 Research
 Open Access
On a class of new means including the generalized SchwabBorchardt mean
 Mustapha Raïssouli^{1, 2}Email authorView ORCID ID profile and
 József Sándor^{3}
https://doi.org/10.1186/s1366001714121
© The Author(s) 2017
 Received: 25 March 2017
 Accepted: 23 May 2017
 Published: 13 June 2017
Abstract
The socalled SchwabBorchardt mean plays an important role in the theory of (bivariate) means. It includes a lot of standard means, such as the logarithmic mean, the first and second Seiffert means and the NeumanSándor mean. In this paper, we investigate an approach which allows us to construct a class of new means. Such class includes the (generalized) SchwabBorchardt mean and other old/new means derived as well.
Keywords
 bivariate mean
 weighted mean
 SchwabBorchardt mean
MSC
 26E60
1 Introduction
Means arise in various contexts and contribute as good tool to solving many scientific problems. It has been proved, throughout a lot of works, that the theory of means is useful from the theoretical point of view as well as for practical purposes.
2 Weighted means
In this section, we state more notions needed later. We begin by the following definition.
Definition 2.1
 (i)
\(m_{\lambda}\) is a mean, in the sense of (1.1), for all fixed \(\lambda\in[0,1]\).
 (ii)
\(m_{0}(a,b)=a\) and \(m_{1}(a,b)=b\), for all \(a,b>0\).
 (iii)
\(m_{\lambda}(a,b)=m_{1\lambda}(b,a)\), for all \(a,b>0\) and every \(\lambda\in[0,1]\).
 (iv)
\(m_{\lambda}\) coincides with m if \(\lambda=1/2\).
A natural question arises from the above: what should be the reasonable weighted means associated to the symmetric means L, T, M and P. For the mean L, there are various weighted Lmeans that have been introduced in the literature; see [15–17] for instance. This understands that a given symmetric mean could have more one weighted mean. For the simplest means A, H, G, S, Q (and more generally \(A_{r}\)) only one weighted mean to each (as far as we know) is known in the literature. In fact, following two distinct points of view we can obtain different weighted means associated to the same symmetric mean, and this according to Definition 2.1 of course. As example, let us consider the symmetric logarithmic mean L. Its weighted mean is not simple to deduce from its explicit form in a and b previously mentioned. However, L has other equivalent forms known in the literature. In this paper (see Example 6.1), we will obtain via our approach a reasonable weighted Lmeans (i.e. satisfying all conditions of Definition 2.1). Another simple example explaining more the latter situation is stated in the following.
Example 2.1
It is worth mentioning that the two chains of inequalities (1.5) and (2.1) look alike while they are of course completely different.
3 Basic notions and preliminary tools
 (\(c_{1}\)):

k is homogeneous of degree 1, i.e. \(k(\alpha a,\alpha b)=\alpha k(a,b)\) for all \(a,b,\alpha>0\).
 (\(c_{2}\)):

\(k(a,b)>0\) for \(a>b\), \(k(a,b)<0\) for \(a< b\) and \(k(a,a)=0\), for all \(a,b>0\).
 (\(c_{3}\)):

The maps \(t\longmapsto k(t,1)\) and \(t\longmapsto t^{1}k(t,1)\) are strictly monotone increasing on \((0,\infty)\).
 (\(c_{4}\)):

The map \(t\longmapsto k(t,1)t^{1}k(t,1)\) is monotone increasing on \([1,\infty)\) and monotone decreasing on \((0,1]\).
 (\(c_{5}\)):

The map \(t\longmapsto k(t,1)\) is continuously differentiable on \((0,1)\cup(1,\infty)\).
In what follows, we denote by \({\mathcal {K}}\) the set of all map \(k:(0,\infty)\times(0,\infty)\longrightarrow{\mathbb {R}}\) satisfying the conditions (\(c_{1}\))(\(c_{5}\)). It is not hard to see that \({\mathcal {K}}\) is a convex cone i.e., for all \(k,h\in{\mathcal {K}}\) and every \(\alpha>0\) we have \(h+k\in{\mathcal {K}}\) and \(\alpha\cdot k\in{\mathcal {K}}\). The following definition may be stated.
Definition 3.1
The next result, which summarizes the elementary properties of \(k_{1}\) and \(k_{2}\), will be needed later.
Proposition 3.1
 (i)
\(k_{1}\) and \(k_{2}\) are with positive values, both continuous on \((0,1)\cup(1,\infty)\).
 (ii)
\(k_{1}(t)>k_{2}(t)\) if \(t>1\), \(k_{1}(t)< k_{2}(t)\) if \(0< t<1\).
 (iii)
For all \(t>0\), we have \(k_{1}(1/t)=t^{2}k_{2}(t)\) and \(k_{2}(1/t)=t^{2}k_{1}(t)\).
 (iv)
The map \(t\longmapsto k_{1}(t)/k_{2}(t)\) is continuous on \((0,\infty)\).
Proof
For the sake of simplicity, we set \(k_{12}=k_{1}/k_{2}\) throughout the following. Proposition 3.1(iv) asserts that the map \(k_{12}:(0,\infty)\longrightarrow(0,\infty)\) is continuous on \((0,\infty )\).
Lemma 3.2
Proof
The conditions (\(c_{1}\)), (\(c_{2}\)) and (\(c_{5}\)) are obviously satisfied while (\(c_{3}\)) and (\(c_{4}\)) follow after an elementary computation. Details are simple and therefore omitted here. □
We notice that \(k_{1}\) and \(k_{2}\) (of the previous lemma) are not continuous at \(t=1\) unless \(1/q^{*}=0\) i.e. \(q=1\). This corresponds to the following simplest example.
Example 3.1
Take \(q=1\) in the previous lemma i.e. k is defined by \(k(a,b)=ab\) for all \(a,b>0\). Simple computation leads to \(k_{1}(t)=1\), \(k_{2}(t)=1/t^{2}\) and \(k_{12}(t)=t^{2}\), for every \(t>0\).
Other examples may be stated as follows.
Example 3.2
Example 3.3
The next lemma will also be needed in the sequel.
Lemma 3.3
Proof
4 General approach
As already pointed out before, the previous section was devoted to listing convenient conditions on the binary map k in the aim that (3.1) defines a mean. In this section we will complete our previous discussion by stating favorable conditions about the function f.
Now, we are in a position to state our first main result stated as follows.
Theorem 4.1
 (i)
\(M_{f,k}\) is a homogeneous mean.
 (ii)If, moreover, f and k are such that$$\begin{aligned}& \forall t>0,\quad f (1/t )=t^{2}f(t), \end{aligned}$$(4.3)then \(M_{f,k}\) is symmetric.$$\begin{aligned}& \forall a,b>0,\quad k(a,b)=k(b,a)\quad (\textit{i.e. } k \textit{ is antisymmetric}), \end{aligned}$$(4.4)
 (iii)
If the function f is continuous then \(M_{f,k}\) is also continuous.
Proof
The reverse of Theorem 4.1(iii) is not always true, i.e. the mean \(M_{f,k}\) could be continuous for not continuous f. The following example explains this situation.
Example 4.1
The following corollary is of great interest for practical purposes.
Corollary 4.2
Proof
Let m be a mean and for \(t>0\) we set \(f(t)=m (k_{1}(t),k_{2}(t) )\) for some \(k\in{\mathcal {K}}\). By (1.1), f satisfies (4.1) and by the previous theorem \(m^{\sigma_{k}}\) is a homogeneous mean. If m is symmetric, Proposition 3.1(iv) implies that \(f(1/t)=t^{2}f(t)\) for all \(t>0\). The symmetry of \(m^{\sigma_{k}}\) follows then from the previous theorem. □
Remark 4.1
(i) The present approach extends that of [18] for a general class of maps k and for means not necessary homogeneous/symmetric/continuous. In fact, the above theorem and corollary give Theorem 2.1 and Corollary 2.2 of [18], respectively, when we consider the simplest k defined by \(k(a,b)=ab\) and m is a symmetric homogeneous continuous mean.
(ii) In what follows, the mean \(m^{\sigma_{k}}\) defined by (4.8) will be called the \(\sigma_{k}\)mean transform of m. For \(k\in{\mathcal {K}}\) defined by (3.2), we write \(m^{\sigma_{q}}\). In particular, \(m^{\sigma_{2}}\) is that with k of Example 4.1. For \(q=1\) in (3.2) i.e. \(k(a,b)=ab\), we simply write \(m^{\sigma}\).
Choosing \(k\in{\mathcal {K}}\) and m particular mean, we can obtain a lot of homogeneous (symmetric or not) means illustrating the above results. As trivial examples, it is not hard to check that \(\min^{\sigma _{k}}=\max\) and \(\max^{\sigma_{k}}=\min\), for all \(k\in{\mathcal {K}}\). To the aim to not lengthen this section, we prefer to present other examples in another section below.
5 Examples and properties of \(m\longmapsto m^{\sigma_{k}}\)
As a first example we present the following in form of result by virtue of its interest.
Proposition 5.1
Let \(k\in{\mathcal {K}}\). Then the relationship \(A_{\lambda}^{\sigma _{k}}=H_{1\lambda}\) holds for all \(\lambda\in[0,1]\). In particular \(A^{\sigma_{k}}=H\).
Proof
It is worth mentioning that in the previous relationship \(A_{\lambda }^{\sigma_{k}}=H_{1\lambda}\), \(k\in{\mathcal {K}}\) is arbitrary. This could be coming from the fact that the weighted arithmetic mean \(A_{\lambda}\) has a linear affine character.
Now we state more examples of interest.
Example 5.1
Let k be as in Example 3.2 whose mean transform is denoted by \(m^{\sigma_{2}}\).
(ii) Let \(m=G_{2/3}:=a^{1/3}b^{2/3}\). By similar arguments as in the previous (i), we simply verify that \(G_{2/3}^{\sigma_{2}}(a,b)=\mathit{SB}(b,a)\) i.e. \(G_{2/3}^{\sigma_{2}}=\mathit{SB}^{T}\), where \(\mathit{SB}^{T}\) denotes the mean transpose of SB defined by \(\mathit{SB}^{T}(a,b)=\mathit{SB}(b,a)\) for all \(a,b>0\).
We will go back again to this situation in section below.
Example 5.2
Further examples in a general context will be discussed in the next sections.
We now give some properties of the meanmap \(m\longmapsto m^{\sigma _{k}}\). The first is stated as follows.
Proposition 5.2
 (i)
Let \(f,g:(0,\infty)\longrightarrow(0,\infty)\) be two functions satisfying (4.1) with \(f\leq g\). Then \(M_{f,k}(a,b)\geq M_{g,k}(a,b)\) for all \(a,b>0\).
 (ii)
Let \(m_{1}\) and \(m_{2}\) be two means such that \(m_{1}< m_{2}\). Then \(m_{1}^{\sigma_{k}}>m_{2}^{\sigma_{k}}\).
Proof
This follows from (4.2) and (4.8), respectively. Details are simple and therefore omitted here. □
Proposition 5.3
Proof
First, we recall that \(k(a,b)>0\) for \(a>b\) and \(k(a,b)<0\) if \(a< b\). Secondly, we have \(k_{1}(t)>k_{2}(t)\) for \(t>1\) and \(k_{1}(t)< k_{2}(t)\) if \(0< t<1\) (see Proposition 3.1). This, with a simple manipulation on (4.8), yields the desired result. Details are simple and therefore omitted here. □
The following example illustrates the previous proposition.
Example 5.3
Now, we can observe the next question: let \(m_{1}\) and \(m_{2}\) be two means and \(k,h\in{\mathcal {K}}\) such that \(m_{1}^{\sigma_{k}}=m_{2}^{\sigma_{h}}\). We ask if this implies that \(m_{1}=m_{2}\) and \(k=h\). Proposition 5.1 shows that it is not true, since \(A_{\lambda}^{\sigma_{k}}=H_{1\lambda}=A_{\lambda}^{\sigma_{h}}\) for all \(k,h\in{\mathcal {K}}\). However, the next result may be stated.
Proposition 5.4
Let \(m_{1}\) and \(m_{2}\) be two continuous homogeneous means such that \({m_{1}^{\sigma_{k}}=m_{2}^{\sigma_{k}}}\) for some \(k\in{\mathcal {K}}\). Assume that the map \(k_{12}:(0,\infty)\longrightarrow(0,\infty)\) is onto. Then we have \(m_{1}=m_{2}\).
Proof
Example 5.4
Let k be defined as in (3.2). Then \(k_{12}(t)=t^{q+1}\) which is onto for \(q>0\). It follows that if \(m_{1}\) and \(m_{2}\) are as in the previous proposition, with \(m_{1}^{\sigma_{q}}=m_{2}^{\sigma_{q}}\) for some \(q>0\) then \(m_{1}=m_{2}\).
6 Application 1: power mean including SB
As already pointed out before, this section displays various applications of the above theoretical approach for constructing some new power means including, among other, the SchwabBorchardt mean. The next result, giving us a lot of power homogeneous means, is of great interest.
Theorem 6.1
Proof
Now, let us present the following example of interest.
Example 6.1
Now, taking \(q=2\) in the above theorem we obtain the following corollary.
Corollary 6.2
Proof
If \(q=2\) then \(q^{*}=2\). Making the change of variables \(s=\operatorname{argch} t\) and \(s=\arccos t\) in the two integrals of (6.1), respectively, we obtain the desired result after an elementary manipulation. Details are simple and therefore omitted here. □
Now, choosing particular values for q, λ in the above, we can state the following interesting examples.
Example 6.2
(i) Taking \(\lambda=1/3\in[0,1]\) in (6.4), we find (after a simple computation) the SchwabBorchardt mean i.e. \(X_{2,1/3}=\mathit{SB}\). Since \(G_{\lambda}(a,b)=G_{1\lambda}(b,a)\), we have \(X_{q,\lambda}(a,b)=X_{q,1\lambda}(b,a)\) for fixed \(q>0\). It follows that \(X_{2,2/3}(a,b)=\mathit{SB}(b,a)\).
(ii) Theorem 6.1, with Remark 4.1, can be formulated as follows: For all \(\lambda\in[0,1]\) and \(q>0\), we have \(G_{\lambda }^{\sigma_{q}}=X_{q,\lambda}\). In particular, Example 6.1 can be formulated as \(G_{\lambda}^{\sigma}=L_{1\lambda}\). Also, the previous (i) means that \(G_{1/3}^{\sigma_{2}}=\mathit{SB}\) and \(G_{2/3}^{\sigma_{2}}=\mathit{SB}^{T}\).
Now, let us observe another interesting special situation given in the following example.
Example 6.3
Example 6.4
We can then see \(X_{2,\lambda}\) defined by (6.4) as weighted mean associated to the symmetric mean \(X_{2,1/2}\) given through (6.6). It seems that explicit computation of \(\alpha(x)\) and so that of \(X_{2,1/2}(a,b)\), for all \(a,b>0\), in terms of elementary functions is impossible.
7 Application 2: weighted means of T and M
In this section we give more application of our present approach. In particular, weighted mean associated to the symmetric means T will be investigated. We preserve the same notations as previously and we start with the following central result.
Theorem 7.1
Proof
Generally, \(Y_{q,\lambda}\) previously introduced cannot be explicitly computed, except for few particular values of q, such as \(q=1\) and \(q=1/2\). The case \(q=1\), which corresponds to \(k(a,b)=ab\), is presented in the following corollary.
Corollary 7.2
Proof
It is suitable to give more justification for our above weighted means. The following result is another reason of such suggestion.
Proposition 7.3
Proof
Now, about the weighted mean of P. This needs a long discussion which will be developed in Section 10 below. Other Lweighted means will be discussed there, by analogy with those of T and M previously investigated.
8 Generated function
Let \(k\in{\mathcal {K}}\) with \(k=\langle k_{1},k_{2}\rangle\) and m be a homogeneous mean. We start this section by stating the following definition.
Definition 8.1
Since \(x\longmapsto k(x,1)\) is continuously differentiable on \((0,1)\cup (1,\infty)\), so is \(x\longmapsto F_{m}^{k}(x)\) provided that \(x\longmapsto m(x,1)\) is as well.
Example 8.1
Example 8.2
Example 8.3
Let k be defined by \(k(a,b)=\sqrt{a^{2}b^{2}}\) if \(a\geq b\) and \(k(a,b)=\sqrt{b^{2}a^{2}}\) if \(a\leq b\).
We left to the reader the task for computing \(F_{H_{\lambda}}^{k}\) in a similar manner.
Now we state the following result.
Proposition 8.1
Proof
It is a simple exercise whose details are omitted here. □
In order to state an application of the above, we introduce more notation. If \(f,g:(0,\infty)\longrightarrow(0,\infty)\) are such that \(f(x)< g(x)\) for all \(x>0\) with \(x\neq1\) then we write \(f\prec g\). With this, we have the following.
Proposition 8.2
 (i)
If \(F_{m_{1}}^{k}=F_{m_{2}}^{k}\) for some \(k\in{\mathcal {K}}\) then we have \(m_{1}=m_{2}\).
 (ii)
If \(F_{m_{1}}^{k}\prec F_{m_{2}}^{k}\) for some \(k\in{\mathcal {K}}\) then \(m_{1}>m_{2}\).
Proof
This follows immediately from (8.2). Details are simple. □
The assertion (i) of the previous proposition means that the map \(m\longmapsto F_{m}^{k}\), for fixed \(k\in{\mathcal {K}}\), is onetoone (on the set of homogeneous means). It is also possible to show that the map \(k\longmapsto F_{m}^{k}\), for fixed homogeneous mean m, is onetoone. Assertion (ii) is more interesting and can be used for showing some mean inequalities. In particular, the chain of inequalities (7.4) can be proved here in a simple and fast way as explained in the following.
Theorem 8.3
Proof
Another example of applications is given in the following result.
Proposition 8.4
Proof
First, we notice that this double inequality was already proved in the literature; see [1] for instance. Our aim here is to prove it again by using our new approach, in a fast way.
9 Inverse transform of \(m\longmapsto m^{\sigma_{k}}\)
This section displays the inverse meanmap of \(m\longmapsto m^{\sigma _{k}}\). The main result here is stated as follows.
Theorem 9.1
Proof
As consequence of the above theorem we obtain the following result which is of interest in practical purposes. For the sake of simplicity we adopt the notation \(x^{1/q}=\sqrt [q]{x}\) for all \(x,q>0\).
Corollary 9.2
Proof
Here we have \(k_{12}(t)=t^{q+1}\) and \(k_{2}\) is explicitly given in Example 3.1. The desired result follows after simple computation and reduction. □
In particular, if \(k(a,b)=ab\) i.e. \(q=1\), \(1/q^{*}=0\) and \(F_{m}\) denotes the generated function of m, then the binary map: \(r_{m}(a,b)=aF_{m}(\sqrt{a/b})\) for all \(a,b>0\), \(a\neq b\), defines a homogeneous mean with \(r_{m}^{\sigma}=m\). Further, \(r_{m}\) is symmetric (resp. continuous) provided that m is as well. This particular situation corresponds to that developed in [18].
Corollary 9.3
Let \((m,k)\in\Omega\). Then there exists one and only one mean \(r=r_{m,k}\in{\mathcal {M}}_{\mathrm{hc}}\) such that \(r^{\sigma_{k}}=m\). We then write \(r=m^{\sigma_{k}}\).
Proof
The existence follows from the previous theorem, since \(R_{m,k}^{\sigma _{k}}=m\). The uniqueness is an immediate consequence of Proposition 5.3. Details are simple and therefore omitted here for the reader. □
Under the hypotheses of Corollary 9.3 and combining the above results, the unique mean \(r\in{\mathcal {M}}_{\mathrm{hc}}\) such that \(r^{\sigma _{k}}=m\) is given by \(r=R_{m,k}\), where \(R_{m,k}\) is defined by (9.1). We can then write \(m^{\sigma_{k}}=R_{m,k}\) and \(R_{m,k}\) will be called the \(\sigma_{k}\)inverse mean of m. We then have \((m^{\sigma_{k}} )^{\sigma_{k}}=m\) and \((m^{\sigma_{k}} )^{\sigma_{k}}=m\) for every \((m,k)\in\Omega\).
The following example illustrates the previous results.
Example 9.1
(i) Following Proposition 5.1, we have \(A_{\lambda}^{\sigma _{k}}=H_{1\lambda}\). By Corollary 9.3 we then deduce that \(H_{\lambda}^{\sigma_{k}}=A_{1\lambda}=G^{2}/H_{\lambda}\) for all \(\lambda \in[0,1]\) and every \(k\in{\mathcal {K}}\) satisfying the hypotheses of the previous corollary.
(ii) Example 6.2 asserts that \(G_{\lambda}^{\sigma_{q}}=X_{\lambda ,q}\) for all \(\lambda\in(0,1)\) and \(q>0\). We can then write \(X_{\lambda ,q}^{\sigma_{q}}=G_{\lambda}\) and in particular, \(\mathit{SB}^{\sigma _{2}}=G_{1/3}\).
(iii) Theorem 7.1 asserts that \(Y_{\lambda,q}^{\sigma _{q}}=H_{\lambda}\) for every \(\lambda\in(0,1)\) and \(q>0\). In particular, (7.1) yields \(T_{\lambda}^{\sigma}=H_{1\lambda}=G^{2}/A_{\lambda }\) for each \(\lambda\in(0,1)\).
Other examples of interest are given in the following result.
Theorem 9.4
Proof
Remark 9.1
The equalities of the previous theorem can be linked by nice and simple relationships which can be used for obtaining inequalities between the involved weighted means. For more details, see Section 11 below.
Now, we state the next result which is also of interest.
Corollary 9.5
Let \(m_{1},m_{2}\in{\mathcal {M}}_{\mathrm{hc}}\). Let \(k\in{\mathcal {K}}\) be such that \((m_{1},k)\in\Omega\) and \((m_{2},k)\in\Omega\). If \(m_{1}^{\sigma _{k}}< m_{2}^{\sigma_{k}}\) then we have \(m_{1}>m_{2}\).
Proof
This follows immediately from the definition of \(m\longmapsto m^{\sigma _{k}}\) with Proposition 5.2(iii). □
We can show again all inequalities of (7.4) by using the previous corollary. This is explained in the following example.
Example 9.2
We left to the reader the task for verifying the other inequalities in a similar manner.
Another example of application is given in the following.
Example 9.3
10 About the Pweighted mean
As already pointed out before, this section deals with the weighted mean of P. We will see that we can introduce more one Pweighted means following different point of view. We also introduce other Lweighted means.
Proposition 10.1
Proof
Now, we can state the following result giving more justification to our previous suggestion.
Proposition 10.2
We have \(L_{\lambda}< P_{\lambda}< A_{\lambda}\) for all \(\lambda\in(0,1)\).
Proof

By Proposition 8.2(ii), it is sufficient to show that the double inequalityholds for all \(x>0\) with \(x\neq1\). According to Example 8.1 and Proposition 10.1, we have to prove that$$F_{A_{\lambda}}(x)< F_{P_{\lambda}}(x)< F_{L_{\lambda}}(x) $$holds for all \(x>0\) with \(x\neq1\). The left side of this double inequality as well as its right side is reduced to \(x^{1\lambda }<(1\lambda)x+\lambda\) which is the wellknown Young (or weighted arithmeticgeometric) inequality. The desired double inequality is proved.$$\frac{1}{ ((1\lambda)x+\lambda )^{2}}< \frac{x^{\lambda 1}}{(1\lambda)x+\lambda}< x^{2(\lambda1)} $$

Following Corollary 9.5, it is sufficient to show thatBy Theorem 9.4 and Proposition 10.1, this is equivalent to$$A_{\lambda}^{\sigma}< P_{\lambda}^{\sigma}< L_{\lambda}^{\sigma}. $$which, in its two sides, is reduced to \(G_{\lambda}< S_{\lambda}\), so finishing the proof. □$$\frac{G^{2}}{S_{\lambda}}< \frac{G^{2}}{ (G_{\lambda}S_{\lambda} )^{1/2}}< \frac{G^{2}}{G_{\lambda}}, $$
For the weighted means \(T_{\lambda}\) and \(M_{\lambda}\) we have seen that \(T_{\lambda}=\mathit{SB}(A_{\lambda},Q_{\lambda})\) and \(M_{\lambda }=\mathit{SB}(Q_{\lambda},A_{\lambda})\). Analogous relation for \(P_{\lambda}\) seems to be not obvious. We then put the following as open problem.
Problem 1
Prove or disprove that \(P_{\lambda}\) defined by (10.1) satisfies \(P_{\lambda}=\mathit{SB} (G_{\lambda},A_{\lambda} )\). Similar question can be posed for \(L_{\lambda}=\mathit{SB} (A_{\lambda },G_{\lambda} )\).
Now, we can ask what is the more reasonable Pweighted mean among the two previous ones. In fact, it depends on what we want to do and what we want to have. For instance, if we desire to conserve the inequalities \(L_{\lambda}< P_{\lambda}< A_{\lambda}\), by analogy with \(L< P< A\), then the Pweighted mean given by (10.1) is more convenient, since that given by (10.2) does not satisfy the previous double inequality (we omit its verification here).
11 Inequalities involving the previous weighted means
As pointed out before, we present here some inequalities involving three means among the previous weighted means. For this purpose, we need a list of theoretical results which we will state in what follows. We begin by the first proposition.
Proposition 11.1
Proof
This follows immediately from Theorem 9.4 and Proposition 10.1. Details are simple and therefore omitted here. □
Now, we state the following result.
Theorem 11.2
 (i)Pointwise convexity: for all \(\alpha\in(0,1)\) and any means \(m_{1}\) and \(m_{2}\) we haveIf, moreover, \(m_{1}\neq m_{2}\) are comparable the previous mean inequality becomes strict.$$\bigl((1\alpha)m_{1}+\alpha m_{2} \bigr)^{\sigma_{k}} \leq(1\alpha)m_{1}^{\sigma _{k}}+\alpha m_{2}^{\sigma_{k}}. $$
 (ii)Pointwise geometric strict concavity: for all \(\alpha\in(0,1)\) and any means \(m_{1}\neq m_{2}\) one has$$ \bigl(m_{1}^{1\alpha}m_{2}^{\alpha} \bigr)^{\sigma_{k}}> \bigl(m_{1}^{\sigma_{k}} \bigr)^{1\alpha} \bigl(m_{2}^{\sigma_{k}} \bigr)^{\alpha}. $$(11.1)
Proof
It is similar to those of Theorem 4.2 and Theorem 4.4 of [18], pages 9697, with some precautions. Details are omitted here to the aim to not lengthen the present paper. □
In applications, the following corollary is of interest.
Corollary 11.3
Proof
Now, we will illustrate the previous statements with the following example.
Example 11.1
(iii) We leave to the reader the routine task for obtaining more mean inequalities in a similar way to previously.
We end this paper by stating the following remark.
Remark 11.1
The mean inequalities obtained in Example 11.1, for \(\lambda\in (0,1)\), justify again that \(L_{\lambda}\), \(M_{\lambda}\), \(T_{\lambda}\) and \(P_{\lambda}\) are reasonable weighted means of L, M, T and P, respectively. This is so because, for \(\lambda=1/2\), they yield the known mean inequalities \(M^{2}>AT\), \(T^{2}>MQ\) and \(P^{2}>AL\), already proved in [2].
Declarations
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Authors’ Affiliations
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