Sharp bounds for Seiffert and Neuman-Sándor means in terms of generalized logarithmic means

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1 Chu et al. Journal of Inequalities and Applications 2013, 2013:10 R E S E A R C H Open Access Sharp bounds for Seiffert and Neuman-Sándor means in terms of generalized logarithmic means Yu-Ming Chu 1*, Bo-Yong Long 2,Wei-MingGong 1 and Ying-Qing Song 1 * Correspondence: chuyuming2005@yahoo.com.cn 1 School of Mathematics and Computation Sciences, Hunan City University, Yiyang, , China Full list of author information is available at the end of the article Abstract In this paper, we prove three sharp inequalities as follows: P(a, b)>l 2 (a, b), T(a, b)>l 5 (a, b)andm(a, b)>l 4 (a, b)foralla, b >0witha b.here,l r (a, b), M(a, b), P(a, b) andt(a, b) are therth generalized logarithmic, Neuman-Sándor, first and second Seiffert means of a andb, respectively. MSC: 26E60 Keywords: generalized logarithmic mean; Neuman-Sándor mean; first Seiffert mean; second Seiffert mean 1 Introduction The Neuman-Sándor mean M(a, b) 1] and the first and second Seiffert means P(a, b) 2] and T(a, b)3] oftwopositivenumbersa and b are defined by ab a b, 2 arcsinh( M(a, b)= ab a+b ), a, a = b, ab P(a, b)= 4 arctan(, a b, a/b)π a, a = b (1.1) (1.2) and ab a b, 2 arctan( T(a, b)= ab a+b ), a, a = b, (1.3) respectively. Recently, these means, M, P and T, have been the subject of intensive research. In particular, many remarkable inequalities for M, P and T can be found in the literature 1, 48]. The power mean M p (a, b)oforderr of two positive numbers a and b is defined by ( ap +b p ) 1/p, p 0, 2 M p (a, b)= ab, p = Chu et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

2 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 2 of 13 The main properties for M p (a, b)aregivenin9]. In particular, the function p M p (a, b) (a b) is continuous and strictly increasing on R. The arithmetic-geometric mean AG(a, b)oftwopositivenumbersa and b is defined as the common limit of sequences {a n } and {b n },whicharegivenby a 0 = a, b 0 = b, a n+1 = a n + b n, b n+1 = a n b n. 2 Let H(a, b) =2ab/(a + b), G(a, b) = ab, L(a, b) =(b a)/(log b log a), I(a, b) = 1/e(b b /a a ) 1/(ba), A(a, b)=(a + b)/2, and S(a, b)= (a 2 + b 2 )/2 be the harmonic, geometric, logarithmic, identric, arithmetic and root-square means of two positive numbers a and b with a b, respectively. Then it is well known that the inequalities H(a, b)=m 1 (a, b)<g(a, b)=m 0 (a, b)<l(a, b)<ag(a, b)<i(a, b) < A(a, b)=m 1 (a, b)<m(a, b)<t(a, b)<s(a, b)=m 2 (a, b) hold for all a, b >0witha b. For r >0therth generalized logarithmic mean L r (a, b)oftwopositivenumbersa and b is defined by L r (a, b)=l 1/r( a r, b r) b r a r r(log blog a) = ]1/r, a b, a, a = b. (1.4) It is not difficult to verify that L r (a, b) is continuous and strictly increasing with respect to r (0, + )forfieda, b >0witha b. In 2, 3] Seiffert proved that the double inequalities L(a, b) <P(a, b) <I(a, b) and A(a, b) <T(a, b) <S(a, b) hold for all a, b >0witha b. The following bounds for the first Seiffert mean P(a, b) in terms of power mean were presented by Jagers in 10]: M 1/2 < P(a, b)<m 2/3 (a, b) for all a, b >0witha b. Hästö 11, 12] proved that the function T(1, )/M p (1, ) isincreasingon(0,+ ) if p 1 and found the sharp lower power mean bound for the Seiffert mean P(a, b) as follows: P(a, b)>m log 2/ log π (a, b) for all a, b >0witha b.

3 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 3 of 13 In 13] the authors presented the following best possible Lehmer mean bounds for the Seiffert means P(a, b)andt(a, b): L 1/6 (a, b)<p(a, b)<l 0 (a, b) and L 0 (a, b)<t(a, b)<l 1/3 (a, b) for all a, b >0witha b. Here, L p (a, b)=(a p+1 + b p+1 )/(a p + b p ) is the Lehmer mean of a and b. In 14, 15] the authors proved that the inequalities and α 1 S(a, b)+(1α 1 )A(a, b)<t(a, b)<β 1 S(a, b)+(1β 1 )A(a, b), S α 2 (a, b)a 1α 2 (a, b)<t(a, b)<s β 2 (a, b)a 1β 2 (a, b) α 3 T(a, b)+(1α 3 )G(a, b)<a(a, b)<β 3 T(a, b)+(1β 3 )G(a, b) hold for all a, b >0witha b if and only if α 1 (4 π)/( 21)π], β 1 2/3, α 2 2/3, β 2 42log π/ log 2, α 3 3/5 and β 3 π/4. For all a, b >0witha b, the following inequalities can be found in 16, 17]: and L(a, b)=l 1 (a, b)<ag(a, b)<l 3/2 (a, b)<m 1/2 (a, b). Neuman and Sándor 1]establishedthat P(a, b) <A(a, b) <M(a, b) <T(a, b) π 2 P(a, b)>a(a, b)>arcsinh(1)m(a, b)>π T(a, b) 4 for all a, b >0witha b. In particular, the Ky Fan inequalities G(a, b) L(a, b) P(a, b) A(a, b) M(a, b) T(a, b) < < < < < G(a, b ) L(a, b ) P(a, b ) A(a, b ) M(a, b ) T(a, b ) hold for all 0 < a, b 1/2 with a b, a =1a and b =1b. It is the aim of this paper to find the best possible generalized logarithmic mean bounds for the Neuman-Sándor and Seiffert means. 2 Lemmas In order to establish our main results, we need three lemmas, which we present in this section. Lemma 2.1 The inequality ( 4 ) 1 1/2 2 1 < 4 log 4 arctan π (2.1) holds for all >1.

4 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 4 of 13 Proof Let ( f ()= log 2 ) 1 2. (2.2) 4 arctan π Then f ()canberewrittenas f ()= ( 4 1)f 1 () 4(4 arctan π) 2 log, (2.3) f 1 ()=(4arctan π) 2 4(2 1)log Simple computations lead to f 1 (1) = 0, (2.4) f f 2(), 2 (2.5) and 4 log f 2 ()=8arctan π, f 2 (1) = 0, (2.6) f 2 3() ( 2 +1), 2 (2.7) f 3 ()=4 ( 2 1 ) log , f 3 (1) = 0, (2.8) f 3 ()=8log , f 3 (1) = 0, (2.9) f 3 ()=8log , f 3 (1) = 0 (2.10) f 3 ()=8 3 ( +1)( 1)( ) <0 (2.11) for >1. Inequality (2.11)impliesthatf 3 () is strictly decreasing in 1, + ), then equation (2.10) leads to the conclusion that f 3 () is strictly decreasing in 1, + ). From equations (2.4)-(2.9) and the monotonicity of f 3 (), we clearly see that for >1. f 1 ()<0 (2.12)

5 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 5 of 13 Therefore, inequality (2.1) followsfromequations (2.2) and(2.3) togetherwithinequal- ity (2.12). Lemma 2.2 The inequality ( 5 ) 1 1/5 1 < 5 log 2 arctan 1 +1 (2.13) holds for all >1. Proof Let g()= 1 ( 5 ) ( 5 log 1 log 5 log Then simple computations lead to 1 2 arctan 1 +1 ). (2.14) g(1) = 0, (2.15) g ()= 5( )log + 5 1]g 1 () 5( 5 1)arctan( 1 +1 ) log, (2.16) 5( 5 1)log 1 g 1 ()= arctan (1 + 2 )5( )log + 5 1] +1, g 1 (1) = 0, (2.17) g 1 ()= 3 g 2 () (1 + 2 ) 2 5( )log + 5 1] 2, (2.18) g 2 ()=50 ( ) log 2 5 ( ) log +4 ( ), g 2 (1) = 0, (2.19) g 2 ()=50( ) log 2 +5 ( ) log , g 2 (1) = 0, (2.20) g 2 ()=100( ) log ( ) log ,

6 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 6 of 13 g 2 (1) = 0, (2.21) g 2 ()=600( ) log ( ) log +1, , , , , g 2 (1) = 0, (2.22) g (4) 2 ()=2g 3(), (2.23) g 3 ()=600 ( ) log ( , ) log +1, , , ,010 1, , , g 3 (1) = 0, (2.24) g 3 () = 3,000(6 +1)log2 +10 ( 3, , ,400 +1, ) log +2, , , ,060 +8,510 2, , , , ,845 8 < 3,000(6 +1)( 1)log +10 ( 3, , ,400 +1, ) log +2, , , ,060 +8,510 2, , , , ,845 8 =10 ( 3, , , , ) log +2, , , , ,510 2, , , , , (2.25) Let g 4 ()=10 ( 3, , , , ) log +2, , , , ,510 2, , , , ,845 8.

7 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 7 of 13 Then g 4 (1) = 1,040 < 0, (2.26) g 4 ()=109 g 5 (), (2.27) g 5 ()= ( 15, , , , ,520 +5,040 ) log 2, , , , ,014 2,106 < ( 15, , , , , ,040 11) log 2, , , , ,014 2,106 = ( 11, ) log 7, ,014 2,106 <0 (2.28) for >1. Equation (2.27) and inequality (2.28)leadtotheconclusionthatg 4 ()isstrictlydecreasing in 1, + ). Then equation (2.26)impliesthat g 4 () < 0 (2.29) for >1. Inequalities (2.25)and(2.29)implythatg 3 ()<0.Thenequation(2.24)showsthat g 3 ()<0 (2.30) for >1. From equations (2.17)-(2.23) and inequality (2.30),we clearly see that g 1 ()<0 (2.31) for >1. Therefore, inequality (2.13) follows easily from equations (2.14)-(2.16) andinequal- ity (2.31). Lemma 2.3 The inequality ( ) 1 arcsinh 4 ( 1)4 log <0 (2.32) +1 4( 4 1) holds for all >1.

8 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 8 of 13 Proof Let ( )] 1 h()=log arcsinh 4 log +1 Then simple computations lead to ( 1) 4 log 4( 4 1) ]. (2.33) h(1) = 0, (2.34) h ()= 4(2 + +1)log + 4 1]h 1 () ( 4 1)arcsinh( 1 +1 ) log, (2.35) 4 2( 4 1)log h 1 ()= ( + 1)4( )log + 4 1] 1+ 2 ( ) 1 arcsinh, +1 h 1 (1) = 0, (2.36) 2 h 2 1 ()= h 2 () ( + 1)(1 + 2 ) 3/2 4( )log + 4 1], (2.37) 2 h 2 ()=16 ( ) log 2 4 ( 4 1 )( ) log +3 ( 4 1 ) 2( 1+ 2 ), h 2 (1) = 0, (2.38) h 2 ()=2h 3(), (2.39) h 3 ()=8 ( ) log 2 2 ( ) log , h 3 (1) = 0, (2.40) h 3 ()=h 4(), (2.41) h 4 ()=16 ( ) log 2 4 ( ) log , h 4 (1) = 0, (2.42) h 4 ()=h 5(), (2.43)

9 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 9 of 13 h 5 ()=80(9 +4)log 2 4 ( ) log , h 5 (1) = 0, (2.44) h 5 ()=720log2 +4 ( ) log , h 5 (1) = 0, (2.45) h 5 ()=49 h 6 (), (2.46) h 6 ()=8 ( ) log ( 2 1 )( ) <0 (2.47) for all >1. Equations (2.45) and(2.46) together with inequality (2.47) implythath 5 () isstrictly decreasing in 1, + ). Then equation (2.44) leadsto h 5 ()<0 (2.48) for all >1. From equations (2.36)-(2.43) and inequality (2.48),we clearly see that h 1 ()<0 (2.49) for all >1. Therefore, inequality (2.32) follows from equations (2.33)-(2.35) and inequality (2.49). 3 Main results Theorem 3.1 The inequality P(a, b)>l 2 (a, b) holds for all a, b >0with a b, and L 2 (a, b) is the best possible lower generalized logarithmic mean bound for the first Seiffert mean P(a, b).

10 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 10 of 13 Proof From (1.2)and(1.4), we clearly see that both P(a, b)andl r (a, b) are symmetric and homogenous of degree one. Without loss of generality, we assume that b =1anda = 2 >1. Then (1.2)and(1.4)leadto L 2 ( 2,1 ) P ( 2,1 ) = ( 4 ) 1 1/ log 4 arctan π. (3.1) Therefore, P( 2,1)>L 2 ( 2, 1) follows from Lemma 2.1 and equation (3.1). Net, we prove that L 2 (a, b) is the best possible lower generalized logarithmic mean bound for the first Seiffert mean P(a, b). For any ɛ >0and >0,from(1.2)and(1.4), one has (1 + ) 2+ɛ ] 1 1/(2+ɛ) L 2+ɛ (1 +,1)P(1 +,1)= (2 + ɛ) log(1 + ) 4 arctan 1+ π. (3.2) Letting 0 and making use of Taylor epansion, we get (1 + ) 2+ɛ ] 1 1/(2+ɛ) (2 + ɛ) log(1 + ) 4 arctan 1+ π = 1+ 2+ɛ (2 + ɛ)(1 + 2ɛ) o ( 2)] 1/(2+ɛ) o( 3 ) = ɛ o ( 2) ] o ( 2) ] = ɛ o ( 2). (3.3) Equations (3.2) and(3.3) implythatforanyɛ >0,thereeistsδ 1 = δ 1 (ɛ) >0suchthat L 2+ɛ (1 +,1)>P(1 +,1)for (0, δ 1 ). Remark 3.1 It follows from (1.2)and(1.4)that P(,1) lim + L λ (,1) = lim λ 1/λ (1 1/) log 1/λ =+ (3.4) + π(1 λ ) 1/λ for all λ >0. Equation (3.4)impliesthatλ >0suchthatL λ (a, b)>p(a, b) for all a, b >0doesnoteist. Theorem 3.2 The inequality T(a, b)>l 5 (a, b) holds for all a, b >0with a b, and L 5 (a, b) is the best possible lower generalized logarithmic mean bound for the second Seiffert mean T(a, b). Proof Without loss of generality, we assume that b =1anda = >1.Then(1.3) and(1.4) lead to L 5 (,1)T(,1)= ( 5 ) 1 1/5 5 log 1 2 arctan (3.5)

11 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 11 of 13 Therefore, T(,1)>L 5 (, 1) follows from Lemma 2.2 and equation (3.5). Net, we prove that L 5 (a, b) is the best possible lower generalized logarithmic mean bound for the second Seiffert mean T(a, b). For any ɛ >0and >0,from(1.3)and(1.4), one has (1 + ) 5+ɛ ] 1 1/(5+ɛ) L 5+ɛ (1 +,1)T(1 +,1)= (5 + ɛ) log(1 + ) 2 arctan. (3.6) 2+ Letting 0 and making use of Taylor epansion, we have (1 + ) 5+ɛ ] 1 1/(5+ɛ) (5 + ɛ) log(1 + ) 2 arctan 2+ = 1+ 5+ɛ (5 + ɛ)(7 + 2ɛ) o ( 2)] 1/(5+ɛ) o( 3 ) = ɛ o ( 2) ] o ( 2) ] = ɛ o ( 2). (3.7) Equations (3.6) and(3.7) implythatforanyɛ >0,thereeistsδ 2 = δ 2 (ɛ) >0suchthat L 5+ɛ (1 +,1)>T(1 +,1)for (0, δ 2 ). Remark 3.2 It follows from (1.3)and(1.4)that T(,1) lim + L μ (,1) = lim 2μ 1/μ (1 1/) log 1/μ =+ (3.8) + π(1 μ ) 1/μ for all μ >0. Equation (3.8)impliesthatμ >0suchthatL μ (a, b)>t(a, b) for all a, b >0doesnoteist. Theorem 3.3 The inequality M(a, b)>l 4 (a, b) holds for all a, b >0with a b, and L 4 (a, b) is the best possible lower generalized logarithmic mean bound for the Neuman-Sándor mean M(a, b). Proof Without loss of generality, we assume that b =1anda = >1.Then(1.1) and(1.4) lead to L 4 4 (,1)M4 (,1) = log ( 1) 4 16 arcsinh 4 ( 1 = arcsinh 4 ( 1 +1 ) log +1 ) arcsinh 4 ( 1 +1 ) ( ] 1)4 log. (3.9) 4( 4 1) Therefore, M(,1)>L 4 (, 1) follows from Lemma 2.3 and equation (3.9).

12 Chuet al. Journal of Inequalities and Applications 2013, 2013:10 Page 12 of 13 Net, we prove that L 4 (a, b) is the best possible lower generalized logarithmic mean bound for the Neuman-Sándor mean M(a, b). For any ɛ >0and >0,from(1.1)and(1.4), one has (1 + ) 4+ɛ ] 1 1/(4+ɛ) L 4+ɛ (1 +,1)T(1 +,1)= (4 + ɛ) log(1 + ) 2 arcsinh( (3.10) ). 2+ Letting 0 and making use of Taylor epansion, we have (1 + ) 4+ɛ ] 1 1/(4+ɛ) (4 + ɛ) log(1 + ) 2 arcsinh( 2+ ) = 1+ 4+ɛ (4 + ɛ)(5 + 2ɛ) o ( 2)] 1/(4+ɛ) o( 3 ) = ɛ o ( 2) ] o ( 2) ] = ɛ o ( 2). (3.11) Equations (3.10) and(3.11) implythatforanyɛ >0,thereeistsδ 3 = δ 3 (ɛ)>0suchthat L 4+ɛ (1 +,1)>M(1 +,1)for (0, δ 3 ). Remark 3.3 It follows from (1.1)and(1.4)that M(,1) lim + L ν (,1) = lim ν 1/ν (1 1/) log 1/ν =+ (3.12) + 2 arcsinh(1)(1 ν ) 1/ν for all ν >0. Equation (3.12) impliesthatν >0suchthatL ν (a, b) >M(a, b) for all a, b >0doesnot eist. Competing interests The authors declare that they have no competing interests. Authors contributions YMC provided the main idea and carried out the proof of Remarks in this article. BYL carried out the proof of Lemma 2.1 and Theorem 3.1 in this article. WMG carried out the proof of Lemma 2.2 and Theorem 3.2 in this article. YQS carried out the proof of Lemma 2.3 and Theorem 3.3 in this article. All authors read and approved the final manuscript. Author details 1 School of Mathematics and Computation Sciences, Hunan City University, Yiyang, , China. 2 College of Mathematics Science, Anhui University, Hefei, , China. Acknowledgements This research was supported by the Natural Science Foundation of China under Grants and , the Natural Science Foundation of Hunan Province under Grant 09JJ6003 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T Received: 15 June 2012 Accepted: 18 December 2012 Published: 7 January 2013 References 1. Neuman, E, Sándor, J: On the Schwab-Borchardt mean. Math. Pannon. 14(2), (2003) 2. Seiffert, H-J: Problem 887. Nieuw Arch. Wiskd. 11, 176 (1993) 3. Seiffert, H-J: Aufgabe β16. Die Wurzel 29, (1995) 4. Chu, Y-M, Qiu, Y-F, Wang, M-K: Sharp power mean bounds for combination of Seiffert and geometric means. Abstr. Appl. Anal. 2010, Article ID (2010)

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