# A Sharp Double Inequality between Harmonic and Identric Means

Authors: Yu-Ming Chu; Miao-Kun Wang; Zi-Kui Wang
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We find the greatest valuep 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 a≠b. Here, H(a,b), and I(a,b) denote the harmonic and identric means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

The classical harmonic meanH(a,b) and identric mean I(a,b) of two positive numbers a and b are defined by(1.1)H(a,b)=2aba+b,(1.2)I(a,b)={1e(bbaa)1/(b-a),a≠b,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 [1–17].LetMp(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 a≠b, respectively. Then it is well-known that (1.3)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 a≠b.Long and Chu [18] 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 a≠b and α,β>0 with α+β<1.In [19], the authors proved that the double inequality (1.4)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 a≠b if and only if α≤2/π and β≥5/6.The following sharp bounds forI,(LI)1/2, and (L+I)/2 in terms of power means are presented in [20]: (1.5)M2/3(a,b)<I(a,b)<Mlog⁡2(a,b),M0(a,b)<L(a,b)I(a,b)<M1/2(a,b),Mlog⁡2/(1+log⁡2)(a,b)<L(a,b)+I(a,b)2<M1/2(a,b)for alla,b>0 with a≠b.Alzer and Qiu [21] proved that the inequalities (1.6)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 a≠b if and only if α≤2/3 and β≥2/e=0.73575, and so forth.For fixeda,b>0 with a≠b and x∈[0,1/2], let (1.7)f(x)=H(xa+(1-x)b,xb+(1-x)a).Then it is not difficult to verify thatf(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 a≠b. The main purpose of this paper is to answer these questions. Our main result is  Theorem 1.1.Theorem 1.1.
Ifp,q∈(0,1/2), then the double inequality
(1.8)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 a≠b if and only if p≤(1-1-2/e)/2 and q≥(6-6)/12.


## 2. Proof of Theorem1.1

Proof of Theorem1.1.
Letλ=(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
(2.1)I(a,b)<H(λa+(1-λ)b,λb+(1-λ)a),(2.2)I(a,b)>H(μa+(1-μ)b,μb+(1-μ)a),
hold for all a,b>0 with a≠b.
Without loss of generality, we assume thata>b. Let t=a/b>1 and r∈(0,1/2), then from (1.1) and (1.2)   one has
(2.3)log⁡H(ra+(1-r)b,rb+(1-r)a)-log⁡I(a,b)=log⁡{r(1-r)t2+[r2+(1-r)2]t+r(1-r)}-log⁡(t+1)-tlog⁡tt-1+1+log⁡2.
Let(2.4)g(t)=log⁡{r(1-r)t2+[r2+(1-r)2]t+r(1-r)}-log⁡(t+1)-tlog⁡tt-1+1+log⁡2.
Then simple computations lead to
(2.5)g(1)=0,(2.6)lim⁡t→+∞g(t)=log⁡[r(1-r)]+1+log⁡2,(2.7)g′(t)=g1(t)(t-1)2,
where
(2.8)g1(t)=log⁡t-(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)],(2.9)g1(1)=0,(2.10)lim⁡t→+∞g1(t)=+∞,(2.11)g1′(t)=g2(t)t(t+1)2[r(1-r)t2+(2r2-2r+1)t+r(1-r)]2,
where
(2.12)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,(2.13)g2(1)=0,(2.14)lim⁡t→+∞g2(t)=+∞,(2.15)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,(2.16)g2′(1)=0,(2.17)lim⁡t→+∞g2′(t)=+∞,(2.18)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),(2.19)g2′′(1)=-2(24r2-24r+5),(2.20)lim⁡t→+∞g2′′(t)=+∞,(2.21)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),(2.22)g2′′′(1)=-12(24r2-24r+5),(2.23)lim⁡t→∞⁡g2′′′(t)=∞,(2.24)g2(4)(t)=360r2(1-r)2t2+120(2r4-4r3-2r2+4r-1)t-24(17r4-34r3+25r2-8r+1),(2.25)g2(4)(1)=48(4r4-8r3-10r2+14r-3),(2.26)lim⁡t→+∞g2(4)(t)=+∞,(2.27)g2(5)(t)=720r2(1-r)2t+120(2r4-4r3-2r2+4r-1),(2.28)g2(5)(1)=120(8r4-16r3+4r2+4r-1).
We divide the proof into two cases.Case 1 (r=λ=(6-6)/12).
Then  (2.19), (2.22), (2.25), and (2.28) lead to
(2.29)g2′′(1)=0,(2.30)g2′′′(1)=0,(2.31)g2(4)(1)=133>0,(2.32)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
(2.33)g(t)>0
for t∈(1,+∞).
Therefore, inequality (2.1) follows from   (2.3) and (2.4) together with inequality (2.33).
Case 2 (r=μ=(1-1-2/e)/2).
Then  (2.19), (2.22), (2.25), and (2.28) lead to
(2.34)g2′′(1)=-2e(5e-12)<0,(2.35)g2′′′(1)=-12e(5e-12)<0,(2.36)g2(4)(1)=-48e2(3e2-7e-1)<0,(2.37)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
(2.38)lim⁡t→+∞g(t)=log⁡[r(1-r)]+1+log⁡2=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
(2.39)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 parameterλ=(6-6)/12 is the best possible parameter in (0,1/2) such that inequality (2.1) holds for all a,b>0 with a≠b. 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
(2.40)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 parameterμ=(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 a≠b. 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
(2.41)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).



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*Source: 657935-2011-10-12.xml*