# Sharp Bounds for Toader Mean in terms of Arithmetic and Second Contraharmonic Means

Authors: Wei-Mao Qian; Ying-Qing Song; Xiao-Hui Zhang; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We present the best possible parametersλ1,μ1∈R and λ2,μ2∈1/2,1 such that double inequalities λ1C(a,b)+1-λ1A(a,b)<T(a,b)<μ1C(a,b)+1-μ1A(a,b), Cλ2a+1-λ2b,λ2b+1-λ2a<T(a,b)<Cμ2a+1-μ2b,μ2b+1-μ2a hold for all a,b>0 with a≠b, where A(a,b)=(a+b)/2, C(a,b)=a3+b3/a2+b2 and T(a,b)=2∫0π/2a2cos2θ+b2sin2θdθ/π are the arithmetic, second contraharmonic, and Toader means of a and b, respectively.


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## Body

## 1. Introduction

Fora,b>0 the Toader mean T(a,b) [1], second contraharmonic mean C(a,b), and arithmetic mean A(a,b) of a and b are given by(1)Ta,b=2π∫0π/2a2cos2⁡θ+b2sin2⁡θdθ=2aE1-b/a2π,a>b,2bE1-a/b2π,a<b,a,a=b,(2)Ca,b=a3+b3a2+b2,Aa,b=a+b2, respectively, where E(r)=∫0π/21-r2sin2⁡t1/2dt(r∈(0,1)) is the complete elliptic integral of the second kind. The Toader mean T(a,b) is well known in mathematical literature for many years; it satisfies(3)Ta,b=REa2,b2,T1,r=2πE1-r2 for all a,b>0 and 0<r<1, where (4)REa,b=1π∫0∞at+b+bt+att+a3/2t+b3/2dt stands for the symmetric complete elliptic integral of the second kind (see [2–4]); therefore it cannot be expressed in terms of the elementary transcendental functions.Recently, the Toader meanT(a,b) has been the subject of intensive research. In particular, many remarkable inequalities for the Toader mean can be found in the literature [5–9].Letp∈R, q∈[0,1], and a,b>0. Then the pth power mean Mp(a,b), pth Gini mean Gp(a,b), pth Lehmer mean Lp(a,b), and qth generalized Seiffert mean Sq(a,b) are defined by (5)
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, respectively. It is well known that Mp(a,b), Gp(a,b), Lp(a,b), and Sq(a,b) are continuous and strictly increasing with respect to p∈R and q∈[0,1] for fixed a,b>0 with a≠b, respectively.Vuorinen [10] conjectured that inequality (6)M3/2a,b<Ta,b holds for all a,b>0 with a≠b. This conjecture was proved by Qiu and Shen [11] and Barnard et al. [12], respectively.Alzer and Qiu [13] presented a best possible upper power mean bound for the Toader mean as follows: (7)Ta,b<Mlog⁡2/log⁡π-log⁡2a,b for all a,b>0 with a≠b.In [14, 15], the authors found the best possible parameters α,β∈[0,1] and λ,μ∈R such that double inequalities Sα(a,b)<T(a,b)<Sβ(a,b) and Gλ(a,b)<T(a,b)<Gμ(a,b) hold for all a,b>0 with a≠b.Chu and Wang [16] proved that double inequality(8)Lpa,b<Ta,b<Lqa,b holds for all a,b>0 with a≠b if and only if p≤0 and q≥1/4.Inequality (8) leads to(9)Aa,b=L0a,b<Ta,b<L1/4a,b<L2a,b=Ca,b for all a,b>0 with a≠b.Leta,b>0 with a≠b be fixed and f(x)=C[xa+(1-x)b,xb+(1-x)a]. Then it is not difficult to verify that f(x) is continuous and strictly increasing on [1/2,1]. Note that(10)f12=Aa,b<Ta,b<Ca,b=f1.Motivated by inequalities (9) and (10), it is natural to ask what are the best possible parameters, λ1,μ1∈R and λ2,μ2∈(1/2,1), such that double inequalities (11)λ1Ca,b+1-λ1Aa,b<Ta,b<μ1Ca,b+1-μ1Aa,b,Cλ2a+1-λ2b,λ2b+1-λ2a<Ta,b<Cμ2a+1-μ2b,μ2b+1-μ2a hold for all a,b>0 with a≠b? The main purpose of this paper is to answer this question.


## 2. Main Results

In order to prove our main results we need some basic knowledge and two lemmas, which we present in this section.Forr∈(0,1) the complete elliptic integral K(r) of the first kind is defined by (12)Kr=∫0π/21-r2sin2⁡t-1/2dt.We clearly see that(13)K0+=E0+=π2,E1-=1, and K(r) and E(r) satisfy formulas (see [17, Appendix E, p. 474-475]) (14)dKrdr=Er-1-r2Krr1-r2,dErdr=Er-Krr,E2r1+r=2Er-1-r2Kr1+r.Lemma 1 (see [17, Theorem 1.25]).
Let-∞<a<b<∞, f,g:[a,b]→(-∞,∞) be continuous on [a,b] and differentiable on (a,b), and g′(x)≠0 on (a,b). If f′(x)/g′(x) is increasing (decreasing) on (a,b), then so are (15)fx-fagx-ga,fx-fbgx-gb. If f′(x)/g′(x) is strictly monotone, then the monotonicity in the conclusion is also strict.Lemma 2 (see [17, Theorem 3.21]).
(1) Functionr↦[E(r)-(1-r2)K(r)]/r2 is strictly increasing from (0,1) to (π/4,1).
(2) Functionr↦(1-r2)λK(r) is strictly decreasing from (0,1) to (0,π/2) if λ≥1/4.Theorem 3.
Double inequality(16)λ1Ca,b+1-λ1Aa,b<Ta,b<μ1Ca,b+1-μ1Aa,b holds for all a,b>0 with a≠b if and only if λ1≤1/8 and μ1≥4/π-1=0.2732….Proof.
SinceA(a,b), T(a,b), and C(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b>0. Let r=(a-b)/(a+b)∈(0,1). Then (1) and (2) lead to(17)Ta,b=2aπE2r1+r=2Aa,bπ2Er-1-r2Kr,(18)Ca,b=Aa,b1+3r21+r2.
We clearly see that inequality (16) is equivalent to(19)λ1<Ta,b-Aa,bCa,b-Aa,b<μ1.
It follows from (17) and (18) that(20)Ta,b-Aa,bCa,b-Aa,b=1π2Er-1-r2Kr-π/2r2/1+r2.
Let(21)fr=2Er-1-r2Kr-π/2r2/1+r2,f1r=2Er-1-r2Kr-π2,f2r=r21+r2.Then simple computations lead to(22)f10+=f20+=0,f1′rf2′r=1+r222Er-1-r2Krr2.
From Lemmas1 and 2 together with (21) and (22) we know that f(r) is strictly increasing on (0,1) and(23)f0+=f1′0+f2′0+=π8,f1-=4-π.
Therefore, Theorem3 follows from (19)–(21) and (23) together with the monotonicity of f(r).Theorem 4.
Letλ2, μ2∈(1/2,1). Then double inequality (24)Cλ2a+1-λ2b,λ2b+1-λ2a<Ta,b<Cμ2a+1-μ2b,μ2b+1-μ2aholds for all a,b>0 with a≠b if and only if λ2≤1/2+2/8=0.6767… and μ2≥1/2+(4-π)/(3π-4)/2=0.6988….Proof.
Letλ2∗=1/2+2/8 and μ2∗=1/2+(4-π)/(3π-4)/2. We first prove that(25)Ta,b>Cλ2∗a+1-λ2∗b,λ2∗b+1-λ2∗a,(26)Ta,b<Cμ2∗a+1-μ2∗b,μ2∗b+1-μ2∗afor all a,b>0 with a≠b.
Without loss of generality, we assume thata>b. Let r=(a-b)/(a+b)∈(0,1) and p∈(1/2,1). Then (2) leads to(27)Cpa+1-pb,pb+1-pa=Aa,b32p-12r2+12p-12r2+1.
It follows from (17) and (27) that(28)Ta,b-Cpa+1-pb,pb+1-pa=Aa,b2π2Er-1-r2Kr+22p-12r2+1-3.Let(29)gr=2π2Er-1-r2Kr+22p-12r2+1-3,(30)g1r=1rg′r.Then making use of Lemma 2 and simple computations lead to(31)g0=0,(32)g1=4π+22p-12+1-3,(33)g1r=2πEr-1-r2Krr2-42p-122p-12r2+12,(34)g10=12-42p-12,(35)g11=2π-42p-122p-12+12.
We divide the proof into two cases.
Case 1. Consider p=λ2∗=1/2+2/8. Then (34) becomes(36)g10=0.It follows from Lemma 2(1) and (33) together with (36) that(37)g1r>0 for all r∈(0,1).
Therefore, inequality (25) follows easily from (28)–(31) and (37).
Case 2. Consider p=μ2∗=1/2+(4-π)/(3π-4)/2. Then (32), (34), and (35) lead to(38)g1=0,(39)g10=-36-11π6π-8<0,(40)g11=3π2-14π+16π2>0.It follows from Lemma 2(1), (33), (39), and (40) that there exists r0∈(0,1) such that g1(r)<0 for r∈(0,r0) and g1(r)>0 for r∈(r0,1). Then (30) leads to the conclusion that g(r) is strictly decreasing on (0,r0] and strictly increasing on (r0,1].
Therefore, inequality (26) follows easily from (28), (29), (31), (38), and the piecewise monotonicity of g(r).
Next, we prove thatλ2=λ2∗=1/2+2/8 is the best possible parameter on (1/2,1) such that inequality (41)Ta,b>Cλ2a+1-λ2b,λ2b+1-λ2aholds for all a,b>0 with a≠b.
Indeed, ifλ2∗=1/2+2/8<p<1, then (34) leads to g1(0)<0 and there exists δ1∈(0,1) such that(42)g1r<0 for r∈(0,δ1).
Equations (28)–(31) and inequality (42) imply that (43)Ta,b<Cpa+1-pb,pb+1-pafor (a-b)/(a+b)∈(0,δ1).
Finally, we prove thatμ2=μ2∗=1/2+(4-π)/(3π-4)/2 is the best possible parameter on (1/2,1) such that double inequality (44)Ta,b<Cμ2a+1-μ2b,μ2b+1-μ2afor all a,b>0 with a≠b.
In fact, if1/2<p<μ2∗=1/2+(4-π)/(3π-4)/2, then (32) leads to g(1)>0 and there exists δ2∈(0,1) such that(45)gr>0for r∈(1-δ2,1).
Equations (28) and (29) together with inequality (45) imply that (46)Ta,b>Cpa+1-pb,pb+1-pafor (a-b)/(a+b)∈(1-δ2,1).Letr∈(0,1), a=1, b=1-r2, λ1=1/8, μ1=4/π-1, λ2=1/2+2/8, and μ2=1/2+(4-π)/(3π-4)/2. Then Theorems 3 and 4 lead to Corollary 5 as follows.Corollary 5.
Double inequalities(47)π1+1-r23/2162-r2+7π1+1-r21/232<Er<4-π1+1-r23/222-r2+π2-11+1-r21/2,π232-21r2+211-r21/2+111-r23/236-18r2+281-r21/2<Er<π23π-4-3π-2r2+3π-21-r21/2+21-r23/2π2-r2+4π-21-r21/2hold for all r∈(0,1).



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