Research Article Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means

Transcription

1 Abstract and Applied Analysis Volume 01, Article ID 486, 1 pages Research Article Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means Tie-Hong Zhao, 1 Yu-Ming Chu, Yun-Liang Jiang, and Yong-Min Li 4 1 Department of Mathematics, Hangzhou Normal University, Hangzhou 1006, China School of Mathematics and Computation Sciences, Hunan City University, Yiyang 41000, China School of Information & Engineering, Huzhou Teachers College, Huzhou 1000, China 4 School of Automation, Southeast University, Nanjing 10096, China Correspondence should be addressed to Yu-Ming Chu; chuyuming005@yahoo.com.cn Received 19 January 01; Accepted 1 February 01 Academic Editor: Khalil Ezzinbi Copyright 01 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 I α 1 (a, b)q 1 α 1 (a, b) < M(a, b) < I β 1 (a, b)q 1 β 1 (a, b), I α (a, b)c 1 α (a, b) < M(a, b) < I β (a, b)c 1 β (a, b) hold for all a, b > 0 with a =bif and only if α 1 1/, β 1 log[ log(1 + )]/(1 log ), α 5/7,and β log[ log(1+ )],I(a, b), M(a, b), Q(a, b),andc(a, b) are the identric, Neuman-Sándor, quadratic, and contraharmonic means of a and b,respectively. 1. Introduction For p Rand 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),andpth power mean M p (a, b) are defined by M (a, b) = I (a, b) = 1 e ( bb 1/(b a) a a ), a b sinh 1 [(a b) / (a+b)], 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) = ab/(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)/,andt(a, b) = (a b)/[ 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 Q (a, b) = a +b, C(a, b) = a +b a+b, {( +b p 1/p ), M p (a, b) = p=0, { { ab, p = 0, (1) H (a, b) =M 1 (a, b) <G(a, b) =M 0 (a, b) <L(a, b) <P(a, b) <I(a, b) <A(a, b) <M 1 (a, b) <M(a, b) <T(a, b) <Q(a, b) =M (a, b) <C(a, b) () respectively, sinh 1 (x) = log(x+ 1+x ) is the inverse hyperbolic sine function. hold for all a, b > 0 with a =b.

2 Abstract and Applied Analysis Neuman and Sándor [1, 8]establishedthat A (a, b) A (a, b) <M(a, b) < log (1 + ), π T (a, b) <M(a, b) <T(a, b), 4 M (a, b) < A (a, b) P (a, b), A (a, b) T (a, b) <M(a, b) < A (a, b) +T (a, b), A (a, b) +Q(a, b) M (a, b) < for all a, b > 0 with a =b. Let 0<a,b 1/with a =b, a =1 a,andb =1 b. ThentheKyFaninequalities 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) M (a, b) T (a, b) < < M(a,b ) T(a,b ) 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[0] and Zhao et al. [1]proved that the inequalities αq (a, b) + (1 α) A (a, b) <M(a, b) <βq(a, b) +(1 β)a(a, b), λc (a, b) + (1 λ) A (a, b) <M(a, b) <μc(a, b) +(1 μ)a(a, b), α 1 H (a, b) +(1 α 1 )Q(a, b) <M(a, b) <β 1 H (a, b) +(1 β 1 )Q(a, b), α C (a, b) +(1 α )Q(a, b) <M(a, b) <β C (a, b) +(1 β )Q(a, b) hold for all a, b > 0 with a =bif and only if α [1 log(1 + )]/[( 1)log(1 + )], β 1/, λ [1 log(1 + )]/ log(1 + ), μ 1/6, α 1 /9, β 1 1 1/[ log(1 + )], α 1/,andβ 1 1/[ log(1 + )]. In [], Chu and Long gave the best possible constants p, q, α, andβ such that the double inequalities M p (a, b) < M(a, b) < M q (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 I(a,b ) =(a a b b ) 1/(a+b), (6) I (a, b) () (5) which has been investigated in [ 5]. Alzer [6]provedthat the inequalities A (a, b) G (a, b) < I (a, b) L (a, b) < I (a, b) +L(a, b) < A (a, b) +G(a, b) (7) hold for all a, b > 0 with a =b. The following sharp bounds for I, (IL) 1/,and(I + L)/ in terms of the power mean and the convex combination of arithmetic and geometric means are given in [7]as M / (a, b) <I(a, b) <M log (a, b), M 0 (a, b) < I (a, b) L (a, b) <M 1/ (a, b), M log /(1+log ) (a, b) I (a, b) +L(a, b) < A (a, b) + 1 G (a, b) <M 1/ (a, b), <I(a, b) < e A (a, b) +(1 )G(a, b) e for all a, b > 0 with a =b. Chu et al. [8] presented the optimal constants α 1,β 1,α, and β such that the double inequalities α 1 Q (a, b) +(1 α 1 )A(a, b) < π/ π a cos θ+b sin θdθ 0 <β 1 Q (a, b) +(1 β 1 )A(a, b), Q α (a, b) A 1 α (a, b) < π/ π a cos θ+b sin θdθ 0 <Q β (a, b) A 1 β (a, b) hold for all a, b > 0 with a =b. The aim of this paper is to find the best possible constants α 1,β 1,α and β such that the double inequalities I α 1 (a, b) Q 1 α 1 (a, b) <M(a, b) <I β 1 (a, b) Q 1 β 1 (a, b), (8) (9) I α (a, b) C 1 α (a, b) <M(a, b) <I β (a, b) C 1 β (a, b) (10) hold for all a, b > 0 with a =b. All numerical computations are carried out using mathematica software.. Lemmas In order to prove our main results, we need several lemmas, which we present in this section.

3 Abstract and Applied Analysis Lemma 1. The double inequality x+ x 15 < 1+x sinh 1 (x) <x+ x x7 105 (11) holds Proof. To prove Lemma 1,itsufficestoprovethat η 1 (x) = 1+x sinh 1 (x) (x+ x )>0, (1) 15 η (x) = 1+x sinh 1 (x) (x+ x x7 10<0 From the expressions of η 1 (x) and η (x),weget η 1 (x) = xη 1 (x) 1+x, η 1 (0) =η (0) =0, (1) η (x) = xη (x) 1+x, (14) η 1 (x) = sinh 1 (x) (x x ) 1+x, η (x) = sinh 1 (x) (x x + 8x5 1 1+x, η 1 (0) =η (0) =0, (15) η 8x 4 1 (x) = >0, (16) 1+x η 16x 6 (x) = <0, (17) 5 1+x Therefore, inequality (1) followsfrom(14) (16), and inequality (1)followsfrom(14) (17). Lemma. Let Proof. To prove inequalities (19) and(0), it suffices to show that ι 1 (x) for x (0, 1),and ι (x) := log [ (1+x)1+x (1 x) 1 x ] x xlog (1 x ) ( x + x5 5 + x7 >0 := log [ (1+x)1+x (1 x) 1 x ] x xlog (1 x ) ( x + x5 5 + x7 7 +x9 )<0 (1) () for x (0, /4). From (1)and(), one has ι 1 (0 + )=ι (0 + )=0, () for x (0, 1),and ι 1 (x) = x8 1 x >0 (4) ι 9x8 (x) = 1 x (7 9 x ) <0 (5) for x (0, /4). Therefore, inequality (1) follows from () and(4), and inequality () follows from ()and(5). Lemma. Let 1 Φ 1 (x) = 1+x sinh 1 (x) 1 x(1+x ). (6) Then the double inequality x 4x 45 + x5 <Φ 1 (x) < x 4x x5 (7) 5 holds for x (0, 0.7). Proof. To prove inequality (7), it suffices to show that φ 1 (x) =x 1+x sinh 1 (x) L (x) = log [ (1+x)1+x (1 x) 1 x ] x xlog (1 x ). (18) x(1+x ) sinh 1 (x) ( x 4x 45 + x5 ) (8) Then L (x) > x + x5 5 + x7 7 for x (0, 1),and (19) L (x) < x + x5 5 + x7 7 +x9 (0) for x (0, /4). >0, φ (x) =x 1+x sinh 1 (x) <0 for x (0, 0.7). x(1+x ) sinh 1 (x) ( x 4x x5 (9)

4 4 Abstract and Applied Analysis First, we prove inequality (8). From the expression of φ 1 (x),wehave φ 1 (0) =0, φ 1 (0.7) = 0.00, (0) φ 1 (x) = φ 1 (x) = 10x + 8x + x 5 45x 7 Equation ()leadsto xφ 1 (x) 90 1+x, (1) (60 16x 69x x 6 ) 1+x sinh 1 (x). () φ 1 (0.6) =.017, φ 1 (0.7) =.551, φ xφ 1 (x) = 1 (x) () 1+x, φ 1 (x) = 56x 09x + 4x x 7 Note that + (8 4x + 75x x 6 ) 1+x sinh 1 (x). (4) 60 16x 69x x 6 >0 (5) for x (0, 0.6],and φ 1 (x) > 56x 09x + 4x x 7 + (8 4x + 75x x 6 ) (x+ x 1=x 15 ( x x x x 8 ) > x 15 [ (0.6) (0.6) 4 ] = x 975 >0 (8) for x [0.6, 0.7). From (), (7), and (8), we clearly see that there exists x 1 (0.6, 0.7) such that φ 1 (x) > 0 for x (0,x 1) and φ 1 (x) < 0 for x (x 1, 0.7). Then(1) leadstotheconclusionthat φ 1 (x) is strictly increasing on (0, x 1 ] and strictly decreasing on [x 1,0.7). Therefore, inequality (8) follows from (0) and the piecewise monotonicity of φ 1 (x). Next, we prove inequality (9). From the expression of φ (x),weget φ (0) =0, φ (x) = xφ (x) 45 1+x, (9) φ (x) = x (18x6 +x 4 x 0) for x (0, 0.6],and +(15 4x +x 4 + 7x 6 ) (40) 8 4x + 75x x 6 >0 (6) for x [0.6, 0.7). It follows from ()and(4) (6)together with Lemma 1 that φ 1 (x) > 10x + 8x + x 5 45x 7 1+x sinh 1 (x). It follows from Lemma 1 and (40)that φ (x) > x (18x6 +x 4 x 0) +(15 4x +x 4 + 7x 6 )(x+ x 1 (41) (60 16x 69x x 6 )(x+ x ) = x5 ( x 60x 4 ) x5 = 10078x [ ] >0 (7) = x5 15 ( x + 708x 4 88x 6 )>0 for x (0, 0.7). Therefore, inequality (9) follows from (9) together with (41). Lemma 4. Let Φ (x) = 1 1+x sinh 1 (x) 1 x x(1+x ). (4)

5 Abstract and Applied Analysis 5 Then the double inequality for x (0, 0.66),and 5x 79x x5 10 <Φ (x) < 5x 79x x5 5 holds for x (0, /4). (4) 4 108x + 1x x 6 > ( 4 ) + 1 (0.66) 4 (5) Proof. To prove Lemma 4,itsufficestoprovethat + 96 (0.66) 6 = > 0 φ 1 (x) := x 1+x (1 x ) sinh 1 (x) x(1+x ) sinh 1 (x) ( 5x 79x x5 10 )>0, φ (x) := x 1+x (1 x ) sinh 1 (x) (44) x(1+x ) sinh 1 (x) ( 5x 79x x5 <0 (45) for x (0, /4). We first prove inequality (44). From the expression of φ 1 (x),weobtain φ 1 (0) =0, φ 1 ( ) = > 0, (46) 4 φ 1 (x) = φ 1 (x) = 10x + 8x + 59x 5 99x 7 xφ 1 (x) 90 1+x, (47) (60 16x 177x x 6 ) 1+x sinh 1 (x). (48) Equation (48)leadsto φ 1 (0.66) = 6.0 > 0, φ 1 ( ) = < 0, 4 (49) φ xφ 1 (x) = 1 (x) 1+x, (50) φ 1 (x) = 56x 705x + 86x x 7 Note that + 14 (4 108x + 1x x 6 ) 1+x sinh 1 (x) x 177x x 6 >60 16 (0.66) 177 (0.66) 4 = > 0 (51) (5) for x [0.66, /4). It follows from Lemma 1,(48), and (51) (5)that φ 1 (x) > 10x + 8x + 59x 5 99x 7 (60 16x 177x x 6 ) (x+ x x7 10 = x5 105 [ x 40x x x 6 (1 x )] > x5 105 [ (0.66) 40 (0.66) 4 ] = x > 0 (54) for x (0, 0.66),and φ 1 (x) > 56x 705x + 86x x (4 108x + 1x x 6 )(x+ x 1 = x 15 [ x + 169x x x 6 (1 x )] > x 15 [ (0.66) +169 (0.66) 4 ] = x > 0 (55) for x [0.66, /4). From (50) and (55), we know that φ 1 (x) is strictly decreasing on [0.66, /4), and this in conjunction with (49) and(54) leadstotheconclusionthat there exists x 1 (0.66, /4) such that φ 1 (x) > 0 for x (0,x 1 ) and φ 1 (x) < 0 for x (x 1,/4).Then(47) implies that φ 1 (x) is strictly increasing on (0, x 1 ] and strictly decreasing on [x 1,/4). Therefore, inequality (44) follows from (46) and the piecewise monotonicity of φ 1 (x).

6 6 Abstract and Applied Analysis Next, we prove inequality (45). From the expression of φ (x) one has φ (0) =0, φ (x) = xφ (x) 45 1+x, (56) = x 5 5 (1 + x ) (1 x )>0, Υ 1 (x) ( 4x 4x 5 + 8x5 < 1 x (x + x5 5 + x7 7 +x9 ) φ (x) = 60x 4x +x x 7 +4(15 4x +x x 6 ) (57) + x 1+x (4x 4x 5 + 8x5 = x7 (1 x ) (1+x ) <0 (61) 1+x sinh 1 (x). It follows from Lemma 1 and (5)that φ (x) > 60x 4x +x x 7 +4(15 4x +x x 6 )(x+ x 1 for x (0, 0.7). Therefore, Lemma 5 follows easily from (61). Lemma 6. Let L(x) be defined as in Lemma and L (x) Υ (x) = x + x 1+x. (6) Then the double inequality = x5 15 ( x + 16x 4 196x 6 )>0 (58) for x (0, /4). Therefore, inequality (45) followsfrom(56) togetherwith (58). Lemma 5. Let L(x) be defined as in Lemma and Υ 1 (x) = Then the double inequality L (x) x + x 1+x. (59) 4x 4x 5 + 4x5 5 <Υ 1 (x) < 4x 4x 5 + 8x5 7 holds for x (0, 0.7). Proof. From Lemma,onehas Υ 1 (x) ( 4x 4x 5 + 4x5 > 1 x (x + x5 5 + x7 + x 1+x ( 4x 4x 5 + 4x5 (60) 7x 9x 5 + 7x5 5 <Υ (x) < 7x 9x x5 7 holds for x (0, /4). Proof. It follows from Lemma that Υ (x) ( 7x 9x 5 + 7x5 > 1 x (x + x5 5 + x7 = + x 1+x (7x 9x 5 + 7x5 44x 5 5 (1 + x ) (1 x )>0, Υ (x) ( 7x 9x x5 < 1 x (x + x5 5 + x7 7 +x9 ) + x 1+x (7x 9x x5 = x7 ( x ) (1+x ) <0 for x (0, /4). Therefore, Lemma 6 follows from (64). (6) (64)

7 Abstract and Applied Analysis 7 Lemma 7. The inequality x 1+x >[sinh 1 (x)] (65) holds Proof. Let Then ζ (x) = x 1+x [sinh 1 (x)]. (66) ζ (x) = ζ (0) =0, ζ 1 (x) (1 + x ), (67) / ζ 1 (x) =x ( + x ) [ 1+x sinh 1 (x)]. (68) It follows from Lemma 1 and (68)that ζ 1 (x) >x ( + x ) (x+ x x7 10 =x 6 [ ( x x6 67 (1 x )+ x6 75 ]>0 (69) Therefore, Lemma 7 follows from (67)together with(69). Lemma 7 and x>sinh 1 (x) give ω 1 (x) < 0 and ω 1 (x) = x [sinh 1 (x)] [ x 1+x (sinh 1 (x)) ]>0 This in turn implies that (7) [ ω 1 (x) 1+x ] = ω 1 (x) (1 + x ) xω 1 (x) (1 + x ) >0 (74) On the other hand, from the expression of ω (x),weget ω (1) = > 0, ω x (x) = (1 + x ) + ω (x) / [sinh 1 (x)], (75) ω (x) = x 1+x sinh 1 (x), ω (0) =0, ω (x) = x <0 (1 + x / ) (76) From (75) (76), we clearly see that ω (x) < 0 and ω (x) > 0 This in turn implies that [ ω (x) (1 + x ) ] / = ω (x) (1 + x ) / x 1+x ω (x) (1 + x ) (77) Lemma 8. Let μ 1 (x) = 1+x (x + x ) 1 (1 + x )[sinh 1 (x)] x (1 + x ) / sinh 1 (x). Then μ 1 (x) < 0. for x [0.7, 1). (70) <0 Equation (7) together with inequalities (74) and(77) lead to the conclusion that μ 1 (x) ω 1 (1) ω + (0.7) [1 + (0.7) ] / (78) Proof. Let = < 0. Then ω 1 (x) = 1 x 1 [sinh 1 (x)], ω (x) = 1+x x sinh 1 (x). (71) μ 1 (x) = ω 1 (x) 1+x + ω (x) (1 + x ). / (7) for x [0.7,1). Lemma 9. Let μ (x) = 1+4x x 4 1 (x+x ) (1 + x )[sinh 1 (x)] x (1 + x ) / sinh 1 (x). Then μ (x) < 0.51 for x [0.65, 1). (79)

8 8 Abstract and Applied Analysis Proof. Let τ 1 (x) = 1 x 1 [sinh 1 (x)] =μ 1 (x), (80) τ (x) = x 1+x x sinh 1 (x), then μ (x) = τ 1 (x) 1+x + τ (x) (1 + x ). / (81) From (74), we clearly see that [ τ 1 (x) 1+x ] =[ ω 1 (x) 1+x ] >0 (8) On the other hand, from the expression of τ (x) together with Lemma 1,weget τ (1) = > 0, τ (x) = 1 sinh 1 (x) xτ (x) (1 + x ) / [sinh 1 (x)], τ (x) =(5+x )[sinh 1 (x)] (1+x ), τ τ (x) =x[sinh 1 (x)] (0.65) = 0.60, +[ 5+x 1+x 1+x sinh 1 (x) x]>0 (8) From (8), we clearly see that τ (x) < 0 and τ (x) > 0 for x [0.65, 1). This in turn implies that [ τ (x) (1 + x ) ] = τ (x) (1 + x ) / x 1+x τ (x) / (1 + x ) <0 (84) for x [0.65, 1). Equation (81) together with inequalities (8) and(84) lead to the conclusion that μ (x) τ 1 (1) τ + (0.65) = 0.50 < 0.51 [1 + (0.65) ] / (85) for x [0.65, 1). Lemma 10. Let L(x) be defined as in Lemma and (1+x 4 ) L (x) ] 1 (x) = (1 x )(1+x ) x. (86) Then ] 1 (x) > 1. for x [0.7, 1). Proof. Differentiating ] 1 (x) yields ] L (x) 1 (x) = x 4 +8x 0x 4 6x 8 x(1 x ) (1 + x ). (87) It follows from (19)and(87)that ] 1 (x) > 1 x [ ( + x 5 + x4 +8x 0x 4 6x 8 (1 x ) (1 + x ) ] = x ( x 97x x 6 + 6x 8 + 6x x 1 ) 5(1 x ) (1 + x ) > x 5(1 x ) (1 + x ) [ (0.7) x 4 (x ] x > 5(1 x ) (1 + x ) >0 (88) for x [0.7,1). Therefore, ] 1 (x) ] 1 (0.7) = 1.14 > 1. for x [0.7, 1) follows from (88). Lemma 11. Let L(x) be defined as in Lemma and x +x 4 L (x) ] (x) = (1 x )(1+x ) x. (89) Then ] (x) > 1.8 for x [0.65, 1). Proof. Differentiating ] (x) yields ] L (x) (x) = x 4 (1+7x 17x 4 +5x 6 4x 8 ) x(1 x ) (1 + x ). (90) It follows from (19) and(90) together with the monotonicity of the function 561x 7x 4 on [0.65, 1) that ] (x) > 1 x [ ( + x 5 + x4 (1+7x 17x 4 +5x 6 4x 8 ) ) 7 (1 x ) (1 + x ) ] = x ( x 7x x 6 + 6x 8 + 6x x 1 ) 5(1 x ) (1 + x ) > x [ (0.65) 7 (0.65) (0.65) 6 ] 5(1 x ) (1 + x ) x 7. = 5(1 x ) (1 + x ) >0 (91) for x [0.65, 1). Equation (91) leads to the conclusion that ] (x) ] (0.65) = 1.89 > 1.8 for x [0.65, 1).

9 Abstract and Applied Analysis 9 Lemma 1. Let Φ 1 (x) and Υ 1 (x) be defined, respectively, as in Lemmas and 5, andθ 1 (x; p) = Φ 1 (x) pυ 1 (x). Then Θ 1 (x; p) is strictly decreasing on [0.7, 1) if p>1/6. Proof. Differentiating Θ 1 (x; p) with respect to x and making use of Lemmas 8 and 10,weget dθ 1 (x; p) dx =Φ 1 (x) pυ 1 (x) =μ 1 (x) p] 1 (x) < 0. 1 (9) 6 1. = 0 for x [0.7, 1) and p>1/6. This in turn implies that Θ 1 (x; p) is strictly decreasing on [0.7, 1) if p>1/6. Lemma 1. Let Φ (x) and Υ (x) be defined, respectively, as in Lemmas 4 and 6, andθ (x; q) = Φ (x) qυ (x). Then Θ (x; q) is strictly decreasing on [0.65, 1) if q>/5. Proof. Differentiating Θ (x; q) with respect to x and making use of Lemmas 9 and 11,wehave dθ 1 (x; q) dx =Φ (x) qυ (x) =μ (x) q] (x) < 0.51 (9) = 0.04 < 0 for x [0.65, 1) and q > /5. This in turn implies that Θ (x; q) is strictly decreasing on [0.65, 1) if q>/5.. Main Results Theorem 14. The double inequality I α 1 (a, b) Q 1 α 1 (a, b) <M(a, b) <I β 1 (a, b) Q 1 β 1 (a, b) (94) we assume that a>b.letp (0, 1), x = (a b)/(a + b),and λ 1 = log[ log(1 + )]/(1 log ).Thenx (0, 1),and I (a, b) A (a, b) = 1 [(1+x)1+x 1/x e (1 x) 1 x ], M (a, b) A (a, b) = x sinh 1 (x), log [Q (a, b)] log [M (a, b)] log [Q (a, b)] log [I (a, b)] = lim x 0 + lim x 1 Q (a, b) A (a, b) = 1+x, (95) log 1+x log x+log [sinh 1 (x)] log 1+x log [(1+x) 1+x /(1 x) 1 x ]/(x) +1, log 1+x log x+log [sinh 1 (x)] log 1+x log [(1+x) 1+x /(1 x) 1 x ]/(x) +1 (96) = 1, (97) log 1+x log x+log [sinh 1 (x)] log 1+x log [(1+x) 1+x /(1 x) 1 x ]/(x) +1 =λ 1. (98) 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)] = p x log [(1+x)1+x ] p 1 x (1 x) +(1 p)log 1+x x log [ sinh 1 (x) ]:=D p (x). Equation (99)leadsto D p (0 + )=0, (99) D p (1 )=log [ log (1 + )] p (1 log ), D λ1 (1 )=0, (100) holds for all a, b > 0 with a =b if and only if β 1 log[ log(1 + )]/(1 log ) = 0.7 and α 1 1/. D p (x) = 1+px x+x + 1 L (x) 1+x sinh 1 (x) x =Φ 1 (x) pυ 1 (x) =Θ 1 (x; p), (101) Proof. Since I(a, b), M(a, b), andq(a, b) are symmetric and homogeneous of degree one, then without loss of generality, L(x), Φ 1 (x), Υ 1 (x), and Θ 1 (x; p) are defined as in Lemmas,, 5,and1,respectively.

10 10 Abstract and Applied Analysis It follows from (101) together with Lemmas and 5 that D 1/ (x) < x 4x x5 5 1 (4x 4x 5 + 4x5 (10) = x 5 (8 9 x )<0 for x (0, 0.7). Moreover,weseeclearly,fromLemma 1, that D 1/ (x) is strictly decreasing on [0.7, 1) and so D 1/ (x) < D 1/ (0.7) = < 0 for x [0.7, 1). Thisin conjunction with (100)and(10)impliesthat D 1/ (x) <0 (10) On the other hand, (101) andlemmas and 5 together with the monotonicity of the function (17 18λ 1 )x /45 + (7 16λ 1 )x 4 /14 on (0, 0.7) lead to D λ 1 (x) > x 4x 45 + x5 λ 1 ( 4x 4x 5 + 8x5 =x[ (1 λ 1) >x[ (1 λ 1) λ 1 14 (17 18λ 1) 45 (17 18λ 1) 45 (0.7) 4 ] = ( λ 1)x >0 x λ 1 x 4 ] 14 (0.7) (104) for x (0, 0.7). It follows from Lemma 1 that D λ 1 (x) is strictly decreasing on [0.7, 1).Notethat D λ 1 (0.7) = 0.09 > 0, D λ 1 (1 )=. (105) From (104) and(105) together with the monotonicity of D λ 1 (x) on [0.7, 1), we clearly see that there exists c 1 (0.7, 1) such that D λ1 (x) is strictly increasing on (0, c 1 ] and strictly decreasing on [c 1,1). This in conjunction with (100) implies that D λ1 (x) >0 (106) Equation (99) together with inequalities (10) and(106) gives rise to M (a, b) >I 1/ (a, b) Q 1/ (a, b), (107) M (a, b) <I λ 1 (a, b) Q 1 λ 1 (a, b). Therefore, Theorem 14 follows from (107) togetherwith the following statements. (i) If α 1 <1/,then(96)and(97) imply that there exists δ 1 (0, 1) such that M(a, b) < I α 1 (a, b)q 1 α 1 (a, b) for all a, b > 0 with (a b)/(a + b) (0, δ 1 ). (ii) If β 1 >λ 1,then(96) and(98) imply that there exists δ (0, 1) such that M(a, b) > I β 1 (a, b)q 1 β 1 (a, b) for all a, b > 0 with (a b)/(a + b) (1 δ,1). Theorem 15. The double inequality I α (a, b) C 1 α (a, b) <M(a, b) <I β (a, b) C 1 β (a, b) (108) holds for all a, b > 0 with a =b ifandonlyifα 5/7and β log[ log(1 + )] = Proof. We will follow the same idea in the proof of Theorem 14. SinceI(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.letq (0, 1), λ = log[ log(1 + )], and x = (a b)/(a + b).thenx (0, 1). Making use of (95)togetherwithC(a, b)/a(a, b) = 1 + x gives log [C (a, b)] log [M (a, b)] log [C (a, b)] log [I (a, b)] lim x 0 + lim x 1 = log (1 + x ) log x+log [sinh 1 (x)] log (1 + x ) log [(1+x) 1+x /(1 x) 1 x ]/(x) +1, log (1 + x ) log x+log [sinh 1 (x)] log (1 + x ) log [(1+x) 1+x /(1 x) 1 x ]/(x) +1 (109) = 5 7, (110) log (1 + x ) log x+log [sinh 1 (x)] log (1 + x ) log [(1+x) 1+x /(1 x) 1 x ]/(x) +1 =λ. (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)] = q log [(1+x)1+x x (1 x) 1 x ] q+(1 q)log (1 + x ) x log [ sinh 1 (x) ]:=E q (x). (11)

11 Abstract and Applied Analysis 11 Equation (11)leadsto E q (0 + )=0, E q (1 )=log [ log (1 + )] q, E λ (1 )=0, (11) such that E λ (x) is strictly increasing on (0, c ] and strictly decreasing on [c,1). This in conjunction with (11) implies that E λ (x) >0 (119) E q (x) = 1 x +qx 1 L (x) x+x + 1+x sinh 1 (x) x =Φ (x) qυ (x) =Θ (x; q), (114) L(x), Φ (x), Υ (x), and Θ (x; q) are defined as in Lemmas, 4, 6,and1,respectively. It follows from Lemmas 4, 6, and1 together with (114) that E 5/7 (x) <( 5x 79x x5 5 7 (7x 9x 5 + 7x5 = 4x 5 (7 6 x )<0 (115) for x (0, 0.65) and E 5/7 (x) is strictly decreasing on [0.65, 1). Thus, we have E 5/7 (x) < E 5/7 (0.65) = for x [0.65, 1). This in conjunction with (11)and(115)impliesthat E 5/7 (x) <0 (116) On the other hand, Lemmas 4, 6, and1 together with (114)leadto E λ (x) >( 5x 79x x5 10 ) λ ( 7x 9x x5 =x[ 5 7λ >x[ 5 7λ 79 81λ λ λ (0.65) 4 ] x 150λ 77 x 4 ] 70 (0.65) = λ x> (117) for x (0, 0.65) and E λ (x) is strictly decreasing on [0.65, 1). Note that E λ (0.65) = , E λ (1 ) =. (118) From (117)and(118) together with the monotonicity of E λ (x) on [0.65, 1), we clearly see that there exists c (0.65, 1) Equation (11) together with inequalities (116) and(119) lead to the conclusion that M (a, b) >I 5/7 (a, b) C /7 (a, b), M (a, b) <I λ (a, b) C 1 λ (a, b). (10) Therefore, Theorem 15 follows from (10) together with the following statements. (i) If α <5/7,then(109)and(110) imply that there exists δ (0, 1) such that M(a, b) < I α (a, b)c 1 α (a, b) for all a, b > 0 with (a b)/(a + b) (0, δ ). (ii) If β >λ,then(109)and(111) imply that there exists δ 4 (0, 1) such that M(a, b) > I β (a, b)c 1 β (a, b) for all a, b > 0 with (a b)/(a + b) (1 δ 4,1). Acknowledgments ThisresearchwassupportedbytheNaturalScienceFoundation of China under Grants , , and and the Natural Science Foundation of Zhejiang Province under Grants Z and LY1F001. References [1] E. Neuman and J. Sándor, On the Schwab-Borchardt mean, Mathematica Pannonica,vol.14,no.,pp.5 66,00. [] J. Chen and B. Hu, The identric mean and the power mean inequalities of Ky Fan type, Facta Universitatis, no.4,pp.15 18, [] J. Sándor, A note on some inequalities for means, Archiv der Mathematik,vol.56,no.5,pp ,1991. [4] J. Sándor, On certain inequalities for means, Mathematical Analysis and Applications,vol.189,no.,pp , [5] J. Sándor, On refinements of certain inequalities for means, Archivum Mathematicum,vol.1,no.4,pp.79 8,1995. [6] J. Sándor, Two inequalities for means, International Mathematics and Mathematical Sciences,vol.18,no.,pp.61 6, [7] T. Trif, On certain inequalities involving the identric mean in n variables, Universitatis Babeş-Bolyai,vol.46,no.4,pp , 001. [8] E. Neuman and J. Sándor, On the Schwab-Borchardt mean. II, Mathematica Pannonica,vol.17,no.1,pp.49 59,006. [9] L. Zhu, New inequalities for means in two variables, Mathematical Inequalities & Applications, vol.11,no.,pp.9 5, 008. [10] L. Zhu, Some new inequalities for means in two variables, Mathematical Inequalities & Applications,vol.11,no.,pp , 008.

12 1 Abstract and Applied Analysis [11] O. Kouba, New bounds for the identric mean of two arguments, Inequalities in Pure and Applied Mathematics, vol.9,no.,article71,6pages,008. [1] M.-K. Wang, Y.-M. Chu, and Y.-F. Qiu, Some comparison inequalities for generalized Muirhead and identric means, Inequalities and Applications, vol.010,articleid 9560, 10 pages, 010. [1] Y.-M. Chu, M.-K. Wang, and Z.-K. Wang, A sharp double inequality between harmonic and identric means, Abstract and Applied Analysis, vol. 011, Article ID 65795, 7 pages, 011. [14] Y.-F. Qiu, M.-K. Wang, Y.-M. Chu, and G.-D. Wang, Two sharp inequalities for Lehmer mean, identric mean and logarithmic mean, Mathematical Inequalities, vol. 5, no., pp , 011. [15] S. Gao, Inequalities for the Seiffert s means in terms of the identric mean, Mathematical Sciences,vol.10,no.1-, pp. 1,011. [16] M.-K. Wang, Z.-K. Wang, and Y.-M. Chu, An optimal double inequality between geometric and identric means, Applied Mathematics Letters,vol.5,no.,pp ,01. [17] Y.-M. Chu, S.-W. Hou, and Z.-H. Shen, Sharp bounds for Seiffert mean in terms of root mean square, Inequalities and Applications,vol.01,article11,6pages,01. [18] Y.-M. Chu and S.-W. Hou, Sharp bounds for Seiffert mean in terms of contraharmonic mean, Abstract and Applied Analysis, vol.01,articleid45175,6pages,01. [19]Y.M.Li,B.Y.Long,andY.M.Chu, Sharpboundsforthe Neuman-Sándor mean in terms of generalized logarithmic mean, Mathematical Inequalities, vol. 6, no. 4, pp , 01. [0] E. Neuman, A note on a certain bivariate mean, Mathematical Inequalities,vol.6,no.4,pp.67 64,01. [1] T. H. Zhao, Y. M. Chu, and B. Y. Liu, Optimal bounds for Neuman-Sándor mean in terms of the convex combinations of harmonic, geometric, quadratic, and contraharmonic means, Abstract and Applied Analysis, vol. 01, Article ID 065, 9 pages, 01. [] Y. M. Chu and B. Y. Long, Bounds of the Neuman-Sándor mean using power and identric means, Abstract and Applied Analysis,vol.01,ArticleID8591,6pages,01. [] J. Sándor, On the identric and logarithmic means, Aequationes Mathematicae,vol.40,no.-,pp.61 70,1990. [4] J.Sándor, On certain identities for means, Universitatis Babeş- Bolyai,vol.8,no.4,pp.7 14,199. [5] J. Sándor and I. Raşa, Inequalities for certain means in two arguments, Nieuw Archief voor Wiskunde, vol. 15, no. 1-, pp , [6] H. Alzer, Ungleichungen für Mittelwerte, Archiv der Mathematik,vol.47,no.5,pp.4 46,1986. [7] H. Alzer and S.-L. Qiu, Inequalities for means in two variables, Archiv der Mathematik,vol.80,no.,pp.01 15,00. [8] Y. M. Chu, M. K. Wang, and S. L. Qiu, Optimal combinations bounds of root-square and arithmetic means for Toader mean, The Proceedings of the Indian Academy of Sciences, vol.1,no. 1, pp , 01.

13 Advances in Operations Research Advances in Decision Sciences Mathematical Problems in Engineering Algebra Probability and Statistics The Scientific World Journal International Differential Equations Submit your manuscripts at International Advances in Combinatorics Mathematical Physics Complex Analysis International Mathematics and Mathematical Sciences Stochastic Analysis Abstract and Applied Analysis International Mathematics Discrete Dynamics in Nature and Society Discrete Mathematics Applied Mathematics Function Spaces Optimization