# Sharp Bounds for Seiffert Mean in Terms of Contraharmonic Mean

Authors: Yu-Ming Chu; Shou-Wei Hou
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We find the greatest valueα and the least value β in (1/2,1) such that the double inequality C(αa+(1-α)b,αb+(1-α)a)<T(a,b)<Cβa+1-βb,βb+(1-βa) holds for all a,b>0 with a≠b. Here, T(a,b)=(a-b)/[2 arctan((a-b)/(a+b))]  and Ca,b=(a2+b2)/(a+b) are the Seiffert and contraharmonic means of a and b, respectively.


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the Seiffert mean T(a,b) and contraharmonic mean C(a,b) are defined by
(1.1)T(a,b)=a-b2arctan((a-b)/(a+b)),(1.2)C(a,b)=a2+b2a+b,respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities and properties for these means can be found in the literature [1–12].LetA(a,b)=(a+b)/2, G(a,b)=ab, S(a,b)=(a2+b2)/2, and let Mp(a,b)=((ap+bp)/2)1/p(p≠0) and M0(a,b)=ab be the arithmetic, geometric, square root, and pth power means of two positive numbers a and b, respectively. Then it is well known that Mp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b, and the inequalities (1.3)G(a,b)=M0(a,b)<A(a,b)=M1(a,b)<S(a,b)=M2(a,b)<C(a,b)
hold for all a,b>0 with a≠b.Seiffert [12] proved that the double inequality (1.4)A(a,b)=M1(a,b)<T(a,b)<M2(a,b)=S(a,b)
holds for all a,b>0 with a≠b.Hästö [13] proved that the function T(1,x)/Mp(1,x) is increasing in (0,∞) if p≤1.In [14], the authors found the greatest value p and the least value q such that the double inequality Hp(a,b)<T(a,b)<Hq(a,b) holds for all a,b>0 with a≠b. Here, Hk(a,b)=((ak+(ab)k/2+bk)/3)1/k(k≠0), and H0(a,b)=ab is the kth power-type Heron mean of a and b.Wang et al. [15] answered the question: what are the best possible parameters λ and μ such that the double inequality Lλ(a,b)<T(a,b)<Lμ(a,b) holds for all a,b>0 with a≠b, where Lr(a,b)=(ar+1+br+1)/(ar+br) is the rth Lehmer mean of a and b.In [16, 17], the authors proved that the inequalities (1.5)α1T(a,b)+(1-α1)G(a,b)<A(a,b)<β1T(a,b)+(1-β1)G(a,b),α2S(a,b)+(1-α2)A(a,b)<T(a,b)<β2S(a,b)+(1-β2)A(a,b),Sα3(a,b)A1-α3(a,b)<T(a,b)<Sβ3(a,b)A1-β3(a,b)
hold for all a,b>0 with a≠b if and only if α1≤3/5, β1≥π/4, α2≤(4-π)/[(2-1)π], β2≥2/3, α3≤2/3 and β3≥4-2logπ/log2.For fixeda,b>0 with a≠b, let x∈[1/2,1] and (1.6)J(x)=C(xa+(1-x)b,xb+(1-x)a).Then it is not difficult to verify thatJ(x) is continuous and strictly increasing in [1/2,1]. Note that J(1/2)=A(a,b)<T(a,b) and J(1)=C(a,b)>T(a,b). Therefore, it is natural to ask what are the greatest value α and the least value β in (1/2,1) such that the double inequality (1.7)C(αa+(1-α)b,αb+(1-α)a)<T(a,b)<C(βa+(1-β)b,βb+(1-β)a)
holds for all a,b>0 with a≠b. The main purpose of this paper is to answer this question. Our main result is the following Theorem 1.1.Theorem 1.1.
Ifα,β∈(1/2,1), then the double inequality
(1.8)C(αa+(1-α)b,αb+(1-α)a)<T(a,b)<C(βa+(1-β)b,βb+(1-β)a)
holds for all a,b>0 with a≠b if and only if α≤(1+4/π-1)/2 and β≥(3+3)/6.


## 2. Proof of Theorem1.1

Proof of Theorem1.1.
Letλ=(1+4/π-1)/2 and μ=(3+3)/6. We first proof that the inequalities
(2.1)T(a,b)>C(λa+(1-λ)b,λb+(1-λ)a),(2.2)T(a,b)<C(μa+(1-μ)b,μb+(1-μ)a)
hold for all a,b>0 with a≠b.
From (1.1) and (1.2) we clearly see that both T(a,b) and C(a,b) are symmetric and homogenous of degree 1. Without loss of generality, we assume that a>b. Let t=a/b>1 and p∈(1/2,1), then from (1.1) and (1.2) one has
(2.3)C(pa+(1-p)b,pb+(1-p)a)-T(a,b)=b[pt+(1-p)]2+[(1-p)t+p]22(t+1)arctan((t-1)/(t+1))×{2arctan(t-1t+1)-t2-1[pt+(1-p)]2+[(1-p)t+p]2}.
Let
(2.4)f(t)=2arctan(t-1t+1)-t2-1[pt+(1-p)]2+[(1-p)t+p]2.
Then simple computations lead to
(2.5)f(1)=0,(2.6)lim⁡t→+∞f(t)=π2-1p2+(1-p)2,(2.7)f′(t)=2f1(t){[pt+(1-p)]2+[(1-p)t+p]2}2(t2+1),
where
(2.8)f1(t)=(4p4-8p3+10p2-6p+1)t4-2(2p-1)2(2p2-2p+1)t3+2(12p4-24p3+18p2-6p+1)t2-2(2p-1)2(2p2-2p+1)t+4p4-8p3+10p2-6p+1,(2.9)f1(1)=0.
Letf2(t)=f1′(t)/2, f3(t)=f2′(t)/2, f4(t)=f3′(t)/3. Then from (2.8) we get
(2.10)f2(t)=2(4p4-8p3+10p2-6p+1)t3-3(2p-1)2(2p2-2p+1)t2+2(12p4-24p3+18p2-6p+1)t-(2p-1)2(2p2-2p+1),(2.11)f2(1)=0,(2.12)f3(t)=3(4p4-8p3+10p2-6p+1)t2-3(2p-1)^2(2p2-2p+1)t+12p4-24p3+18p2-6p+1,(2.13)f3(1)=6p2-6p+1,(2.14)f4(t)=2(4p4-8p3+10p2-6p+1)t-(2p-1)2(2p2-2p+1),(2.15)f4(1)=6p2-6p+1.
We divide the proof into two cases.
Case 1 (p=λ=(1+4/π-1)/2).
Then (2.6), (2.13), and (2.15) lead to
(2.16)lim⁡t→+∞f(t)=0,(2.17)f3(1)=-2(π-3)π<0,(2.18)f4(1)=-2(π-3)π<0.
Note that(2.19)4p4-8p3+10p2-6p+1=4+2π-π2π2>0.
It follows from (2.8), (2.10), (2.12), (2.14), and (2.19) that
(2.20)lim⁡t→+∞f1(t)=+∞,(2.21)lim⁡t→+∞f2(t)=+∞,(2.22)lim⁡t→+∞f3(t)=+∞,(2.23)lim⁡t→+∞f4(t)=+∞.
From (2.14) and inequality (2.19), we clearly see that f4(t) is strictly increasing in [1,+∞). Then (2.18) and (2.23) lead to the conclusion that there exists t0>1 such that f4(t)<0 for t∈[1,t0) and f4(t)>0 for t∈(t0,+∞). Hence, f3(t) is strictly decreasing in [1,t0] and strictly increasing in [t0,+∞).
It follows from (2.17) and (2.22) together with the piecewise monotonicity of f3(t) that there exists t1>t0>1 such that f2(t) is strictly decreasing in [1,t1] and strictly increasing in [t1,+∞).
From (2.11) and (2.21) together with the piecewise monotonicity of f2(t), we conclude that there exists t2>t1>1 such that f1(t) is strictly decreasing in [1,t2] and strictly increasing in [t2,+∞).
Equations (2.7), (2.9), and (2.20) together with the piecewise monotonicity of f1(t) imply that there exists t3>t2>1 such that f(t) is strictly decreasing in [1,t3] and strictly increasing in [t3,+∞).
Therefore, inequality (2.1) follows from (2.3)–(2.5) and (2.16) together with the piecewise monotonicity of f(t).Case 2 (p=μ=(3+3)/6).
Then (2.8) leads to
(2.24)f1(t)=(t-1)49>0
for t>1.
Inequality (2.24) and (2.7) imply that f(t) is strictly increasing in [1,+∞). Therefore, inequality (2.2) follows from (2.3)–(2.5) together with the monotonicity of f(t).
From inequalities (2.1) and (2.2) together with the monotonicity of J(x)=C(xa+(1-x)b,xb+(1-x)a) in [1/2,1], we know that inequality (1.8) holds for all α≤(1+4/π-1)/2, β≥(3+3)/6, and all a,b>0 with a≠b.
Next, we prove thatλ=(1+4/π-1)/2 is the best possible parameter in [1/2,1] such that inequality (2.1) holds for all a,b>0 with a≠b.
For any1>p>λ=(1+4/π-1)/2, from (2.6) one has
(2.25)lim⁡t→+∞f(t)=π2-1p2+(1-p)2>0.
Equations (2.3) and (2.4) together with inequality (2.25) imply that for any 1>p>λ=(1+4/π-1)/2 there exists T0=T0(p)>1 such that
(2.26)C(pa+(1-p)b,pb+(1-p)a)>T(a,b)
for a/b∈(T0,+∞).
Finally, we prove thatμ=(3+3)/6 is the best possible parameter such that inequality (2.2) holds for all a,b>0 with a≠b.
For any1/2<p<μ=(3+3)/6, from (2.13) one has
(2.27)f3(1)=6p2-6p+1<0.
From inequality (2.27) and the continuity of f3(t), we know that there exists δ=δ(p)>0 such that
(2.28)f3(t)<0
for t∈(1,1+δ).
Equations (2.3)–(2.5), (2.7), (2.9), and (2.11) together with inequality (2.28) imply that for any 1/2<p<μ=(3+3)/6 there exists δ=δ(p)>0 such that
(2.29)T(a,b)>C(pa+(1-p)b,pb+(1-p)a)
for a/b∈(1,1+δ).



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*Source: 425175-2011-12-27.xml*