Optimal bounds for Neuman-Sándor mean in terms of the convex combination of the logarithmic and the second Seiffert means

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1 Chen et al. Journal of Inequalities and Applications (207) 207:25 DOI 0.86/s R E S E A R C H Open Access Optimal bounds for Neuman-Sándor mean in terms of the conve combination of the logarithmic and the second Seiffert means Jing-Jing Chen, Jian-Jun Lei and Bo-Yong Long * * Correspondence: longboyong@ahu.edu.cn School of Mathematical Science, Anhui University, Hefei, 2060, China Abstract In the article, we prove that the double inequality αl(a, b) +( α)t(a, b) <NS(a, b) <βl(a, b) +( β)t(a, b) holds for a, b >0witha bif and only if α / and β π/[ log( + 2)], where NS(a, b), L(a, b) andt(a, b) denote the Neuman-Sándor, logarithmic and second Seiffert means of two positive numbers a and b, respectively. MSC: 26E60 Keywords: Neuman-Sándor mean; logarithmic mean; the second Seiffert mean Introduction For a, b >0witha b, the Neuman-Sándor mean NS(a, b) [], the second Seiffert mean T(a, b)[2], and the logarithmic mean L(a, b)[] aredefinedby NS(a, b)= T(a, b)= L(a, b)= a b 2 sinh [(a b)/(a + b)], (.) a b 2 tan [(a b)/(a + b)], (.2) a b log a log b, respectively. It can be observed that the logarithmic mean L(a, b)canberewrittenas(see as []) L(a, b)= a b 2 tanh [(a b)/(a + b)], (.) where sinh ()=log( ), tanh ()=log ( + )/( ) andtan ()=arctan(), are the inverse hyperbolic sine, inverse hyperbolic tangent, and inverse tangent, respectively. The Author(s) 207. This article is distributed under the terms of the Creative Commons Attribution.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page 2 of Recently, the means NS, T, L and other means have been the subject of etensive research. In particular, many remarkable inequalities for the Neuman-Sándor, second Seiffert and logarithmic means can be found in the literature [2 6]. Let P(a, b) =(a b)/(2 sin [(a b)/(a + b)]), S(a, b)= (a 2 + b 2 )/2, A(a, b) =(a + b)/2, I(a, b) =/e(b b /a a ) /(b a), G(a, b) = ab, andh(a, b) =2ab/(a + b) denote the first Seiffert, root-square, arithmetic, identric, geometric, and the harmonic means of two positive numbers a and b with a b, respectively. Then it is well known that the inequality S(a, b)>t(a, b)>ns(a, b)>a(a, b)>i(a, b)>p(a, b)>l(a, b)>g(a, b)>h(a, b) holds for a, b >0witha b. In [7]and[8], the authors proved that the double inequalities S(a, b) α A α (a, b)<ns(a, b)<s(a, b) β A β (a, b), α S(a, b)+( α )G(a, b)<ns(a, b)<β S(a, b)+( β )G(a, b) hold for all a, b >0witha b if and only if α /, 2(log(2 + 2) log )/ log 2 β, α 2/ and β /[ 2 log( + 2)]. In [9], it was showed that the inequality P α 2 (a, b)t α 2 (a, b)<ns(a, b)<p β 2 (a, b)t β 2 (a, b) holds for all a, b >0witha b if and only if α 2 > / and ( ) log( + 2) β 2 log / log 2= π Let L p (a, b)=(a p+ + b p+ )/(a p + b p ) be the Lehmer mean of two positive numbers a and b with a b.in[0], the authors proved the double inequality L α (a, b)<ns(a, b)<l β (a, b) holds for all a, b >0witha b if and only if α = is the unique solution of the equation (p +) /p =2log( + 2), and β =2. Let ( ap +b p ) /p, p 0, 2 M p (a, b)= ab, p =0, be the pth power means of two positive numbers a and b with a b. In[20], the authors proved the sharp double inequality M log 2/(log π log 2) (a, b)<t(a, b)<m 5/ (a, b) holds.

3 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page of Gao [2] proved the optimal double inequality I(a, b)<t(a, b)< 2e I(a, b) π holds for all a, b >0witha b. Yang [22]provedtheinequality A /(p) p (a, b)g /(p) (a, b)<l(a, b)<a /(q) q (a, b)g /(q) (a, b) holds for all a, b >0witha b if and only if p / 5and0<q /. And the inequality M 0 (a, b)<l(a, b)<m / (a, b) was proved by Lin in [2]. In [2], the authors present bounds for L in terms of G and A G 2/ (a, b)a / (a, b)<l(a, b)< 2 G(a, b)+ A(a, b) for all a, b >0witha b. The purpose of this paper is to answer the following questions: What are the least value α and the greatest value β such that αl(a, b) +( α)t(a, b) <NS(a, b) <βl(a, b) +( β)t(a, b) holds for all a, b >0witha b? 2 Lemmas It is well known that, for (0, ), tanh ()= = 2n + 2n+, (2.) n=0 tan ()= = ( ) n 2n + 2n+. (2.2) To establish our main result, we need several lemmas as follows. n=0 Lemma 2. ([25]) Let H()= sinh +2 (sinh ). 2 Then H() is strictly increasing on (0,).Moreover, the inequality H()< 9 (2.)

4 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page of holds for any (0, 0.76) and the inequality H()> 7 90 (2.) holds for any (0, ). Lemma 2.2 Let S()=/tanh /[( 2 )(tanh ) 2 ]. Then S()< 2 5 (2.5) for any (0, ) and S()> 2 5 (2.6) for any (0, 0.76). Proof Let G()= ( 2)( tanh ) 2 [S() ] 5. Then direct computation leads to G(0) = 0, (2.7) G ()= g() tanh, (2.8) where g()=( )tanh It follows that g ()= 2 g (), (2.9) where g ()=( 2 5 6)( 2 )tanh Considering (2.), we have g ()< ( )( 2)( ) = ( ) 5 < 2( ) <0, (2.0) for (0, ). Thus, (2.9) and(2.0) aswellasg(0) = 0 imply g()<0for (0, ). Therefore, combining (2.7) and(2.8), we get G() < 0for (0,). It means inequality (2.5) holds. Let Q()= ( 2)( tanh ) 2 [S()+ 2 ]

5 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page 5 of Direct computation leads to where Q(0) = 0, (2.) Q ()= 2 q () tanh, (2.2) q ()= ( ) tanh. When (0, 0.7], considering (2.)andthefact =( 2 6 )+( )>0,wecanget q ()> ( )( ) = ( > ) >0. When (0.7, 0.76), direct computation leads to q (0.76) = > 0, (2.) q ()=q 2()/ ( 2), (2.) where q 2 ()= ( ) tanh. Considering (2.) and the fact <2( )<0,wecanget q 2 ()< ( )( ) = <2 ( ) + 2( 6 2 2) < 0. (2.5) Thus, (2.)-(2.5)implythat q ()>0 (2.6) holds for any (0.7, 0.76). Therefore, Q() >0for (0, 0.76) follows from (2.), (2.2) and(2.6). That means inequality (2.6)holds. Lemma 2. Let T()=/tan /[( + 2 )(tan ) 2 ]. Then T()< (2.7)

6 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page 6 of for any (0, ) and T()> (2.8) for any (0, 0.76). Proof Let 2 M()= [T() + 2 ] ( )( tan ) 2. Differentiating M(), we have M ()=[t()tan ]/2, where t()= ( ) tan. For (0, ), we have < <0.Thusfrom(2.2), we can get t()< ( )( ) = < <0. Therefore M () <0for (0, ). Considering the fact M(0) = 0, we get M() <0for (0, ). So the inequality (2.7)holds. Let N()= [T() ] 5 (+ 2 )( tan ) 2. 7 Differentiating N(), we have N ()=n()tan,where n()= ( ) tan. Because of ( ) >0 for (0, 0.76), it follows that n()> ( ) = 5 7 >0. Considering the fact N(0) =0, the inequality (2.8)holds.

7 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page 7 of Lemma 2. The function f () =λs()+( λ)t() H() is strictly decreasing on (0.76,), where λ = π/[ log( + 2)]= and H(), S() and T() are defined as in Lemmas 2., 2.2 and 2., respectively. Proof Direct computation leads to S tanh ()=2 ( 2 ) 2 (tanh ), S ϕ() ()=2 ( 2 ) (tanh ), where ϕ()=(+ 2 )tanh (tanh ) 2. It follows that ϕ ()= R() 2, where R()= ( 2 )(tanh ) 2 (6 +2)tanh +6 2.From(2.), we can get R()< ( 2)( + = <0. ) 2 ( 6 +2 )( + ) +6 2 Thus ϕ() is strictly decreasing on (0.76, ). Considering the fact ϕ(0.76) = < 0, we have ϕ() < 0 for any (0.76, ). In other words, S () is strictly decreasing on (0.76, ). Let φ()=λs()+( λ)t(). It was proved that T () is strictly decreasing on (0.7, ) in Lemma 5 of [26]. Thus, from the monotonicity of S ()andt (), we have φ ()<λs (0.76) + ( λ)t (0.76)= <0 for any (0.76, ). That is to say, φ() is strictly decreasing on (0.76, ). Considering the monotonicity of H() in Lemma 2., the proof is completed. Lemma 2.5 We have λ 28 27λ λ >0 5 for (0, 0.76), where λ = π/[ log( + 2)]= Proof Let η()= λ 28 27λ λ. Then it is easy to verify that η()is decreasingon(0,μ), where μ = 5 60 log( + 2) 27π π 28log( + = ) Considering η(0.76) = >0, wehaveη()>0for (0, 0.76).

8 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page 8 of Main results Theorem. The double inequality αl(a, b) +( α)t(a, b) <NS(a, b) <βl(a, b) +( β)t(a, b) holds for any a, b >0with a bifandonlyifα / and β π log( + 2) = Proof Because NS(a, b), L(a, b), T(a, b) are symmetric and homogeneous of degree, without loss of generality, we can assume that a > b and := (a b)/(a + b) (0, ). Let p (0, ) and λ = π/[ log( + 2)] = Then by (.)-(.), direct computation leads to Let NS(a, b) A(a, b) = sinh, L(a, b) A(a, b) = tanh, T(a, b) A(a, b) = tan. tl(a, b)+( t)t(a, b) M(a, b) F t ()= A(a, b) = t tanh +( t) tan sinh. (.) Then it follows that F ( 0 + ) =0, (.2) F λ ( 0 + ) = F λ ( ) = 0. (.) Differentiating F t (), we have [ ] F t ()=t tanh 2 (tanh ) 2 [ +( t) [ tan sinh := ts()+( t)t() H(), (tan ) 2 ] (sinh ) 2 ] where H(),S() and T() are defined as in Lemmas 2.-2.,respectively.

9 Chenetal. Journal of Inequalities and Applications (207) 207:25 Page 9 of On one hand, from inequalities (2.), (2.5)and(2.6), we clearly see that F ()= S()+ T() H() < ( 2 ) 5 + ( ) ( ) 90 = <0 for any (0, ). It leads to F ()<F (0) = 0 (.) for any (0, ). Thus, from (.) it follows that NS(a, b)> L(a, b)+ T(a, b) for all a, b >0witha b. Considering L(a, b)<ns(a, b)<t(a, b), we can get NS(a, b) >αl(a, b) +( α)t(a, b) (.5) for all α / and a, b >0witha b. On the other hand, from inequalities (2.), (2.6)and(2.7), we have ( 2 F λ ()> λ [ λ = 28 ) ( 2 +( λ) ] 27λ λ 5 for (0, 0.76). According to Lemma 2.5,wehave ) ( ) 9 F λ ()>0 (.6) for (0, 0.76). Lemma 2. shows that F λ () is strictly decreasing on (0.76, ). This fact and F λ (0.76) = > 0 together with F λ ( )= imply that there eists 0 (0.76, ) such that F λ () is strictly increasing on (0, 0 ] and strictly decreasing on [ 0,). Equations (.) and(.) together with the piecewise monotonicity of F λ () leadtothe conclusion that NS(a, b) <λl(a, b) +( λ)t(a, b) for all a, b >0witha b. Considering L(a, b)<m(a, b)<t(a, b), we can get NS(a, b) <βl(a, b) +( β)t(a, b) (.7) holds for β λ and all a, b >0witha b.

10 Chenetal.Journal of Inequalities and Applications (207) 207:25 Page 0 of Finally, we prove that L(a, b)/ + T(a, b)/ and λl(a, b)+( λ)t(a, b)arethebestpossible lower and upper mean bound for the Neuman-Sándor mean M(a, b). For any ɛ, ɛ 2 >0,lett = / ɛ, t 2 = λ + ɛ 2.Thenonecanget ( ) F t ()= ɛ tanh + ( ) + ɛ tan F t2 ()=(λ + ɛ 2 ) tanh +( λ ɛ 2) tan Let 0 + and 2, then the Taylor epansion leads to sinh, (.8) sinh. (.9) F t ( )= 2 ɛ 2 + O( ), (.0) F t2 ( 2 )= ɛ 2 /π + O( 2 ). (.) Equations (.8)and(.0)implythatifα < /, then, for any ɛ >0,thereeistsσ (0, ) such that NS(a, b) < (/ ɛ )L(a, b) + (/ ɛ )T(a, b) for all a, b with (a b)/(a + b) (0, σ ). Equations (.9)and(.)implythatifβ > λ,then,foranyɛ 2 >0,thereeistsσ 2 (0, ) such that NS(a, b) >(λ + ɛ 2 )L(a, b)+( λ ɛ 2 )T(a, b) for all a, b with (a b)/(a + b) ( σ 2,). Conclusion In the article, we give the sharp upper and lower bounds for Neuman-Sándor mean in terms of the linear conve combination of the logarithmic and second Seiffert means. Acknowledgements This research was supported by Foundations of Anhui Educational Committee (KJ207A029) and Anhui University (J , J00090, Y000228), China. Competing interests The authors declare that they have no competing interests. Authors contributions All the authors worked jointly. All the authors read and approved the final manuscript. Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Received: 26 May 207 Accepted: September 207 References. Neuman, E, Sándor, J: On the Schwab-Borchardt mean. Math. Pannon. (2), (200) 2. Seiffert, HJ: Aufgabe β 6. Die Wurzel 29, (995). Carlson, BC: The logarithmic mean. Am. Math. Mon. 79, (972). Chu, Y-M, Long, B-Y: Bounds of the Neuman-Sandor mean using power and identric means. Abstr. Appl. Anal. 20, Article ID 8259 (20) 5. Chu, Y-M, Long, B-Y, Gong, W-M, Song, Y-Q: Sharp bounds for Seiffert and Neuman-Sandor means in terms of generalized logarithmic means. J. Inequal. Appl. 20, ArticleID0 (20) 6. Chu, Y-M, Qian, W-M, Wu, L-M, Zhang, X-H: Opimal bounds for the first and second Seiffert means in terms of geometric, arithmetic and contraharmonc means. J. Inequal. Appl. 205, ArticleID (205) 7. Chu, Y-M, Wang, M-K, Gong, W-M: Two sharp double inequalities for Seiffert mean. J. Inequal. Appl. 20, Article ID (20) 8. Chu, Y-M, Wang, M-K, Wang, Z-K: Best possible inequalities among harmonic, logarithmic and Seiffert means. Math. Inequal. Appl. 5(2),5-22 (202) 9. Chu, Y-M, Zhao, T-H, Liu, B-Y: Optimal bounds for Neuman-Sándor mean in terms of the conve combination of logarithmic and quadratic or contra-harmonic means. J. Math. Inequal. 8(2), (20)

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