# Optimal Two Parameter Bounds for the Seiffert Mean

Authors: Hui Sun; Ying-Qing Song; Yu-Ming Chu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We obtain sharp bounds for the Seiffert mean in terms of a two parameter family of means. Our results generalize and extend the recent bounds presented in the Journal of Inequalities and Applications (2012) and Abstract and Applied Analysis (2012).


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the Seiffert mean T(a,b), root mean square S(a,b), and contraharmonic mean C(a,b) are defined by
(1)T(a,b)=a-b2arctan[(a-b)/(a+b)],(2)S(a,b)=a2+b22,(3)C(a,b)=a2+b2a+b,
respectively. It is well known that the inequalities
(4)T(a,b)<S(a,b)<C(a,b)
hold for all a,b>0 with a≠b.Recently,T(a,b), S(a,b), and C(a,b) have been the subject of intensive research. In particular, many remarkable inequalities and properties for these means can be found in the literature [1–8].Forα,β,λ,μ∈(1/2,1), very recently Chu et al. [9, 10] proved that the inequalities
(5)S(αa+(1-α)b,αb+(1-α)a)<T(a,b)<S(βa+(1-β)b,βb+(1-β)a),(6)C(λa+(1-λ)b,λb+(1-λ)a)<T(a,b)<C(μa+(1-μ)b,μb+(1-μ)a)
hold for all a,b>0 with a≠b if and only if α≤(1+16/π2-1)/2, β≥(3+6)/6, λ≤(1+4/π-1)/2, and μ≥(3+3)/6.Lett∈(1/2,1), p≥1/2, and
(7)Qt,p(a,b)=Cp(ta+(1-t)b,tb+(1-t)a)A1-p(a,b),
where A(a,b)=(a+b)/2 is the classical arithmetic mean of a and b. Then from (2), (3), and (7) we clearly see that
(8)Qt,1/2(a,b)=S(ta+(1-t)b,tb+(1-t)a),Qt,1(a,b)=C(ta+(1-t)b,tb+(1-t)a),
and Qt,p(a,b) is strictly increasing with respect to t∈(1/2,1) for fixed a,b>0 with a≠b.It is natural to ask what are the greatest valuet1=t1(p) and the least value t2=t2(p) in (1/2,1) such that the double inequality
(9)Qt1,p(a,b)<T(a,b)<Qt2,p(a,b)
holds for all a,b>0 with a≠b and p≥1/2. The aim of this paper is to answer this question; our main result is the following Theorem 1.Theorem 1.
Ift1,t2∈(1/2,1) and p∈[1/2,∞), then the double inequality
(10)Qt1,p(a,b)<T(a,b)<Qt2,p(a,b)
holds for all a,b>0 with a≠b if and only if t1≤1/2+[(4/π)1/p-1]/2 and t2≥1/2+3p/(6p).Remark 2.
If we takep=1/2 and p=1 in Theorem 1, then inequality (10) reduces to inequalities (5) and (6), respectively.


## 2. Proof of Theorem1

In order to prove Theorem1 we need two lemmas, which we present in this section.Lemma 3 (see [11, Theorem  1.25]).
For-∞<a<b<∞, let f,g:[a,b]→ℝ be continuous on [a,b] and differentiable on (a,b); let g′(x)≠0 on (a,b). If f′(x)/g′(x) is increasing (decreasing) on (a,b), then so are
(11)f(x)-f(a)g(x)-g(a),f(x)-f(b)g(x)-g(b).
If f′(x)/g′(x) is strictly monotone, then the monotonicity in the conclusion is also strict.Lemma 4.
Letu∈[0,1], p≥1/2 and
(12)fu,p(x)=plog(1+ux2)-logx+logarctanx.
Then(1)
fu,p(x)>0 for x∈(0,1) if and only if 3pu≥1;(2)
fu,p(x)<0 for x∈(0,1) if and only if 1+u≤(4/π)1/p.Proof.
By (12) and simple computations one has
(13)limx→0fu,p(x)=0,(14)fu,p(1)=plog(1+u)+log(π4),(15)fu,p′(x)=2pux1+ux2+1(1+x2)arctanx-1x=(u[(2p-1)x2(1+x2)arctanx+x3]000-[(1+x2)arctanx-x])00×(x(1+x2)(1+ux2)arctanx)-1=(2p-1)x2(1+x2)arctanx+x3x(1+x2)(1+ux2)arctanx[u-g(x)],
where
(16)g(x)=(1+x2)arctanx-x(2p-1)x2(1+x2)arctanx+x3.
Letg1(x)=arctanx-x/(1+x2) and g2(x)=(2p-1)x2arctanx+x3/(1+x2). Then g2(x) is strictly increasing in (0,1):
(17)g(x)=g1(x)g2(x),g1(0)=g2(0)=0,(18)g1′(x)g2′(x)=1(2p-1)[(1+x2)2arctanx]/x+px2+p+1.
It is not difficult to verify that the functionx→[(1+x2)2arctanx]/x is strictly increasing from (0,1) onto (1,π); hence (18) implies that g1′(x)/g2′(x) is strictly decreasing in (0,1). Therefore, g(x) is strictly decreasing in (0,1) follows from Lemma 3 and (17) together with the monotonicity of g2(x) and g1′(x)/g2′(x). Moreover, making use of l'Hôpital's rule we get
(19)limx→0g(x)=13p,(20)g(1)=π-2(2p-1)π+2.
We divide the proof into three cases.
Case  1 (u≥1/(3p)). Then from (15) and (19) together with the monotonicity of g(x) we conclude that fu,p(x) is strictly increasing in (0,1). Therefore fu,p(x)>0 for x∈(0,1) follows from (13) and the monotonicity of fu,p(x).
Case  2 (u≤(π-2)/[(2p-1)π+2]). Then from (15) and (20) together with the monotonicity of g(x) we clearly see that fu,p(x) is strictly decreasing in (0,1). Therefore fu,p(x)<0 for x∈(0,1) follows from (13) and the monotonicity of fu,p(x).
Case  3 ((π-2)/[(2p-1)π+2]<u<1/(3p)). Then from (15), (19), and (20) together with the monotonicity of g(x) we know that there exists x0∈(0,1) such that fu,p(x) is strictly decreasing in (0,x0) and strictly increasing in (x0,1). Therefore, fu,p(x)<0 in (0,1) if and only if fu,p(1)≤0 follows from (13) and the piecewise monotonicity of fu,p(x), which (14) gives immediately 1+u≤(4/π)1/p.Proof of Theorem1.
Since bothQt,p(a,b) and T(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) and (7) we get
(21)log(Qt,p(a,b)T(a,b))=log(Qt,p(a,b)A(a,b))-log(T(a,b)A(a,b))=plog[1+(1-2t)2x2]-logx+logarctanx.
Therefore, Theorem1 follows from Lemma 4 and (21).



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*Source: 438971-2013-08-01.xml*