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# A Note on Mixed-Mean Inequalities

*Journal of Inequalities and Applications*
**volume 2010**, Article number: 509323 (2010)

## Abstract

We give a simpler proof of a result of Holland concerning a mixed arithmetic-geometric mean inequality. We also prove a result of mixed-mean inequality involving the symmetric means.

## 1. Introduction

Let be the generalized weighted power means: , where , , and , with . Here denotes the limit of as . Unless specified, we always assume that . When there is no risk of confusion, we will write for and we also define .

The celebrated Hardy's inequality (see [1, Theorem ] ) asserts that, for ,

Among the many different proofs of Hardy's inequality as well as its generalizations and extensions in the literature, one novel approach is via the mixed-mean inequalities (see, e.g., [2, Theorem ]). By mixed-mean inequalities, we will mean the following inequalities:

where are two matrices with nonnegative entries and the above inequality being meant to hold for any vector with nonnegative entries. Here and, when , we want the inequality above to be reversed.

The meaning of mixed mean becomes more clear when are weighted mean matrices. Here we say that a matrix is a weighted mean matrix if for and

Now we focus our attention to the case of (1.2) for being weighted mean matrices given in (1.3). In this case, for fixed , we define , . Then we have the following mixed-mean inequalities of Nanjundiah [3] (see also [4]).

Theorem 1.1.

Let and . If, for , , then

with equality holding if and only if .

A very elegant proof of Theorem 1.1 for the case is given by Kedlaya in [5]. In fact, the following Popoviciu-type inequalities were established in [5] (see also [4, Theorem ]).

Theorem 1.2.

Let . If, for , , then

with equality holding if and only if .

It is easy to see that the case of Theorem 1.1 follows from Theorem 1.2. As was pointed out by Kedlaya that the method used in [5] can be applied to establish both Popoviciu-type and Rado-type inequalities for mixed means for a general pair , the details were worked out in [6] and the following Rado-type inequalities were established in [6].

Theorem 1.3.

Let and . If, for , , then

with equality holding if and only if and the above inequality reverses when .

A different proof of Theorem 1.1 for the case was given in [7] and Bennett used essentially the same approach in [8, 9] to study (1.2) for the cases being lower triangular matrices, namely, if . Among other things, he showed [8] that inequalities (1.2) hold when are Hausdorff matrices.

In [10], Holland further improved the condition in Theorem 1.3 for the case by proving the following.

Theorem 1.4.

Let . If, for , , then

with equality holding if and only if .

It is our goal in this paper to first give a simpler proof of the above result by modifying Holland's own approach. This is done in the next section and, in Section 3, we will prove a result of mixed-mean inequality involving the symmetric means.

## 2. A Proof of Theorem 1.4

First, we recast (1.7) as

We now note that

We may assume that , and the case for some will follow by continuity. Thus on dividing on both sides of (2.1) and using (2.2), we can recast (2.1) as

We now express that , with to recast (2.3) as

We now set , , to further recast the above inequality as

It now follows from the assumption of Theorem 1.4 that

so that by the arithmetic-geometric mean inequality we have

Similarly, we have

Now it is easy to see that inequality (2.5) follows on adding inequalities (2.7) and (2.8), and this completes the proof of Theorem 1.4.

## 3. A Discussion on Symmetric Means

Let ; we recall that the th symmetric function of and its mean are defined by

It is well known that, for fixed of dimension , is a nonincreasing function of for with (with weights , ). In view of the mixed-mean inequalities for the generalized weighted power means (Theorem 1.1), it is natural to ask whether similar results hold for the symmetric means. Of course one may have to adjust the notion of such mixed means in order for this to make sense for all . For example, when , , the notion of is not even defined. From now on, we will only focus on the extreme cases of the symmetric means; namely, or . In these cases it is then natural to define , and, on recasting , we see that it is also natural for us to define (note that this is not consistent with our definition of above).

We now prove a mixed-mean inequality involving and . We first note the following result of Marcus and Lopes [11] (see also [12, pages 33–35]).

Theorem 3.1.

Let and for . Then

with equality holding if and only if or there exists a constant such that .

We also need the following lemma of C. D. Tarnavas and D. D. Tarnavas [6].

Lemma 3.2.

Let be a convex function and suppose that for , that . Then

The equality holds if and only if or when is strictly convex. When is concave, then the above inequality is reversed.

We now apply Lemma 3.2 to obtain the following.

Lemma 3.3.

For and , ,

with equality holding in both cases if and only if or .

Proof.

The case yields an identity, so we may assume that here. Write . Note that , and now Lemma 3.2 with implies that . On expanding , we obtain

Hence,

Using , we obtain

So by (3.6),

which is just what we want.

We now prove the following mixed-mean inequality involving the symmetric means.

Theorem 3.4.

Let and define . Then

with equality holding if and only if . It follows that

with equality holding if and only if .

Proof.

It suffices to prove (3.9) here. We may assume that here and we will use the idea in [6]. Lemma 3.3 implies that

where the last inequality follows from Theorem 3.1 for the case . It is easy to see that the above inequality is equivalent to (3.9) and this completes the proof.

Now we let and define with here. Then it is interesting to see whether the following inequality holds or not:

We note here that, if the above inequality holds, then it is easy to deduce from it via the approach in [2] the following Hardy-type inequality:

where we define . We now end this paper by proving the following result.

Theorem 3.5.

Let and . Then

where one defines .

Proof.

We follow an approach of Knopp [13, 14] here (see also [15]). For , we define

It is easy to check by partial summation that

Certainly, we have and, for , we apply the inequality to the numbers to see that

We now show by induction that for ; equivalently, this is

Note first that the above inequality holds when and suppose now that it holds for some . Then by induction,

Now using , we have

It is easy to see that the last expression above is no less than when and this proves inequality (3.18) for the case . This completes the proof of the theorem.

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Gao, P. A Note on Mixed-Mean Inequalities.
*J Inequal Appl* **2010**, 509323 (2010). https://doi.org/10.1155/2010/509323

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DOI: https://doi.org/10.1155/2010/509323