥 ) = [ ( 1 + 𝑥 ) 3 𝛼 − 1 − 𝜖 − 1 ] [ 𝛼 + ( 𝛼 / 2 ) 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 ] 2 − 3 𝛼 + 𝜖 − ( 3 𝛼 − 1 − 𝜖 ) 𝑥 .
Let 𝑥 → 0 ; making use of Taylor expansion we get 𝑓 5 1 ( 𝑥 ) = 2 4 𝜖 ( 3 𝛼 − 1 − 𝜖 ) ( 2 − 3 𝛼 + 𝜖 ) 𝑥 3 𝑥 + 𝑜 3 . ( 2 . 3 0 )
Equations (2.29) and (2.30) imply that for any 𝛼 ∈ ( 1 / 2 , 2 / 3 ) and any 𝜖 ∈ ( 0 , 3 𝛼 − 1 ) , there exists 0 < 𝛿 5 = 𝛿 5 ( 𝜖 , 𝛼 ) < 1 , such that 𝐿 3 𝛼 − 2 − 𝜖 ( 1 , 1 + 𝑥 ) < 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) for 𝑥 ∈ ( 0 , 𝛿 5 ) .Case 6. 𝛼 ∈ ( 2 / 3 , 1 ) . For any 𝜖 ∈ ( 0 , 3 𝛼 − 2 ) and 𝑥 ∈ ( 0 , 1 ) , we get [ ] 𝛼 𝐴 ( 1 , 1 + 𝑥 ) + ( 1 − 𝛼 ) 𝐺 ( 1 , 1 + 𝑥 ) 3 𝛼 − 2 − 𝜖 − 𝐿 3 𝛼 − 2 − 𝜖 ( 1 , 1 + 𝑥 ) 3 𝛼 − 2 − 𝜖 = 𝛼 𝛼 + 2 𝑥 + ( 1 − 𝛼 ) ( 1 + 𝑥 ) 1 / 2 3 𝛼 − 2 − 𝜖 − ( 1 + 𝑥 ) 3 𝛼 − 𝜖 − 1 − 1 = 𝑓 ( 3 𝛼 − 1 − 𝜖 ) 𝑥 6 ( 𝑥 ) , ( 3 𝛼 − 1 − 𝜖 ) 𝑥 ( 2 . 3 1 ) where 𝑓