Abstract and Applied Analysis
Volume 2016 (2016), Article ID 1249604, 8 pages
http://dx.doi.org/10.1155/2016/1249604

Research Article

On Some Inequalities Involving Three or More Means

Mustapha Raïssouli1,2 and Mohamed Chergui³

¹Department of Mathematics, Science Faculty, Taibah University, P.O. Box 30097, Al Madinah Al Munawwarah 41477, Saudi Arabia
²Department of Mathematics, Science Faculty, Moulay Ismail University, 50060 Meknes, Morocco
³Department of Mathematics, Centre Régional des Métiers de l’Education et de la Formation, (CRMEF), 14000 Kenitra, Morocco

Received 15 April 2016; Accepted 10 July 2016

Academic Editor: Alberto Fiorenza

Copyright © 2016 Mustapha Raïssouli and Mohamed Chergui. 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.

Abstract.

We investigate some results about mean-inequalities involving a large number of bivariate means. As application, we derive a lot of inequalities between four or more means among the standard means known in the literature.

1. Introduction

By (bivariate) mean we understand a map between positive real numbers satisfying the following double inequality:

As usual, we define continuous (resp., symmetric/homogeneous) means in the habitual way. The standard examples of such means are given in the following:

with , and they are known as the arithmetic mean, geometric mean, harmonic mean, contraharmonic mean, logarithmic mean, first Seiffert mean [1], second Seiffert mean [2], and Neuman-Sándor mean [3], respectively. Other examples of means (not needed here) can be found in the literature; see [4], for instance, and the references cited therein.

The two means,

were introduced in [5] (see page 9 and page 24, resp.). These two means are included in the so-called Seiffert type means discussed in [6].

As usual, we identify a mean with its value at by setting for the sake of simplicity. If and are two means, we write for meaning that for all with . Two different means and are called comparable if or holds. As it is well-known, means , , , , , , and are mutually comparable with . Concerning means and we have . However, the means and are not comparable; see Section 5.

The notation refers to the dual mean of defined by for all . As it is well-known, if is symmetric and homogeneous then so is , with which we briefly write as .

For over the last years, mean-theory has been the subject of intensive research. It has been proved, throughout a large number of works, that mean-theory is useful from theoretical point of view and for practical purposes. The most interesting subject in mean-theory is to investigate mean-inequalities that occur in a primordial place in the literature. For instance, many inequalities involving trigonometric/hyperbolic functions can be derived from mean-inequalities in a simple and fast way. As far as we know, published results about comparison of the previous standard means concern mean-inequalities involving only two or three means. Inspired by an approach, recently introduced by the first author in [7], we present here some results about inequalities involving a large number of means. In particular, inequalities involving four or more means among the standard means , , , , , and are discussed. Inequalities involving the means and are also obtained.

2. Background Material

In this section, we state some results that will be needed throughout the following. We begin by recalling the following result; see [7].

Theorem 1. Let be a continuous homogeneous symmetric mean. Then the binary map defined by and

for all with is a continuous homogeneous symmetric mean (called the integral mean-transform of ).

As example, we have , , and . The mean-map is pointwisely strictly decreasing; that is, if and are two continuous homogeneous symmetric means such that , then we have .

As pointed out in [7], a continuous homogeneous symmetric mean will be called regular mean, for the sake of simplicity. A regular mean will be called -regular if the map is continuously differentiable on and the function (called the generated function of ) defined by

for all , with , satisfies

for all . The previous means are regular and -regular, except which is not -regular (see [7, Examples  3.1, 3.2, 3.3]).

The following result has also been proved in [7].

Theorem 2. Let be a -regular mean with its generated function . Then the binary map defined by

for all is a regular mean with .

If we denote by and the sets of all regular means and -regular means, respectively, then the mean-map is a bijection from into and we can write (see [7])

With this, the following result, which will be needed throughout the following, has been established in [7].

Theorem 3. The following relationships hold:

where the notation refers to the dual mean of .

If, for , we set

it is easy to see that if then we have [7]

Finally, we recall the following result; see, for example, [7].

Theorem 4. The following relationships are met:

Using the previous result, it is easy to see that, for all , we have

which, with (11), immediately implies (simultaneously and in a fast way) the known chains and , respectively. The previous inequalities are graphically illustrated in Figure 1.

Figure 1: Curves of .

3. Some Needed Results

In this section we present some results that will be needed for obtaining a lot of mean-inequalities involving four or more means among the previously mentioned means. We preserve the same notations as in the previous sections. We begin by stating the following result.

Theorem 5. Let be regular means and such that . Then one has

If moreover two different means among are comparable then (14) is strict.

Proof. First, it is easy to see that is a regular mean. By definition, for all , we have

This, with the linearity of integral, gives

Taking inverse-sides of (16) and using the fact that the real-valued function is (strictly) convex on , generalized Jensen’s inequality asserts that

Now, if two means among are different and comparable, say for all with and some , then the strict convexity of implies that previous Jensen’s inequality is strict and the proof is complete.

For practical purposes, the following corollary is of interest.

Corollary 6. Let and be -regular means. Assume that, for some such that , the inequality

holds true for all . Then we have

Proof. Since , with , then (18) can be written as follows:

This, with (14) and the fact that the mean-map is pointwisely strictly decreasing, yields

which implies the desired result.

The following example shows that the reverse statement of the previous corollary is not always true.

Example 7. It is easy to see that the inequality

holds for all with equality if and only if . This, with the fact that for , yields

Writing this inequality in the form

we then obtain (by virtue of Theorem 3)

This, with the known inequality (see, e.g., [8, 9]), justifies our claim.

Theorem 8. Let be regular means and such that . Then there holds

If moreover two means among are different then (26) is strict.

Proof. Clearly, is a regular mean. Without loss of generality, we can assume that . By definition of the integral transform-mean we have

According to generalized Hölder’s inequality for integrals, the latter inequality yields

Since the last inequality becomes

or again

Taking inverses of the two sides of this latter inequality we find

Now, if the means are not all equal then we can see that Hölder’s inequality is here strict, thus completing the proof.

The following corollary is of interest for practical purposes.

Corollary 9. Let and be -regular means. Assume that, for some such that , the inequality

holds true for all . Then we have

If moreover there exist two means among which are different then (33) is strict.

Proof. In similar way as the proof of Corollary 6, we deduce the desired result from the previous theorem. Detail is simple and therefore omitted here.

We end this section by stating the following remark.

Remark 10. It is worth mentioning that the mean-inequality (33) is strict even if (32) is an equality, provided two means among are different.

4. Inequalities Involving , , , , , and

This section is devoted to investigating some applications for mean-inequalities. As already pointed before, we give mean-inequalities involving some means among those recalled in the previous introduction. Concerning mean-inequalities in a geometric combination involving only three means,

have been proved in [10, 11]; see also [7]. Other inequalities in geometric combination, such as , can be found in the literature. Here refers to the quadratic mean; that is, . In a parallel way, mean-inequalities in a (convex) combination sum were the subject of many published papers. See [12] and the related references cited therein.

The following result gives more mean-inequalities involving only three means among , , , , , and .

Theorem 11. The following inequalities hold true:

Proof. Let us prove the inequality . Using Theorem 4, simple observation leads to the fact that

hold for all . This, with Corollary 9, yields the first mean-inequality. The other mean-inequalities can be proved in the same way. Detail is simple and therefore omitted here with the aim of not lengthening this paper.

Remark 12. In [13] the authors showed that if and only if . This means that the mean-inequality (previously mentioned) is the best possible.
Otherwise, it is worth mentioning that certain inequalities of the above theorem can be deduced from those of (34). As example, from , that is, , one has , since this latter inequality is equivalent to . However, our aim here is to show how to apply the same procedure, described by the present approach, for proving old/new mean-inequalities in a simple and fast way.

Similarly, we can find a lot of mean-inequalities involving four means among the standard means , , , , , and . For instance, we cite the following.

Theorem 13. The following mean-inequalities are satisfied:

Proof. We show and , for example. The proofs of the other inequalities are analogue.
For all , we can easily verify that (by Theorem 4)

By Corollary 9 we immediately obtain the first inequality. For the second one, we have

which by the same reason as previous implies . Now, getting in this latter inequality we then obtain the second desired inequality.

We left to the reader the routine task for obtaining more inequalities in an analogous way. Similarly, inequalities involving five means can be obtained as well. The following result lists some of them.

Theorem 14. The following mean-inequalities hold:

Proof. We show and , for instance. For all , we can easily see that (always by Theorem 4)

which, with Corollary 9, yields . Getting , we deduce the first desired inequality. The second one can be similarly obtained from the following:

Concerning mean-inequalities involving all the six means , and , we have the following result.

Theorem 15. The three following mean-inequalities hold:

Proof. We just show the first inequality, for example. In similar way as previous, we have (for all )

from which we get . Replacing by we obtain the desired inequality.

We end this section by stating the following interesting result.

Theorem 16. For every real number we have

Proof. By Theorem 4, simple computation leads to

for all . This, with Corollary 9, yields the desired mean-inequality.

Remark 17. Inequality (45) contains some known mean-inequalities shown in the literature by different ways. In fact, taking it yields which is the last inequality in (34). Writing it in the equivalent form

and then letting we find again ; see (34). For , it gives .

5. Inequalities Involving and

As already pointed before, the means and satisfy the following chain of inequalities:

We notice that means and are not comparable. In fact, in [14] the authors proved that the double inequality

takes place if and only if and , where , and , is the th-power mean. Therefore, and are not comparable because .

For recent developments about means and we refer the reader to [14–18]. As far as we know, mean-inequalities involving three or more means and including and/or are not investigated in the literature yet. Adopting the previous approach, we can give a lot of mean-inequalities involving and . We start with the case of convex arithmetic combination.

Theorem 18. The following inequalities hold:

Proof. For all , we have (by Theorem 4)

since the real-function is strictly concave on . A simple verification leads to

for all . This, with , gives

which, with Corollary 6, yields the first inequality of (50). The second inequality of (50) can be proved in a similar way.

Now, we give some inequalities in convex geometric combination between three means including and .

Theorem 19. The following mean-inequalities hold:

Proof. We will show the inequalities and , for instance. For all , it is easy to see that

These, with Corollary 9, yield the desired inequalities.

Inequalities involving four or more means including and can be obtained in a similar way as previous. As examples, we cite the following:

Other more interesting examples are given in the following result.

Theorem 20. The two inequalities

hold for each , while

is valid for every .

Proof. According to Theorem 4 we have (for all )

since and for all . This, with Corollary 9, gives (57). Inequalities (58) and (59) can be proved in a similar way.

For particular cases of the parameter , we can find again some mean-inequalities among those stated in the previous results. The following example explains this latter situation.

Example 21. (1) With , (57) gives which is the first inequality of (54). Writing it in the form

and getting we find which is the last inequality in (54).
(2) With , (59) yields while with it gives . Writing (59) in the equivalent form

and then letting we obtain .
Other mean-inequalities can be deduced from (58) in a similar way. We left it to the reader.

It is clear that whenever we have a mean-inequality involving means we can deduce from it a mean-inequality involving only means, provided the involved means are mutually comparable. The following example explains more precisely such situation.

Example 22. Let us consider the first inequality of Theorem 15; that is, . This, with , implies that .

As it is well-known, mean-inequalities could be good tool for proving some inequalities involving trigonometric/hyperbolic functions whose proofs in a direct way present serious difficulties. Let us observe this situation in the following example.

Example 23. Let us consider (45); that is,

(i) Let and set ,. Simple computation leads to

Substituting these in the previous mean-inequality we obtain

(ii) Let and ,. We then have

which when substituted in the previous mean-inequality yield

Now, take . Inequality (65) implies

while (67) yields

which are well-known inequalities. We left to the reader the routine task for deducing trigonometric/hyperbolic inequalities from mean-inequalities (57), (58), and (59).

Competing Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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