On Some Intermediate Mean Values

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Volume 2013, Article ID 283127,7pages http://dx.doi.org/10.1155/2013/283127

Research Article

On Some Intermediate Mean Values

Slavko Simic

Mathematical Institute SANU, Kneza Mihaila 36, 11000 Belgrade, Serbia

Correspondence should be addressed to Slavko Simic; [email protected]

Received 25 June 2012; Revised 9 December 2012; Accepted 16 December 2012

Academic Editor: Mowaffaq Hajja

Copyright © 2013 Slavko Simic. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We give a necessary and sufficient mean condition for the quotient of two Jensen functionals and define a new classΛ_𝑓,𝑔(𝑎, 𝑏)of mean values where𝑓, 𝑔are continuously differentiable convex functions satisfying the relation𝑓󸀠󸀠(𝑡) = 𝑡𝑔󸀠󸀠(𝑡),𝑡 ∈ R+. Then we asked for a characterization of𝑓, 𝑔such that the inequalities𝐻(𝑎, 𝑏) ≤ Λ_𝑓,𝑔(𝑎, 𝑏) ≤ 𝐴(𝑎, 𝑏)or𝐿(𝑎, 𝑏) ≤ Λ_𝑓,𝑔(𝑎, 𝑏) ≤ 𝐼(𝑎, 𝑏)hold for each positive𝑎, 𝑏, where𝐻, 𝐴, 𝐿, 𝐼are the harmonic, arithmetic, logarithmic, and identric means, respectively. For a subclass ofΛ with𝑔󸀠󸀠(𝑡) = 𝑡𝑠, 𝑠 ∈R, this problem is thoroughly solved.

1. Introduction

It is said that the mean𝑃is intermediate relating to the means 𝑀and𝑁, 𝑀 ≤ 𝑁if the relation

𝑀 (𝑎, 𝑏) ≤ 𝑃 (𝑎, 𝑏) ≤ 𝑁 (𝑎, 𝑏) (1)

holds for each two positive numbers𝑎,𝑏. It is also well known that

min{𝑎, 𝑏} ≤ 𝐻 (𝑎, 𝑏) ≤ 𝐺 (𝑎, 𝑏)

≤ 𝐿 (𝑎, 𝑏) ≤ 𝐼 (𝑎, 𝑏) ≤ 𝐴 (𝑎, 𝑏) ≤ 𝑆 (𝑎, 𝑏) ≤ max{𝑎, 𝑏} ,

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where

𝐻 = 𝐻 (𝑎, 𝑏) := 2(1_𝑎+1 𝑏)

−1 ;

𝐺 = 𝐺 (𝑎, 𝑏) := √𝑎𝑏; 𝐿 = 𝐿 (𝑎, 𝑏) := 𝑏 − 𝑎 log 𝑏 −log 𝑎;

𝐼 = 𝐼 (𝑎, 𝑏) :=(𝑏

𝑏_/𝑎𝑎₎1/(𝑏−𝑎)

𝑒 ;

𝐴 = 𝐴 (𝑎, 𝑏) := 𝑎 + 𝑏₂ ; 𝑆 = 𝑆 (𝑎, 𝑏) := 𝑎𝑎/(𝑎+𝑏)𝑏𝑏/(𝑎+𝑏) (3)

are the harmonic, geometric, logarithmic, identric, arith-metic, and Gini mean, respectively.

An easy task is to construct intermediate means related to two given means𝑀and𝑁with𝑀 ≤ 𝑁. For instance, for an arbitrary mean𝑃, we have that

𝑀 (𝑎, 𝑏) ≤ 𝑃 (𝑀 (𝑎, 𝑏) , 𝑁 (𝑎, 𝑏)) ≤ 𝑁 (𝑎, 𝑏). (4)

The problem is more difficult if we have to decide whether the given mean is intermediate or not. For example, the relation

𝐿 (𝑎, 𝑏) ≤ 𝑆𝑠(𝑎, 𝑏) ≤ 𝐼 (𝑎, 𝑏) (5)

holds for each positive𝑎and𝑏if and only if0 ≤ 𝑠 ≤ 1, where the Stolarsky mean𝑆_𝑠is defined by (cf [1])

𝑆_𝑠(𝑎, 𝑏) := ( 𝑏𝑠− 𝑎𝑠 𝑠 (𝑏 − 𝑎))

1/(𝑠−1)

. (6)

Also,

𝐺 (𝑎, 𝑏) ≤ 𝐴𝑠(𝑎, 𝑏) ≤ 𝐴 (𝑎, 𝑏) (7)

holds if and only if0 ≤ 𝑠 ≤ 1, where the H¨older mean of order 𝑠is defined by

𝐴𝑠(𝑎, 𝑏) := (𝑎 𝑠_{+ 𝑏}𝑠

2 )

1/𝑠

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An inverse problem is to find best possible approximation of a given mean𝑃by elements of an ordered class of means 𝑆. A good example for this topic is comparison between the logarithmic mean and the class𝐴_𝑠of H¨older means of order 𝑠. Namely, since𝐴₀=lim_{𝑠 → 0}𝐴_𝑠= 𝐺and𝐴₁= 𝐴, it follows from (2) that

𝐴₀≤ 𝐿 ≤ 𝐴₁. (9)

Since𝐴_𝑠is monotone increasing in𝑠, an improving of the above is given by Carlson [2]:

𝐴₀≤ 𝐿 ≤ 𝐴_1/2. (10)

Finally, Lin showed in [3] that

𝐴₀≤ 𝐿 ≤ 𝐴_1/3 (11)

is the best possible approximation of the logarithmic mean by the means from the class𝐴_𝑠.

Numerous similar results have been obtained recently. For example, an approximation of Seiffert’s mean by the class 𝐴𝑠is given in [4,5].

In this paper we will give best possible approximations for a whole variety of elementary means (2) by the class𝜆_𝑠 defined below (seeTheorem 5).

Let 𝑓, 𝑔 be twice continuously differentiable (strictly) convex functions onR+. By definition (cf [6], page 5),

𝑓 (𝑎, 𝑏) := 𝑓 (𝑎) + 𝑓 (𝑏) − 2𝑓 (𝑎 + 𝑏₂ ) > 0, 𝑎 ̸= 𝑏,

𝑓 (𝑎, 𝑏) = 0,

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if and only if𝑎 = 𝑏.

It turns out that the expression

Λ_𝑓,𝑔(𝑎, 𝑏) := 𝑓 (𝑎, 𝑏)

𝑔 (𝑎, 𝑏) = 𝑓 (𝑎) + 𝑓 (𝑏) − 2𝑓 ((𝑎 + 𝑏) /2)𝑔 (𝑎) + 𝑔 (𝑏) − 2𝑔 ((𝑎 + 𝑏) /2) (13)

represents a mean of two positive numbers𝑎,𝑏; that is, the relation

min{𝑎, 𝑏} ≤ Λ_𝑓,𝑔(𝑎, 𝑏) ≤max{𝑎, 𝑏} (14)

holds for each𝑎, 𝑏 ∈R+, if and only if the relation

𝑓󸀠󸀠(𝑡) = 𝑡𝑔󸀠󸀠(𝑡) (15)

holds for each𝑡 ∈R+.

Let𝑓, 𝑔 ∈ 𝐶∞(0, ∞)and denote byΛthe set{(𝑓, 𝑔)}of convex functions satisfying the relation (15). There is a natural question how to improve the bounds in (14); in this sense we come upon the following intermediate mean problem.

Open Question. Under what additional conditions on𝑓, 𝑔 ∈

Λ, the inequalities

𝐻 (𝑎, 𝑏) ≤ Λ𝑓,𝑔(𝑎, 𝑏) ≤ 𝐴 (𝑎, 𝑏), (16)

or, more tightly,

𝐿 (𝑎, 𝑏) ≤ Λ𝑓,𝑔(𝑎, 𝑏) ≤ 𝐼 (𝑎, 𝑏), (17)

hold for each𝑎, 𝑏 ∈R+?

As an illustration, consider the function𝑓_𝑠(𝑡)defined to be

𝑓𝑠(𝑡) = { { { { { { {

𝑡𝑠_{− 𝑠𝑡 + 𝑠 − 1}

𝑠 (𝑠 − 1) , 𝑠 (𝑠 − 1) ̸= 0; 𝑡 −log𝑡 − 1, 𝑠 = 0; 𝑡log𝑡 − 𝑡 + 1, 𝑠 = 1.

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Since

𝑓_𝑠󸀠(𝑡) = { { { { { { { { { { { { { { {

𝑡𝑠−1_{− 1}

𝑠 − 1 , 𝑠 (𝑠 − 1) ̸= 0; 1 − l_𝑡, 𝑠 = 0;

log𝑡, 𝑠 = 1,

𝑓󸀠󸀠

𝑠 (𝑡) = 𝑡𝑠−2, 𝑠 ∈R, 𝑡 > 0,

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it follows that 𝑓_𝑠(𝑡) is a twice continuously differentiable convex function for𝑠 ∈R,𝑡 ∈R+.

Moreover, it is evident that(𝑓_𝑠+1, 𝑓_𝑠) ∈ Λ.

We will give in the sequel a complete answer to the above question concerning the means

𝑓_𝑠+1(𝑎, 𝑏)

𝑓_𝑠(𝑎, 𝑏) := 𝜆𝑠(𝑎, 𝑏) (20)

defined by 𝜆𝑠(𝑎, 𝑏)

= { { { { { { { { { { { { { { { { { { { { { { { { {

𝑠 − 1 𝑠 + 1

𝑎𝑠+1_{+ 𝑏}𝑠+1_{− 2((𝑎 + 𝑏) /2)}𝑠+1

𝑎𝑠_{+ 𝑏}𝑠_{− 2((𝑎 + 𝑏) /2)}𝑠 , 𝑠 ∈R/ {−1, 0, 1} ;

2log((𝑎 + 𝑏) /2) −log𝑎 −log𝑏

1/2𝑎 + 1/2𝑏 − 2/ (𝑎 + 𝑏) , 𝑠 = −1;

𝑎log𝑎 + 𝑏log𝑏 − (𝑎 + 𝑏)log((𝑎 + 𝑏) /2)

2log((𝑎 + 𝑏) /2) −log𝑎 −log𝑏 , 𝑠 = 0;

(𝑏 − 𝑎)2

4 (𝑎log𝑎 + 𝑏log𝑏 − (𝑎 + 𝑏)log((𝑎 + 𝑏) /2)), 𝑠 = 1.

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Those means are obviously symmetric and homogeneous of order one.

As a consequence we obtain some new intermediate mean values; for instance, we show that the inequalities

𝐻 (𝑎, 𝑏) ≤ 𝜆−1(𝑎, 𝑏) ≤ 𝐺 (𝑎, 𝑏) ≤ 𝜆0(𝑎, 𝑏) ≤ 𝐿 (𝑎, 𝑏)

≤ 𝜆₁(𝑎, 𝑏) ≤ 𝐼 (𝑎, 𝑏) (22)

hold for arbitrary𝑎, 𝑏 ∈R+. Note that

𝜆₋₁ = 2𝐺2_{𝐴 − 𝐻}log(𝐴/𝐺); 𝜆₀= 𝐴log(𝑆/𝐴) log(𝐴/𝐺);

𝜆₁= 1 2

𝐴 − 𝐻 log(𝑆/𝐴).

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2. Results

We prove firstly the following

Theorem 1. Let𝑓, 𝑔 ∈ 𝐶2(𝐼)with𝑔󸀠󸀠 > 0. The expression

Λ_𝑓,𝑔(𝑎, 𝑏)represents a mean of arbitrary numbers𝑎, 𝑏 ∈ 𝐼if

and only if the relation(15)holds for𝑡 ∈ 𝐼.

Remark 2. In the same way, for arbitrary𝑝, 𝑞 > 0, 𝑝 + 𝑞 = 1,

it can be deduced that the quotient

Λ_𝑓,𝑔(𝑝, 𝑞; 𝑎, 𝑏) := 𝑝𝑓 (𝑎) + 𝑞𝑓 (𝑏) − 𝑓 (𝑝𝑎 + 𝑞𝑏) 𝑝𝑔 (𝑎) + 𝑞𝑔 (𝑏) − 𝑔 (𝑝𝑎 + 𝑞𝑏) (24) represents a mean value of numbers𝑎, 𝑏if and only if (15) holds.

A generalization of the above assertion is the next.

Theorem 3. Let 𝑓, 𝑔 : 𝐼 → R be twice continuously

differentiable functions with𝑔󸀠󸀠 > 0on𝐼and let𝑝 = {𝑝_𝑖},

𝑖 = 1, 2, . . . , ∑ 𝑝_𝑖= 1be an arbitrary positive weight sequence. Then the quotient of two Jensen functionals

Λ_𝑓,𝑔(𝑝, 𝑥) := ∑𝑛1𝑝𝑖𝑓 (𝑥𝑖) − 𝑓 (∑𝑛1𝑝𝑖𝑥𝑖)

∑𝑛₁𝑝_𝑖𝑔 (𝑥_𝑖) − 𝑔 (∑𝑛₁𝑝_𝑖𝑥_𝑖), 𝑛 ≥ 2, (25)

represents a mean of an arbitrary set of real numbers

𝑥₁, 𝑥₂, . . . , 𝑥_𝑛∈ 𝐼if and only if the relation

𝑓󸀠󸀠(𝑡) = 𝑡𝑔󸀠󸀠(𝑡) (26)

holds for each𝑡 ∈ 𝐼.

Remark 4. It should be noted that the relation𝑓󸀠󸀠(𝑡) = 𝑡𝑔󸀠󸀠(𝑡)

determines𝑓in terms of𝑔in an easy way. Precisely,

𝑓 (𝑡) = 𝑡𝑔 (𝑡) − 2𝐺 (𝑡) + 𝑐𝑡 + 𝑑, (27)

where𝐺(𝑡) := ∫₁𝑡𝑔(𝑢)𝑑𝑢and𝑐and𝑑are constants.

Our results concerning the means 𝜆_𝑠(𝑎, 𝑏), 𝑠 ∈ R are included in the following.

Theorem 5. For the class of means𝜆_𝑠(𝑎, 𝑏)defined above, the

following assertions hold for each𝑎, 𝑏 ∈R+.

(1)The means𝜆_𝑠(𝑎, 𝑏)are monotone increasing in𝑠;

(2)𝜆_𝑠(𝑎, 𝑏) ≤ 𝐻(𝑎, 𝑏)for each𝑠 ≤ −4;

(3)𝐻(𝑎, 𝑏) ≤ 𝜆_𝑠(𝑎, 𝑏) ≤ 𝐺(𝑎, 𝑏)for−3 ≤ 𝑠 ≤ −1;

(4)𝐺(𝑎, 𝑏) ≤ 𝜆_𝑠(𝑎, 𝑏) ≤ 𝐿(𝑎, 𝑏)for−1/2 ≤ 𝑠 ≤ 0;

(5)there is a number𝑠₀∈ (1/12, 1/11)such that𝐿(𝑎, 𝑏) ≤

𝜆_𝑠(𝑎, 𝑏) ≤ 𝐼(𝑎, 𝑏)for𝑠₀≤ 𝑠 ≤ 1;

(6)there is a number𝑠₁ ∈ (1.03, 1.04)such that𝐼(𝑎, 𝑏) ≤

𝜆_𝑠(𝑎, 𝑏) ≤ 𝐴(𝑎, 𝑏)for𝑠₁≤ 𝑠 ≤ 2;

(7)𝐴(𝑎, 𝑏) ≤ 𝜆_𝑠(𝑎, 𝑏) ≤ 𝑆(𝑎, 𝑏)for each2 ≤ 𝑠 ≤ 5;

(8)there is no finite𝑠such that the inequality𝑆(𝑎, 𝑏) ≤

𝜆_𝑠(𝑎, 𝑏)holds for each𝑎, 𝑏 ∈R+. The above estimations are best possible.

3. Proofs

3.1. Proof ofTheorem 1. We prove firstly the necessity of the

condition (15).

SinceΛ_𝑓,𝑔(𝑎, 𝑏)is a mean value for arbitrary𝑎, 𝑏 ∈ 𝐼; 𝑎 ̸= 𝑏, we have

min{𝑎, 𝑏} ≤ Λ_𝑓,𝑔(𝑎, 𝑏) ≤max{𝑎, 𝑏} . (28)

Hence

lim

𝑏 → 𝑎Λ𝑓,𝑔(𝑎, 𝑏) = 𝑎. (29)

From the other hand, due to l’Hospital’s rule we obtain

lim

𝑏 → 𝑎Λ𝑓,𝑔(𝑎, 𝑏) =𝑏 → 𝑎lim(

𝑓󸀠(𝑏) − 𝑓󸀠((𝑎 + 𝑏) /2) 𝑔󸀠_{(𝑏) − 𝑔}󸀠_{((𝑎 + 𝑏) /2)})

= lim 𝑏 → 𝑎(

2𝑓󸀠󸀠(𝑏) − 𝑓󸀠󸀠((𝑎 + 𝑏) /2) 2𝑔󸀠󸀠_{(𝑏) − 𝑔}󸀠󸀠_{((𝑎 + 𝑏) /2)})

= 𝑓󸀠󸀠(𝑎) 𝑔󸀠󸀠_(𝑎).

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Comparing (29) and (30) the desired result follows.

Suppose now that (15) holds and let𝑎 < 𝑏. Since𝑔󸀠󸀠(𝑡) > 0 𝑡 ∈ [𝑎, 𝑏]by the Cauchy mean value theorem there exists 𝜉 ∈ ((𝑎 + 𝑡)/2, 𝑡)such that

𝑓󸀠_{(𝑡) − 𝑓}󸀠_{((𝑎 + 𝑡) /2)} 𝑔󸀠_{(𝑡) − 𝑔}󸀠_{((𝑎 + 𝑡) /2)} =

𝑓󸀠󸀠_(𝜉)

𝑔󸀠󸀠_(𝜉) = 𝜉. (31) But,

𝑎 ≤ 𝑎 + 𝑡₂ < 𝜉 < 𝑡 ≤ 𝑏, (32)

and, since𝑔󸀠is strictly increasing,𝑔󸀠(𝑡)−𝑔󸀠((𝑎+𝑡)/2) > 0, 𝑡 ∈ [𝑎, 𝑏].

Therefore, by (31) we get

𝑎 (𝑔󸀠(𝑡) − 𝑔󸀠(𝑎 + 𝑡

2 )) ≤ 𝑓󸀠(𝑡) − 𝑓󸀠( 𝑎 + 𝑡

2 )

≤ 𝑏 (𝑔󸀠(𝑡) − 𝑔󸀠(𝑎 + 𝑡 2 )) .

(33)

Finally, integrating (33) over𝑡 ∈ [𝑎, 𝑏]we obtain the assertion fromTheorem 1.

3.2. Proof ofTheorem 3. We will give a proof of this assertion

by induction on𝑛.

ByRemark 2, it holds for𝑛 = 2.

Next, it is not difficult to check the identity 𝑛

∑ 1

𝑝_𝑖𝑓 (𝑥_𝑖) − 𝑓 (∑𝑛 1

𝑝_𝑖𝑥_𝑖)

= (1 − 𝑝_𝑛) (𝑛−1∑ 1

𝑝_𝑖󸀠𝑓 (𝑥_𝑖) − 𝑓 (𝑛−1∑ 1

𝑝_𝑖󸀠𝑥_𝑖))

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where

𝑇 :=𝑛−1∑ 1

𝑝_𝑖󸀠𝑥_𝑖; 𝑝_𝑖󸀠:= 𝑝𝑖

(1 − 𝑝_𝑛), 𝑖 = 1, 2, . . . , 𝑛 − 1;

𝑛−1 ∑ 1 𝑝

󸀠 𝑖 = 1.

(35)

Therefore, by induction hypothesis andRemark 2, we get 𝑛

∑ 1

𝑝_𝑖𝑓 (𝑥_𝑖) − 𝑓 (∑𝑛 1

𝑝_𝑖𝑥_𝑖)

≤max{𝑥₁, 𝑥₂, . . . 𝑥_𝑛−1} (1 − 𝑝_𝑛)

× (𝑛−1∑ 1

𝑝󸀠_𝑖𝑔 (𝑥_𝑖) − 𝑔 (𝑛−1∑ 1

+max{𝑇, 𝑥_𝑛} [(1 − 𝑝_𝑛) 𝑔 (𝑇) + 𝑝_𝑛𝑔 (𝑥_𝑛) −𝑔 ((1 − 𝑝_𝑛) 𝑇 + 𝑝_𝑛𝑥_𝑛)] ≤max{𝑥₁, 𝑥₂, . . . , 𝑥_𝑛}

× ((1 − 𝑝_𝑛) (𝑛−1∑ 1

𝑝_𝑖󸀠𝑔 (𝑥_𝑖) − 𝑔 (𝑛−1∑ 1

+ [(1 − 𝑝_𝑛) 𝑔 (𝑇) + 𝑝𝑛𝑔 (𝑥𝑛) − 𝑔 ((1 − 𝑝𝑛) 𝑇 + 𝑝𝑛𝑥𝑛)] )

=max{𝑥₁, 𝑥₂, . . . , 𝑥_𝑛} ( 𝑛 ∑

1𝑝𝑖𝑔 (𝑥𝑖) − 𝑔 ( 𝑛 ∑

1𝑝𝑖𝑥𝑖)) . (36) The inequality

min{𝑥₁, 𝑥₂, . . . , 𝑥_𝑛} ≤ Λ_𝑓,𝑔(𝑝, 𝑥) (37)

can be proved analogously.

For the proof of necessity, put𝑥₂ = 𝑥₃ = ⋅ ⋅ ⋅ = 𝑥_𝑛 and proceed as inTheorem 1.

Remark 6. It is evident from (15) that if𝐼 ⊆R+then𝑓has to

be also convex on𝐼. Otherwise, it shouldn’t be the case. For example, the conditions ofTheorem 3are satisfied with𝑓(𝑡) = 𝑡3_/3_,_{𝑔(𝑡) = 𝑡}2_{, 𝑡 ∈}_R_{. Hence, for an arbitrary sequence}_{𝑥

𝑖}𝑛1 of real numbers, we obtain

min{𝑥₁, 𝑥₂, . . . , 𝑥_𝑛} ≤ ∑ 𝑛

1𝑝𝑖𝑥3𝑖 − (∑𝑛1𝑝𝑖𝑥𝑖)3 3 (∑𝑛₁𝑝_𝑖𝑥2

𝑖 − (∑𝑛1𝑝𝑖𝑥𝑖)2) ≤max{𝑥₁, 𝑥₂, . . . , 𝑥_𝑛} .

(38)

Because the above inequality does not depend on 𝑛, a probabilistic interpretation of the above result is contained in the following.

Theorem 7. For an arbitrary probability law𝐹 of random

variable𝑋with support on(−∞, +∞), one has

(𝐸𝑋)3+ 3 (min𝑋) 𝜎_𝑋2 ≤ 𝐸𝑋3≤ (𝐸𝑋)3+ 3 (max𝑋) 𝜎2_𝑋. (39)

3.3. Proof of Theorem 5, Part (1). We will prove a general

assertion of this type. Namely, for an arbitrary positive sequencex = {𝑥_𝑖} and an associated weight sequencep = {𝑝_𝑖},𝑖 = 1, 2, . . ., denote

𝜒_𝑠(p,x)

:= { { { { { { { { { { { { {

∑ 𝑝_𝑖𝑥𝑠

𝑖− (∑ 𝑝𝑖𝑥𝑖)𝑠

𝑠 (𝑠 − 1) , 𝑠 ∈R/ {0, 1} ;

log(∑ 𝑝_𝑖𝑥_𝑖) − ∑ 𝑝_𝑖 log 𝑥_𝑖, 𝑠 = 0;

∑ 𝑝_𝑖𝑥_𝑖log𝑥_𝑖− (∑ 𝑝_𝑖𝑥_𝑖)log(∑ 𝑝_𝑖𝑥_𝑖) , 𝑠 = 1.

(40)

For𝑠 ∈R,𝑟 > 0we have

𝜒𝑠(p,x) 𝜒𝑠+𝑟+1(p,x) ≥ 𝜒𝑠+1(p,x) 𝜒𝑠+𝑟(p,x) , (41)

which is equivalent to

Theorem 8. The sequence{𝜒_𝑠+1(p,x)/𝜒_𝑠(p,x)} is monotone

increasing in𝑠,𝑠 ∈R.

This assertion follows applying the result from [7, Theo-rem 2] which states the following.

Lemma 9. For−∞ < 𝑎 < 𝑏 < 𝑐 < +∞, the inequality

(𝜒_𝑏(p,x))𝑐−𝑎≤ (𝜒_𝑎(p,x))𝑐−𝑏(𝜒_𝑐(p,x))𝑏−𝑎 (42)

holds for arbitrary sequencesp,x.

Putting there𝑎 = 𝑠,𝑏 = 𝑠 + 1,𝑐 = 𝑠 + 𝑟 + 1and𝑎 = 𝑠, 𝑏 = 𝑠 + 𝑟,𝑐 = 𝑠 + 𝑟 + 1, we successively obtain

(𝜒_𝑠+1(p,x))𝑟+1≤ (𝜒_𝑠(p,x))𝑟𝜒_𝑠+𝑟+1(p,x),

(𝜒𝑠+𝑟(p,x))𝑟+1≤ 𝜒𝑠(p,x) (𝜒𝑠+𝑟+1(p,x))𝑟.

(43)

Since 𝑟 > 0, multiplying those inequalities we get the relation (41), that is, the proof ofTheorem 8.

The part (1) ofTheorem 5follows for𝑝₁= 𝑝₂= 1/2. A general way to prove the rest ofTheorem 5is to use an easy-checkable identity

𝜆_𝑠(𝑎, 𝑏)

𝐴 (𝑎, 𝑏) = 𝜆𝑠(1 + 𝑡, 1 − 𝑡) , (44)

(5)

Since0 < 𝑎 < 𝑏, we get0 < 𝑡 < 1. Also,

𝐻 (𝑎, 𝑏)

𝐴 (𝑎, 𝑏) = 1 − 𝑡2; 𝐺 (𝑎, 𝑏)𝐴 (𝑎, 𝑏) = √1 − 𝑡2; 𝐿 (𝑎, 𝑏)

𝐴 (𝑎, 𝑏) =

2𝑡

log(1 + 𝑡) −log(1 − 𝑡); 𝐼 (𝑎, 𝑏)

𝐴 (𝑎, 𝑏)

=exp((1 + 𝑡)log(1 + 𝑡) − (1 − 𝑡)log(1 − 𝑡)

2𝑡 − 1) ;

𝑆 (𝑎, 𝑏) 𝐴 (𝑎, 𝑏)

=exp(1

2((1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)log(1 − 𝑡))) . (45)

Therefore, we have to compare some one-variable inequalities and to check their validness for each𝑡 ∈ (0, 1).

For example, we will prove that the inequality

𝜆𝑠(𝑎, 𝑏) ≤ 𝐿 (𝑎, 𝑏) (46)

holds for each positive𝑎, 𝑏if and only if𝑠 ≤ 0.

Since𝜆_𝑠(𝑎, 𝑏)is monotone increasing in𝑠, it is enough to prove that

𝜆₀(𝑎, 𝑏)

𝐿 (𝑎, 𝑏) ≤ 1. (47)

By the above formulae, this is equivalent to the assertion that the inequality

𝜙 (𝑡) ≤ 0 (48)

holds for each𝑡 ∈ (0, 1), with

𝜙 (𝑡) := log(1 + 𝑡) −_2𝑡log(1 − 𝑡)

× ((1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)log(1 − 𝑡)) +log(1 + 𝑡) +log(1 − 𝑡).

(49)

We will prove that the power series expansion of 𝜙(𝑡) have non-positive coefficients. Thus the relation (48) will be proved.

Since

log(1 + 𝑡) −log(1 − 𝑡)

2𝑡 =

∞ ∑ 0

𝑡2𝑘 2𝑘 + 1;

log(1 + 𝑡) +log(1 − 𝑡) = −𝑡2 ∞ ∑ 0

𝑡2𝑘 𝑘 + 1;

(1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)log(1 − 𝑡)

= 𝑡2∑∞ 0

𝑡2𝑘 (𝑘 + 1) (2𝑘 + 1),

(50)

we get

𝜙 (𝑡) 𝑡2

=∑∞ 𝑛=0

(−_{𝑛 + 1}1 +∑𝑛 𝑘=0

1

(2𝑛 − 2𝑘 + 1) (𝑘 + 1) (2𝑘 + 1)) 𝑡2𝑛

=∑∞ 0𝑐𝑛𝑡

2𝑛_.

(51)

Hence,

𝑐₀= 𝑐₁= 0; 𝑐₂= −₉₀1 , (52)

and, after some calculation, we get

𝑐_𝑛= 2

(𝑛 + 1) (2𝑛 + 3)((𝑛 + 2) 𝑛 ∑

1 1

2𝑘 + 1− (𝑛 + 1) 𝑛 ∑

1 1 2𝑘) ,

𝑛 > 1. (53)

Now, one can easily prove (by induction, e.g.) that

𝑑_𝑛:= (𝑛 + 2)∑𝑛 1

1

2𝑘 + 1 − (𝑛 + 1) 𝑛 ∑

1 1

2𝑘 (54)

is a negative real number for𝑛 ≥ 2. Therefore𝑐_𝑛≤ 0, and the proof of the first part is done. For0 < 𝑠 < 1we have

𝜆_𝑠(𝑎, 𝑏) 𝐿 (𝑎, 𝑏) − 1

=(1 − 𝑠) ((1 + 𝑡)

𝑠+1_{+ (1 − 𝑡)}𝑠+1_{− 2)}_log_{((1 + 𝑡) / (1 − 𝑡))}

2𝑡 (1 + 𝑠) (2 − (1 + 𝑡)𝑠_{− (1 − 𝑡)}𝑠₎ −1

= 1₆𝑠𝑡2+ 𝑂 (𝑡4) (𝑡 󳨀→ 0) .

(55)

Therefore,𝜆_𝑠(𝑎, 𝑏) > 𝐿(𝑎, 𝑏)for𝑠 > 0 and sufficiently small𝑡 := (𝑏 − 𝑎)/(𝑏 + 𝑎).

Similarly, we will prove that the inequality

𝜆𝑠(𝑎, 𝑏) ≤ 𝐼 (𝑎, 𝑏) (56)

holds for each𝑎,𝑏;0 < 𝑎 < 𝑏if and only if𝑠 ≤ 1. As before, it is enough to consider the expression

𝐼 (𝑎, 𝑏)

𝜆₁(𝑎, 𝑏) = 𝑒𝜇(𝑡)𝜈 (𝑡) := 𝜓 (𝑡) , (57)

with

𝜇 (𝑡) = (1 + 𝑡)log(1 + 𝑡) − (1 − 𝑡)_2𝑡 log(1 − 𝑡)− 1;

𝜈 (𝑡) = (1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)_𝑡₂ log(1 − 𝑡).

(6)

It is not difficult to check the identity

𝜓󸀠(𝑡) = −𝑒𝜇(𝑡)_𝑡𝜙 (𝑡)₃ . (59)

Hence by (48), we get𝜓󸀠(𝑡) > 0, that is,𝜓(𝑡)is monotone increasing for𝑡 ∈ (0, 1).

Therefore

𝐼 (𝑎, 𝑏)

𝜆₁(𝑎, 𝑏) ≥𝑡 → 0lim+𝜓 (𝑡) = 1. (60)

By monotonicity it follows that𝜆_𝑠(𝑎, 𝑏) ≤ 𝐼(𝑎, 𝑏)for𝑠 ≤ 1. For𝑠 > 1, (𝑏 − 𝑎)/(𝑏 + 𝑎) = 𝑡, we have

𝜆_𝑠(𝑎, 𝑏) − 𝐼 (𝑎, 𝑏) = (1

6(𝑠 − 1) 𝑡2+ 𝑂 (𝑡4)) 𝐴 (𝑎, 𝑏) (𝑡 󳨀→ 0+) .

(61)

Hence,𝜆_𝑠(𝑎, 𝑏) > 𝐼(𝑎, 𝑏)for𝑠 > 1and𝑡sufficiently small. From the other hand,

lim 𝑡 → 1−[

𝜆_𝑠(𝑎, 𝑏) 𝐼 (𝑎, 𝑏) − 1] =

𝑒 (𝑠 − 1) (2𝑠+1_{− 2)}

2 (𝑠 + 1) (2𝑠_{− 2)} − 1 := 𝜏 (𝑠) . (62)

Examining the function𝜏(𝑠), we find out that it has the only real zero at𝑠₀≈ 1.0376and is negative for𝑠 ∈ (1, 𝑠₀).

Remark 10. Since𝜓(𝑡)is monotone increasing, we also get

𝐼 (𝑎, 𝑏)

𝜆1(𝑎, 𝑏) ≤𝑡 → 1lim−𝜓 (𝑡) = 4log2

𝑒 . (63)

Hence

1 ≤ 𝐼 (𝑎, 𝑏) 𝜆1(𝑎, 𝑏)≤

4log2

𝑒 . (64)

A calculation gives4log2/𝑒 ≈ 1.0200.

Note also that

𝜆₂(𝑎, 𝑏) ≡ 𝐴 (𝑎, 𝑏) . (65)

Therefore, applying the assertion from the part 1, we get

𝜆_𝑠(𝑎, 𝑏) ≤ 𝐴 (𝑎, 𝑏), 𝑠 ≤ 2;

𝜆_𝑠(𝑎, 𝑏) ≥ 𝐴 (𝑎, 𝑏), 𝑠 ≥ 2. (66)

Finally, we give a detailed proof of the part 7.

We have to prove that𝜆_𝑠(𝑎, 𝑏) ≤ 𝑆(𝑎, 𝑏)for𝑠 ≤ 5. Since 𝜆_𝑠(𝑎, 𝑏)is monotone increasing in𝑠, it is sufficient to prove that the inequality

𝜆5(𝑎, 𝑏) ≤ 𝑆 (𝑎, 𝑏) (67)

holds for each𝑎, 𝑏 ∈R+.

Therefore, by the transformation given above, we get

log𝜆5 𝐴

=log[2 3

(1 + 𝑡)6_{+ (1 − 𝑡)}6_{− 2} (1 + 𝑡)5+ (1 − 𝑡)5− 2]

=log[2 15

15 + 15𝑡2+ 𝑡4 2 + 𝑡2 ]

≤log[1 + 𝑡 2_{+ 𝑡}4_/4

1 + 𝑡2_/2 ] =log(1 + 𝑡2

2)

= 𝑡₂2 −𝑡₈4 +₂₄𝑡6 − ⋅ ⋅ ⋅

≤ 𝑡₂2 +₁₂𝑡4 +₃₀𝑡6 + ⋅ ⋅ ⋅

= 1

2((1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)log(1 − 𝑡))

=log 𝑆 𝐴,

(68)

and the proof is done.

Further, we have to show that𝜆_𝑠(𝑎, 𝑏) > 𝑆(𝑎, 𝑏)for some positive𝑎,𝑏whenever𝑠 > 5.

Indeed, since

(1 + 𝑡)𝑠+ (1 − 𝑡)𝑠_{− 2 = (𝑠2)𝑡}2_{+ (𝑠4)𝑡}4+ 𝑂 (𝑡6) , (69)

for𝑠 > 5and sufficiently small𝑡, we get

𝜆_𝑠

𝐴 =

𝑠 − 1 𝑠 + 1

(𝑠+1

2 ) 𝑡2+ (𝑠+14 ) 𝑡4+ 𝑂 (𝑡6) (𝑠

2) 𝑡2+ (𝑠4) 𝑡4+ 𝑂 (𝑡6)

= 1 + (𝑠 − 1) (𝑠 − 2) 𝑡

2_{/12 + 𝑂 (𝑡}4₎

1 + (𝑠 − 2) (𝑠 − 3) 𝑡2_{/12 + 𝑂 (𝑡}4₎

= 1 + (𝑠 6−

1

3) 𝑡2+ 𝑂 (𝑡4) .

(70)

Similarly,

𝑆

𝐴 =exp( 1

2((1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)log(1 − 𝑡)))

=exp(𝑡 2

2 + 𝑂 (𝑡4)) = 1 + 𝑡2

2 + 𝑂 (𝑡4) .

(71)

Hence,

1

𝐴(𝜆𝑠− 𝑆) = 1

6(𝑠 − 5) 𝑡2+ 𝑂 (𝑡4) , (72)

(7)

As for the part 8, applying the above transformation we obtain

𝜆_𝑠(𝑎, 𝑏) 𝑆 (𝑎, 𝑏)

= 𝑠 − 1 𝑠 + 1

(1 + 𝑡)𝑠+1+ (1 − 𝑡)𝑠+1− 2 (1 + 𝑡)𝑠_{+ (1 − 𝑡)}𝑠_{− 2}

×exp(−1

2((1 + 𝑡)log(1 + 𝑡) + (1 − 𝑡)log(1 − 𝑡))) , (73)

where0 < 𝑎 < 𝑏,𝑡 = (𝑏 − 𝑎)/(𝑏 + 𝑎). Since for𝑠 > 5,

lim 𝑡 → 1−

𝜆_𝑠

𝑆 =

𝑠 − 1 𝑠 + 1

2𝑠− 1

2𝑠_{− 2}, (74)

and the last expression is less than one, it follows that the inequality𝑆(𝑎, 𝑏) < 𝜆_𝑠(𝑎, 𝑏)cannot hold whenever 𝑏/𝑎 is sufficiently large.

The rest of the proof is straightforward.

Acknowledgment

The author is indebted to the referees for valuable sugges-tions.

References

[1] K. B. Stolarsky, “Generalizations of the logarithmic mean,”

Mathematics Magazine, vol. 48, pp. 87–92, 1975.

[2] B. C. Carlson, “The logarithmic mean,”The American

Mathe-matical Monthly, vol. 79, pp. 615–618, 1972.

[3] T. P. Lin, “The power mean and the logarithmic mean,” The

American Mathematical Monthly, vol. 81, pp. 879–883, 1974.

[4] P. A. H¨ast¨o, “Optimal inequalities between Seiffert’s mean and power means,”Mathematical Inequalities & Applications, vol. 7, no. 1, pp. 47–53, 2004.

[5] Z.-H. Yang, “Sharp bounds for the second Seiert mean in terms of power mean,”http://arxiv.org/abs/1206.5494.

[6] G. H. Hardy, J. E. Littlewood, and G. P¨olya, Inequalities, Cambridge University Press, Cambridge, UK, 1978.

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