AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation65793510.1155/2011/657935657935Research ArticleA Sharp Double Inequality between Harmonic and Identric MeansChuYu-Ming1WangMiao-Kun1WangZi-Kui2DošlýOndřej1Department of MathematicsHuzhou Teachers CollegeHuzhou 313000China2Department of MathematicsHangzhou Normal UniversityHangzhou 310012Chinahztc.edu.cn201112102011201131052011060820112011Copyright © 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.

We find the greatest value p and the least value q in (0,1/2) such that the double inequality H(pa+(1-p)b,pb+(1-p)a)<I(a,b)<H(qa+(1-q)b,qb+(1-q)a) holds for all a,b>0 with ab. Here, H(a,b), and I(a,b) denote the harmonic and identric means of two positive numbers a and b, respectively.

1. Introduction

The classical harmonic mean H(a,b) and identric mean I(a,b) of two positive numbers a and b are defined byH(a,b)=2aba+b,I(a,b)={1e(bbaa)1/(b-a),ab,a,a=b, respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for H and I can be found in the literature .

Let Mp(a,b)=[(ap+bp)/2]1/p, L(a,b)=(a-b)/(loga-logb), G(a,b)=ab, A(a,b)=(a+b)/2, and P(a,b)=(a-b)/[4arctan(a/b)-π] be the pth power, logarithmic, geometric, arithmetic, and Seiffert means of two positive numbers a and b with ab, 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)<P(a,b)<I(a,b)<A(a,b)=M1(a,b)<max{a,b} for all a,b>0 with ab.

Long and Chu  answered the question: what are the greatest value p and the least value q such that Mp(a,b)<Aα(a,b)Gβ(a,b)H1-α-β(a,b)<Mq(a,b) for all a,b>0 with ab and α,β>0 with α+β<1.

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.

The following sharp bounds for I,  (LI)1/2, and (L+I)/2 in terms of power means are presented in : M2/3(a,b)<I(a,b)<Mlog2(a,b),M0(a,b)<L(a,b)I(a,b)<M1/2(a,b),Mlog2/(1+log2)(a,b)<L(a,b)+I(a,b)2<M1/2(a,b)

for all a,b>0 with ab.

Alzer and Qiu  proved that the inequalities αA(a,b)+(1-α)G(a,b)<I(a,b)<βA(a,b)+(1-β)G(a,b) hold for all positive real numbers a and b with ab if and only if α2/3 and β2/e=0.73575, and so forth.

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

Then it is not difficult to verify that f(x) is continuous and strictly increasing in [0,1/2]. Note that f(0)=H(a,b)<I(a,b) and f(1/2)=A(a,b)>I(a,b). Therefore, it is natural to ask what are the greatest value p and the least value q in (0,1/2) such that the double inequality H(pa+(1-p)b,pb+(1-p)a)<I(a,b)<H(qa+(1-q)b,qb+(1-q)a) holds for all a,b>0 with ab. The main purpose of this paper is to answer these questions. Our main result is Theorem 1.1.

Theorem 1.1.

If p,q(0,1/2), then the double inequality H(pa+(1-p)b,pb+(1-p)a)<I(a,b)<H(qa+(1-q)b,qb+(1-q)a) holds for all a,b>0 with ab if and only if p(1-1-2/e)/2 and q(6-6)/12.

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=(6-6)/12 and μ=(1-1-2/e)/2. Then from the monotonicity of the function f(x)=H(xa+(1-x)b,xb+(1-x)a) in [0,1/2] we know that to prove inequality (1.8) we only need to prove that inequalities I(a,b)<H(λa+(1-λ)b,λb+(1-λ)a),I(a,b)>H(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 r(0,1/2), then from (1.1) and (1.2) one has logH(ra+(1-r)b,rb+(1-r)a)-logI(a,b)=log{r(1-r)t2+[r2+(1-r)2]t+r(1-r)}-log(t+1)-tlogtt-1+1+log2.

Let g(t)=log{r(1-r)t2+[r2+(1-r)2]t+r(1-r)}-log(t+1)-tlogtt-1+1+log2. Then simple computations lead to g(1)=0,limt+g(t)=log[r(1-r)]+1+log2,g(t)=g1(t)(t-1)2, where g1(t)=logt-(t-1)[(2r2-2r+1)t2+4r(1-r)t+2r2-2r+1](t+1)[r(1-r)t2+(2r2-2r+1)t+r(1-r)],g1(1)=0,limt+g1(t)=+,g1(t)=g2(t)t(t+1)2[r(1-r)t2+(2r2-2r+1)t+r(1-r)]2, where g2(t)=r2(1-r)2t6+(2r4-4r3-2r2+4r-1)t5-(17r4-34r3+25r2-8r+1)t4+4(7r4-14r3+13r2-6r+1)t3-(17r4-34r3+25r2-8r+1)t2+(2r4-4r3-2r2+4r-1)t+r2(1-r)2,g2(1)=0,limt+g2(t)=+,g2(t)=6r2(1-r)2t5+5(2r4-4r3-2r2+4r-1)t4-4(17r4-34r3+25r2-8r+1)t3+12(7r4-14r3+13r2-6r+1)t2-2(17r4-34r3+25r2-8r+1)t+2r4-4r3-2r2+4r-1,g2(1)=0,limt+g2(t)=+,g2′′(t)=30r2(1-r)2t4+20(2r4-4r3-2r2+4r-1)t3-12(17r4-34r3+25r2-8r+1)t2+24(7r4-14r3+13r2-6r+1)t-2(17r4-34r3+25r2-8r+1),g2′′(1)=-2(24r2-24r+5),limt+g2′′(t)=+,g2′′′(t)=120r2(1-r)2t3+60(2r4-4r3-2r2+4r-1)t2-24(17r4-34r3+25r2-8r+1)t+24(7r4-14r3+13r2-6r+1),g2′′′(1)=-12(24r2-24r+5),limtg2′′′(t)=,g2(4)(t)=360r2(1-r)2t2+120(2r4-4r3-2r2+4r-1)t-24(17r4-34r3+25r2-8r+1),g2(4)(1)=48(4r4-8r3-10r2+14r-3),limt+g2(4)(t)=+,g2(5)(t)=720r2(1-r)2t+120(2r4-4r3-2r2+4r-1),g2(5)(1)=120(8r4-16r3+4r2+4r-1).

We divide the proof into two cases. Case 1 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M111"><mml:mi>r</mml:mi><mml:mo>=</mml:mo><mml:mi>λ</mml:mi><mml:mo>=</mml:mo><mml:mo stretchy="false">(</mml:mo><mml:mn>6</mml:mn><mml:mo>-</mml:mo><mml:msqrt><mml:mn>6</mml:mn></mml:msqrt><mml:mo stretchy="false">)</mml:mo><mml:mo>/</mml:mo><mml:mn>12</mml:mn></mml:math></inline-formula>).

Then (2.19), (2.22), (2.25), and (2.28) lead to g2′′(1)=0,g2′′′(1)=0,g2(4)(1)=133>0,g2(5)(1)=653>0.

From (2.27) we clearly see that g2(5)(t) is strictly increasing in [1,+), then inequality (2.32) leads to the conclusion that g2(5)(t)>0 for t[1,+), hence g2(4)(t) is strictly increasing in [1,+).

It follows from inequality (2.31) and the monotonicity of g2(4)(t) that g2′′′(t) is strictly increasing in [1,+). Then (2.30) implies that g2′′′(t)>0 for t[1,+), so g2′′(t) is strictly increasing in [1,+).

From (2.29) and the monotonicity of g2′′(t) we clearly see that g2(t) is strictly increasing in [1,+).

From (2.5), (2.7), (2.9), (2.11), (2.13), (2.16), and the monotonicity of g2(t) we conclude that g(t)>0 for t(1,+).

Therefore, inequality (2.1) follows from (2.3) and (2.4) together with inequality (2.33).

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M135"><mml:mi>r</mml:mi><mml:mo>=</mml:mo><mml:mi>μ</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>2</mml:mn><mml:mo>/</mml:mo><mml:mi>e</mml:mi></mml:msqrt><mml:mo stretchy="false">)</mml:mo><mml:mo>/</mml:mo><mml:mn>2</mml:mn></mml:math></inline-formula>).

Then (2.19), (2.22), (2.25), and (2.28) lead to g2′′(1)=-2e(5e-12)<0,g2′′′(1)=-12e(5e-12)<0,g2(4)(1)=-48e2(3e2-7e-1)<0,g2(5)(1)=120e2(2+2e-e2)>0. From (2.27) and (2.37) we know that g2(4)(t) is strictly increasing in [1,+). Then (2.26) and (2.36) lead to the conclusion that there exists t1>1 such that g2(4)(t)<0 for t[1,t1) and g2(4)(t)>0 for t(t1,+), hence g2′′′(t) is strictly decreasing in [1,t1] and strictly increasing in [t1,+).

It follows from (2.23) and (2.35) together with the piecewise monotonicity of g2′′′(t) that there exists t2>t1>1 such that g2′′(t) is strictly decreasing in [1,t2] and strictly increasing in [t2,+). Then (2.20) and (2.34) lead to the conclusion that there exists t3>t2>1 such that g2(t) is strictly decreasing in [1,t3] and strictly increasing in [t3,+).

From (2.16) and (2.17) together with the piecewise monotonicity of g2(t) we clearly see that there exists t4>t3>1 such that g2(t)<0 for t(1,t4) and g2(t)>0 for t(t4,+). Therefore, g2(t) is strictly decreasing in [1,t4] and strictly increasing in [t4,+). Then (2.11)–(2.14) lead to the conclusion that there exists t5>t4>1 such that g1(t) is strictly decreasing in [1,t5] and strictly increasing in [t5,+).

It follows from (2.7)–(2.10) and the piecewise monotonicity of g1(t) that there exists t6>t5>1 such that g(t) is strictly decreasing in [1,t6] and strictly increasingin [t6,+).

Note that (2.6) becomes limt+g(t)=log[r(1-r)]+1+log2=0 for r=μ=(1-1-2/e)/2.

From (2.5) and (2.38) together with the piecewise monotonicity of g(t) we clearly see that g(t)<0 for t(1,+).

Therefore, inequality (2.2) follows from (2.3) and (2.4) together with inequality (2.39).

Next, we prove that the parameter λ=(6-6)/12 is the best possible parameter in (0,1/2) such that inequality (2.1) holds for all a,b>0 with ab. In fact, if r<λ=(6-6)/12, then (2.19) leads to g2′′(1)=-2(24r2-24r+5)<0. From the continuity of g2′′(t) we know that there exists δ>0 such that g2′′(t)<0 for t(1,1+δ).

It follows from (2.3)–(2.5), (2.7), (2.9), (2.11), (2.13), and (2.16) that I(a,b)>H(ra+(1-r)b,rb+(1-r)a) for a/b(1,1+δ).

Finally, we prove that the parameter μ=(1-1-2/e)/2 is the best possible parameter in (0,1/2) such that inequality (2.2) holds for all a,b>0 with ab. In fact, if (1-1-2/e)/2=μ<r<1/2, then (2.6) leads to limt+g(t)>0. Hence, there exists T>1 such that g(t)>0 for t(T,+).

Therefore, H(ra+(1-r)b,rb+(1-r)a)>I(a,b) for a/b(T,+), follows from (2.3) and (2.4) together with inequality (2.41).

Acknowledgments

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

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