JAMJournal of Applied Mathematics1687-00421110-757XHindawi Publishing Corporation26123710.1155/2011/261237261237Research ArticleAn Optimal Double Inequality between Seiffert and Geometric MeansChuYu-Ming1WangMiao-Kun1WangZi-Kui2ButcherJ. C.1Department of MathematicsHuzhou Teachers CollegeHuzhou 313000Chinahutc.zj.cn2Department of MathematicsHangzhou Normal UniversityHangzhou 310012Chinahztc.edu.cn201123112011201130062011141020112011Copyright © 2011 Yu-Ming Chu 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 α,β(0,1/2) we prove that the double inequality G(αa+(1α)b,αb+(1α)a)<P(a,b)<G(βa+(1β)b,βb+(1β)a) holds for all a,b>0 with ab if and only if α(114/π2)/2 and β(33)/6. Here, G(a,b) and P(a,b) denote the geometric and Seiffert means of two positive numbers a and b, respectively.

1. Introduction

For a,  b>0 with ab the Seiffert mean P(a,b) was introduced by Seiffert  as follows:P(a,b)=a-b4arctana/b-π.

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature .

Let H(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(a-b)/(loga-logb), I(a,b)=1/e(bb/aa)1/(b-a), A(a,b)=(a+b)/2, C(a,b)=(a2+b2)/(a+b), and Mp(a,b)=[(ap+bp)/2]1/p(p0) and M0(a,b)=ab be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and pth power means of two different positive numbers a and b, respectively. Then it is well known that min{a,b}<H(a,b)=M-1(a,b)<G(a,b)=M0(a,b)<L(a,b)<I(a,b)<A(a,b)=M1(a,b)<C(a,b)<max{a,b} for all a,b>0 with ab.

For all a,b>0 with ab, Seiffert  established that L(a,b)<P(a,b)<I(a,b); Jagers  proved that M1/2(a,b)<P(a,b)<M2/3(a,b) and M2/3(a,b) is the best possible upper power mean bound for the Seiffert mean P(a,b); Seiffert  established that P(a,b)>A(a,b)G(a,b)/L(a,b) and P(a,b)>2A(a,b)/π; Sándor  presented that (A(a,b)+G(a,b))/2<P(a,b)<A(a,b)(A(a,b)+G(a,b))/2 and A2(a,b)G(a,b)3<P(a,b)<(G(a,b)+2A(a,b))/3; Hästö  proved that P(a,b)>Mlog2/logπ(a,b) and Mlog2/logπ(a,b) is the best possible lower power mean bound for the Seiffert mean P(a,b).

Very recently, Wang and Chu  found the greatest value α and the least value β such that the double inequality Aα(a,b)H1-α(a,b)<P(a,b)<Aβ(a,b)H1-β(a,b) holds for a,b>0 with ab; For any α(0,1), Chu et al.  presented the best possible bounds for Pα(a,b)G1-α(a,b) in terms of the power mean; In  the authors proved that the double inequality αA(a,b)+(1-α)H(a,b)<P(a,b)<βA(a,b)+(1-β)H(a,b) holds for all a,b>0 with ab if and only if α2/π and β5/6; Liu and Meng  proved that the inequalities α1C(a,b)+(1-α1)G(a,b)<P(a,b)<β1C(a,b)+(1-β1)G(a,b),α2C(a,b)+(1-α2)H(a,b)<P(a,b)<β2C(a,b)+(1-β2)H(a,b) hold for all a,b>0 with ab if and only if α12/9, β11/π, α21/π and β25/12.

For fixed a,b>0 with ab and x[0,1/2], let g(x)=G(xa+(1-x)b,xb+(1-x)a).

Then it is not difficult to verify that g(x) is continuous and strictly increasing in [0,1/2]. Note that g(0)=G(a,b)<P(a,b) and g(1/2)=A(a,b)>P(a,b). Therefore, it is natural to ask what are the greatest value α and least value β in (0,1/2) such that the double inequality G(αa+(1-α)b,αb+(1-α)a)<P(a,b)<G(βa+(1-β)b,βb+(1-β)a) holds for all a,b>0 with ab. The main purpose of this paper is to answer these questions. Our main result is the following Theorem 1.1.

Theorem 1.1.

If α,β(0,1/2), then the double inequality G(αa+(1-α)b,αb+(1-α)a)<P(a,b)<G(βa+(1-β)b,βb+(1-β)a) holds for all a,b>0 with ab if and only if α(1-1-4/π2)/2 and β(3-3)/6.

2. Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>Proof of Theorem <xref ref-type="statement" rid="thm1.1">1.1</xref>.

Let λ=(1-1-4/π2)/2 and μ=(3-3)/6. We first prove that inequalities P(a,b)>G(λa+(1-λ)b,λb+(1-λ)a),P(a,b)<G(μa+(1-μ)b,μb+(1-μ)a) hold for all a,b>0 with ab.

Without loss of generality, we assume that a>b. Let t=a/b>1 and p(0,1/2), then from (1.1) one has logG(pa+(1-p)b,pb+(1-p)a)-logP(a,b)=12log  [(pt2+(1-p))((1-p)t2+p)]-logt2-14arctant-π. Let f(t)=12log  [(pt2+(1-p))((1-p)t2+p)]-logt2-14arctant-π, then simple computations lead to f(1)=0,limt+f(t)=12log  [p(1-p)]+logπ,f(t)=t(t2+1)(t2-1)(4arctant-π)(pt2+(1-p))((1-p)t2+p)f1(t), where f1(t)=4(t2-1)(pt2+1-p)[(1-p)t2+p]t(t2+1)2-4arctant+π.f1(1)=0,limt+f1(t)=+,f1(t)=4f2(t2)t2(t2+1)4, where f2(t)=p(1-p)t5-(3p-2)(3p-1)t4+2(5p2-5p+1)t3+2(5p2-5p+1)t2-(3p-2)(3p-1)t+p(1-p).

Note that f2(1)=0,limt+f2(t)=+,f2(t)=5p(1-p)t4-4(3p-2)(3p-1)t3+6(5p2-5p+1)t2+4(5p2-5p+1)t-(3p-2)(3p-1),f2(1)=0,limt+f2(t)=+,f2(t)=20p(1-p)t3-12(3p-2)(3p-1)t2+12(5p2-5p+1)t+4(5p2-5p+1),f2(t)=-8(6p2-6p+1),limt+f2(t)=+,f3(t)=60p(1-p)t2-24(3p-2)(3p-1)t+12(5p2-5p+1),f2(1)=-36(6p2-6p+1),limt+f2(t)=+,f2(4)(t)=120p(1-p)t-24(3p-2)(3p-1),f2(4)(1)=-48(7p2-7p+1),limt+f2(4)(t)=+.

We divide the proof into two cases.

Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M115"><mml:mi>p</mml:mi><mml:mo>=</mml:mo><mml:mi>λ</mml:mi><mml:mo>=</mml:mo><mml:mo stretchy="false">(</mml:mo><mml:mn>1</mml:mn><mml:mo>-</mml:mo><mml:msqrt><mml:mn>1</mml:mn><mml:mo>-</mml:mo><mml:mn>4</mml:mn><mml:mo>/</mml:mo><mml:msup><mml:mrow><mml:mi>π</mml:mi></mml:mrow><mml:mrow><mml:mn>2</mml:mn></mml:mrow></mml:msup></mml:msqrt><mml:mo stretchy="false">)</mml:mo><mml:mo>/</mml:mo><mml:mn>2</mml:mn></mml:math></inline-formula>).

Then (2.6), (2.18), (2.21), and (2.24) become limt+f(t)=0,f2(1)=-8(π2-6)π2<0,f2(1)=-36(π2-6)π2<0,f2(4)(1)=-48(π2-7)π2<0.

From (2.23) we clearly see that f2(4)(t) is strictly increasing in [1,+), then (2.25) and inequality (2.29) lead to the conclusion that there exists λ1>1 such that f2(4)(t)<0 for t[1,λ1) and f2(4)(t)>0 for t(λ1,+). Thus, f2(t) is strictly decreasing in [1,λ1] and strictly increasing in [λ1,+).

It follows from (2.22) and inequality (2.28) together with the piecewise monotonicity of f2(t) that there exists λ2>λ1>1 such that f2(t) is strictly decreasing in [1,λ2] and strictly increasing in [λ2,+). Then (2.19) and inequality (2.27) lead to the conclusion that there exists λ3>λ2>1 such that f2(t) is strictly decreasing in [1,λ3] and strictly increasing in [λ3,+).

From (2.15) and (2.16) together with the piecewise monotonicity of f2(t) we know that there exists λ4>λ3>1 such that f2(t) is strictly decreasing in [1,λ4] and strictly increasing in [λ4,+). Then (2.11)–(2.13) lead to the conclusion that there exists λ5>λ4>1 such that f1(t) is strictly decreasing in [1,λ5] and strictly increasing in [λ5,+).

It follows from (2.7)–(2.10) and the piecewise monotonicity of f1(t) that there exists λ6>λ5>1 such that f(t) is strictly decreasing in [1,λ6] and strictly increasing in [λ6,+).

Therefore, inequality (2.1) follows from (2.3)–(2.5) and the piecewise monotonicity of f(t).

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M154"><mml:mi>p</mml:mi><mml:mo>=</mml:mo><mml:mi>μ</mml:mi><mml:mo>=</mml:mo><mml:mo stretchy="false">(</mml:mo><mml:mn>3</mml:mn><mml:mo>-</mml:mo><mml:msqrt><mml:mn>3</mml:mn></mml:msqrt><mml:mo stretchy="false">)</mml:mo><mml:mo>/</mml:mo><mml:mn>6</mml:mn></mml:math></inline-formula>).

Then (2.18), (2.21) and (2.24) become f2(1)=0,f2(1)=0,f2(4)(1)=8>0.

From (2.23) we clearly see that f2(4)(t) is strictly increasing in [1,+), then inequality (2.32) leads to the conclusion that f2(t) is strictly increasing in [1,+).

Therefore, inequality (2.2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), (2.15), and inequalities (2.30) and (2.31) together with the monotonicity of f2(t).

Next, we prove that λ=(1-1-4/π2)/2 is the best possible parameter such that inequality (2.1) holds for all a,b>0 with ab. In fact, if (1-1-4/π2)/2=λ<p<1/2, then (2.6) leads to limt+f(t)=12log  [p(1-p)]+logπ>0.

Inequality (2.33) implies that there exists T=T(p)>1 such that f(t)>0 for t(T,+).

It follows from (2.3) and (2.4) together with inequality (2.34) that P(a,b)<G(pa+(1-p)b,pb+(1-p)a) for a/b(T2,+).

Finally, we prove that μ=(3-3)/6 is the best possible parameter such that inequality (2.2) holds for all a,b>0 with ab. In fact, if 0<p<μ=(3-3)/6, then from (2.18) we get f2(1)<0, which implies that there exists δ>0 such that f2(t)<0 for t[1,1+δ).

Therefore, P(a,b)>G(pa+(1-p)b,pb+(1-p)a) for a/b(1,(1+δ)2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), and (2.15) together with inequality (2.35).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grant 11071069 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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