# Some Inequalities for Bounding Toader Mean

Authors: Wen-Hui Li; Miao-Miao Zheng
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

By finding linear relations among differences between two special means, the authors establish some inequalities for bounding Toader mean in terms of the arithmetic, harmonic, centroidal, and contraharmonic means.


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## Body

## 1. Introduction

It is well known that the quantities(1)A(a,b)=a+b2,G(a,b)=ab,H(a,b)=2aba+b,C¯(a,b)=2(a2+ab+b2)3(a+b),C(a,b)=a2+b2a+b,S(a,b)=a2+b22,Mp(a,b)={(ap+ap2)1/p,p≠0ab,p=0
are, respectively, called in the literature the arithmetic, geometric, harmonic, centroidal, contraharmonic, root-square means, and the power mean of order p of two positive numbers a and b. In [1], Toader introduced a mean
(2)T(a,b)=2π∫0π/2‍a2cos2θ+b2sin2θdθ={2aπℰ(1-(ba)2),a>b,2bπℰ(1-(ba)2),a<b,a,a=b,
where
(3)ℰ=ℰ(r)=∫0π/2‍1-r2sin2θdθ,ℰ′=ℰ′(r)=ℰ(r′),ℰ(0)=π2,ℰ(1)=1,
for r∈[0,1] and r′=1-r2 is Legendre's complete elliptic integral of the second kind; see [2] and [3, pages 40–46].In [4], Vuorinen conjectured that
(4)M3/2(a,b)<T(a,b)
for all a,b>0 with a≠b. This conjecture was verified in [5, 6], respectively. Later in [7], it was presented that
(5)T(a,b)<M(ln2)/ln(π/2)(a,b)
for all a,b>0 with a≠b. The constants 3/2 and ln2/ln(π/2)=1.53… which appeared in (4) and (5) are the best possible.Utilizing inequalities (4) and (5) and using the fact that the power mean Mp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b may conclude that
(6)A(a,b)=M1(a,b)<T(a,b)<M2(a,b)=S(a,b)
for all a,b>0 with a≠b. In [8, Theorem 3.1], it was demonstrated that the double inequality
(7)αS(a,b)+(1-α)A(a,b)<T(a,b)<βS(a,b)+(1-β)A(a,b)
holds for all a,b>0 with a≠b if and only if
(8)α≤12,β≥4-π(2-1)π.Recently in [9, Theorems 1.1 to 1.3], it was shown that the double inequalities
(9)α1C¯(a,b)+(1-α1)A(a,b)<T(a,b)<β1C¯(a,b)+(1-β1)A(a,b),(10)α2A(a,b)+1-α2C¯(a,b)<1T(a,b)<β2A(a,b)+1-β2C¯(a,b)
hold for all a,b>0 with a≠b if and only if
(11)α1≤34,β1≥12π-3,α2≤π-3,β2≥14.The equation (4.4) in [10, page 1013] reads that
(12)2[A(a,b)-H(a,b)]=32[C¯(a,b)-H(a,b)]=C(a,b)-H(a,b)=(a-b)2a+b.
Motivated by (12), we further find that
(13)6[C¯(a,b)-A(a,b)]=3[C(a,b)-C¯(a,b)]=2[C(a,b)-A(a,b)]=(a-b)2a+b.
It is not difficult to see that the double inequality (9) can be rearranged as
(14)α1<T(a,b)-A(a,b)C¯(a,b)-A(a,b)<β1.
Therefore, replacing the denominator in (14) by one of differences in (12) and (13) yields
(15)12α1C(a,b)-12α1C¯(a,b)+A(a,b)<T(a,b)<12β1C(a,b)-12β1C¯(a,b)+A(a,b),(16)13α1C(a,b)+(1-13α1)A(a,b)<T(a,b)<13β1C(a,b)+(1-13β1)A(a,b),(17)(1+13α1)A(a,b)-13α1H(a,b)<T(a,b)<(1+13β1)A(a,b)-13β1H(a,b),(18)14α1C¯(a,b)-14α1H(a,b)+A(a,b)<T(a,b)<14β1C¯(a,b)-14β1H(a,b)+A(a,b),(19)16α1C(a,b)-16α1H(a,b)+A(a,b)<T(a,b)<16β1C(a,b)-16β1H(a,b)+A(a,b),
where α1 and β1 satisfy (11). On the other hand, the arithmetic mean A(a,b) in the numerator of (14) can also be replaced by the harmonic, contraharmonic, or centroidal means.For our own convenience, we denote the difference of means in (12) and (13) by
(20)MCH(a,b)=(a-b)2a+b.
The quantity MCH(a,b) is nonnegative and convex on (0,∞)×(0,∞). See [11, Theorem 2.1].Now we naturally pose the following problem.Problem 1.
What are the best constantsα and β such that the double inequality
(21)α<T(a,b)-H(a,b)MCH(a,b)<β
holds for all positive numbers a and b with a≠b?The main purposes of this paper are to answer the previous problem, to provide an alternative proof for inequalities (14) to (19), and, finally, to remark the connection between Toader mean and the complete elliptic integral of the second kind.


## 2. Lemmas

To attain our main purposes, we need the following lemmas.For0<r<1, denote r′=1-r2. Legendre's complete elliptic integrals of the first kind may be defined in [12, 13] by
(22)𝒦=𝒦(r)=∫0π/2‍11-r2sin2θdθ,𝒦′=𝒦′(r)=𝒦(r′),𝒦(0)=π2,𝒦(1)=∞.Lemma 2 (see [14, Appendix E, pages 474-475]).
For0<r<1 and r′=1-r2, one has
(23)d𝒦dr=ℰ-(r′)2𝒦r(r′)2,dℰdr=ℰ-𝒦r,d(ℰ-(r′)2𝒦)dr=r𝒦,ℰ(2r1+r)=2ℰ-(r′)2𝒦1+r.Lemma 3 (see [14, Theorem 1.25]).
For-∞<a<b<∞, let f,g:[a,b]→ℝ be continuous on [a,b], differentiable on (a,b), and g′(x)≠0 on (a,b). If f′(x)/g′(x) is (strictly) increasing (or (strictly) decreasing, resp.) on (a,b), so are the functions
(24)f(x)-f(a)g(x)-g(a),f(x)-f(b)g(x)-g(b).Lemma 4 (see [14, Theorem 3.21]).
The function(25)h(r)=ℰ-(r′)2𝒦r2
is strictly increasing and convex from (0,1) onto (π/4,1).Lemma 5.
The function(26)f(r)=2ℰ-(r′)2𝒦r2
is strictly decreasing on (0,1) and satisfies
(27)limr⟶0+f(r)=∞,limr⟶1-f(r)=2.Proof.
By the first three formulas in Lemma2, simple computations lead to
(28)f′(r)=(r′)2𝒦-3ℰr3≜f1(r)r3,f1′(r)=(r′)2𝒦+𝒦-2ℰr≜f2(r)r,f2′(r)=r(r′)2[ℰ-(r′)2𝒦]≜r3(r′)2h(r),
where the function h(r) is defined by (25) in Lemma 4.
From(29)f1′(0)=f2(0)=f2′(0)=0,f1(1)=-3,
it follows that
(30)f2′(r)>0,f2(r)>0,f1′(r)>0,f1(r)<0,f′(r)<0.
Hence, the function f(r) is strictly decreasing on (0,1). Further, by easily obtained limits in (27), the proof of Lemma 5 is complete.


## 3. Some Inequalities for Bounding Toader Mean

Now, we are in a position to give an affirmative solution to Problem1 and to provide an alternative proof for inequalities (14) to (19).Theorem 6.
The double inequality(31)αMCH(a,b)+H(a,b)<T(a,b)<βMCH(a,b)+H(a,b)
holds for all a,b>0 with a≠b if and only if
(32)α≤58=0.625,β≥2π=0.636….Proof.
Without loss of generality, assume thata>b>0. Let t=b/a. Then, t∈(0,1) and
(33)T(a,b)-H(a,b)MCH(a,b)=(2/π)ℰ(1-t2)-(2t/(1+t))(1-t)2/(1+t).
Let r=(1-t)/(1+t). Then, r∈(0,1), and by the last formula in Lemma 2,
(34)T(a,b)-H(a,b)MCH(a,b)=(2/π)ℰ(2r/(1+r))+r-12r2/(1+r)=f1(r)f2(r)+12,
where
(35)f1(r)=2π[2ℰ-(r′)2𝒦]-1,f2(r)=2r2.
By the middle two formulas in Lemma 2, a straightforward calculation leads to
(36)f1(0)=f2(0)=0,f1′(r)=2πℰ-(r′)2𝒦r,f2′(r)=4r.
So, we have
(37)f1′(r)f2′(r)=12πℰ-(r′)2𝒦r2=h(r)2π.
Combining this with Lemmas 3 and 4 reveals that the function f1(r)/f2(r) is strictly increasing on (0,1). Moreover, using L’Hôpital’s rule, we obtain
(38)limr⟶0+f1(r)f2(r)+12=limr⟶0+12πℰ-(r′)2𝒦r2+12=58,limr⟶1-f1(r)f2(r)+12=limr⟶1-(2/π)[2ℰ-(r′)2𝒦]-12r2+12=limr⟶1-ℰπr2+limr⟶1-h(r)π=2π.
The proof of Theorem 6 is thus complete.Theorem 7.
For alla,b>0 with a≠b,
(
1
) the double inequality
(39)α1MCH(a,b)+A(a,b)<T(a,b)<β1MCH(a,b)+A(a,b)
holds if and only if(40)α1≤18=0.125,β1≥2π-12=0.136…,
(
2
) the double inequality
(41)α2MCH(a,b)+C(a,b)<T(a,b)<β2MCH(a,b)+C(a,b)
holds if and only if(42)α2≤-38=-0.375,β2≥2π-1=-0.363…,
(
3
) the double inequality
(43)α3MCH(a,b)+C¯(a,b)<T(a,b)<β3MCH(a,b)+C¯(a,b)
holds if and only if(44)α3≤-124=-0.041…,β3≥2π-23=-0.030….Proof.
From the identities in (12), it follows that
(45)H(a,b)=A(a,b)-12MCH(a,b)=C(a,b)-MCH(a,b)=C¯(a,b)-23MCH(a,b).
Substituting these into the inequality (31) in Theorem 6 acquires inequalities (39) to (43) in Theorem 7.Theorem 8.
The inequality(46)T(a,b)>λMCH(a,b)
holds for all a,b>0 with a≠b if and only if λ≤2/π.Proof.
Without loss of generality, assume thata>b>0. Let t=b/a. Then, t∈(0,1) and
(47)T(a,b)MCH(a,b)=(2/π)ℰ(1-t2)(1-t)2/(1+t).
Let r=(1-t)/(1+t). Then, r∈(0,1), and by the last formula in Lemma 2,
(48)T(a,b)MCH(a,b)=(2/π)ℰ(2r/(1+r))2r2/(1+r)=1π2ℰ-(r′)2𝒦r2.
Therefore, from Lemma 5, it follows that function T(a,b)/MCH(a,b) is strictly decreasing and
(49)limr⟶1-T(a,b)MCH(a,b)=limr⟶1-1π2ℰ-(r′)2𝒦r2=2π.
Theorem 8 is thus proved.


## 4. Remarks

Finally, we would like to remark several things, including the connection between Toader mean and the complete elliptic integral of the second kind.Remark 9.
The double inequality (39) is equivalent to (9). Consequently, inequalities (14) to (19) are recovered once again.Remark 10.
The coefficient3/2 in (12) corrects an error which appeared at the corresponding position in (4.4) in [10, page 1013]. Luckily, this error does not influence the correctness of any other conclusions in [10].Remark 11.
We point out that Toader meanT(a,b) satisfies
(50)T(a,b)=RE(a2,b2),
where
(51)RE(x,y)=1π∫0∞‍(xt+x+yt+y)t(t+x)(t+y)dt
is the complete symmetric elliptic integral of the second kind and is a symmetric and homogeneous function; see [15, equation (9.2-3)] and [16, page 250, equation (11)]. Numerous inequalities involving RE, ℰ, and 𝒦 are known in the mathematical literature; see [2, 16–19] and [3, pages 40–46] and closely related references therein. In the past years, the fact that Toader mean T and the elliptic integral RE are the same has been overlooked by several researchers.



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