Index of a bivariate mean and applications
- Mustapha Raïssouli^{1, 2}Email authorView ORCID ID profile
https://doi.org/10.1186/s13660-016-1156-3
© Raïssouli 2016
Received: 13 June 2016
Accepted: 26 August 2016
Published: 15 September 2016
Abstract
Exploring some results of (Raïssouli in J. Math. Inequal. 10(1):83-99, 2016) from another point of view, we introduce here some power-operations for (bivariate) means. As application, we construct some classes of means in one or two parameters including some standard means. We also define a law between means which allows us to obtain, among others, a simple relationship involving the three familiar means, namely the first Seiffert mean, the second Seiffert mean, and the Neuman-Sándor mean. At the end, more examples of interest are discussed and open problems are derived as well.
Keywords
MSC
1 Introduction
In [1], the following result has been established (see Corollary 2.2).
Theorem A
With this, the following result has been proved in [1] (see Theorem 3.2).
Theorem B
For example, we have \(H^{-\sigma}=A\), \(L^{-\sigma}=G\) and \(T^{-\sigma }=H\). See [1] for more examples and detail.
The following result has also been proved there (see Corollary 3.4).
Theorem C
The remainder of this paper will be organized as follows: after this introduction, Section 2 is devoted to a list of lemmas that will be needed throughout the paper. In Section 3 we define \(m^{(q)}\) the mean-power of a mean m of order q, with \(|q|\leq1\). As examples, we obtain \(P=G^{(-1)}\), \(T=H^{(-1)}\), and \(P=A^{(1/2)}\). This allows us to construct, in Section 4, a family of means involving one parameter. Section 5 displays the definition of a new concept, so-called index of a mean, in the aim to define \(m^{(q)}\) when \(|q|>1\). We obtain, among others, \(P^{(2)}=A=G^{(-2)}\) and \(L^{(q)}=L\) for each q real number. In Section 6, we introduce a law ⊙ between means and we study its properties. As a first application, we obtain a simple relationship involving the three means P, M, and T, namely \(M=T^{(1/2)}\odot P\). Further applications are discussed in Section 7 where we construct some families of means involving two parameters and including all the previous means. Finally, Section 8 is focused on giving more examples of applications as well as deriving open problems for future research.
2 Some lemmas needed
As already pointed out, we state here some lemmas that will be needed in the sequel. First, we mention that every homogeneous mean m can be written in the form \(m=Gg (A/G )\) for some function g, since a and b can be both expressed in terms of A and G. The following lemma explains this situation.
Lemma 2.1
Proof
We notice that \(F_{\min}(z)=u(z)\) and \(F_{\max}(z)=l(z)\), where min and max denote the trivial means \((a,b)\longmapsto\min(a,b)\) and \((a,b)\longmapsto\max(a,b)\), respectively.
A function g, defined from \([1,\infty)\) into \((0,\infty)\) and satisfying \(l(z)\leq g(z)\leq u(z)\) for all \(z\geq1\), will be called here an admissible function. The functions l and u are the lower and upper admissible functions, respectively. To obtain other examples of admissible functions g, it is sufficient to take \(g(z)=\eta (l(z),u(z) )\) for some mean η (symmetric, homogeneous, or not). The following example explains the latter situation.
Example 2.2
- (i)
\(g(z):=A (l(z),u(z) )=z=F_{H}(z)\).
- (ii)
\(g(z):=G (l(z),u(z) )=1=F_{L}(z)\).
- (iii)
\(g(z):=H (l(z),u(z) )=1/z=F_{T}(z)\).
Inversely, let g be an admissible function, does \(m\in{\mathcal{M}}_{\sigma}\) exist such that \(g(z)=m^{-\sigma}/G\) whenever \(z=A/G\)? The following result answers affirmatively the latter question.
Lemma 2.3
Proof
If g is as assumed, it is easy to see that \(Gg (A/G )\) is a regular mean. Detail is simple and therefore omitted here. □
Remark that, for all \(z\geq1\), we have \(0< l(z)\leq1\leq u(z)\) and \(l(z)=1/u(z)\). This implies that if g is admissible then its point-wise inverse \(g^{-1}\) i.e. \(g^{-1}(z)=1/g(z)\), is also admissible. More generally, let q be a real number and define the point-wise q-power of g by \(g^{q}(z):= (g(z) )^{q}\), with \(g^{0}(z)=1\), for all \(z\geq1\). If g is admissible, it is easy to see that \(g^{2}\) is in general not admissible. A function g for which \(g^{q}\) is admissible, for some real number q, will be called q-admissible. It is immediate that, if g is q-admissible then so is \(g^{-1}:=1/g\). With this, the following result may be stated.
Lemma 2.4
Let g be an admissible function. Then g is q-admissible whenever \(|q|\leq1\).
Proof
The following lemma will also be needed in some situations below.
Lemma 2.5
Let g be an admissible function and \(q\geq1\). Assume that \(g(z)\leq1\) (resp. \(g(z)\geq1\)) for all \(z\geq1\). If g is q-admissible then g is s-admissible whenever \(1\leq s\leq q\).
Proof
We end this section by stating another needed lemma recited in the following.
Lemma 2.6
Proof
3 Mean-power and mean-iterate
In this section, we will observe the previous results from another point of view in the aim to interpret them in service of means.
Proposition 3.1
We can then state the following definition.
Definition 3.2
Let \(q\in[-1,1]\). The mean \(r_{q}\in{\mathcal{M}}_{\sigma}\) defined by the previous proposition will be called the q-mean-power of m and we write \(r_{q}=m^{(q)}\), with \(m^{(0)}=L\). If \(q=1/n\), with \(n\geq2\) integer, \(m^{(1/n)}\) will be called the n-mean-iterate of m. In particular, \(m^{(1/2)}\) is the mean-root of m and \(m^{(-1)}\) is the mean-inverse of m. Clearly, \(m^{(1)}=m\).
Using (3.1), it is easy to see that \((m^{(q_{1})} )^{(q_{2})}=m^{(q_{1}q_{2})}\) whenever \(q_{1},q_{2}\in[-1,1]\). The following examples illustrate the previous concepts.
Example 3.3
- (i)
For each \(q\in[-1,1]\), \(L^{(q)}=L\).
- (ii)
\(P=G^{(-1)}\) and \(T=H^{(-1)}\).
- (iii)
\(P=A^{(1/2)}\) and \(G=A^{(-1/2)}\).
Example 3.4
Other various examples of interest will be discussed throughout the following sections. Now, we state the following result summarizing the elementary properties of the mean-map \(m\longmapsto m^{(q)}\) for \(q\in(0,1)\).
Proposition 3.5
- (i)
Let \(m_{1},m_{2}\in{\mathcal{M}}_{\sigma}\) be such that \(F_{m_{1}}(z)< F_{m_{2}}(z)\) for all \(z>1\). Then \(m_{1}^{(q)}>m_{2}^{(q)}\) for each \(q\in(0,1)\).
- (ii)
Let \(m\in{\mathcal{M}}_{\sigma}\) be such that \(F_{m}(z)<1\) (resp. \(F_{m}(z)>1\)) for all \(z>1\). Then we have \(m^{(q_{1})}< m^{(q_{2})}< m\) (resp. \(m< m^{(q_{2})}< m^{(q_{1})}\)) whenever \(0\leq q_{1}< q_{2}\leq1\).
Proof
It is straightforward. We therefore omit it for the reader. □
Now, we state the following results which give more information as regards comparison between the previous means.
Proposition 3.6
- (i)
If \(0< q\leq1/4\) then \(G< H^{(q)}\) and \(T^{(q)}< P\).
- (ii)
If \(1/2\leq q<1\) then \(H^{(q)}< G\).
- (iii)
If \(0< q<2/3\) then \(M^{(q)}< A\) and so \(M^{(q/2)}< P\).
Proof
We use (1.4) and Theorem C. We have to compare \(F_{G}(z)=\sqrt {(z+1)/2}\) and \(F_{H^{(q)}}(z):= (F_{H}(z) )^{q}=z^{q}\). Setting \(\Phi(z)=2z^{2q}-z-1\), \(z>1\), and studying the monotonicity of Φ we deduce the desired result about G and \(H^{(q)}\) in a simple way. Similar method for the other mean-inequalities. Detail is simple and therefore omitted here. □
The previous proposition when combined with (3.2), (3.3), and (3.4) yields the following corollary.
Corollary 3.7
Remark 3.8
For some values of q, like \(q\in(1/4,1/2)\), Proposition 3.6 does not give any information as regards comparison of G and \(H^{(q)}\). In fact, \(F_{G}(z)\) and \((F_{H}(z) )^{q}\) are not comparable, because the related function Φ is not monotonic and satisfies \(\Phi(1)=0\), \(\lim_{z\uparrow\infty}\Phi(z)=-\infty\). Of course, we cannot deduce any conclusion as for comparison of G and \(H^{(q)}\), since (1.4) is just an implication. This is the reason why analogous of (3.5) when \(q\notin(0,1/4]\) cannot be stated in a similar manner as previous.
- (a)
- (b)
What is the expression of \(A^{(q)}\), for each \(q\in[-1,1]\), extending the relationship \(A^{(1/2)}=P\)? The answer to this question will be the aim of the next section. We also give the expression of \(T^{(q)}\).
- (c)
State a reciprocal study of the previous one i.e. let \(r\in {\mathcal{M}}_{\sigma}\) and \(q\in[-1,1]\). Does \(m\in{\mathcal{M}}_{\sigma }\) exist such that \(m^{(q)}=r\)? This will be the purpose of Section 5 below.
4 On a family of 1-power means
As pointed before, this section will be devoted to an expression of \(A^{(q)}\) for \(q\in[-1,1]\). Such an expression should coincide with that of P when we take \(q=1/2\), since \(A^{(1/2)}=P\). Precisely, the following result may be stated.
Theorem 4.1
Proof
The above theorem immediately gives (again) \(A^{(1)}=A\) and \(A^{(1/2)}=P\). Taking \(q=1/n\), with \(n\geq1\) integer, in the previous theorem we obtain a sequence of regular means. Such sequence satisfies interesting properties as summarized in the following result.
Proposition 4.2
Proof
The fact that \((P_{n})\) is strictly point-wisely decreasing follows from Proposition 3.5. This, with the fact that \(P_{n}\) is a mean i.e. \(\min(a,b)\leq P_{n}(a,b)\leq\max(a,b)\) for all \(a,b>0\) and \(n\geq 1\), implies that \((P_{n})\) converges point-wisely to a mean, which we call \(P_{\infty}\). We need to prove \(P_{\infty}=L\). Such a result follows from the next lemma and the proof of the theorem will then be completed. □
Lemma 4.3
Proof
By (3.6) we deduce \(\lim_{n\uparrow\infty }P^{(1/n)}(a,b)=L(a,b)\). We can also see this by writing \(P^{(1/n)}= (A^{(1/2)} )^{(1/n)}=A^{(1/2n)}\).
Now, let us compute \(T^{(q)}\) for \(0< q\leq1\). By similar way as for \(A^{(q)}\), we obtain the following result (details are omitted here).
Theorem 4.4
Since \(H=T^{(-1)}\) we deduce \(H^{(q)}=T^{(-q)}\) for each \(|q|\leq1\). The expression of \(H^{(q)}\) follows from (4.3). For another way of computation of \(G^{(q)}\) and \(P^{(q)}\), see the next section (Examples 5.7, 5.9). For \(M^{(q)}\), see Section 7 below.
5 Index of a mean
Let q be a real number. A mean \(m\in{\mathcal{M}}_{\sigma}\) will be called q-coherent if its regularized function \(F_{m}\) is q-admissible. By Lemma 2.4, we deduce that every \(m\in {\mathcal{M}}_{\sigma}\) is q-coherent for each \(|q|\leq1\). We can then introduce the following definition.
Definition 5.1
It is clear that \(\operatorname{ind}(m)\geq1\) for every \(m\in{\mathcal{M}}_{\sigma}\), since m is q-coherent for \(|q|\leq1\). The following result may be stated as well.
Proposition 5.2
- (i)
\(\operatorname{ind}(m^{(-1)})=i\).
- (ii)
\(\operatorname{ind}(m^{(q)})=i/|q|\) for every \(|q|\leq1\) with \(q\neq0\).
- (iii)
There exists \(r\in{\mathcal{M}}_{\sigma}\) such that \(r^{(1/q)}=m\) whenever \(1\leq q\leq i\).
Proof
(i) follows from the definition of index with the fact that \(l(z)=1/u(z)\).
(iii) If \(\operatorname{ind}(m)=i\) then, by definition, \(F_{m}^{q}(z):= (F_{m}(z) )^{q}\) is admissible for each \(1\leq q\leq i\). Then, by Lemma 2.3, there exists \(r\in{\mathcal{M}}_{\sigma}\) such that \((F_{m}(z) )^{q}=F_{r}(z)\) or again \(F_{m}(z)= (F_{r}(z) )^{1/q}:=F_{r^{(1/q)}}(z)\), where \(r^{(1/q)}\) is defined by (3.1). We then deduce \(m=r^{(1/q)}\) and the proof is complete. □
The following example illustrates the previous concepts.
Example 5.3
- (i)
Since \(F_{L}(z)=1\), L is q-coherent for all real number q and so \(\operatorname{ind}(L)=\infty\). This rejoins Proposition 5.2(ii) with \(m^{(0)}=L\), by adopting the convention \(i/0=\infty\).
- (ii)Since \(F_{\min}(z)=u(z)\) we deduce that the largest \(q\geq1\) such that (for all \(z\geq1\))is \(q=1\). This means that \(\operatorname{ind}(\min)=1\). Similarly, we verify that \(\operatorname{ind}(\max)=1\).$$l(z)\leq \bigl(F_{\min}(z) \bigr)^{q}= \bigl(u(z) \bigr)^{q}\leq u(z) $$
The following proposition gives more examples of interest.
Proposition 5.4
- (i)
\(\operatorname{ind}(A)=\operatorname{ind}(H)=\operatorname{ind}(T)=\operatorname{ind}(M)=1\).
- (ii)
\(\operatorname{ind}(G)=\operatorname{ind}(P)=2\).
- (iii)
L is the unique σ-regular mean such that \(\operatorname{ind}(L)=\infty\).
Proof
(i) Let q be the index of A (resp. H, T or M). Lemma 4.3 (with Theorem C) immediately implies that \(q\leq1\). Since \(\operatorname{ind}(m)\geq1\) for every mean \(m\in{\mathcal{M}}_{\sigma}\), we deduce the desired result.
(ii) Since \(P=A^{(1/2)}\) we deduce by Proposition 5.2(ii), \(\operatorname{ind}(P)=2\operatorname{ind}(A)=2\). The relation \(P=G^{(-1)}\), with Proposition 5.2(i), yields \(\operatorname{ind}(G)=\operatorname{ind}(P)=2\).
The following example is also of interest.
Example 5.5
Let \(q\geq1\). Since \(\operatorname{ind}(A)=1\), Proposition 5.2(ii) implies that \(\operatorname{ind}(A^{1/q})=q\), where \(A^{(1/q)}\) is defined by (4.1). Similarly, \(\operatorname{ind}(T^{1/q})=q\), where \(T^{(1/q)}\) is defined by (4.3). If follows that, for each \(q\geq1\), there exists \(m\in {\mathcal{M}}_{\sigma}\) (not unique) such that \(\operatorname{ind}(m)=q\). In another way, the map \(m\longmapsto \operatorname{ind}(m)\) defined from \({\mathcal{M}}_{\sigma}\) into \([1,\infty)\) is surjective but not injective.
Now, we are in a position to state the following definition.
Definition 5.6
Let \(m\in{\mathcal{M}}_{\sigma}\) be such that \(\operatorname{ind}(m)=i\) and \(1\leq q\leq i\). Then the mean \(r\in{\mathcal{M}}_{\sigma}\) defined by Proposition 5.2(ii) will be denoted by \(r:=m^{(q)}\) and called the q-mean-power of m. In particular, if \(i\geq2\) and \(q=2\) then \(m^{(2)}\) is called the mean-square of m.
Definition 3.2 introduces \(m^{(q)}\) when \(|q|\leq1\) for all \(m\in{\mathcal{M}}_{\sigma}\), while Definition 5.6 defines \(m^{(q)}\) when \(1\leq q\leq i\) provided \(\operatorname{ind}(m)=i\). We can then define \(m^{(q)}\) for \(q\leq-i\) when \(\operatorname{ind}(m)=i\). In fact, we write \(m^{(q)}:= (m^{(-1)} )^{(-q)}\), with \(\operatorname{ind}(m)=\operatorname{ind}(m^{(-1)})\).
Now, let us observe the following example explaining the previous discussion.
Example 5.7
- (i)
By \(\operatorname{ind}(L)=\infty\) we then deduce that \(L^{(q)}\) is also defined for each \(q\geq1\). We then have \(L^{(q)}=L\) for every real number q.
- (ii)
Since \(\operatorname{ind}(P)=2\), the relationship \(P=A^{(1/2)}\) is, following the previous definition, equivalent to \(P^{(2)}=A\).
- (iii)
By \(\operatorname{ind}(A)=1\), the relation \(P_{n}:=A^{(1/n)}\) can then be written as \(P_{n}^{(n)}=A\), since \(\operatorname{ind}(A^{(1/n)})=n\) according to Proposition 5.2(ii).
- (iv)Since \(\operatorname{ind}(G)=2\), \(G^{(q)}\) exists (as a mean) for each \(|q|\leq 2\). We can show that (we omit detail here)$$G^{(2)}= \biggl(\frac{A+G}{2} \biggr)^{\sigma}= \frac{2G^{2}L}{G^{2}+AL}. $$
We end this section by stating the following result, which summarizes some elementary properties of the index.
Proposition 5.8
- (i)
\((m^{(q_{1})} )^{(q_{2})}=m^{(q_{1}q_{2})}\) for all \(|q_{1}|\leq1\) and \(|q_{2}|\leq1\).
- (ii)
\((m^{(q_{1})} )^{(q_{2})}=m^{(q_{1}q_{2})}\) for all \(q_{1}\), \(q_{2}\) such that \(|q_{1}|\leq \operatorname{ind}(m)\) and \(|q_{1}q_{2}|\leq \operatorname{ind}(m)\). In particular, \((m^{(1/q)} )^{(q)}=m\) for each \(|q|\geq1\) and \((m^{(q)} )^{(1/q)}=m\) for every \(|q|\leq \operatorname{ind}(m)\).
- (iii)
\(\operatorname{ind}(m^{(q)})=\operatorname{ind}(m)/|q|\) for each \(|q|\leq \operatorname{ind}(m)\).
Proof
It is straightforward and we therefore omit all detail here. □
Now, we present the following example, which illustrates the previous results.
Example 5.9
6 On a law between means
We preserve the same notation as in the above sections. Inspired by the previous study, we will construct here an internal operation (law) between σ-regular means.
If \(\operatorname{ind}(m)\geq i\) then \((m^{(i_{1})},m^{(i_{2})} )\in{\mathcal{C}}\) whenever \(i_{1}+i_{2}=i\). In particular, \((m,m)\in{\mathcal{C}}\) for every \(m\in{\mathcal{M}}_{\sigma}\) such that \(\operatorname{ind}(m)\geq2\).
A triplet \((m_{1},m_{2},m_{3})\) of σ-regular means will be called ⊙-compatible if \((m_{1},m_{2})\in{\mathcal{C}}\), \((m_{2},m_{3})\in{\mathcal{C}}\), \((m_{1}\odot m_{2},m_{3})\in{\mathcal{C}}\) and \((m_{1},m_{2}\odot m_{3})\in {\mathcal{C}}\). If \(\operatorname{ind}(m)\geq3\) then \((m,m,m)\) is ⊙-compatible. More generally, if \(\operatorname{ind}(m)\geq i\) then \((m^{(i_{1})},m^{(i_{2})},m^{(i_{3})} )\) is ⊙-compatible whenever \(i_{1}+i_{2}+i_{3}=i\).
The following result summarizes some properties of the law ⊙.
Proposition 6.1
- (i)
For all \((m_{1},m_{2})\in{\mathcal{C}}\), \((m_{2},m_{1})\in{\mathcal{C}}\) and \(m_{1}\odot m_{2}=m_{2}\odot m_{1}\).
- (ii)
\((m_{1}\odot m_{2})\odot m_{3}=m_{1}\odot(m_{2}\odot m_{3})\) for all ⊙-compatible triplet \((m_{1},m_{2},m_{3})\).
- (iii)
For all \(m\in{\mathcal{M}}_{\sigma}\), \((m,L)\in{\mathcal{C}}\) and \(m\odot L=L\odot m=m\).
- (iv)For all \(m\in{\mathcal{M}}_{\sigma}\), \((m,m^{(-1)})\in{\mathcal{C}}\) andwhere \(m^{(-1)}\) was defined in Definition 3.2 and is given by$$m\odot m^{(-1)}=m^{(-1)}\odot m=L, $$$$m^{(-1)}= \bigl(\bigl(m^{-\sigma}\bigr)^{*} \bigr)^{\sigma}. $$
Proof
Proposition 6.2
With the above, \(({\mathcal{I}}_{2},\odot )\) is an abelian group.
Proof
First, as previously pointed, ⊙ is stable on \({\mathcal{I}}_{2}\). Further, it is easy to verify that if \(m_{1},m_{2},m_{3}\in{\mathcal{I}}_{2}\) then the triplet \((m_{1},m_{2},m_{3})\) is ⊙-compatible. This, with Proposition 6.1(i), (ii), asserts that ⊙ is commutative and associative on \({\mathcal{I}}_{2}\). Now, we have \(L\in{\mathcal{I}}_{2}\) and so L is the unit element of ⊙ on \({\mathcal{I}}_{2}\). Lastly, if \(m\in{\mathcal{I}}_{2}\) then \(m^{(-1)}\in{\mathcal{I}}_{2}\), since \(\operatorname{ind}(m)=\operatorname{ind}(m^{(-1)})\). The proof is completed. □
We now are in a position to state the following result giving a simple and nice relationship between the three familiar means P, T and M in terms of the law ⊙.
Theorem 6.3
Proof
We notice that the relation \(M=T^{(1/2)}\odot P\) is not equivalent to \(M^{(2)}=T\odot P^{(2)}\), since \(\operatorname{ind}(M)=1\) and so M is not 2-coherent. Also, we cannot write \(M=T^{(1/2)}\odot A^{(1/2)}\) in the form \(M=(T\odot A)^{(1/2)}\), since \((T,A)\notin{\mathcal{C}}\) and so \(T\odot A\) is not defined.
Otherwise, the previous theorem is interesting from theoretical point of view as well as in practical purposes. First, it gives M in a short form involving the simplest means A and H. Second, it regroups P, T and M in a unified expression, simple to prove and easy to remember. Third, we can derive some families of means involving two parameters and including all the previous familiar means A, G, H, L, P, T and M. In the next section, we will give some clarification as regards the latter situation.
We end this section by stating the following result.
Theorem 6.4
Proof
If \(m_{2}=m_{1}^{(-1)}\) then inequality (6.1) gives \(0\leq 2/\operatorname{ind}(m_{1})\), since \(m_{1}\odot m_{1}^{(-1)}=L\) and \(\operatorname{ind}(L)=\infty\). This means that (6.1) is not, in general, an equality. It is the best possible in the sense that if \(m_{1}=m_{2}\), it remains an equality. We can then state the following open question.
7 On some families of 2-power means
We start this section by stating the following needed lemma.
Lemma 7.1
Let \(p,q\geq0\) be such that \(p+q\leq1\). Then \(A^{(p)}\) and \(T^{(q)}\) satisfy \((A^{(p)},T^{(q)} )\in{\mathcal{C}}\) i.e. \(A^{(p)}\odot T^{(q)}\) is well defined.
Proof
We can give an expression of \(W_{p,q}\) as recited in the following result.
Theorem 7.2
Proof
If in (7.2) we take \(p=0\) and \(|q|\leq1\), or \(q=0\) and \(|p|\leq 1\), we obtain (after simple computation) equations (4.1) and (4.3), respectively. This means that (7.2) is also valid for the particular situation (7.1). In another way, (7.2) includes all the previous means A, G, H, L, P, T, and M.
In order to give more example of construction of 2-power means, we need the next lemma.
Lemma 7.3
Proof
The previous lemma implies that \(P^{(p)}\odot M^{(q)}\) exists. It further implies, with Theorem 7.2, an expression of \(Z_{p,q}:=P^{(p)}\odot M^{(q)}\) as recited in the following result.
Theorem 7.4
When \(q=0\) and \(0\leq p\leq2\), (7.5) coincides with (5.1). For \(p=0\) and \(0\leq q\leq1\), it immediately yields an expression of \(M^{(q)}\). All the above expressions are uncomputable exactly, except for a few trivial values of p and q.
8 Further examples
As already pointed out before, this section displays some other examples of interest in the aim to illustrate more the previous concepts as well as their related results.
Applying the above theoretical study to the means U and V, we obtain the following result.
Theorem 8.1
- (i)The means U and V are σ-regular, with$$F_{U}(z)=\frac{1}{z}\sqrt{\frac{z+1}{2}} \quad \textit{and} \quad F_{V}(z)=\sqrt{\frac {z+1}{2z}};\quad z=A/G. $$
- (ii)
\(U=G\odot T=G\odot H^{(-1)}=P^{(-1)}\odot T\), \(V=G\odot T^{(1/2)}\) and \(U=V\odot T^{(1/2)}\).
- (iii)
\(\operatorname{ind}(U)=2\) and \(\operatorname{ind}(V)\geq6\).
Proof
(i) We leave it to the reader as an interesting exercise. We can also consult [1] for similar situations.
(ii) follows from (i), with the definition of ⊙.
Proposition 8.2
Proof
A simple verification asserts that \(F_{U}(z)>F_{M}(z)\) for all \(z>1\). By (1.3) again, we then deduce that \(U< M\). Similarly, it is easy to verify that \(F_{P}(z)< F_{V}(z)\) and \(F_{V}(z)<1=F_{L}(z)\), for all \(z>1\). We then deduce \(L< V< P\). Summarizing, the desired inequalities are obtained. □
Since \(\operatorname{ind}(U)=2\) and \(\operatorname{ind}(V)\geq6\), \(U^{(2)}\) and \(V^{(2)}\) both exist (as means). Then the two relations \(V=G\odot T^{(1/2)}\) and \(U=V\odot T^{(1/2)}\) are equivalent to the following ones: \(V^{(2)}=G^{(2)}\odot T\) and \(U^{(2)}=V^{(2)}\odot T\), respectively. We might also be interested by computing the explicit forms of \(U^{(2)}\) and \(V^{(2)}\). The following result answers the latter claim.
Theorem 8.3
Proof
We left to the reader the routine task for finding the integral expression of \(V^{(q)}\), for \(1\leq q\leq6\), and then deduce some writings of the means \(V^{(3)}\), \(V^{(4)}\) and \(V^{(5)}\) in terms of the previous familiar means.
We now put the following as an open question.
Problem 2
Find the exact value of \(\operatorname{ind}(V)\).
Proposition 8.4
We end this section by stating the following open question.
Problem 3
Evaluate \(\operatorname{ind}(R_{1})\) and \(\operatorname{ind}(R_{2})\).
Declarations
Acknowledgements
The author would like to thank the anonymous referees for their encouragements and suggestions.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
References
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