Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 348326, 12 pages http://dx.doi.org/10.1155/2013/348326 Research Article Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means Tie-Hong Zhao,1 Yu-Ming Chu,2 Yun-Liang Jiang,3 and Yong-Min Li4 1 Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China 3 School of Information & Engineering, Huzhou Teachers College, Huzhou 313000, China 4 School of Automation, Southeast University, Nanjing 210096, China 2 Correspondence should be addressed to Yu-Ming Chu; chuyuming2005@yahoo.com.cn Received 19 January 2013; Accepted 1 February 2013 Academic Editor: Khalil Ezzinbi Copyright © 2013 Tie-Hong Zhao 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. We prove that the double inequalities 𝐼𝛼1 (𝑎, 𝑏)𝑄1−𝛼1 (𝑎, 𝑏) < 𝑀(𝑎, 𝑏) < 𝐼𝛽1 (𝑎, 𝑏)𝑄1−𝛽1 (𝑎, 𝑏), 𝐼𝛼2 (𝑎, 𝑏)𝐶1−𝛼2 (𝑎, 𝑏) < 𝑀(𝑎, 𝑏) < 𝐼𝛽2 (𝑎, 𝑏)𝐶1−𝛽2 (𝑎, 𝑏) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼1 ≥ 1/2, 𝛽1 ≤ log[√2 log(1 + √2)]/(1 − log √2), 𝛼2 ≥ 5/7, and 𝛽2 ≤ log[2 log(1+√2)], where 𝐼(𝑎, 𝑏), 𝑀(𝑎, 𝑏), 𝑄(𝑎, 𝑏), and 𝐶(𝑎, 𝑏) are the identric, Neuman-Sándor, quadratic, and contraharmonic means of 𝑎 and 𝑏, respectively. 1. Introduction For 𝑝 ∈ R and 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏, the identric mean 𝐼(𝑎, 𝑏), Neuman-Sándor mean 𝑀(𝑎, 𝑏) [1], quadratic mean 𝑄(𝑎, 𝑏), contraharmonic mean 𝐶(𝑎, 𝑏), and 𝑝th power mean 𝑀𝑝 (𝑎, 𝑏) are defined by 1/(𝑏−𝑎) 1 𝑏𝑏 𝐼 (𝑎, 𝑏) = ( 𝑎 ) 𝑒 𝑎 𝑀 (𝑎, 𝑏) = 𝑄 (𝑎, 𝑏) = √ , 𝑎−𝑏 , 2sinh [(𝑎 − 𝑏) / (𝑎 + 𝑏)] −1 𝑎2 + 𝑏2 , 2 𝐶 (𝑎, 𝑏) = 𝑎2 + 𝑏2 , 𝑎+𝑏 (1) 1/𝑝 𝑎𝑝 + 𝑏𝑝 { { {( ) 2 𝑀𝑝 (𝑎, 𝑏) = { { { √ { 𝑎𝑏, Recently, the identric, Neuman-Sándor, quadratic, and contraharmonic means have attracted the interest of numerous eminent mathematicians. In particular, many remarkable inequalities for these means can be found in the literature [1– 18]. Let 𝐻(𝑎, 𝑏) = 2𝑎𝑏/(𝑎 + 𝑏), 𝐺(𝑎, 𝑏) = √𝑎𝑏, 𝐿(𝑎, 𝑏) = (𝑏 − 𝑎)/(log 𝑏 − log 𝑎), 𝑃(𝑎, 𝑏) = (𝑎 − 𝑏)/(4 arctan √𝑎/𝑏 − 𝜋), 𝐴(𝑎, 𝑏) = (𝑎+𝑏)/2, and 𝑇(𝑎, 𝑏) = (𝑎−𝑏)/[2 arctan((𝑎−𝑏)/(𝑎+ 𝑏))] be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, and second Seiffert means of two distinct positive numbers 𝑎 and 𝑏, respectively. Then it is well known that the inequalities , 𝑝 ≠ 0, 𝑝 = 0, respectively, where sinh−1 (𝑥) = log(𝑥+√1 + 𝑥2 ) is the inverse hyperbolic sine function. 𝐻 (𝑎, 𝑏) = 𝑀−1 (𝑎, 𝑏) < 𝐺 (𝑎, 𝑏) = 𝑀0 (𝑎, 𝑏) < 𝐿 (𝑎, 𝑏) < 𝑃 (𝑎, 𝑏) < 𝐼 (𝑎, 𝑏) < 𝐴 (𝑎, 𝑏) < 𝑀1 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝑄 (𝑎, 𝑏) = 𝑀2 (𝑎, 𝑏) < 𝐶 (𝑎, 𝑏) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. (2) 2 Abstract and Applied Analysis Neuman and Sándor [1, 8] established that 𝐴 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐴 (𝑎, 𝑏) log (1 + √2) which has been investigated in [23–25]. Alzer [26] proved that the inequalities , √𝐴 (𝑎, 𝑏) 𝐺 (𝑎, 𝑏) < √𝐼 (𝑎, 𝑏) 𝐿 (𝑎, 𝑏) 𝜋 𝑇 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) , 4 𝑀 (𝑎, 𝑏) < 𝐴2 (𝑎, 𝑏) , 𝑃 (𝑎, 𝑏) < (3) √𝐴 (𝑎, 𝑏) 𝑇 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) <√ 𝑀 (𝑎, 𝑏) < 𝐴2 (𝑎, 𝑏) + 𝑇2 (𝑎, 𝑏) , 2 𝑀0 (𝑎, 𝑏) < √𝐼 (𝑎, 𝑏) 𝐿 (𝑎, 𝑏) < 𝑀1/2 (𝑎, 𝑏) , for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. Let 0 < 𝑎, 𝑏 ≤ 1/2 with 𝑎 ≠ 𝑏, 𝑎 = 1 − 𝑎, and 𝑏 = 1 − 𝑏. Then the Ky Fan inequalities 𝑇 (𝑎, 𝑏) 𝑀 (𝑎, 𝑏) < < 𝑀 (𝑎 , 𝑏 ) 𝑇 (𝑎 , 𝑏 ) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. The following sharp bounds for 𝐼, (𝐼𝐿)1/2 , and (𝐼 + 𝐿)/2 in terms of the power mean and the convex combination of arithmetic and geometric means are given in [27] as 𝑀2/3 (𝑎, 𝑏) < 𝐼 (𝑎, 𝑏) < 𝑀log 2 (𝑎, 𝑏) , 2𝐴 (𝑎, 𝑏) + 𝑄 (𝑎, 𝑏) 3 𝐺 (𝑎, 𝑏) 𝐿 (𝑎, 𝑏) 𝑃 (𝑎, 𝑏) 𝐴 (𝑎, 𝑏) < < < 𝐺 (𝑎 , 𝑏 ) 𝐿 (𝑎 , 𝑏 ) 𝑃 (𝑎 , 𝑏 ) 𝐴 (𝑎 , 𝑏 ) (4) were presented in [1]. Li et al. [19] found the best possible bounds for the Neuman-Sándor mean in terms of the generalized logarithmic mean 𝐿 𝑟 (𝑎, 𝑏). Neuman [20] and Zhao et al. [21] proved that the inequalities 𝑀log 2/(1+log 2) (𝑎, 𝑏) < for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. Chu et al. [28] presented the optimal constants 𝛼1 , 𝛽1 , 𝛼2 , and 𝛽2 such that the double inequalities 𝛼1 𝑄 (𝑎, 𝑏) + (1 − 𝛼1 ) 𝐴 (𝑎, 𝑏) < 2 𝜋/2 √ 2 2 𝑎 cos 𝜃 + 𝑏2 sin2 𝜃𝑑𝜃 ∫ 𝜋 0 < 𝛽1 𝑄 (𝑎, 𝑏) + (1 − 𝛽1 ) 𝐴 (𝑎, 𝑏) , 𝜆𝐶 (𝑎, 𝑏) + (1 − 𝜆) 𝐴 (𝑎, 𝑏) 𝑄𝛼2 (𝑎, 𝑏) 𝐴1−𝛼2 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝜇𝐶 (𝑎, 𝑏) + (1 − 𝜇) 𝐴 (𝑎, 𝑏) , (5) 𝛼1 𝐻 (𝑎, 𝑏) + (1 − 𝛼1 ) 𝑄 (𝑎, 𝑏) < < 𝑀 (𝑎, 𝑏) < 𝛽1 𝐻 (𝑎, 𝑏) + (1 − 𝛽1 ) 𝑄 (𝑎, 𝑏) , < 𝑀 (𝑎, 𝑏) < 𝛽2 𝐶 (𝑎, 𝑏) + (1 − 𝛽2 ) 𝑄 (𝑎, 𝑏) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼 ≤ [1 − log(1 + √2)]/[(√2 − 1) log(1 + √2)], 𝛽 ≥ 1/3, 𝜆 ≤ [1 − log(1 + √2)]/ log(1 + √2), 𝜇 ≥ 1/6, 𝛼1 ≥ 2/9, 𝛽1 ≤ 1 − 1/[√2 log(1 + √2)], 𝛼2 ≥ 1/3, and 𝛽2 ≤ 1 − 1/[√2 log(1 + √2)]. In [22], Chu and Long gave the best possible constants 𝑝, 𝑞, 𝛼, and 𝛽 such that the double inequalities 𝑀𝑝 (𝑎, 𝑏) < 𝑀(𝑎, 𝑏) < 𝑀𝑞 (𝑎, 𝑏) and 𝛼𝐼(𝑎, 𝑏) < 𝑀(𝑎, 𝑏) < 𝛽𝐼(𝑎, 𝑏) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. The ratio of identric means leads to the weighted geometric mean 1/(𝑎+𝑏) , (9) 2 𝜋/2 √ 2 2 𝑎 cos 𝜃 + 𝑏2 sin2 𝜃𝑑𝜃 ∫ 𝜋 0 < 𝑄𝛽2 (𝑎, 𝑏) 𝐴1−𝛽2 (𝑎, 𝑏) 𝛼2 𝐶 (𝑎, 𝑏) + (1 − 𝛼2 ) 𝑄 (𝑎, 𝑏) = (𝑎𝑎 𝑏𝑏 ) (8) 2 2 < 𝐼 (𝑎, 𝑏) < 𝐴 (𝑎, 𝑏) + (1 − ) 𝐺 (𝑎, 𝑏) 𝑒 𝑒 < 𝑀 (𝑎, 𝑏) < 𝛽𝑄 (𝑎, 𝑏) + (1 − 𝛽) 𝐴 (𝑎, 𝑏) , 𝐼 (𝑎, 𝑏) 𝐼 (𝑎, 𝑏) + 𝐿 (𝑎, 𝑏) < 𝑀1/2 (𝑎, 𝑏) , 2 2 1 𝐴 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏) 3 3 𝛼𝑄 (𝑎, 𝑏) + (1 − 𝛼) 𝐴 (𝑎, 𝑏) 𝐼 (𝑎2 , 𝑏2 ) 𝐼 (𝑎, 𝑏) + 𝐿 (𝑎, 𝑏) 𝐴 (𝑎, 𝑏) + 𝐺 (𝑎, 𝑏) < 2 2 (7) (6) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. The aim of this paper is to find the best possible constants 𝛼1 , 𝛽1 , 𝛼2 and 𝛽2 such that the double inequalities 𝐼𝛼1 (𝑎, 𝑏) 𝑄1−𝛼1 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽1 (𝑎, 𝑏) 𝑄1−𝛽1 (𝑎, 𝑏) , 𝐼𝛼2 (𝑎, 𝑏) 𝐶1−𝛼2 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽2 (𝑎, 𝑏) 𝐶1−𝛽2 (𝑎, 𝑏) (10) hold for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏. All numerical computations are carried out using mathematica software. 2. Lemmas In order to prove our main results, we need several lemmas, which we present in this section. Abstract and Applied Analysis 3 Lemma 1. The double inequality 𝑥+ 𝑥3 2𝑥5 √ 𝑥3 2𝑥5 8𝑥7 − < 1 + 𝑥2 sinh−1 (𝑥) < 𝑥 + − + 3 15 3 15 105 (11) Proof. To prove inequalities (19) and (20), it suffices to show that 𝜄1 (𝑥) := log [ holds for 𝑥 ∈ (0, 1). −( Proof. To prove Lemma 1, it suffices to prove that 𝜂1 (𝑥) = √1 + 𝑥2 sinh−1 (𝑥) − (𝑥 + 𝜂2 (𝑥) = √1 + 𝑥2 sinh−1 (𝑥) − (𝑥 + 𝑥3 2𝑥5 − ) > 0, 3 15 (12) 𝑥3 2𝑥5 8𝑥7 − + )<0 3 15 105 (13) 𝜂1 (0) = 𝜂2 (0) = 0, 𝜂1 𝑥𝜂1∗ (𝑥) , (𝑥) = √1 + 𝑥2 𝜂2 𝑥𝜂2∗ (𝑥) , (𝑥) = √1 + 𝑥2 (14) 2𝑥3 8𝑥5 √ + ) 1 + 𝑥2 , 3 15 (15) (1 + 𝑥)1+𝑥 ] − 2𝑥 − 𝑥 log (1 − 𝑥2 ) (1 − 𝑥)1−𝑥 (22) 2𝑥3 2𝑥5 2𝑥7 + + + 𝑥9 ) < 0 3 5 7 for 𝑥 ∈ (0, 3/4). From (21) and (22), one has 𝜄1 (0+ ) = 𝜄2 (0+ ) = 0, (23) 2𝑥8 >0 1 − 𝑥2 (24) 7 9𝑥8 ( − 𝑥2 ) < 0 1 − 𝑥2 9 (25) for 𝑥 ∈ (0, 3/4). Therefore, inequality (21) follows from (23) and (24), and inequality (22) follows from (23) and (25). Lemma 3. Let 8𝑥4 > 0, 𝜂1∗ (𝑥) = 3√1 + 𝑥2 (16) 16𝑥6 < 0, 5√1 + 𝑥2 (17) := log [ 𝜄2 (𝑥) = − 𝜂1∗ (0) = 𝜂2∗ (0) = 0, 𝜂2∗ (𝑥) = − 𝜄2 (𝑥) for 𝑥 ∈ (0, 1), and 2𝑥3 √ 𝜂1∗ (𝑥) = sinh−1 (𝑥) − (𝑥 − ) 1 + 𝑥2 , 3 𝜂2∗ (𝑥) = sinh−1 (𝑥) − (𝑥 − 2𝑥3 2𝑥5 2𝑥7 + + )>0 3 5 7 𝜄1 (𝑥) = where (21) for 𝑥 ∈ (0, 1), and −( for 𝑥 ∈ (0, 1). From the expressions of 𝜂1 (𝑥) and 𝜂2 (𝑥), we get (1 + 𝑥)1+𝑥 ] − 2𝑥 − 𝑥 log (1 − 𝑥2 ) (1 − 𝑥)1−𝑥 Φ1 (𝑥) = 1 1 − . √1 + 𝑥2 sinh−1 (𝑥) 𝑥 (1 + 𝑥2 ) (26) Then the double inequality for 𝑥 ∈ (0, 1). Therefore, inequality (12) follows from (14)–(16), and inequality (13) follows from (14)–(17). 2𝑥 34𝑥3 𝑥5 2𝑥 34𝑥3 4𝑥5 − + < Φ1 (𝑥) < − + 3 45 2 3 45 5 holds for 𝑥 ∈ (0, 0.7). (27) Proof. To prove inequality (27), it suffices to show that 𝜙1 (𝑥) = 𝑥√1 + 𝑥2 − sinh−1 (𝑥) Lemma 2. Let 𝐿 (𝑥) = log [ (1 + 𝑥)1+𝑥 ] − 2𝑥 − 𝑥 log (1 − 𝑥2 ) . (1 − 𝑥)1−𝑥 (18) − 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) ( 2𝑥 34𝑥3 𝑥5 − + ) (28) 3 45 2 > 0, Then 𝐿 (𝑥) > 2𝑥3 2𝑥5 2𝑥7 + + 3 5 7 (19) − 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) ( for 𝑥 ∈ (0, 1), and 𝐿 (𝑥) < for 𝑥 ∈ (0, 3/4). 𝜙2 (𝑥) = 𝑥√1 + 𝑥2 − sinh−1 (𝑥) 2𝑥3 2𝑥5 2𝑥7 + + + 𝑥9 3 5 7 (20) 2𝑥 34𝑥3 4𝑥5 − + ) 3 45 5 <0 (29) for 𝑥 ∈ (0, 0.7). 4 Abstract and Applied Analysis First, we prove inequality (28). From the expression of 𝜙1 (𝑥), we have 𝜙1 (0) = 0, 𝜙1 (0.7) = 0.0033 ⋅ ⋅ ⋅ , 𝜙1 (𝑥) = 𝑥𝜙1∗ (𝑥) 90√1 + 𝑥2 , for 𝑥 ∈ (0, 0.6], and 𝜙1∗∗ (𝑥) > − 56𝑥 − 309𝑥3 + 422𝑥5 + 675𝑥7 (30) + 2 (28 − 324𝑥2 + 735𝑥4 + 1260𝑥6 ) (31) × (𝑥 + where 𝑥3 2𝑥5 𝑥3 − )= 3 15 15 ×(−14075+25028𝑥2 +56571𝑥4 + 9660𝑥6 −5040𝑥8 ) 𝜙1∗ (𝑥) = 120𝑥 + 8𝑥3 + 23𝑥5 − 45𝑥7 − 2 (60 − 16𝑥2 − 69𝑥4 + 180𝑥6 ) Equation (32) leads to 𝜙1∗ (𝑥) = − 𝜙1∗ (0.7) = −3.551 ⋅ ⋅ ⋅ , (33) 𝑥𝜙1∗∗ (𝑥) , 1 + 𝑥2 where 𝜙1∗∗ (𝑥) = − 56𝑥 − 309𝑥3 + 422𝑥5 + 675𝑥7 + 2 (28 − 324𝑥2 + 735𝑥4 + 1260𝑥6 ) 𝑥3 [−14075 + 25028 × (0.6)2 + 56571 × (0.6)4 ] 15 = 1416676𝑥3 >0 9375 (32) × √1 + 𝑥2 sinh−1 (𝑥) . 𝜙1∗ (0.6) = 3.017 ⋅ ⋅ ⋅ , > (38) for 𝑥 ∈ [0.6, 0.7). From (33), (37), and (38), we clearly see that there exists 𝑥1 ∈ (0.6, 0.7) such that 𝜙1∗ (𝑥) > 0 for 𝑥 ∈ (0, 𝑥1 ) and 𝜙1∗ (𝑥) < 0 for 𝑥 ∈ (𝑥1 , 0.7). Then (31) leads to the conclusion that 𝜙1 (𝑥) is strictly increasing on (0, 𝑥1 ] and strictly decreasing on [𝑥1 , 0.7). Therefore, inequality (28) follows from (30) and the piecewise monotonicity of 𝜙1 (𝑥). Next, we prove inequality (29). From the expression of 𝜙2 (𝑥), we get 𝜙2 (0) = 0, (34) × √1 + 𝑥2 sinh−1 (𝑥) . 𝜙2 (𝑥) = − Note that 2𝑥𝜙2∗ (𝑥) , 45√1 + 𝑥2 (39) where 60 − 16𝑥2 − 69𝑥4 + 180𝑥6 > 0 𝜙2∗ (𝑥) = 𝑥 (18𝑥6 + 𝑥4 − 2𝑥2 − 30) (35) + 2 (15 − 4𝑥2 + 3𝑥4 + 72𝑥6 ) for 𝑥 ∈ (0, 0.6], and 28 − 324𝑥2 + 735𝑥4 + 1260𝑥6 > 0 (40) × √1 + 𝑥2 sinh−1 (𝑥) . (36) It follows from Lemma 1 and (40) that for 𝑥 ∈ [0.6, 0.7). It follows from (32) and (34)–(36) together with Lemma 1 that 𝜙1∗ 3 5 (𝑥) > 120𝑥 + 8𝑥 + 23𝑥 − 45𝑥 + 2 (15 − 4𝑥2 + 3𝑥4 + 72𝑥6 ) (𝑥 + 7 𝑥3 − 2 (60 − 16𝑥 − 69𝑥 + 180𝑥 ) (𝑥 + ) 3 2 4 = 6 = 𝑥5 (515 − 1077𝑥2 − 360𝑥4 ) 3 ≥ 𝑥5 9 81 [515 − 1077 × − 360 × ] 3 25 625 10078𝑥5 = >0 375 𝜙2∗ (𝑥) > 𝑥 (18𝑥6 + 𝑥4 − 2𝑥2 − 30) (37) 𝑥3 2𝑥5 − ) (41) 3 15 𝑥5 (5 + 2476𝑥2 + 708𝑥4 − 288𝑥6 ) > 0 15 for 𝑥 ∈ (0, 0.7). Therefore, inequality (29) follows from (39) together with (41). Lemma 4. Let Φ2 (𝑥) = 1 √1 + 𝑥2 sinh−1 (𝑥) − 1 − 𝑥2 . 𝑥 (1 + 𝑥2 ) (42) Abstract and Applied Analysis 5 for 𝑥 ∈ (0, 0.66), and Then the double inequality 5𝑥 79𝑥3 11𝑥5 5𝑥 79𝑥3 9𝑥5 − + < Φ2 (𝑥) < − + 3 45 10 3 45 5 4 − 108𝑥2 + 213𝑥4 + 396𝑥6 (43) 3 2 > 4 − 108 × ( ) + 213 × (0.66)4 4 holds for 𝑥 ∈ (0, 3/4). + 396 × (0.66)6 = 16.3972 > 0 Proof. To prove Lemma 4, it suffices to prove that 𝜑1 (𝑥) := 𝑥√1 + 𝑥2 − (1 − 𝑥2 ) sinh−1 (𝑥) − 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) ( 5𝑥 79𝑥3 11𝑥5 − + ) > 0, 3 45 10 (44) 𝜑2 (𝑥) := 𝑥√1 + 𝑥2 − (1 − 𝑥2 ) sinh−1 (𝑥) − 𝑥 (1 + 𝑥2 ) sinh−1 (𝑥) ( 5𝑥 79𝑥3 9𝑥5 − + )<0 3 45 5 (45) 3 𝜑1 ( ) = 0.008457 ⋅ ⋅ ⋅ > 0, 4 𝜑1 (𝑥) = 𝑥𝜑1∗ (𝑥) , 90√1 + 𝑥2 for 𝑥 ∈ [0.66, 3/4). It follows from Lemma 1, (48), and (51)–(53) that 𝜑1∗ (𝑥) > 120𝑥 + 8𝑥3 + 59𝑥5 − 99𝑥7 − 2 (60 − 16𝑥2 − 177𝑥4 + 396𝑥6 ) for 𝑥 ∈ (0, 3/4). We first prove inequality (44). From the expression of 𝜑1 (𝑥), we obtain 𝜑1 (0) = 0, × (𝑥 + = 𝑥3 2𝑥5 8𝑥7 − + ) 3 15 105 𝑥5 [46165 − 82573𝑥2 − 32420𝑥4 105 +7584𝑥6 + 6336𝑥6 (1 − 𝑥2 )] (46) > 𝑥5 [46165 − 82573 × (0.66)2 − 32420 × (0.66)4 ] 105 (47) = 𝑥5 × 4044.5917 ⋅ ⋅ ⋅ > 0 105 where 𝜑1∗ (𝑥) = 120𝑥 + 8𝑥3 + 59𝑥5 − 99𝑥7 − 2 𝜑1∗∗ (𝑥) > −56𝑥 − 705𝑥3 + 836𝑥5 + 1485𝑥7 + 14 (4 − 108𝑥2 + 213𝑥4 + 396𝑥6 ) (𝑥+ Equation (48) leads to 3 𝜑1∗ ( ) = −19.299 ⋅ ⋅ ⋅ < 0, 4 (49) 𝑥𝜙∗∗ (𝑥) 𝜑1∗ (𝑥) = − 1 2 , 1+𝑥 = (50) 𝜑1∗∗ (𝑥) = − 56𝑥 − 705𝑥3 + 836𝑥5 + 1485𝑥7 (51) × √1 + 𝑥2 sinh−1 (𝑥) . Note that 60 − 16𝑥2 − 177𝑥4 + 396𝑥6 > 60 − 16 × (0.66)2 − 177 × (0.66)4 = 19.4451 > 0 (52) 𝑥3 2𝑥5 − ) 3 15 𝑥3 [−32975 + 49598𝑥2 + 123369𝑥4 15 +10668𝑥6 + 11088𝑥6 (1 − 𝑥2 )] > 𝑥3 [−32975 + 49598 ×(0.66)2 +123369 × (0.66)4 ] 15 = 𝑥3 × 12038.83 ⋅ ⋅ ⋅ > 0 15 where + 14 (4 − 108𝑥2 + 213𝑥4 + 396𝑥6 ) (54) for 𝑥 ∈ (0, 0.66), and × (60 − 16𝑥2 − 177𝑥4 + 396𝑥6 ) √1 + 𝑥2 sinh−1 (𝑥) . (48) 𝜑1∗ (0.66) = 6.02 ⋅ ⋅ ⋅ > 0, (53) (55) for 𝑥 ∈ [0.66, 3/4). From (50) and (55), we know that 𝜑1∗ (𝑥) is strictly decreasing on [0.66, 3/4), and this in conjunction with (49) and (54) leads to the conclusion that there exists 𝑥1 ∈ (0.66, 3/4) such that 𝜑1∗ (𝑥) > 0 for 𝑥 ∈ (0, 𝑥1 ) and 𝜑1∗ (𝑥) < 0 for 𝑥 ∈ (𝑥1 , 3/4). Then (47) implies that 𝜑1 (𝑥) is strictly increasing on (0, 𝑥1 ] and strictly decreasing on [𝑥1 , 3/4). Therefore, inequality (44) follows from (46) and the piecewise monotonicity of 𝜑1 (𝑥). 6 Abstract and Applied Analysis Next, we prove inequality (45). From the expression of 𝜑2 (𝑥) one has 12 23𝑥5 ( − 𝑥2 ) > 0, 2 35 (1 + 𝑥 ) 23 Υ1 (𝑥) − ( 𝜑2 (0) = 0, 𝜙2 (𝑥) = − = 𝑥𝜙2∗ (𝑥) , 45√1 + 𝑥2 (56) < where 1 2𝑥3 2𝑥5 2𝑥7 ( + + + 𝑥9 ) 2 2𝑥 3 5 7 𝑥 4𝑥 4𝑥3 8𝑥5 − ( − + ) 1 + 𝑥2 3 5 7 + 𝜙2∗ (𝑥) = − 60𝑥 − 4𝑥3 + 2𝑥5 + 81𝑥7 + 4 (15 − 4𝑥2 + 3𝑥4 + 162𝑥6 ) (57) 4𝑥 4𝑥3 8𝑥5 − + ) 3 5 7 =− 𝑥7 (1 − 𝑥2 ) 2 (1 + 𝑥2 ) <0 (61) × √1 + 𝑥2 sinh−1 (𝑥) . for 𝑥 ∈ (0, 0.7). Therefore, Lemma 5 follows easily from (61). It follows from Lemma 1 and (52) that Lemma 6. Let 𝐿(𝑥) be defined as in Lemma 2 and 𝜙2∗ 3 5 (𝑥) > −60𝑥 − 4𝑥 + 2𝑥 + 81𝑥 7 3 𝑥 2𝑥 + 4 (15 − 4𝑥 + 3𝑥 + 162𝑥 ) (𝑥 + − ) 3 15 2 = 4 6 𝑥5 (10 + 11027𝑥2 + 3216𝑥4 − 1296𝑥6 ) > 0 15 (58) for 𝑥 ∈ (0, 3/4). Therefore, inequality (45) follows from (56) together with (58). Lemma 5. Let 𝐿(𝑥) be defined as in Lemma 2 and Υ1 (𝑥) = Υ2 (𝑥) = 5 7𝑥 9𝑥3 7𝑥5 7𝑥 9𝑥3 15𝑥5 − + < Υ2 (𝑥) < − + 3 5 5 3 5 7 Proof. It follows from Lemma 2 that Υ2 (𝑥) − ( (59) 3 4𝑥 4𝑥 4𝑥 4𝑥 4𝑥 8𝑥 − + < Υ1 (𝑥) < − + 3 5 5 3 5 7 5 holds for 𝑥 ∈ (0, 0.7). Proof. From Lemma 2, one has 4𝑥 4𝑥3 4𝑥5 Υ1 (𝑥) − ( − + ) 3 5 5 > 2𝑥3 2𝑥5 2𝑥7 1 𝑥 ( + + )+ 2 2𝑥 3 5 7 1 + 𝑥2 −( 4𝑥 4𝑥3 4𝑥5 − + ) 3 5 5 = (60) 7𝑥 9𝑥3 7𝑥5 − + ) 3 5 5 1 2𝑥3 2𝑥5 2𝑥7 ( + + ) 2𝑥2 3 5 7 Then the double inequality 5 (63) holds for 𝑥 ∈ (0, 3/4). + 3 (62) Then the double inequality > 𝐿 (𝑥) 𝑥 + . 2𝑥2 1 + 𝑥2 𝐿 (𝑥) 2𝑥 + . 2𝑥2 1 + 𝑥2 7𝑥 9𝑥3 7𝑥5 2𝑥 − ( − + ) 1 + 𝑥2 3 5 5 13 44𝑥5 ( − 𝑥2 ) > 0, 35 (1 + 𝑥2 ) 22 7𝑥 9𝑥3 15𝑥5 Υ2 (𝑥) − ( − + ) 3 5 7 < 2𝑥3 2𝑥5 2𝑥7 1 ( + + + 𝑥9 ) 2 2𝑥 3 5 7 + =− 2𝑥 7𝑥 9𝑥3 15𝑥5 − ( − + ) 1 + 𝑥2 3 5 7 𝑥7 (3 − 𝑥2 ) 2 (1 + 𝑥2 ) <0 for 𝑥 ∈ (0, 3/4). Therefore, Lemma 6 follows from (64). (64) Abstract and Applied Analysis 7 Lemma 7 and 𝑥 > sinh−1 (𝑥) give 𝜔1 (𝑥) < 0 and Lemma 7. The inequality 𝑥3 √1 + 𝑥2 > [sinh−1 (𝑥)] 3 (65) 𝜔1 (𝑥) = 2 −1 𝑥3 [sinh (𝑥)] [ 3 𝑥3 3 √1 + − (sinh−1 (𝑥)) ] > 0 𝑥2 (73) holds for 𝑥 ∈ (0, 1). for 𝑥 ∈ (0, 1). This in turn implies that Proof. Let 𝜁 (𝑥) = 𝑥3 3 − [sinh−1 (𝑥)] . √1 + 𝑥2 (66) 2 𝜔1 (𝑥) 𝜔1 (𝑥) (1 + 𝑥 ) − 2𝑥𝜔1 (𝑥) = >0 ] 2 1 + 𝑥2 (1 + 𝑥2 ) 𝜁 (0) = 0, 𝜁1 (𝑥) 3/2 (1 + 𝑥2 ) 𝜔2 (1) = 0.2796 ⋅ ⋅ ⋅ > 0, (67) , 𝜔2 (𝑥) = − where 2𝑥 𝜔2∗ (𝑥) = (68) It follows from Lemma 1 and (68) that (69) 6 37 208 36𝑥 64𝑥 +( + + ) 525 525 175 3675 6 32𝑥 ]>0 735 [ 𝜔2 (𝑥) <0 2 − 1 (1 + 𝑥2 ) [sinh−1 (𝑥)] 𝑥 (1 + 𝑥2 ) 3/2 sinh−1 (𝑥) (75) , 2 (70) . (76) 2 3/2 (1 + 𝑥2 ) <0 ] 3/2 Lemma 8. Let (𝑥 + 𝑥3 ) [sinh (𝑥)] 2 3/2 (1 + 𝑥2 ) = − −1 for 𝑥 ∈ (0, 1). From (75)–(76), we clearly see that 𝜔2 (𝑥) < 0 and 𝜔2 (𝑥) > 0 for 𝑥 ∈ (0, 1). This in turn implies that for 𝑥 ∈ (0, 1). Therefore, Lemma 7 follows from (67) together with (69). 1 + 3𝑥2 𝑥 𝜔2∗ (𝑥) = − 2 𝜇1 (𝑥) = 𝜔2∗ (𝑥) 𝜔2∗ (0) = 0, 𝑥3 2𝑥5 8𝑥7 > 𝑥2 (3 + 2𝑥2 ) − 3(𝑥 + − + ) 3 15 105 × (1 − 𝑥2 ) + + 𝑥 − sinh−1 (𝑥) , √1 + 𝑥2 𝜁1 (𝑥) 2 3/2 𝑥2 ) (1 + where 2 𝜁1 (𝑥) = 𝑥2 (3 + 2𝑥2 ) − 3[√1 + 𝑥2 sinh−1 (𝑥)] . = 𝑥6 [ (74) for 𝑥 ∈ (0, 1). On the other hand, from the expression of 𝜔2 (𝑥), we get Then 𝜁 (𝑥) = [ 𝜔2 (𝑥) (1 + 𝑥2 ) − 3𝑥√1 + 𝑥2 𝜔2 (𝑥) (1 + (77) 3 𝑥2 ) for 𝑥 ∈ (0, 1). Equation (72) together with inequalities (74) and (77) lead to the conclusion that 𝜇1 (𝑥) ≤ Then 𝜇1 (𝑥) < 0.2 for 𝑥 ∈ [0.7, 1). 𝜔1 (1) 𝜔2 (0.7) + 3/2 2 [1 + (0.7)2 ] (78) = 0.167 ⋅ ⋅ ⋅ < 0.2 Proof. Let 𝜔1 (𝑥) = 𝜔2 (𝑥) = 1 1 − , 2 2 𝑥 [sinh−1 (𝑥)] 2 √1 + 𝑥2 − 𝑥 . sinh−1 (𝑥) for 𝑥 ∈ [0.7, 1). (71) 𝜇2 (𝑥) = Then 𝜇1 (𝑥) = 𝜔2 (𝑥) 𝜔1 (𝑥) + . 1 + 𝑥2 (1 + 𝑥2 )3/2 Lemma 9. Let 1 + 4𝑥2 − 𝑥4 (𝑥 + − (72) 2 𝑥3 ) − 1 (1 + 𝑥2 ) [sinh−1 𝑥 (1 + 𝑥2 ) 3/2 sinh−1 (𝑥) Then 𝜇2 (𝑥) < 0.51 for 𝑥 ∈ [0.65, 1). . (𝑥)] 2 (79) 8 Abstract and Applied Analysis Proof. Differentiating ]1 (𝑥) yields Proof. Let 1 1 − = 𝜇1 (𝑥) , 2 2 𝑥 [sinh−1 (𝑥)] 𝜏1 (𝑥) = ]1 (𝑥) = (80) 𝑥 3 − 𝑥2 − , 𝜏2 (𝑥) = 2 √1 + 𝑥 sinh−1 (𝑥) ]1 (𝑥) > 𝜏2 (𝑥) 𝜏1 (𝑥) + . 1 + 𝑥2 (1 + 𝑥2 )3/2 (81) = From (74), we clearly see that [ 𝜏1 (𝑥) 𝜔1 (𝑥) ] = [ ] >0 1 + 𝑥2 1 + 𝑥2 > (82) > 𝜏2 (1) = 0.2796 ⋅ ⋅ ⋅ > 0, 1 −1 sinh (𝑥) − 3/2 𝑥2 ) [sinh−1 (𝑥)] 2 2 𝜏2∗ (0.65) = 0.6033 ⋅ ⋅ ⋅ , + 2[ (83) 𝜏2 (𝑥) ] = 3/2 (1 + 𝑥2 ) − 3𝑥√1 + 𝑥2 𝜏2 (𝑥) 3 (1 + 𝑥2 ) <0 for 𝑥 ∈ [0.65, 1). Equation (81) together with inequalities (82) and (84) lead to the conclusion that 𝜏 (1) 𝜏2 (0.65) 𝜇2 (𝑥) ≤ 1 = 0.503 ⋅ ⋅ ⋅ < 0.51 (85) + 3/2 2 [1 + (0.65)2 ] ]2 for 𝑥 ∈ [0.65, 1). Lemma 10. Let 𝐿(𝑥) be defined as in Lemma 2 and 2 (1 − 𝑥2 ) (1 + 𝑥2 ) Then ]1 (𝑥) > 1.2 for x ∈ [0.7, 1). 349 5 2𝑥 2 3 35(1 − 𝑥2 ) (1 + 𝑥2 ) >0 − 𝐿 (𝑥) . 𝑥3 3 − 2𝑥2 + 3𝑥4 (1 − 𝑥2 ) (1 + 2 𝑥2 ) 𝐿 (𝑥) . 𝑥3 − 2 4 6 8 3𝐿 (𝑥) 2 (1 + 7𝑥 − 17𝑥 + 5𝑥 − 4𝑥 ) − . (𝑥) = 2 3 𝑥4 𝑥(1 − 𝑥2 ) (1 + 𝑥2 ) (89) (90) It follows from (19) and (90) together with the monotonicity of the function 561𝑥2 − 272𝑥4 on [0.65, 1) that ]2 (𝑥) > = > 2 (1 + 𝑥4 ) + [−84 + 316 × (0.7)2 − Proof. Differentiating ]2 (𝑥) yields (84) ]1 (𝑥) = 35(1 − 3 𝑥2 ) Then ]2 (𝑥) > 1.38 for 𝑥 ∈ [0.65, 1). 5 + 𝑥2 √ 1 + 𝑥2 sinh−1 (𝑥) − 𝑥] > 0 1 + 𝑥2 3/2 2𝑥 2 𝑥2 ) (1 ]2 (𝑥) = 2 𝜏2 (𝑥) (1 + 𝑥2 ) 3 Lemma 11. Let 𝐿(𝑥) be defined as in Lemma 2 and for 𝑥 ∈ (0, 1). From (83), we clearly see that 𝜏2 (𝑥) < 0 and 𝜏2 (𝑥) > 0 for 𝑥 ∈ [0.65, 1). This in turn implies that [ 2 35(1 − 𝑥2 ) (1 + 𝑥2 ) for 𝑥 ∈ [0.7, 1). Therefore, ]1 (𝑥) ≥ ]1 (0.7) = 1.214 ⋅ ⋅ ⋅ > 1.2 for 𝑥 ∈ [0.7, 1) follows from (88). , 𝜏2∗ (𝑥) = (5 + 𝑥2 ) [sinh−1 (𝑥)] − (1 + 𝑥2 ) , 𝜏2∗ (𝑥) = 2𝑥[sinh−1 (𝑥)] 2𝑥 (−84 + 316𝑥2 − 97𝑥4 + 68𝑥6 + 26𝑥8 + 36𝑥10 + 15𝑥12 ) (88) 𝑥𝜏2∗ (𝑥) (1 + 2 2𝑥2 2𝑥4 2 + 8𝑥2 − 20𝑥4 − 6𝑥8 1 [3 ( + + )− ] 2 3 𝑥 3 5 7 (1 − 𝑥2 ) (1 + 𝑥2 ) 2 +68𝑥4 (𝑥2 − )] 5 for 𝑥 ∈ (0, 1). On the other hand, from the expression of 𝜏2 (𝑥) together with Lemma 1, we get 𝜏2 (𝑥) = − (87) It follows from (19) and (87) that then 𝜇2 (𝑥) = 3𝐿 (𝑥) 2 + 8𝑥2 − 20𝑥4 − 6𝑥8 − . 2 3 𝑥4 𝑥(1 − 𝑥2 ) (1 + 𝑥2 ) = 2 (1 + 7𝑥2 − 17𝑥4 + 5𝑥6 − 4𝑥8 ) 1 2 2𝑥2 2𝑥4 ] [3 ( + + )− 2 3 𝑥 3 5 7 (1 − 𝑥2 ) (1 + 𝑥2 ) 2𝑥 (−189 + 561𝑥2 − 272𝑥4 + 103𝑥6 + 26𝑥8 + 36𝑥10 + 15𝑥12 ) 2 3 35(1 − 𝑥2 ) (1 + 𝑥2 ) 2𝑥 [−189 + 561 × (0.65)2 − 272 × (0.65)4 + 103 × (0.65)6 ] 2 3 35(1 − 𝑥2 ) (1 + 𝑥2 ) 2𝑥 × 7.23 ⋅ ⋅ ⋅ 2 3 35(1 − 𝑥2 ) (1 + 𝑥2 ) >0 (91) (86) for 𝑥 ∈ [0.65, 1). Equation (91) leads to the conclusion that ]2 (𝑥) ]2 (0.65) = 1.389 ⋅ ⋅ ⋅ > 1.38 for 𝑥 ∈ [0.65, 1). ≥ Abstract and Applied Analysis 9 Lemma 12. Let Φ1 (𝑥) and Υ1 (𝑥) be defined, respectively, as in Lemmas 3 and 5, and Θ1 (𝑥; 𝑝) = Φ1 (𝑥) − 𝑝Υ1 (𝑥). Then Θ1 (𝑥; 𝑝) is strictly decreasing on [0.7, 1) if 𝑝 > 1/6. Proof. Differentiating Θ1 (𝑥; 𝑝) with respect to 𝑥 and making use of Lemmas 8 and 10, we get we assume that 𝑎 > 𝑏. Let 𝑝 ∈ (0, 1), 𝑥 = (𝑎 − 𝑏)/(𝑎 + 𝑏), and 𝜆 1 = log[√2 log(1 + √2)]/(1 − log √2). Then 𝑥 ∈ (0, 1), and 1 (1 + 𝑥)1+𝑥 𝐼 (𝑎, 𝑏) = [ ] 𝐴 (𝑎, 𝑏) 𝑒 (1 − 𝑥)1−𝑥 𝑀 (𝑎, 𝑏) 𝑥 , = 𝐴 (𝑎, 𝑏) sinh−1 (𝑥) 𝑑Θ1 (𝑥; 𝑝) = Φ1 (𝑥) − 𝑝Υ1 (𝑥) = 𝜇1 (𝑥) − 𝑝]1 (𝑥) 𝑑𝑥 1 < 0.2 − × 1.2 = 0 6 (92) lim+ 𝑥→0 𝑄 (𝑎, 𝑏) √ = 1 + 𝑥2 , 𝐴 (𝑎, 𝑏) (95) log √1 + 𝑥2 − log 𝑥 + log [sinh−1 (𝑥)] , log √1 + 𝑥2 − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1 (96) log √1 + 𝑥2 − log 𝑥 + log [sinh−1 (𝑥)] log √1 + 𝑥2 − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1 1 = , 2 Lemma 13. Let Φ2 (𝑥) and Υ2 (𝑥) be defined, respectively, as in Lemmas 4 and 6, and Θ2 (𝑥; 𝑞) = Φ2 (𝑥) − 𝑞Υ2 (𝑥). Then Θ2 (𝑥; 𝑞) is strictly decreasing on [0.65, 1) if 𝑞 > 2/5. Proof. Differentiating Θ2 (𝑥; 𝑞) with respect to 𝑥 and making use of Lemmas 9 and 11, we have , log [𝑄 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)] log [𝑄 (𝑎, 𝑏)] − log [𝐼 (𝑎, 𝑏)] = for 𝑥 ∈ [0.7, 1) and 𝑝 > 1/6. This in turn implies that Θ1 (𝑥; 𝑝) is strictly decreasing on [0.7, 1) if 𝑝 > 1/6. 1/2𝑥 lim− 𝑥→1 (97) log √1 + 𝑥2 − log 𝑥 + log [sinh−1 (𝑥)] log √1 + 𝑥2 − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1 = 𝜆1. 𝑑Θ1 (𝑥; 𝑞) = Φ2 (𝑥) − 𝑞Υ2 (𝑥) = 𝜇2 (𝑥) − 𝑞]2 (𝑥) 𝑑𝑥 < 0.51 − 2 × 1.38 = −0.042 < 0 5 (98) (93) The difference between the convex combination of log[𝐼(𝑎, 𝑏)], log[𝑄(𝑎, 𝑏)] and log[𝑀(𝑎, 𝑏)] is as follows: 𝑝 log [𝐼 (𝑎, 𝑏)] + (1 − 𝑝) log [𝑄 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)] = for 𝑥 ∈ [0.65, 1) and 𝑞 > 2/5. This in turn implies that Θ2 (𝑥; 𝑞) is strictly decreasing on [0.65, 1) if 𝑞 > 2/5. 3. Main Results Theorem 14. The double inequality 𝑝 (1 + 𝑥)1+𝑥 ]−𝑝 log [ 2𝑥 (1 − 𝑥)1−𝑥 + (1 − 𝑝) log √1 + 𝑥2 − log [ 𝑥 ] := 𝐷𝑝 (𝑥) . sinh−1 (𝑥) (99) Equation (99) leads to 𝐷𝑝 (0+ ) = 0, 𝐷𝑝 (1− ) = log [√2 log (1 + √2)] − 𝑝 (1 − log √2) , 𝐼𝛼1 (𝑎, 𝑏) 𝑄1−𝛼1 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽1 (𝑎, 𝑏) 𝑄1−𝛽1 (𝑎, 𝑏) (94) 𝐷𝜆 1 (1− ) = 0, 𝐷𝑝 (𝑥) = − holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛽1 ≤ log[√2 log(1 + √2)]/(1 − log √2) = 0.337 ⋅ ⋅ ⋅ and 𝛼1 ≥ 1/2. Proof. Since 𝐼(𝑎, 𝑏), 𝑀(𝑎, 𝑏), and 𝑄(𝑎, 𝑏) are symmetric and homogeneous of degree one, then without loss of generality, 1 + 𝑝𝑥2 1 𝐿 (𝑥) + − 3 −1 2 √1 + 𝑥 sinh (𝑥) 𝑥+𝑥 2𝑥2 (100) (101) = Φ1 (𝑥) − 𝑝Υ1 (𝑥) = Θ1 (𝑥; 𝑝) , where 𝐿(𝑥), Φ1 (𝑥), Υ1 (𝑥), and Θ1 (𝑥; 𝑝) are defined as in Lemmas 2, 3, 5, and 12, respectively. 10 Abstract and Applied Analysis It follows from (101) together with Lemmas 3 and 5 that 𝐷1/2 (𝑥) < 2𝑥 34𝑥3 4𝑥5 1 4𝑥 4𝑥3 4𝑥5 − + − ( − + ) (102) 3 45 5 2 3 5 5 2𝑥2 8 ( − 𝑥2 ) < 0 5 9 for 𝑥 ∈ (0, 0.7). Moreover, we see clearly, from Lemma 12, (𝑥) is strictly decreasing on [0.7, 1) and so 𝐷1/2 (𝑥) < that 𝐷1/2 𝐷1/2 (0.7) = −0.109 ⋅ ⋅ ⋅ < 0 for 𝑥 ∈ [0.7, 1). This in conjunction with (100) and (102) implies that (i) If 𝛼1 < 1/2, then (96) and (97) imply that there exists 𝛿1 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) < 𝐼𝛼1 (𝑎, 𝑏)𝑄1−𝛼1 (𝑎, 𝑏) for all 𝑎, 𝑏 > 0 with (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 𝛿1 ). (ii) If 𝛽1 > 𝜆 1 , then (96) and (98) imply that there exists 𝛿2 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) > 𝐼𝛽1 (𝑎, 𝑏)𝑄1−𝛽1 (𝑎, 𝑏) for all 𝑎, 𝑏 > 0 with (𝑎 − b)/(𝑎 + 𝑏) ∈ (1 − 𝛿2 , 1). =− 𝐷1/2 (𝑥) < 0 (103) for 𝑥 ∈ (0, 1). On the other hand, (101) and Lemmas 3 and 5 together with the monotonicity of the function −2(17 − 18𝜆 1 )𝑥2 /45 + (7 − 16𝜆 1 )𝑥4 /14 on (0, 0.7) lead to 𝐷𝜆 1 (𝑥) > 2𝑥 34𝑥3 𝑥5 4𝑥 4𝑥3 8𝑥5 − + − 𝜆1 ( − + ) 3 45 2 3 5 7 = 𝑥[ 2 (1 − 2𝜆 1 ) 2 (17 − 18𝜆 1 ) 2 7 − 16𝜆 1 4 − 𝑥 + 𝑥] 3 45 14 > 𝑥[ 2 (1 − 2𝜆 1 ) 2 (17 − 18𝜆 1 ) − × (0.7)2 3 45 + Theorem 15. The double inequality 𝐼𝛼2 (𝑎, 𝑏) 𝐶1−𝛼2 (𝑎, 𝑏) < 𝑀 (𝑎, 𝑏) < 𝐼𝛽2 (𝑎, 𝑏) 𝐶1−𝛽2 (𝑎, 𝑏) (108) holds for all 𝑎, 𝑏 > 0 with 𝑎 ≠ 𝑏 if and only if 𝛼2 ≥ 5/7 and 𝛽2 ≤ log[2 log(1 + √2)] = 0.566 ⋅ ⋅ ⋅. Proof. We will follow the same idea in the proof of Theorem 14. Since 𝐼(𝑎, 𝑏), 𝑀(𝑎, 𝑏), and 𝐶(𝑎, 𝑏) are symmetric and homogeneous of degree one. Without loss of generality, we assume that 𝑎 > 𝑏. Let 𝑞 ∈ (0, 1), 𝜆 2 = log[2 log(1 + √2)], and 𝑥 = (𝑎 − 𝑏)/(𝑎 + 𝑏). Then 𝑥 ∈ (0, 1). Making use of (95) together with 𝐶(𝑎, 𝑏)/𝐴(𝑎, 𝑏) = 1 + 𝑥2 gives log [𝐶 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)] log [𝐶 (𝑎, 𝑏)] − log [𝐼 (𝑎, 𝑏)] = 7 − 16𝜆 1 × (0.7)4 ] 14 (74969 − 218832𝜆 1 ) 𝑥 = >0 180000 lim+ 𝑥→0 (104) 𝐷𝜆 1 (1− ) = −∞. (105) From (104) and (105) together with the monotonicity of 𝐷𝜆 1 (𝑥) on [0.7, 1), we clearly see that there exists 𝑐1 ∈ (0.7, 1) such that 𝐷𝜆 1 (𝑥) is strictly increasing on (0, 𝑐1 ] and strictly decreasing on [𝑐1 , 1). This in conjunction with (100) implies that 𝐷𝜆 1 (𝑥) > 0 (106) for 𝑥 ∈ (0, 1). Equation (99) together with inequalities (103) and (106) gives rise to 𝑀 (𝑎, 𝑏) > 𝐼 1/2 1/2 (𝑎, 𝑏) 𝑄 (𝑎, 𝑏) , 𝑀 (𝑎, 𝑏) < 𝐼𝜆 1 (𝑎, 𝑏) 𝑄1−𝜆 1 (𝑎, 𝑏) . , log (1 + 𝑥2 ) − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1 (109) log (1 + 𝑥2 ) − log 𝑥 + log [sinh−1 (𝑥)] log (1 + 𝑥2 ) − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1 5 = , 7 for 𝑥 ∈ (0, 0.7). It follows from Lemma 12 that 𝐷𝜆 1 (𝑥) is strictly decreasing on [0.7, 1). Note that 𝐷𝜆 1 (0.7) = 0.0229 ⋅ ⋅ ⋅ > 0, log (1 + 𝑥2 ) − log 𝑥 + log [sinh−1 (𝑥)] log (1 + 𝑥2 ) − log 𝑥 + log [sinh−1 (𝑥)] lim 𝑥 → 1− log (1 Therefore, Theorem 14 follows from (107) together with the following statements. + 𝑥2 ) − log [(1 + 𝑥)1+𝑥 /(1 − 𝑥)1−𝑥 ] / (2𝑥) + 1 = 𝜆2. (111) The difference between the convex combination of log[𝐼(𝑎, 𝑏)], log[𝐶(𝑎, 𝑏)] and log[𝑀(𝑎, 𝑏)] is as follows: 𝑞 log [𝐼 (𝑎, 𝑏)] + (1 − 𝑞) log [𝐶 (𝑎, 𝑏)] − log [𝑀 (𝑎, 𝑏)] = (107) (110) 𝑞 (1 + 𝑥)1+𝑥 ] − 𝑞 + (1 − 𝑞) log (1 + 𝑥2 ) log [ 2𝑥 (1 − 𝑥)1−𝑥 − log [ 𝑥 ] := 𝐸𝑞 (𝑥) . sinh−1 (𝑥) (112) Abstract and Applied Analysis 11 Equation (112) leads to 𝐸𝑞 (0+ ) = 0, 𝐸𝑞 (1− ) = log [2 log (1 + √2)] − 𝑞, 𝐸𝜆 2 (1− ) = 0, (113) 𝐸𝑞 (𝑥) =− 1 − 𝑥2 + 2𝑞𝑥2 1 𝐿 (𝑥) (114) + − √1 + 𝑥2 sinh−1 (𝑥) 𝑥 + 𝑥3 2𝑥2 such that 𝐸𝜆 2 (𝑥) is strictly increasing on (0, 𝑐2 ] and strictly decreasing on [𝑐2 , 1). This in conjunction with (113) implies that 𝐸𝜆 2 (𝑥) > 0 for 𝑥 ∈ (0, 1). Equation (112) together with inequalities (116) and (119) lead to the conclusion that 𝑀 (𝑎, 𝑏) > 𝐼5/7 (𝑎, 𝑏) 𝐶2/7 (𝑎, 𝑏) , = Φ2 (𝑥) − 𝑞Υ2 (𝑥) = Θ2 (𝑥; 𝑞) , where 𝐿(𝑥), Φ2 (𝑥), Υ2 (𝑥), and Θ2 (𝑥; 𝑞) are defined as in Lemmas 2, 4, 6, and 13, respectively. It follows from Lemmas 4, 6, and 13 together with (114) that 𝐸5/7 (𝑥) <( 5𝑥 79𝑥3 9𝑥5 5 7𝑥 9𝑥3 7𝑥5 − + )− ( − + ) 3 45 5 7 3 5 5 =− 4𝑥2 37 ( − 𝑥2 ) < 0 5 63 (115) for 𝑥 ∈ (0, 0.65) and 𝐸5/7 (𝑥) is strictly decreasing on [0.65, 1). Thus, we have 𝐸5/7 (𝑥) < 𝐸5/7 (0.65) = −0.117 ⋅ ⋅ ⋅ for 𝑥 ∈ [0.65, 1). This in conjunction with (113) and (115) implies that 𝐸5/7 (𝑥) < 0 (116) for 𝑥 ∈ (0, 1). On the other hand, Lemmas 4, 6, and 13 together with (114) lead to 𝐸𝜆 2 (𝑥) >( 5𝑥 79𝑥3 11𝑥5 7𝑥 9𝑥3 15𝑥5 − + ) − 𝜆2 ( − + ) 3 45 10 3 5 7 = 𝑥[ 5 − 7𝜆 2 79 − 81𝜆 2 2 150𝜆 2 − 77 4 − 𝑥 − 𝑥] 3 45 70 > 𝑥[ 5 − 7𝜆 2 79 − 81𝜆 2 − × (0.65)2 3 45 − = 150𝜆 2 − 77 × (0.65)4 ] 70 113027173 − 197098950𝜆 2 𝑥>0 100800000 (117) for 𝑥 ∈ (0, 0.65) and 𝐸𝜆 2 (𝑥) is strictly decreasing on [0.65, 1). Note that 𝐸𝜆 2 (0.65) = 0.0609 ⋅ ⋅ ⋅ , 𝐸𝜆 2 (1− ) = −∞. (118) From (117) and (118) together with the monotonicity of 𝐸𝜆 2 (𝑥) on [0.65, 1), we clearly see that there exists 𝑐2 ∈ (0.65, 1) (119) 𝑀 (𝑎, 𝑏) < 𝐼𝜆 2 (𝑎, 𝑏) 𝐶1−𝜆 2 (𝑎, 𝑏) . (120) Therefore, Theorem 15 follows from (120) together with the following statements. (i) If 𝛼2 < 5/7, then (109) and (110) imply that there exists 𝛿3 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) < 𝐼𝛼2 (𝑎, 𝑏)𝐶1−𝛼2 (𝑎, 𝑏) for all 𝑎, 𝑏 > 0 with (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (0, 𝛿3 ). (ii) If 𝛽2 > 𝜆 2 , then (109) and (111) imply that there exists 𝛿4 ∈ (0, 1) such that 𝑀(𝑎, 𝑏) > 𝐼𝛽2 (𝑎, 𝑏)𝐶1−𝛽2 (𝑎, 𝑏) for all 𝑎, 𝑏 > 0 with (𝑎 − 𝑏)/(𝑎 + 𝑏) ∈ (1 − 𝛿4 , 1). Acknowledgments This research was supported by the Natural Science Foundation of China under Grants 11171307, 61174076, and 61173123 and the Natural Science Foundation of Zhejiang Province under Grants Z1110551 and LY12F02012. References [1] E. Neuman and J. Sándor, “On the Schwab-Borchardt mean,” Mathematica Pannonica, vol. 14, no. 2, pp. 253–266, 2003. [2] J. Chen and B. 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