AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 684834 10.1155/2012/684834 684834 Research Article A Sharp Double Inequality between Seiffert, Arithmetic, and Geometric Means Gong Wei-Ming 1 Song Ying-Qing 1 Wang Miao-Kun 2 Chu Yu-Ming 2 Diblík Josef 1 College of Mathematics and Computation Science Hunan City University Yiyang 413000 China hncu.net 2 Department of Mathematics Huzhou Teachers College Huzhou 313000 China hutc.zj.cn 2012 2 9 2012 2012 02 07 2012 21 08 2012 2012 Copyright © 2012 Wei-Ming Gong 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.

For fixed s1 and any t1,t2(0,1/2) we prove that the double inequality Gs(t1a+(1-t1)b,t1b+(1-t1)a)A1-s(a,b)<P(a,b)<Gs(t2a+(1-t2)b,t2b+(1-t2)a)A1-s(a,b) holds for all a,b>0 with ab if and only if t1(1-1-(2/π)2/s)/2 and t2(1-1/3s)/2. Here, P(a,b), A(a,b) and G(a,b) denote the Seiffert, arithmetic, and geometric means of two positive numbers a and b, respectively.

1. Introduction

The Seiffert mean P(a,b)  of two distinct positive numbers a and b is defined by (1.1)P(a,b)=a-b4arctan(a/b)-π.

Recently, the Seiffert mean P(a,b) has been the subject of intensive research. In particular, many remarkable inequalities for P(a,b) can be found in the literature . The Seiffert mean P(a,b) can be rewritten as (see [6, (2.4)]) (1.2)P(a,b)=a-b2arcsin((a-b)/(a+b)).

Let A(a,b)=(a+b)/2, G(a,b)=ab and H(a,b)=2ab/(a+b) be the classical arithmetic, geometric, and harmonic means of two positive numbers a and b, respectively. Then it is well known that inequalities H(a,b)<G(a,b)<P(a,b)<A(a,b) hold for all a,b>0 with ab.

For α,β,λ,μ(0,1/2), Chu et al. [18, 19] proved that the double inequalities (1.3)G(αa+(1-α)b,αb+(1-α)a)<P(a,b)<G(βa+(1-β)b,βb+(1-β)a),H(λa+(1-λ)b,λb+(1-λ)a)<P(a,b)<H(μa+(1-μ)b,μb+(1-μ)a) hold for all a,b>0 with ab if and only if α(1-1-4/π2)/2, β(3-3)/6, λ(1-1-2/π)/2 and μ(6-6)/12.

Let t(0,1/2), s1 and (1.4)Qt,s(a,b)=Gs(ta+(1-t)b,tb+(1-t)a)A1-s(a,b), then it is not difficult to verify that (1.5)Qt,1(a,b)=G(ta+(1-t)b,tb+(1-t)a),Qt,2(a,b)=H(ta+(1-t)b,tb+(1-t)a) and Qt,s(a,b) is strictly increasing with respect to t(0,1/2) for fixed a,b>0 with ab.

It is natural to ask what are the largest value t1=t1(s) and the least value t2=t2(s) in (0,1/2) such that the double inequality Qt1,s(a,b)<P(a,b)<Qt2,s(a,b) holds for all a,b>0 with ab and s1. The main purpose of this paper is to answer this question.

2. Main Result

In order to establish our main result we need two lemmas, which we present in the following.

Lemma 2.1.

If s1, then 1/(3s)+(2/π)2/s<1.

Proof.

Consider the following: (2.1)f(s)=13s+(2π)2/s.

Then simple computations lead to (2.2)lims+f(s)=1,(2.3)f(s)=2s2logπ2[(2π)2/s-16log(π/2)]2s2logπ2[(2π)2-16log(π/2)]=24log(π/2)-π23π2s2 for s1.

Computational and numerical experiments show that (2.4)24log(π2)-π2=0.968>0.

Inequalities (2.3) and (2.4) imply that f(s) is strictly increasing in [1,+). Therefore, Lemma 2.1 follows from (2.1) and (2.2) together with the monotonicity of f(s).

Lemma 2.2.

Let 0u1, s1 and (2.5)fu,s(x)=s2log(1-ux2)-logx+log(arcsinx).

Then inequality fu,s(x)>0 holds for all x(0,1) if and only if 3su1, and inequality fu,s(x)<0 holds for all x(0,1) if and only if u+(2/π)2/s1.

Proof.

If u=0, then we clearly see that 3su1, u+(2/π)2/s<1 and f0,s(x)=log[(arcsinx)/x]>0 for all s1 and x(0,1). In the following discussion, we assume that 0<u1.

From (2.5) and simple computations we have (2.6)limx0+fu,s(x)=0,(2.7)fu,s(x)  =11-x2arcsinx-1+u(s-1)x2x(1-ux2)=1+u(s-1)x2x(1-ux2)arcsinxgu,s(x), where (2.8)gu,s(x)=x(1-ux2)1-x2[1+u(s-1)x2]-arcsinx,(2.9)gu,s(0)=0,(2.10)gu,s(x)=x2(1-x2)3/2[1+u(s-1)x2]2hu,s(x), where (2.11)hu,s(x)=u2(s-1)2x4+u(-s2u+us+4s-2)x2+1-3su,(2.12)hu,s(0)=1-3su,(2.13)hu,s(1)=us(1-u)+(1-u)2.

We divide the proof into four cases.

Case 1 (3su1). Then from (2.11) and (2.12) together with the fact that (2.14)-us2+us+4s-2=2(s-1)+s(u+2su+1)+s(1-3su)>0, we clearly see that (2.15)hu,s(0)0, and hu,s(x) is strictly increasing in [0,1].

Equation (2.12) and the monotonicity of hu,s(x) imply that (2.16)hu,s(x)>0 for x(0,1].

Equation (2.10) and inequality (2.16) lead to the conclusion that gu,s(x) is strictly increasing in [0,1). Then from (2.9) we know that (2.17)gu,s(x)>0 for x(0,1).

It follows from (2.7) and inequality (2.17) that fu,s(x) is strictly increasing in (0,1].

Therefore, fu,s(x)>0 for all x(0,1) follows from (2.6) and the monotonicity of fu,s(x).

Case 2 (3su>1). Then (2.12) and the continuity of hu,s(x) imply that there exists 0<λ<1 such that (2.18)hu,s(x)<0 for x[0,λ).

Therefore, fu,s(x)<0 for x(0,λ) follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.18).

Case 3 (u+(2/π)2/s1). Then Lemma 2.1 and (2.12) lead to (2.19)hu,s(0)=1-3su1-3s[1-(2π)2/s]  <0.

We divide the proof into two subcases.

Subcase 3.1 (u=1). Then (2.13) becomes (2.20)hu,s(1)=0.

Let t=x2, then from (2.11) we clearly see that the function hu,s is a quadratic function of variable t. It follows from inequality (2.19) and (2.20) that (2.21)hu,s(x)<0 for all x[0,1).

Therefore, fu,s(x)<0 for x(0,1) follows easily from (2.6), (2.7), (2.9) and (2.10) together with inequality (2.21).

Subcase 3.2 (0<u<1). Then from (2.5), (2.8), and (2.13) we have (2.22)fu,s(1)=log[π2(1-u)s/2]0,(2.23)limx1-gu,s(x)=+,(2.24)hu,s(1)>0.

From (2.11), (2.19), and (2.24) we clearly see that there exists 0<λ1<1 such that hu,s(x)<0 for x[0,λ1) and hu,s(x)>0 for x(λ1,1]. Then (2.10) implies that gu,s(x) is strictly decreasing in [0,λ1] and strictly increasing in [λ1,1).

From (2.9) and (2.23) together with the piecewise monotonicity of gu,s(x) we clearly see that there exists 0<λ2<1 such that gu,s(x)<0, for x(0,λ2) and gu,s(x)>0 for x(λ2,1). Then (2.7) implies that fu,s(x) is strictly decreasing in (0,λ2] and strictly increasing in [λ2,1].

Therefore, fu,s(x)<0 for x(0,1) follows from (2.6) and (2.22) together with the piecewise monotonicity of fu,s(x).

Case 4 (u+(2/π)2/s<1). Then (2.5) leads to (2.25)fu,s(1)=log[π2(1-u)s/2]>0.

From inequality (2.25) and the continuity of fu,s(x) we know that there exists 0<μ<1 such that fu,s(x)>0 for x(μ,1].

Theorem 2.3.

If t1,t2(0,1/2) and s1, then the double inequality (2.26)Qt1,s(a,b)<P(a,b)<Qt2,s(a,b) holds for all a,b>0 with ab if and only if t1(1-1-(2/π)2/s)/2 and t2(1-1/3s)/2.

Proof.

Since both Qt,s(a,b) and P(a,b) are symmetric and homogeneous of degree 1. Without loss of generality, we assume that a>b. Let x=(a-b)/(a+b)(0,1). Then from (1.2) and (1.4) we have (2.27)log(Qt,s(a,b)P(a,b))=log(Qt,s(a,b)A(a,b))-log(P(a,b)A(a,b))=s2log[1-(1-2t)2x2]-logx+log(arcsinx).

Therefore, Theorem 2.3 follows easily from Lemma 2.2 and (2.27).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grant 11071069 and the Natural Science Foundation of Hunan Province under Grant 09JJ6003.

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