Research Article Some Inequalities for Bounding Toader Mean · 2019. 7. 31. · Toader mean and the...

Hindawi Publishing CorporationJournal of Function Spaces and ApplicationsVolume 2013, Article ID 394194, 5 pageshttp://dx.doi.org/10.1155/2013/394194

Research ArticleSome Inequalities for Bounding Toader Mean

Wen-Hui Li and Miao-Miao Zheng

Department of Mathematics, School of Science, Tianjin Polytechnic University, Tianjin 300387, China

Correspondence should be addressed to Wen-Hui Li; [email protected]

Received 8 May 2013; Revised 11 June 2013; Accepted 16 June 2013

Academic Editor: L. E. Persson

Copyright © 2013 W.-H. Li and M.-M. Zheng. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.

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

1. Introduction

It is well known that the quantities

𝐴 (𝑎, 𝑏) =𝑎 + 𝑏

2, 𝐺 (𝑎, 𝑏) = √𝑎𝑏 ,

𝐻 (𝑎, 𝑏) =2𝑎𝑏

𝑎 + 𝑏, 𝐶 (𝑎, 𝑏) =

2 (𝑎2+ 𝑎𝑏 + 𝑏

2)

3 (𝑎 + 𝑏),

𝐶 (𝑎, 𝑏) =𝑎2+ 𝑏2

𝑎 + 𝑏, 𝑆 (𝑎, 𝑏) = √

𝑎2+ 𝑏2

2,

𝑀𝑝(𝑎, 𝑏) =

{{

{{

{

(𝑎𝑝+ 𝑎𝑝

2)

1/𝑝

, 𝑝 ̸= 0

√𝑎𝑏, 𝑝 = 0

(1)

are, respectively, called in the literature the arithmetic, geo-metric, harmonic, centroidal, contraharmonic, root-squaremeans, and the power mean of order 𝑝 of two positivenumbers 𝑎 and 𝑏. In [1], Toader introduced a mean

𝑇 (𝑎, 𝑏) =2

𝜋∫

𝜋/2

0

√𝑎2cos2𝜃 + 𝑏2sin2𝜃 𝑑𝜃

=

{{{{{{{{

{{{{{{{{

{

2𝑎

𝜋E(√1 − (

𝑏

𝑎)

2

) , 𝑎 > 𝑏,

2𝑏

𝜋E(√1 − (

𝑏

𝑎)

2

) , 𝑎 < 𝑏,

𝑎, 𝑎 = 𝑏,

(2)

where

E = E(𝑟) = ∫𝜋/2

0

√1 − 𝑟2sin2𝜃 𝑑𝜃,

E= E(𝑟) = E (𝑟

) , E (0) =

𝜋

2, E (1) = 1,

(3)

for 𝑟 ∈ [0, 1] and 𝑟 = √1 − 𝑟2 is Legendre’s complete ellipticintegral of the second kind; see [2] and [3, pages 40–46].

In [4], Vuorinen conjectured that

𝑀3/2

(𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) (4)

for all 𝑎, 𝑏 > 0with 𝑎 ̸= 𝑏.This conjecturewas verified in [5, 6],respectively. Later in [7], it was presented that

𝑇 (𝑎, 𝑏) < 𝑀(ln 2)/ ln(𝜋/2) (𝑎, 𝑏) (5)

for all𝑎, 𝑏 > 0with𝑎 ̸= 𝑏.The constants 3/2 and ln 2/ln(𝜋/2) =1.53 . . . which appeared in (4) and (5) are the best possible.

Utilizing inequalities (4) and (5) and using the factthat the power mean 𝑀

𝑝(𝑎, 𝑏) is continuous and strictly

increasing with respect to 𝑝 ∈ R for fixed 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏may conclude that

𝐴 (𝑎, 𝑏) = 𝑀1(𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝑀

2(𝑎, 𝑏) = 𝑆 (𝑎, 𝑏) (6)

for all 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏. In [8, Theorem 3.1], it wasdemonstrated that the double inequality

𝛼𝑆 (𝑎, 𝑏) + (1 − 𝛼)𝐴 (𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) < 𝛽𝑆 (𝑎, 𝑏) + (1 − 𝛽)𝐴 (𝑎, 𝑏)

(7)

2 Journal of Function Spaces and Applications

holds for all 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏 if and only if

𝛼 ≤1

2, 𝛽 ≥

4 − 𝜋

(√2 − 1) 𝜋

. (8)

Recently in [9, Theorems 1.1 to 1.3], it was shown that thedouble inequalities

𝛼1𝐶 (𝑎, 𝑏) + (1 − 𝛼

1) 𝐴 (𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) < 𝛽1𝐶 (𝑎, 𝑏) + (1 − 𝛽

1) 𝐴 (𝑎, 𝑏) ,

(9)

𝛼2

𝐴 (𝑎, 𝑏)+

1 − 𝛼2

𝐶 (𝑎, 𝑏)

<1

𝑇 (𝑎, 𝑏)<

𝛽2

𝐴 (𝑎, 𝑏)+

1 − 𝛽2

𝐶 (𝑎, 𝑏)(10)

hold for all 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏 if and only if

𝛼1≤3

4, 𝛽1≥12

𝜋− 3, 𝛼

2≤ 𝜋 − 3, 𝛽

2≥1

4. (11)

The equation (4.4) in [10, page 1013] reads that

2 [𝐴 (𝑎, 𝑏) − 𝐻 (𝑎, 𝑏)] =3

2[𝐶 (𝑎, 𝑏) − 𝐻 (𝑎, 𝑏)]

= 𝐶 (𝑎, 𝑏) − 𝐻 (𝑎, 𝑏) =(𝑎 − 𝑏)

2

𝑎 + 𝑏.

(12)

Motivated by (12), we further find that

6 [𝐶 (𝑎, 𝑏) − 𝐴 (𝑎, 𝑏)] = 3 [𝐶 (𝑎, 𝑏) − 𝐶 (𝑎, 𝑏)]

= 2 [𝐶 (𝑎, 𝑏) − 𝐴 (𝑎, 𝑏)] =(𝑎 − 𝑏)

2

𝑎 + 𝑏.

(13)

It is not difficult to see that the double inequality (9) can berearranged as

𝛼1<𝑇 (𝑎, 𝑏) − 𝐴 (𝑎, 𝑏)

𝐶 (𝑎, 𝑏) − 𝐴 (𝑎, 𝑏)

< 𝛽1. (14)

Therefore, replacing the denominator in (14) by one ofdifferences in (12) and (13) yields

1

2𝛼1𝐶 (𝑎, 𝑏) −

1

2𝛼1𝐶 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) <1

2𝛽1𝐶 (𝑎, 𝑏) −

1

2𝛽1𝐶 (𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) ,

(15)

1

3𝛼1𝐶 (𝑎, 𝑏) + (1 −

1

3𝛼1)𝐴 (𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) <1

3𝛽1𝐶 (𝑎, 𝑏) + (1 −

1

3𝛽1)𝐴 (𝑎, 𝑏) ,

(16)

(1 +1

3𝛼1)𝐴 (𝑎, 𝑏) −

1

3𝛼1𝐻(𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) < (1 +1

3𝛽1)𝐴 (𝑎, 𝑏) −

1

3𝛽1𝐻(𝑎, 𝑏) ,

(17)

1

4𝛼1𝐶 (𝑎, 𝑏) −

1

4𝛼1𝐻(𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) <1

4𝛽1𝐶 (𝑎, 𝑏) −

1

4𝛽1𝐻(𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) ,

(18)

1

6𝛼1𝐶 (𝑎, 𝑏) −

1

6𝛼1𝐻(𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

< 𝑇 (𝑎, 𝑏) <1

6𝛽1𝐶 (𝑎, 𝑏) −

1

6𝛽1𝐻(𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) ,

(19)

where𝛼1and𝛽1satisfy (11). On the other hand, the arithmetic

mean𝐴(𝑎, 𝑏) in the numerator of (14) can also be replaced bythe harmonic, contraharmonic, or centroidal means.

For our own convenience, we denote the difference ofmeans in (12) and (13) by

𝑀𝐶𝐻

(𝑎, 𝑏) =(𝑎 − 𝑏)

2

𝑎 + 𝑏. (20)

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 constants 𝛼 and 𝛽 such that thedouble inequality

𝛼 <𝑇 (𝑎, 𝑏) − 𝐻 (𝑎, 𝑏)

𝑀𝐶𝐻

(𝑎, 𝑏)< 𝛽 (21)

holds for all positive numbers 𝑎 and 𝑏 with 𝑎 ̸= 𝑏?

The main purposes of this paper are to answer the previ-ous problem, to provide an alternative proof for inequalities(14) to (19), and, finally, to remark the connection betweenToader mean and the complete elliptic integral of the secondkind.

Journal of Function Spaces and Applications 3

2. Lemmas

To attain our main purposes, we need the following lemmas.For 0 < 𝑟 < 1, denote 𝑟 = √1 − 𝑟2. Legendre’s complete

elliptic integrals of the first kind may be defined in [12, 13] by

K = K(𝑟) = ∫𝜋/2

0

1

√1 − 𝑟2sin2𝜃𝑑𝜃,

K= K(𝑟) = K(𝑟

) , K(0) =

𝜋

2, K(1) = ∞.

(22)

Lemma 2 (see [14, Appendix E, pages 474-475]). For 0 < 𝑟 <1 and 𝑟 = √1 − 𝑟2 , one has

𝑑K

𝑑𝑟=

E − (𝑟)2

K

𝑟(𝑟)2

,𝑑E

𝑑𝑟=E −K

𝑟,

𝑑 (E − (𝑟)2

K)

𝑑𝑟= 𝑟K, E(

2√𝑟

1 + 𝑟) =

2E − (𝑟)2

K

1 + 𝑟.

(23)

Lemma 3 (see [14, Theorem 1.25]). For −∞ < 𝑎 < 𝑏 < ∞,let 𝑓, 𝑔 : [𝑎, 𝑏] → R be continuous on [𝑎, 𝑏], differentiableon (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 thefunctions

𝑓 (𝑥) − 𝑓 (𝑎)

𝑔 (𝑥) − 𝑔 (𝑎),

𝑓 (𝑥) − 𝑓 (𝑏)

𝑔 (𝑥) − 𝑔 (𝑏). (24)

Lemma 4 (see [14, Theorem 3.21]). The function

ℎ (𝑟) =

E − (𝑟)2

K

𝑟2(25)

is strictly increasing and convex from (0, 1) onto (𝜋/4, 1).

Lemma 5. The function

𝑓 (𝑟) =

2E − (𝑟)2

K

𝑟2(26)

is strictly decreasing on (0, 1) and satisfies

lim𝑟→0

+

𝑓 (𝑟) = ∞, lim𝑟→1

−

𝑓 (𝑟) = 2. (27)

Proof. By the first three formulas in Lemma 2, simple com-putations lead to

𝑓(𝑟) =

(𝑟)2

K − 3E

𝑟3≜𝑓1(𝑟)

𝑟3,

𝑓

1(𝑟) =

(𝑟)2

K +K − 2E

𝑟≜𝑓2(𝑟)

𝑟,

𝑓

2(𝑟) =

𝑟

(𝑟)2[E − (𝑟

)2

K] ≜𝑟3

(𝑟)2ℎ (𝑟) ,

(28)

where the function ℎ(𝑟) is defined by (25) in Lemma 4.

From

𝑓

1(0) = 𝑓

2(0) = 𝑓

2(0) = 0, 𝑓

1(1) = −3, (29)

it follows that

𝑓

2(𝑟) > 0, 𝑓

2(𝑟) > 0, 𝑓

1(𝑟) > 0,

𝑓1(𝑟) < 0, 𝑓

(𝑟) < 0.

(30)

Hence, the function 𝑓(𝑟) is strictly decreasing on (0, 1).Further, by easily obtained limits in (27), the proof ofLemma 5 is complete.

3. Some Inequalities for BoundingToader Mean

Now, we are in a position to give an affirmative solution toProblem 1 and to provide an alternative proof for inequalities(14) to (19).

Theorem 6. The double inequality

𝛼𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐻 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝛽𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐻 (𝑎, 𝑏)

(31)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏 if and only if

𝛼 ≤5

8= 0.625, 𝛽 ≥

2

𝜋= 0.636 . . . . (32)

Proof. Without loss of generality, assume that 𝑎 > 𝑏 > 0. Let𝑡 = 𝑏/𝑎. Then, 𝑡 ∈ (0, 1) and

𝑇 (𝑎, 𝑏) − 𝐻 (𝑎, 𝑏)

𝑀𝐶𝐻

(𝑎, 𝑏)=

(2/𝜋)E (√1 − 𝑡2 ) − (2𝑡/ (1 + 𝑡))

(1 − 𝑡)2/ (1 + 𝑡)

.

(33)

Let 𝑟 = (1− 𝑡)/(1+ 𝑡). Then, 𝑟 ∈ (0, 1), and by the last formulain Lemma 2,

𝑇 (𝑎, 𝑏) − 𝐻 (𝑎, 𝑏)

𝑀𝐶𝐻

(𝑎, 𝑏)=(2/𝜋)E (2√𝑟 / (1 + 𝑟)) + 𝑟 − 1

2𝑟2/ (1 + 𝑟)

=𝑓1(𝑟)

𝑓2(𝑟)

+1

2,

(34)

where

𝑓1(𝑟) =

2

𝜋[2E − (𝑟

)2

K] − 1, 𝑓2(𝑟) = 2𝑟

2. (35)

By the middle two formulas in Lemma 2, a straightforwardcalculation leads to

𝑓1(0) = 𝑓

2(0) = 0, 𝑓

1(𝑟) =

2

𝜋

E − (𝑟)2

K

𝑟, 𝑓

2(𝑟) = 4𝑟.

(36)

So, we have

𝑓

1(𝑟)

𝑓

2(𝑟)

=1

2𝜋

E − (𝑟)2

K

𝑟2=ℎ (𝑟)

2𝜋. (37)

4 Journal of Function Spaces and Applications

Combining this with Lemmas 3 and 4 reveals that thefunction𝑓

1(𝑟)/𝑓2(𝑟) is strictly increasing on (0, 1). Moreover,

using L’Hôpital’s rule, we obtain

lim𝑟→0

+

𝑓1(𝑟)

𝑓2(𝑟)

+1

2= lim𝑟→0

+

1

2𝜋

E − (𝑟)2

K

𝑟2+1

2=5

8,

lim𝑟→1

−

𝑓1(𝑟)

𝑓2(𝑟)

+1

2= lim𝑟→1

−

(2/𝜋) [2E − (𝑟)2

K] − 1

2𝑟2+1

2

= lim𝑟→1

−

E

𝜋𝑟2+ lim𝑟→1

−

ℎ (𝑟)

𝜋=2

𝜋.

(38)

The proof of Theorem 6 is thus complete.

Theorem 7. For all 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏,(1) the double inequality

𝛼1𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐴 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝛽1𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐴 (𝑎, 𝑏)

(39)

holds if and only if

𝛼1≤1

8= 0.125, 𝛽

1≥2

𝜋−1

2= 0.136 . . . , (40)

(2) the double inequality

𝛼2𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐶 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝛽2𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐶 (𝑎, 𝑏)

(41)


𝛼2≤ −

3

8= −0.375, 𝛽

2≥2

𝜋− 1 = −0.363 . . . , (42)

(3) the double inequality

𝛼3𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐶 (𝑎, 𝑏) < 𝑇 (𝑎, 𝑏) < 𝛽3𝑀𝐶𝐻

(𝑎, 𝑏) + 𝐶 (𝑎, 𝑏)

(43)


𝛼3≤ −

1

24= −0.041 . . . , 𝛽

3≥2

𝜋−2

3= −0.030 . . . .

(44)

Proof. From the identities in (12), it follows that

𝐻(𝑎, 𝑏) = 𝐴 (𝑎, 𝑏) −1

2𝑀𝐶𝐻

(𝑎, 𝑏)

= 𝐶 (𝑎, 𝑏) − 𝑀𝐶𝐻

(𝑎, 𝑏) = 𝐶 (𝑎, 𝑏) −2

3𝑀𝐶𝐻

(𝑎, 𝑏) .

(45)

Substituting these into the inequality (31) in Theorem 6acquires inequalities (39) to (43) inTheorem 7.

Theorem 8. The inequality

𝑇 (𝑎, 𝑏) > 𝜆𝑀𝐶𝐻

(𝑎, 𝑏) (46)

holds for all 𝑎, 𝑏 > 0 with 𝑎 ̸= 𝑏 if and only if 𝜆 ≤ 2/𝜋.

Proof. Without loss of generality, assume that 𝑎 > 𝑏 > 0. Let𝑡 = 𝑏/𝑎. Then, 𝑡 ∈ (0, 1) and

𝑇 (𝑎, 𝑏)

𝑀𝐶𝐻

(𝑎, 𝑏)=

(2/𝜋)E (√1 − 𝑡2 )

(1 − 𝑡)2/ (1 + 𝑡)

. (47)

Let 𝑟 = (1− 𝑡)/(1+ 𝑡). Then, 𝑟 ∈ (0, 1), and by the last formulain Lemma 2,

𝑇 (𝑎, 𝑏)

𝑀𝐶𝐻

(𝑎, 𝑏)=(2/𝜋)E (2√𝑟 / (1 + 𝑟))

2𝑟2/ (1 + 𝑟)=1

𝜋

2E − (𝑟)2

K

𝑟2.

(48)

Therefore, from Lemma 5, it follows that function 𝑇(𝑎, 𝑏)/𝑀𝐶𝐻(𝑎, 𝑏) is strictly decreasing and

lim𝑟→1

−

𝑇 (𝑎, 𝑏)

𝑀𝐶𝐻

(𝑎, 𝑏)= lim𝑟→1

−

1

𝜋

2E − (𝑟)2

K

𝑟2=2

𝜋. (49)

Theorem 8 is thus proved.

4. Remarks

Finally, we would like to remark several things, including theconnection between Toader mean and the complete ellipticintegral of the second kind.

Remark 9. The double inequality (39) is equivalent to (9).Consequently, inequalities (14) to (19) are recovered onceagain.

Remark 10. The coefficient 3/2 in (12) corrects an error whichappeared at the corresponding position in (4.4) in [10, page1013]. Luckily, this error does not influence the correctness ofany other conclusions in [10].

Remark 11. We point out that Toader mean 𝑇(𝑎, 𝑏) satisfies

𝑇 (𝑎, 𝑏) = 𝑅𝐸(𝑎2, 𝑏2) , (50)

where

𝑅𝐸(𝑥, 𝑦) =

1

𝜋∫

∞

0

(𝑥

𝑡 + 𝑥+

𝑦

𝑡 + 𝑦)

𝑡

√(𝑡 + 𝑥) (𝑡 + 𝑦)

𝑑𝑡

(51)

is the complete symmetric elliptic integral of the secondkind and is a symmetric and homogeneous function; see[15, equation (9.2-3)] and [16, page 250, equation (1.6)].Numerous inequalities involving 𝑅

𝐸,E, andK 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 𝑅

𝐸are

the same has been overlooked by several researchers.

Acknowledgments

The authors would like to thank the anonymous referees fortheir valuable comments on the original version of this paperand Professor Feng Qi for his kind help in the whole processof composing this paper.

Journal of Function Spaces and Applications 5

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