# Bounds for Combinations of Toader Mean and Arithmetic Mean in Terms of Centroidal Mean

Authors: Wei-Dong Jiang
Journal: The Scientific World Journal
 (2013)

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## Abstract

The authors find the greatest valueλ and the least value μ, such that the double inequality C¯(λa+(1-λb),λb+(1-λ)a)<αA(a,b)+(1-α)T(a,b)<C¯(μa+(1-μ)b,μb+(1-μ)a) holds for all α∈(0,1) and a,b>0 with a≠b, where C¯(a,b)=2(a2+ab+b2)/3(a+b), A(a,b)=(a+b)/2, and Ta,b=2/π∫0π/2a2cos2θ+b2sin2θdθ denote, respectively, the centroidal, arithmetic, and Toader means of the two positive numbers a and b.


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## Body

## 1. Introduction

In [1], Toader introduced a mean
(1)T(a,b)=2π∫0π/2a2cos2θ+b2sin2θdθ={2aℰ1-(b/a)2π,a>b,2bℰ1-(a/b)2π,a<b,a,a=b,
where
(2)ℰ=ℰ(r)=∫0π/2(1-r2sin2θ)1/2dθ,
for r∈[0,1] is the complete elliptic integral of the second kind.In recent years, there have been plenty of literature, such as [2–6], dedicated to the Toader mean.Forp∈ℝ and a,b>0, the centroidal mean C¯(a,b) and pth power mean Mp(a,b) are, respectively, defined by
(3)C¯(a,b)=2(a2+ab+b2)3(a+b),Mp(a,b)={(ap+ap2)1/p,p≠0,ab,p=0.In [7], Vuorinen conjectured that
(4)M3/2(a,b)<T(a,b),
for all a,b>0 with a≠b. This conjecture was verified by Qiu and Shen [8] and by Barnard et al. [9], respectively.In [10], Alzer and Qiu presented a best possible upper power mean bound for the Toader mean as follows:
(5)T(a,b)<Mlog2/log(π/2)(a,b),
for all a,b>0 with a≠b.Chu et al. [5] proved that the double inequality
(6)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≤3/4 and β≥1/2+4π-π2/(2π).Very recently, Hua and Qi [11] proved that the double inequality
(7)αC¯(a,b)+(1-α)A(a,b)<T(a,b)<βC¯(a,b)+(1-β)A(a,b)
is valid for all a,b>0 with a≠b if and only if α≤3/4 and β≥(12/π)-3. Where A(a,b)=(a+b)/2 denote the arithmetic mean.For positive numbersa,b>0 with a≠b, let
(8)J(x)=C¯(xa+(1-x)b,xb+(1-x)a)
be on [1/2,1]. It is not difficult to directly verify that J(x) is continuous and strictly increasing on [1/2,1].The main purpose of the paper is to find the greatest valueλ and the least value μ, such that the double inequality C¯(λa+(1-λb),λb+(1-λ)a)<αA(a,b)+(1-α)T(a,b)<C¯(μa+(1-μ)b,μb+(1-μ)a) holds for all α∈(0,1) and a,b>0 with a≠b. As applications, we also present new bounds for the complete elliptic integral of the second kind.


## 2. Preliminaries and Lemmas

In order to establish our main result, we need several formulas and Lemmas below.For0<r<1 and r′=1-r2, Legendre’s complete elliptic integrals of the first and second kinds are defined in [12, 13] by
(9)𝒦=𝒦(r)=∫0π/2(1-r2sin2θ)-1/2dθ,𝒦'=𝒦′(r)=𝒦(r′),𝒦(0)=π2,𝒦(1)=∞,ℰ=ℰ(r)=∫0π/2‍(1-r2sin2θ)1/2dθ,ℰ'=ℰ′(r)=ℰ(r′),ℰ(0)=π2,ℰ(1)=1,
respectively.For0<r<1, the formulas
(10)d𝒦dr=ℰ-r′2𝒦rr′2,dℰdr=ℰ-𝒦r,d(ℰ-r′2𝒦)dr=r𝒦,d(𝒦-ℰ)dr=rℰr′2,ℰ(2r1+r)=2ℰ-r′2𝒦1+r
were presented in [14, Appendix E, pages 474-475].Lemma 1 (see [14, Theorem 3.21(1), 3.43 exercises 13(a)]).
The function(ℰ-r′2𝒦)/r2 is strictly increasing from (0,1) to (π/4,1), and the function 2ℰ-r′2𝒦 is increasing from (0,1) to (π/2,2).Lemma 2.
Letu,α∈(0,1) and
(11)fu,α(r)=13ur2-(1-α)(2π(2ℰ(r)-(1-r2)𝒦(r))-1).
Then, fu,α>0, for all r∈(0,1) if and only if u≥3(1-α)(4/π-1), and fu,α<0, for all r∈(0,1) if and only if u≤3(1-α)/4.Proof.
From (11), one has
(12)fu,α(0+)=0,(13)fu,α(1-)=13u-(1-α)(4π-1),(14)fu,α′(r)=23r[u-3(1-α)g(r)],
where g(r)=(1/π)((ℰ-r′2𝒦)/r2).
We divide the proof into four cases.
Case 1 (
u
≥
3
(
1
-
α
)
/
π
).  From (14) and Lemma 1 together with the monotonicity of g(r), we clearly see that fu,α(r) is strictly increasing on (0,1). Therefore, fu,α(r)>0, for all r∈(0,1).
Case 2 (
u
≤
3
(
1
-
α
)
/
4
). From (14) and Lemma 1 together with the monotonicity of g(r), we obtain that fu,α(r) is strictly decreasing on (0,1). Therefore, fu,α(r)<0, for all r∈(0,1).
Case 3 (
3
(
1
-
α
)
/
4
<
u
≤
3
(
1
-
α
)
(
4
/
π
-
1
)
). From (13) and (14) together with the monotonicity of g(r), we see that there exists λ∈(0,1), such that fu,α(r) is strictly increasing in (0,λ] and strictly decreasing in [λ,1) and
(15)fu,α(1-)≤0.
Therefore, making use of (12) and inequality (15) together with the piecewise monotonicity of fu,α(r) leads to the conclusion that there exists 0<λ<η<1, such that fu,α(r)>0 for r∈(0,η) and fu,α(r)<0 for r∈(η,1).
Case 4 (
3
(
1
-
α
)
(
4
/
π
-
1
)
≤
u
<
3
(
1
-
α
)
/
π
). Equation (13) leads to
(16)fu,α(1-)≥0.
From (13) and (14) together with the monotonicity of g(r), we clearly see that there exists λ∈(0,1), such that fu,α(r) is strictly increasing in (0,λ] and strictly decreasing in [λ,1). Therefore, fu,α(r)>0 for r∈(0,1) follows from (12) and (16) together with the piecewise monotonicity of fu,α(r).


## 3. Main Results

Now, we are in a position to state and prove our main results.Theorem 3.
Ifα∈(0,1) and λ,μ∈(1/2,1), then the double inequality
(17)C¯(λa+(1-λ)b,λb+(1-λ)a)<αA(a,b)+(1-α)T(a,b)<C¯(μa+(1-μ)b,μb+(1-μ)a)
holds for all a,b>0 with a≠b if and only if
(18)λ≤12+3(1-α)4,μ≥12(1+3(1-α)(4π-1)).Proof.
SinceA(a,b), T(a,b), and C¯(a,b) are symmetric and homogeneous of degree one, without loss of generality, we assume that a>b. Let p∈(1/2,1), t=b/a∈(0,1), and r=(1-t)/(1+t). Then,
(19)C¯(pa+(1-p)b,pb+(1-p)a)-αA(a,b)-(1-α)T(a,b)=a23((p+(1-p)ba)2+(p+(1-p)ba)(pba+1-p)+(pba+1-p)2)(1+ba)-1-αa1+(b/a)2-(1-α)2aπℰ(1-(ba)2)=a{23((pt+1-p)2(pt+1-p)2(p+(1-p)t)2+(p+(1-p)t)(pt+1-p)+(pt+1-p)2)(1+t)-1-α1+t2-(1-α)2πℰ(1-t2)}=a{(1-2p)2r2+33(1+r)-α11+r-(1-α)2π2ℰ-r′2𝒦1+r}=a1+r[13(1-2p)2r2+1-α-(1-α)2π(2ℰ-r′2𝒦)].
Therefore, Theorem 3 follows easily from Lemma 2 and (19).Letα=1/4,  λ=7/8,  μ=(1/2)(1+(34/π-1/2)). Then, from Theorem 3, we get new bounds for the complete elliptic integral ℰ(r) of the second kind in terms of elementary functions as follows.Corollary 4.
Forr∈(0,1) and r′=1-r2, one has
(20)π2[5+6r′+5r′28(1+r′)]<ℰ(r)<π[r′+(2/π)(1-r′)21+r′].


## 4. Remarks

Remark 5.
In the recent past, the complete elliptic integrals have attracted the attention of numerous mathematicians. In [4], it was established that
(21)π2[121+r′22+1+r′4]<ℰ(r)<π2[4-π(2-1)π1+r′22+(2π-4)(1+r′)2(2-1)π],
for all r∈(0,1).
Guo and Qi [15] proved that
(22)π2-12log(1+r)1-r(1-r)1+r<ℰ(r)<π-12+1-r24rlog1+r1-r,
for all r∈(0,1).
Yin and Qi [16] presented that
(23)π26+21-r2-3r222≤ℰ(r)≤π210-21-r2-5r222,
for all r∈(0,1).
It was pointed out in [4] that the bounds in (21) for ℰ(r) are better than the bounds in (22) for some r∈(0,1).Remark 6.
The lower bound in (20) for ℰ(r) is better than the lower bound in (21). Indeed,
(24)5+6x+5x28(1+x)-[121+x22+1+x4]=3x2+2x+3-22(1+x2)(1+x)8(1+x),(3x2+2x+3)2-(22(1+x2)(1+x))2=(1-x)4>0,
for all x∈(0,1).Remark 7.
The following equivalence relations forx∈(0,1) show that the lower bound in (20) for ℰ(r) is better than the lower bound in (23):
(25)5+6x+5x28(1+x)>6+2x-3(1-x2)22⟺(5x2+6x+5)2>8(x+1)2(3x2+2x+3)⟺(x-1)4>0.



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*Source: 842542-2013-10-22.xml*