# A Sharp Double Inequality between Seiffert, Arithmetic, and Geometric Means

Authors: Wei-Ming Gong; Ying-Qing Song; Miao-Kun Wang; Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

For fixeds≥1 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 a≠b 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.


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## Body

## 1. Introduction

The Seiffert meanP(a,b) [1] of two distinct positive numbers a and b is defined by
(1.1)P(a,b)=a-b4arctan(a/b)-π.Recently, the Seiffert meanP(a,b) has been the subject of intensive research. In particular, many remarkable inequalities for P(a,b) can be found in the literature [2–17]. 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)).LetA(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 a≠b.Forα,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 a≠b 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.Lett∈(0,1/2), s≥1 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 a≠b.It is natural to ask what are the largest valuet1=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 a≠b and s≥1. 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.
Ifs≥1, 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 s≥1.
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.
Let0≤u≤1, s≥1 and
(2.5)fu,s(x)=s2log(1-ux2)-logx+log(arcsinx).
Then inequalityfu,s(x)>0 holds for all x∈(0,1) if and only if 3su≤1, and inequality fu,s(x)<0 holds for all x∈(0,1) if and only if u+(2/π)2/s≥1.Proof.
Ifu=0, then we clearly see that 3su≤1, u+(2/π)2/s<1 and f0,s(x)=log[(arcsinx)/x]>0 for all s≥1 and x∈(0,1). In the following discussion, we assume that 0<u≤1.
From (2.5) and simple computations we have
(2.6)limx→0+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 (3su≤1). 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/s≥1). Then Lemma 2.1 and (2.12) lead to
(2.19)hu,s(0)=1-3su≤1-3s[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.
Lett=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)limx→1-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.
Ift1,t2∈(0,1/2) and s≥1, then the double inequality
(2.26)Qt1,s(a,b)<P(a,b)<Qt2,s(a,b)
holds for all a,b>0 with a≠b if and only if t1≤(1-1-(2/π)2/s)/2 and t2≥(1-1/3s)/2.Proof.
Since bothQt,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, Theorem2.3 follows easily from Lemma 2.2 and (2.27).



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*Source: 684834-2012-09-02.xml*