Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means

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1 Gong et al. Journal of Inequalities and Applications 01, 01:55 R E S E A R C H Open Access Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means Wei-Ming Gong 1, Xu-Hui Shen and Yu-Ming Chu 1* * Correspondence: chuyuming005@16.com 1 School of Mathematics and Computation Science, Hunan City University, Yiyang, 41000, China Full list of author information is available at the end of the article Abstract In this paper, we find the least value α and the greatest value β such that the double inequality P α (a, bq 1 α (a, b<m(a, b<p β (a, bq 1 β (a, b holds true for all a, b >0witha b,wherep(a, b, M(a, bandq(a, b are the first Seiffert, Neuman-Sándor and quadratic means of a and b, respectively. MSC: 6E60 Keywords: Neuman-Sándor mean; first Seiffert mean; quadratic mean 1 Introduction Let u, v and w be the bivariate means such that u(a, b <w(a, b <v(a, b for all a, b >0 with a b. The problems of finding the best possible parameters α and β such that the inequalities αu(a, b+(1 αv(a, b<w(a, b<βu(a, b+(1 βv(a, bandu α (a, bv 1 α (a, b< w(a, b<u β (a, bv 1 β (a, b hold for all a, b >0witha bhave attracted the interest of many mathematicians. For a, b >0witha b, the first Seiffert mean P(a, b [1], the Neuman-Sándor mean M(a, b[], the quadratic mean Q(a, b aredefinedby a b P(a, b= 4 arctan( a/b π, M(a, b= a b sinh 1 ( a b a + b Q(a, b=, a+b, (1.1 respectively. In here, sinh 1 (=log( is the inverse hyperbolic sine function. Recently, the means P, M and Q have been the subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [ 14]. The first Seiffert mean P(a, b canbe rewritten as(see[, Eq. (.4] P(a, b= a b arcsin[(a b/(a + b]. (1. 01 Gong 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 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page of 1 Let H(a, b = ab/(a + b, L(a, b =(b a/(log b log a, A(a, b =(a + b/, T(a, b = (a b/[ arctan((a b/(a+b] and C(a, b=(a +b /(a+b be the harmonic, logarithmic, arithmetic, second Seiffert and contra-harmonic means of a and b, respectively. Then it is known that the inequalities H(a, b <L(a, b <P(a, b <A(a, b <M(a, b <T(a, b <Q(a, b <C(a, b hold for all a, b >0witha b. Neumanand Sándor [, 15] proved that the inequalities π 4 log(1 + T(a, b<m(a, b< A(a, b log(1 +, T (a, b Q (a, b<m(a, b< T (a, b Q(a, b, H ( T(a, b, A(a, b < M(a, b<l ( A(a, b, Q(a, b, T(a, b>h ( M(a, b, Q(a, b, M(a, b< A (a, b P(a, b, A/ (a, bq 1/ (a, b<m(a, b< A(a, bt(a, b<m(a, b< A (a, b+t (a, b, A(, y A(1,1 y < M(, y M(1,1 y < T(, y T(1,1 y, A(a, b+q(a, b, 1 A(1,1 y 1 A(, y < 1 M(1,1 y 1 M(, y < 1 T(1,1 y 1 T(, y, A(, ya(1,1 y <M(, ym(1,1 y <T(, yt(1,1 y hold for all a, b >0and, y (0, 1/] with a b and y. Li et al. [16] proved that the double inequality L p0 (a, b<m(a, b<l (a, b holdsforall a, b >0witha b, wherel p (a, b=[(b p+1 a p+1 /((p +1(b a] 1/p (p 1, 0, L 0 (a, b= 1/e(b b /a a 1/(b a and L 1 (a, b =(b a/(log b log a isthepth generalized logarithmic mean of a and b, andp 0 = istheuniquesolutionoftheequation(p +1 1/p = log(1 +. In [1], Neuman proved that the double inequalities Q α (a, ba 1 α (a, b<m(a, b<q β (a, ba 1 β (a, b (1. and C λ (a, ba 1 λ (a, b<m(a, b<c μ (a, ba 1 μ (a, b (1.4 hold for all a, b >0witha b if α 1/, β [log( + log ]/ log, λ 1/6 and μ [log( + log ]/ log. Jiang and Qi [17, 18] gave the best possible parameters α, β, t 1 and t in (0, 1/ such that the inequalities Q ( αa +(1 αb, αb +(1 αa < M(a, b<q ( βa +(1 βb, βb +(1 βa, Q t1,p(a, b<m(a, b<q t,p(a, b

3 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page of 1 hold for all a, b >0witha b and p 1/, where Q t,p (a, b =C p (ta +(1 tb, tb +(1 taa 1 p (a, b. Inspired by inequalities (1. and(1.4, in this paper, we present the optimal upper and lower bounds for the Neuman-Sándor mean M(a, b in terms of the geometric conve combinations of the first Seiffert mean P(a, b and the quadratic mean Q(a, b. All numerical computations are carried out using Mathematica software. Lemmas In order to establish our main result, we need several lemmas, which we present in this section. Lemma.1 Thedoubleinequality < 1+ sinh 1 (< (.1 holds Proof To show inequality (.1, it suffices to prove that ω 1 (= 1+ sinh 1 ( ( + 5 >0 (. 15 and ω (= 1+ sinh 1 ( ( <0 (. 105 From the epressions of ω 1 (andω (, we get ω 1 (0 = ω (0 = 0, (.4 ω 1 (= ω 1 ( 1+, ω (= ω ( 1+, (.5 where ω1 (=sinh 1 ( ( 1+, ω (=sinh 1 ( ( , 15 ω1 (0 = ω (0 = 0, (.6 ω1 8 4 (= >0 (.7 1+ and ω (= <0 (.8 1+

4 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page 4 of 1 Therefore, inequality (. follows from (.4-(.7, and inequality (. follows from (.4-(.6and(.8. Lemma. The inequality 1+ > [ sinh 1 ( ] holds Proof Let (0, 1, then from (1.wehave M(1 +,1 >A / (1 +,1 Q 1/ (1 +,1. (.9 Therefore, Lemma. followsfrom (.9. Lemma. The inequality 1 arcsin(> 5 (.10 holds for (0, 0.7, and the inequality 1 arcsin(< 5 15 (.11 holds for (0,1,where arcsin( is the inverse sine function. Proof Let ϕ 1 (= 1 arcsin( + + 5, (.1 ϕ (= 1 arcsin( (.1 Then simple computations lead to ϕ 1 (0 = ϕ (0 = 0, (.14 ϕ 1 (= ϕ 1 ( 1, ϕ (= ϕ ( 1, (.15 where ( ϕ1 (= arcsin(, ( ϕ (= + 1 arcsin(. Note that ϕ1 (0 = ϕ (0 = 0, ϕ 1 (0.7= , (.16

5 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page 5 of 1 ϕ1 (= (9 0, (.17 1 ϕ (= 8 4 <0 (.18 1 From (.17 we clearly see that ϕ 1 ( is strictly increasing on (0, 5/10] and strictly decreasing on [ 5/10, 0.7. This in conjunction with (.16impliesthat ϕ1 (>0 (.19 for (0, 0.7. Therefore, inequality (.10 follows from (.1, (.14, (.15 and(.19, and inequality (.11 follows from (.1and(.14-(.16togetherwith(.18. Lemma.4 Let (= Then the inequality 1 1+ sinh 1 ( 1 (1 +. (> (.0 holds for (0, 0.7, and (< (.1 holds Proof To show inequalities (.0and(.1, it suffices to prove that φ 1 (:= ( 1+ [ ( ] sinh 1 ( ( for (0, 0.7, and = 1+ sinh 1 ( ( 1+ ( sinh 1 ( >0 (. φ (:= ( 1+ [ ( ] sinh 1 ( ( = 1+ sinh 1 ( ( 1+ ( sinh 1 ( <0 (.

6 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page 6 of 1 From the epressions of φ 1 (andφ (, one has where φ 1 (0 = φ (0 = 0, (.4 φ 1 (= φ 1 (, φ (= φ (, (.5 φ 1 (=( 1, ( , sinh 1 (, (.6 φ (=( Note that + ( sinh 1 (. ( ,75 8 > ( ( ,75 (0.7 8 = > 0 (.8 for (0, 0.7. Lemma.1 and equations (.6-(.8leadto φ1 (>( 1, ( ,75 8( for (0, 0.7, and = 7 ( 157,11 + 1,151,00 +07, , ,600 8 >0 ( φ (>( ( ( = 5 ( 5+, >0 (.0 15 Therefore, inequality (.followsfrom (.4, (.5and(.9,and inequality (. follows from (.4, (.5and(.0. Lemma.5 Let ϒ(= Then the inequality 1 ( arcsin(. ϒ(> (.1

7 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page 7 of 1 holds for (0, 0.7, and ϒ(< (. holds Proof Let ɛ 1 (:= ( 1+ [ ( ] 4 1 arcsin( ϒ( = 1 arcsin( ( 1+ + ( 1+ ( 4 1 arcsin( (. and ɛ (:= ( 1+ [ ( ] 4 1 arcsin( ϒ( = 1 arcsin( ( 1+ + ( 1+ ( 4 1 arcsin( An easy calculation gives rise to (.4 where ɛ 1 (0 = ɛ (0 = 0, (.5 ɛ 1 (= 90(1 ɛ 1 (, ɛ (= 45(1 ɛ (, (.6 ɛ1 (= ( ( , arcsin(, (.7 ɛ (= ( 1 ( ( arcsin(. (.8 Note that ,15 8 > (0.7 1,15 (0.7 8 = > 0 (.9 for (0, 0.7.

8 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page 8 of 1 It follows from (.10, (.7and(.9that ɛ1 (> ( ( ,15 8( 5 [ ( = 5 1, ,047 ( 1 ] ,15 8 >0 (.40 for (0, 0.7. We claim that ɛ (<0 (.41 Indeed, let q( = ,thenq( = , q( = and ( q (= 4 [8 + 10, ] < Therefore, there eists unique 0 = (0, 1 such that q(>0for (0, 0 andq( 0for[ 0, 1. This in conjunction with (.11and(.8leadsto ɛ (< ( 1 ( ( ( 5 15 [ ( = 5 1,897,05, ,46,644 +,619, ( 1 ( ] <0 5,8 for (0, 0 andɛ ( (1 ( <0for [ 0,1. Therefore, inequality (.1 followsfrom (., (.5, (.6 and(.40, andinequality (. follows from (.-(.6and(.41. Lemma.6 Let μ(= 1+ ( + 1 (1 + [sinh 1 (] (1 + / sinh 1 (. Then μ(<0.for [0.7, 1. Proof Let μ 1 (= 1 1 [sinh 1 (], μ (= 1+ sinh 1 (. Then μ(= μ 1( 1+ + μ (. (.4 (1 + /

9 Gonget al. Journal of Inequalities and Applications 01, 01:55 Page 9 of 1 Lemma. together with > sinh 1 (givesμ 1 (<0and [ μ 1 (= [sinh 1 (] This in turn implies that [ μ1 ( ( sinh 1 ( ] >0 ] = μ 1 ((1 + μ 1 ( (1 + >0 (.4 On the other hand, from the epression of μ (, we get where μ (1= >0, (.44 μ (= (1 + + μ ( / [sinh 1 (], (.45 μ (= 1+ sinh 1 (, (.46 μ (0 = 0, (.47 μ (= <0 (.48 (1 + / From (.44-(.48 we clearly see that μ (<0andμ (>0for (0, 1. This in turn implies that [ μ ( (1 + / ] = μ ((1 + / 1+ μ ( (1 + <0 (.49 Equation (.4 and inequalities (.4and(.49 lead to the conclusion that μ( μ 1(1 + μ (0.7 = < 0. [1 + (0.7 ] / for [0.7, 1. Lemma.7 Let ν(= 1+ ( (1 arcsin ( (1 / arcsin(. Then ν(< 1.48for [0.7, 1. Proof Differentiating ν(yields ν (= ( + arcsin(ν 1 (+(1 ν ( (1 5/ (1 + arcsin, (.50 (

10 Gong et al. Journal of Inequalities and Applications 01, 01:55 Page 10 of 1 where ν 1 (= 1 ( 1+ arcsin(, (.51 ν (= ( [ 1 arcsin( ] ( +. (.5 Equation (.51leads to ν 1 (0.7 = , (.5 ν 1 (= arcsin( 1 <0 (.54 for [0.7, 1. Therefore, ν 1 (<0 (.55 for [0.7, 1 follows from (.5and(.54. It follows from (.5and(.11that ν (< ( ( ( + = 5 ( <0 (.56 7 for [0.7, 1. Equation (.50 together with inequalities (.55and(.56 leads to the conclusion that ν( is strictly decreasing on [0.7, 1. This in turn implies that ν( ν(0.7 = < 1.48 for [0.7, 1. Lemma.8 Let λ 0 =[log(log(1 + + log ]/[ log π log ]= ,and (= (+λ 0 ϒ(, where ( and ϒ( are defined as in Lemmas.4 and.5, respectively. Then the function ( is strictly decreasing on [0.7, 1. Proof Let μ( and ν( bedefinedasinlemmas.6and.7, respectively. Then differentiating (yields (= (+λ 0 ϒ (=μ(+λ 0 ν(< λ 0 = <0 for [0.7, 1. This in turn implies that ( is strictly decreasing on [0.7, 1. Main result Theorem.1 The double inequality P α (a, bq 1 α (a, b<m(a, b<p β (a, bq 1 β (a, b

11 Gong et al. Journal of Inequalities and Applications 01, 01:55 Page 11 of 1 holds for all a, b >0with a bifandonlyifα 1/ and β [ log(log(1 + + log ]/[ log π log ]= Proof Since P(a, b, M(a, b andq(a, b are symmetric and homogeneous of degree 1, without loss of generality, we assume that a > b. Letp (0, 1, λ 0 =[log(log(1 + + log ]/[ log π log ] and =(a b/(a + b. Then (0, 1, P(a, b A(a, b = arcsin(, M(a, b A(a, b = sinh 1 (, Q(a, b A(a, b = 1+, log[q(a, b] log[m(a, b] log[q(a, b] log[p(a, b] = log(1 + log +log[sinh 1 (] log(1 + log +log[arcsin(] (.1 and log(1 + log +log[sinh 1 (] lim 0 + log(1 + log +log[arcsin(] = 1, (. log(1 + log +log[sinh 1 (] lim 1 log(1 + log +log[arcsin(] = λ 0. (. The difference between the conve combination of log[p(a, b], log[q(a, b] and log[m(a, b] is given by p log [ P(a, b ] +(1 p log [ Q(a, b ] log [ M(a, b ] [ ] = p log + 1 p arcsin( log( 1+ [ ] log sinh 1 := D p (. (.4 ( Equation (.4leads to D p ( 0 + =0, D p ( 1 = log [ log(1 + ] p log( π D p (= p 1 arcsin(, D λ0 ( 1 =0, (.5 (1 p ( = (+pϒ(, (.6 1+ sinh 1 ( where (andϒ( are defined as in Lemmas.4 and.5,respectively. From Lemmas.4 and.5, we clearly see that D 1/ (= (+ 1 ϒ( < = for (0, 1, and ( ( <0 (.7 D λ 0 (= (+λ 0 ϒ( > λ ( 4

12 Gong et al. Journal of Inequalities and Applications 01, 01:55 Page 1 of 1 [ (1 λ0 = 4(1 λ ( λ ] := F λ0 (>0 (.8 for (0, 0.7. Note that F λ0 (0 = (1 λ 0 / > 0, F λ0 (0.7 = > 0 (.9 and [( F λ 0 (= 0 1,508,85λ 0 +,4,16 + 4,05,160λ 0 8,07,5λ ] 0 9,450 89,0,500 <0 (.10 for (0, 0.7. Inequalities (.8-(.10lead to the conclusionthat D λ 0 (>0 (.11 for (0, 0.7. It follows from Lemma.8 and (.6 thatd λ 0 ( is strictly decreasing in [0.7, 1. Then from (.11 andd λ 0 (0.7 = together with D λ 0 (1 =, we know that there eists (0.7, 1 such that D λ0 ( is strictly increasing on (0, ] and strictly decreasing on [, 1. This in conjunction with (.5impliesthat D λ0 (>0 (.1 Equations (.4, (.5, (.7and(.1lead to the conclusion that M(a, b<p λ 0 (a, bq 1 λ 0 (a, b (.1 and M(a, b>p 1/ (a, bq 1/ (a, b. (.14 Therefore, Theorem.1 followsfrom (.1 and(.14together with thefollowingstatements: Ifα < 1/,then(.1and(. imply that there eists δ 1 (0, 1 such that M(a, b<p α (a, bq 1 α (a, b for all a, b >0with (a b/(a + b (0, δ 1. Ifβ > λ 0,then(.1and(. imply that there eists δ (0, 1 such that M(a, b>p β (a, bq 1 β (a, b for all a, b >0with (a b/(a + b (1 δ,1. Competing interests The authors declare that they have no competing interests.

13 Gong et al. Journal of Inequalities and Applications 01, 01:55 Page 1 of 1 Authors contributions W-MG provided the main idea and carried out the proof of Theorem.1. X-HS carried out the proof of Lemmas Y-MC carried out the proof of Lemmas All authors read and approved the final manuscript. Author details 1 School of Mathematics and Computation Science, Hunan City University, Yiyang, 41000, China. College of Nursing, Huzhou Teachers College, Huzhou, 1000, China. Acknowledgements This research was supported by the Natural Science Foundation of China under Grants and , and the Natural Science Foundation of Zhejiang Province under Grants LY1H and LY1A Received: 1 July 01 Accepted: 18 October 01 Published: Nov 01 References 1. Seiffert, H-J: Problem 887. Nieuw Arch. Wiskd. 11(, 176 (199. Neuman, E, Sándor, J: On the Schwab-Borchardt mean. Math. Pannon. 14(,5-66 (00. Chu, Y-M, Hou, S-W, Shen, Z-H: Sharp bounds for Seiffert mean in terms of root mean square. J. Inequal. Appl. 01, 11 (01 4. Chu, Y-M, Wang, M-K: Refinements of the inequalities between Neuman-Sándor, arithmetic, contra-harmonic and quadratic means. arxiv:109.90v1 [math.ca] 5. Chu, Y-M, Wang, M-K, Qiu, S-L: Optimal combinations bounds of root-square and arithmetic means for Toader mean. Proc. Indian Acad. Sci. Math. Sci. 1(1, (01 6. Chu, Y-M, Wang, M-K, Wang, Z-K: Best possible inequalities among harmonic, geometric, logarithmic and Seiffert means. Math. Inequal. Appl. 15(, (01 7. Hästö, PA: A monotonicity property of ratios of symmetric homogeneous means. JIPAM. J. Inequal. Pure Appl. Math. (5, Article ID 71 (00 8. Hästö, PA: Optimal inequalities between Seiffert s mean and power mean. Math. Inequal. Appl. 7(1,47-5 ( Gao, S-Q: Inequalities for the Seiffert s means in terms of the identric mean. J. Math. Sci. Adv. Appl. 10(1-, -1 ( Gong, W-M, Song, Y-Q, Wang, M-K, Chu, Y-M: A sharp double inequality between Seiffert, arithmetic, and geometric means. Abstr. Appl. Anal. 01,Article ID ( Jiang, W-D: Some sharp inequalities involving reciprocals of the Seiffert and other means. J. Math. Inequal. 6(4, (01 1. Liu, H, Meng, J-X: The optimal conve combination bounds for Seiffert s mean. J. Inequal. Appl. 011, Article ID ( Neuman, E: A note on a certain bivariate mean. J. Math. Inequal. 6(4, ( Neuman, E, Sándor, J: On certain means of two arguments and their etension. Int. J. Math. Math. Sci. 16, ( Neuman, E, Sándor, J: On the Schwab-Borchardt mean II. Math. Pannon. 17(1, ( Li, Y-M, Long, B-Y, Chu, Y-M: Sharp bounds for the Neuman-Sándor mean in terms of generalized logarithmic mean. J. Math. Inequal. 6(4, ( Jiang, W-D, Qi, F: Sharp bounds for Neuman-Sándor mean in terms of the root-mean-square. arxiv:101.67v1 [math.ca] 18. Jiang, W-D, Qi, F: Sharp bounds in terms of a two-parameter family of means for Neuman-Sándor mean. Preprint /109-4X Cite this article as: Gong et al.: Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means. Journal of Inequalities and Applications 01, 01:55