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(1r2sin2θ)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.
For p∈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+(1α)b,αb+(1α)a) <T(a,b) <C(βa+(1β)b,βb+(1β)a)
holds for all a,b>0 with a≠b if and only if α≤3/4 and β≥1/2+4ππ2/(2π).
Very recently, Hua and Qi [11] proved that the double inequality
(7)αC¯(a,b)+(1α)A(a,b) <T(a,b) <βC¯(a,b)+(1β)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 numbers a,b>0 with a≠b, let
(8)J(x)=C¯(xa+(1x)b,xb+(1x)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 value λ and the least value μ, such that the double inequality C¯(λa+(1λb),λb+(1λ)a)<αA(a,b)+(1α)T(a,b)<C¯(μa+(1μ)b,μb+(1μ)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.
For 0<r<1 and r′=1r2, Legendre’s complete elliptic integrals of the first and second kinds are defined in [12, 13] by
(9)𝒦=𝒦(r)=∫0π/2(1r2sin2θ)1/2dθ,𝒦'=𝒦′(r)=𝒦(r′),𝒦(0)=π2, 𝒦(1)=∞,ℰ=ℰ(r)=∫0π/2(1r2sin2θ)1/2dθ,ℰ'=ℰ′(r)=ℰ(r′),ℰ(0)=π2, ℰ(1)=1,
respectively.
For 0<r<1, the formulas
(10)d𝒦dr=ℰr′2𝒦rr′2, dℰdr=ℰ𝒦r,d(ℰr′2𝒦)dr=r𝒦, d(𝒦ℰ)dr=rℰr′2,ℰ(2r1+r)=2ℰr′2𝒦1+r
were presented in [14, Appendix E, pages 474475].
Lemma 1 (see [<xref reftype="bibr" rid="B3">14</xref>, Theorem 3.21(1), 3.43 exercises 13(a)]).
The function (ℰr′2𝒦)/r2 is strictly increasing from (0,1) to (π/4,1), and the function 2ℰr′2𝒦 is increasing from (0,1) to (π/2,2).
Lemma 2.
Let u,α∈(0,1) and
(11)fu,α(r)=13ur2 (1α)(2π(2ℰ(r)(1r2)𝒦(r))1).
Then, fu,α>0, for all r∈(0,1) if and only if u≥3(1α)(4/π1), and fu,α<0, for all r∈(0,1) if and only if u≤3(1α)/4.
Proof.
From (11), one has
(12)fu,α(0+)=0,(13)fu,α(1)=13u(1α)(4π1),(14)fu,α′(r)=23r[u3(1α)g(r)],
where g(r)=(1/π)((ℰr′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.
If α∈(0,1) and λ,μ∈(1/2,1), then the double inequality
(17)C¯(λa+(1λ)b,λb+(1λ)a) <αA(a,b)+(1α)T(a,b) <C¯(μa+(1μ)b,μb+(1μ)a)
holds for all a,b>0 with a≠b if and only if
(18)λ≤12+3(1α)4,μ≥12(1+3(1α)(4π1)).
Proof.
Since A(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=(1t)/(1+t). Then,
(19)C¯(pa+(1p)b,pb+(1p)a) αA(a,b)(1α)T(a,b) =a23((p+(1p)ba)2 +(p+(1p)ba)(pba+1p) +(pba+1p)2)(1+ba)1 αa1+(b/a)2 (1α)2aπℰ(1(ba)2) =a{23((pt+1p)2(pt+1p)2(p+(1p)t)2 +(p+(1p)t)(pt+1p) +(pt+1p)2)(1+t)1 α1+t2(1α)2πℰ(1t2)} =a{(12p)2r2+33(1+r)α11+r (1α)2π2ℰr′2𝒦1+r} =a1+r[13(12p)2r2+1α (1α)2π(2ℰr′2𝒦)].
Therefore, Theorem 3 follows easily from Lemma 2 and (19).
Let α=1/4, λ=7/8, μ=(1/2)(1+(34/π1/2)). 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.
For r∈(0,1) and r′=1r2, one has
(20)π2[5+6r′+5r′28(1+r′)]<ℰ(r)<π[r′+(2/π)(1r′)21+r′].