Table of Contents Author Guidelines Submit a Manuscript
International Journal of Mathematics and Mathematical Sciences
VolumeΒ 2009, Article IDΒ 293539, 11 pages
http://dx.doi.org/10.1155/2009/293539
Research Article

On a Class of Ky Fan-Type Inequalities

Division of Mathematical Sciences, School of Physical and Mathematical Sciences, Nanyang Technological University, Singapore 637371

Received 19 August 2009; Accepted 15 December 2009

Academic Editor: Sever SilvestruΒ Dragomir

Copyright Β© 2009 Peng Gao. 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 study one class of Ky Fan-type inequalities, which has ties with the original Ky Fan inequality. Our result extends the known ones.

1. Introduction

Let 𝑀𝑛,π‘Ÿ(𝐱;πͺ) be the generalized weighted means: 𝑀𝑛,π‘Ÿ(𝐱;πͺ)=(βˆ‘π‘›π‘–=1π‘žπ‘–π‘₯π‘Ÿπ‘–)1/π‘Ÿ, where 𝑀𝑛,0(𝐱;πͺ) denotes the limit of 𝑀𝑛,π‘Ÿ(𝐱;πͺ) as π‘Ÿβ†’0+. Here 𝐱=(π‘₯1,…,π‘₯𝑛), πͺ=(π‘ž1,…,π‘žπ‘›) with π‘žπ‘–>0(1≀𝑖≀𝑛) satisfying βˆ‘π‘›π‘–=1π‘žπ‘–=1. In this paper, we always assume 0<π‘₯1≀π‘₯2≀⋯≀π‘₯𝑛. To any given 𝐱 and 𝑑β‰₯0, we set π±ξ…ž=(1βˆ’π‘₯1,…,1βˆ’π‘₯𝑛),𝐱𝑑=(π‘₯1+𝑑,…,π‘₯𝑛+𝑑).

We define 𝐴𝑛(𝐱;πͺ)=𝑀𝑛,1(𝐱;πͺ),𝐺𝑛(𝐱;πͺ)=𝑀𝑛,0(𝐱;πͺ),𝐻𝑛(𝐱;πͺ)=𝑀𝑛,βˆ’1(𝐱;πͺ), and we shall write 𝑀𝑛,π‘Ÿ for 𝑀𝑛,π‘Ÿ(𝐱;πͺ), 𝑀𝑛,π‘Ÿ,𝑑 for 𝑀𝑛,π‘Ÿ(𝐱𝑑;πͺ), and π‘€ξ…žπ‘›,π‘Ÿ for 𝑀𝑛,π‘Ÿ(π±ξ…ž;πͺ) if π‘₯𝑛<1 and similarly for other means when there is no risk of confusion. We further denote πœŽπ‘›=βˆ‘π‘›π‘–=1π‘žπ‘–(π‘₯π‘–βˆ’π΄π‘›)2.

When π‘₯𝑛<1, we define

Ξ”π‘Ÿ,𝑠,𝛼,𝑑=ξ€·π‘€ξ…žπ›Όπ‘›,π‘Ÿ,π‘‘βˆ’π‘€ξ…žπ›Όπ‘›,𝑠,𝑑/𝛼𝑀𝛼𝑛,π‘Ÿ,π‘‘βˆ’π‘€π›Όπ‘›,𝑠,𝑑,/𝛼(1.1) where we set 𝑀0𝑛,π‘Ÿ/0=ln𝑀𝑛,π‘Ÿ and we shall write Ξ”π‘Ÿ,𝑠,𝛼 for Ξ”π‘Ÿ,𝑠,𝛼,0 and Ξ”π‘Ÿ,𝑠 for Ξ”π‘Ÿ,𝑠,1. In order to include the case of equality for various inequalities in our discussions, for any given inequality, we define 0/0 to be the number which makes the inequality an equality. The author [1, Theorem 2.1] has shown the following (in fact, only the case 𝛼=1 is shown there but one can easily extend the result to all 𝛼≀2 following the method there).

Theorem 1.1. For π‘Ÿ>𝑠 and 𝛼≀2, the following inequalities are equivalent: π‘Ÿβˆ’π‘ 2π‘₯12βˆ’π›ΌπœŽπ‘›β‰₯𝑀𝛼𝑛,π‘Ÿβˆ’π‘€π›Όπ‘›,𝑠𝛼β‰₯π‘Ÿβˆ’π‘ 2π‘₯𝑛2βˆ’π›ΌπœŽπ‘›,ξ‚΅π‘₯(1.2)𝑛1βˆ’π‘₯𝑛2βˆ’π›Όβ‰₯Ξ”π‘Ÿ,𝑠,𝛼β‰₯ξ‚΅π‘₯11βˆ’π‘₯1ξ‚Ά2βˆ’π›Ό,(1.3) where in (1.3) one requires π‘₯𝑛<1.

In fact, one can further show that (see [2]) the two inequalities in Theorem 1.1 are equivalent to

ξ‚΅π‘₯𝑛𝑑+π‘₯1ξ‚Ά2βˆ’π›Όβ‰₯Ξ”π‘Ÿ,𝑠,𝛼,𝑑β‰₯ξ‚΅π‘₯1𝑑+π‘₯𝑛2βˆ’π›Ό(1.4) being valid for all 𝑑β‰₯0. We point out here that when inequality (1.2) holds for some π‘Ÿ,𝑠, one can often expect for a better result than (1.4), namely,

ξ‚΅π‘₯𝑛𝑑+π‘₯𝑛2βˆ’π›Όβ‰₯Ξ”π‘Ÿ,𝑠,𝛼,𝑑β‰₯ξ‚΅π‘₯1𝑑+π‘₯1ξ‚Ά2βˆ’π›Ό.(1.5)

We note that inequality (1.2) does not hold for all pairs π‘Ÿ,𝑠 (see [1]). Cartwright and Field [3] first proved the validity of (1.2) for π‘Ÿ=1,𝑠=0,𝛼=1. For other extensions and refinements of (1.2), see [2, 4–8]. When 𝛼=1, inequality (1.3) is commonly referred as the additive Ky Fan’s inequality. We refer the reader to the survey article [9] and the references therein for an account of Ky Fan's inequality.

In this paper, we will focus on the special case 𝛼=0 of (1.2), which has ties with the following result of Ky Fan that initiated the study of the whole subject.

Theorem 1.2 (see [10,page 5]). For π‘₯π‘–βˆˆ(0,1/2], Ξ”1,0,0≀1, with equality holding if and only if π‘₯1=β‹…β‹…β‹…=π‘₯𝑛.

A nice result of Wang and Chen [11] determines all the pairs π‘Ÿ,𝑠 with π‘Ÿ>𝑠 such that Ξ”π‘Ÿ,𝑠,0≀1 is satisfied when π‘₯π‘–βˆˆ(0,1/2]. Their result is contained in the following.

Theorem 1.3. For π‘Ÿ>𝑠,π‘₯π‘–βˆˆ(0,1/2], Ξ”π‘Ÿ,𝑠,0≀1 holds if and only if |π‘Ÿ+𝑠|≀3,2𝑠/𝑠β‰₯2π‘Ÿ/π‘Ÿ when 𝑠>0, 𝑠2π‘ β‰€π‘Ÿ2π‘Ÿ when π‘Ÿ<0.

We note here that Theorem 1.2 follows from the left-hand side inequality of (1.3) for the case π‘Ÿ=1,𝑠=0,and𝛼=0, which in turn is a consequence of the above mentioned result of Cartwright and Field. In fact, we have the following result which is contained implicitly in [12].

Theorem 1.4. If either side of inequality (1.2) holds for π‘Ÿ,𝑠,𝛼≀2, then the same side inequality of (1.2) also holds for π‘Ÿ,𝑠 and any 𝛽≀𝛼. Moreover, the above assertion also holds when applied to (1.3) or (1.4).

On combining the above result with the result of Cartwright and Field we see that (1.2) holds for π‘Ÿ=1,𝑠=𝛼=0 and consequently (1.3) holds for π‘Ÿ=1,𝑠=𝛼=0 in virtue of Theorem 1.1.

Now, it is natural to be motivated by the result of Wang and Chen, in view of the discussions above, to ask whether one can determine all the pairs π‘Ÿ,𝑠 with π‘Ÿ>𝑠 such that either one of the inequalities (1.2)–(1.4) holds for 𝛼=0. It is our goal in this paper to investigate such a problem. Before we proceed, we would like to summarize the known results in this area. On taking 𝑙=2,𝑑=1 in [5, Proposition 2.3], we deduce with the help of Theorem 1.4 that (1.2) holds for βˆ’1≀𝑠≀1,π‘ β‰€π‘Ÿβ‰€1+𝑠,𝛼=0. On the other hand, [5, Corollary 3.2] combined with Theorem 1.4 implies that (1.3) holds for 𝛼=0,0β‰€π‘ β‰€π‘Ÿβ‰€1 and π‘Ÿβˆ’1≀𝑠≀1,0β‰€π‘Ÿβ‰€2. We also observe that if (1.2) holds for π‘Ÿ>𝑠 and 𝑠>π‘ ξ…ž, then it also holds for π‘Ÿ>π‘ ξ…ž. As (1.2) and (1.3) are equivalent, we conclude that when 𝛼=0, (1.2) holds for any π‘Ÿ>𝑠,0β‰€π‘Ÿβ‰€2,βˆ’1≀𝑠≀1.

2. The Main Theorem

Lemma 2.1. Let π‘Ÿ>𝑠, 𝐼1=(0,1],𝐼2=[1,+∞) and let 𝐸 denote the region 𝐸={(π‘ž1,π‘ž2)βˆ£π‘ž1β‰₯0,π‘ž2β‰₯0,π‘ž1+π‘ž2=1}. Define π·π‘Ÿ,𝑠𝑑;π‘ž1,π‘ž2ξ€Έ=π‘‘π‘Ÿβˆ’1βˆ’π‘‘π‘ βˆ’1ξ€·π‘ž+(π‘Ÿβˆ’π‘ )(1βˆ’π‘‘)1+π‘ž2π‘‘π‘Ÿπ‘žξ€Έξ€·1+π‘ž2𝑑𝑠.(2.1) Then for 𝑠β‰₯0, π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 holds for all (𝑑,π‘ž1,π‘ž2)∈𝐼1×𝐸 if and only if 𝑠≀1 and π‘Ÿ+𝑠≀3 and π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 holds for all (𝑑,π‘ž1,π‘ž2)∈𝐼2×𝐸 if and only if π‘Ÿβ‰€2 and π‘Ÿ+𝑠≀3.
For 𝑠<0, if π‘Ÿβ‰€0, then π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 holds for all (𝑑,π‘ž1,π‘ž2)∈𝐼1×𝐸 if and only if βˆ’1β‰€π‘Ÿβ‰€0 and βˆ’3β‰€π‘Ÿ+𝑠≀0 and π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 holds for all (𝑑,π‘ž1,π‘ž2)∈𝐼2×𝐸 if and only if 𝑠β‰₯βˆ’2 and βˆ’3β‰€π‘Ÿ+𝑠≀0.
For 𝑠<0<π‘Ÿ, π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 holds for all (𝑑,π‘ž1,π‘ž2)∈𝐼2×𝐸 if and only if π‘Ÿβ‰€2 and π‘Ÿ+𝑠β‰₯0 or 𝑠β‰₯βˆ’2 and π‘Ÿ+𝑠≀0.

Proof. When 𝑠β‰₯0, in order for π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 to hold for all (𝑑,π‘ž1,π‘ž2)∈𝐼1×𝐸, one just needs to check the case π‘ž1=1,π‘ž2=0. In this case we can rewrite π·π‘Ÿ,𝑠(𝑑;1,0) as 𝑓(𝑑)=π‘‘π‘Ÿβˆ’1βˆ’π‘‘π‘ βˆ’1+(π‘Ÿβˆ’π‘ )(1βˆ’π‘‘).(2.2) Note that 𝑓(1)=π‘“ξ…ž(1)=0; hence in order for 𝑓(𝑑)≀0 to hold for all 0<𝑑≀1, it is necessary that π‘“ξ…žξ…ž(1)≀0. Note that π‘“ξ…žξ…ž(𝑑)=(π‘Ÿβˆ’1)(π‘Ÿβˆ’2)π‘‘π‘Ÿβˆ’3βˆ’(π‘ βˆ’1)(π‘ βˆ’2)π‘‘π‘ βˆ’3 and from this one checks easily that π‘“ξ…žξ…ž(1)≀0 is equivalent to π‘Ÿ+𝑠≀3. On the other hand, on taking 𝑑→0+, we see that one needs to have 𝑠≀1 in order for 𝑓(𝑑)≀0 to hold for all 0<𝑑≀1. Now, it also follows from 𝑠≀1 that 𝑑3βˆ’π‘ π‘“ξ…žξ…ž(𝑑)=(π‘Ÿβˆ’1)(π‘Ÿβˆ’2)π‘‘π‘Ÿβˆ’π‘ βˆ’(π‘ βˆ’1)(π‘ βˆ’2)≀max{(π‘Ÿβˆ’1)(π‘Ÿβˆ’2)βˆ’(π‘ βˆ’1)(π‘ βˆ’2),βˆ’(π‘ βˆ’1)(π‘ βˆ’2)}≀0. Hence one deduces via Taylor expansion of 𝑓(𝑑) at 1 that 𝑓(𝑑)≀0 for all 0<𝑑≀1.
Similarly, when 𝑠β‰₯0, in order for π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 to hold for all (𝑑,π‘ž1,π‘ž2)∈𝐼2×𝐸, one just needs to check 𝑓(𝑑)≀0 for 𝑑β‰₯1. As 𝑓(1)=π‘“ξ…ž(1)=0, certainly it is necessary to have π‘“ξ…žξ…ž(1)≀0 and lim𝑑→+βˆžπ‘“(𝑑)≀0. These imply that π‘Ÿβ‰€2 and π‘Ÿ+𝑠≀3 and one checks easily that these conditions are also sufficient.
As a consequence of the above discussion, one can deduce the assertion of the lemma for the case 𝑠<0 and π‘Ÿβ‰€0 by noting that π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)=π‘‘π‘Ÿ+π‘ π·βˆ’π‘ ,βˆ’π‘Ÿ(𝑑;π‘ž2,π‘ž1).
It remains to treat the case π‘Ÿ>0>𝑠. We let 𝑔(π‘ž)=(1βˆ’π‘ž+π‘žπ‘‘π‘Ÿ)(1βˆ’π‘ž+π‘žπ‘‘π‘ ), and note that π‘”ξ…žξ…ž(π‘ž)=2(π‘‘π‘Ÿβˆ’1)(π‘‘π‘ βˆ’1)≀0. It follows from this that in order for π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 to hold for (𝑑,π‘ž1,π‘ž2)∈𝐼2×𝐸, it suffices to check the cases π‘ž2=0,1. When π‘Ÿ+𝑠β‰₯0, we only need to check the case π‘ž2=0 and in this case one can discuss similarly to the case 𝑠β‰₯0 above to conclude the assertion of the lemma. We just point out here that as 𝑠<0<π‘Ÿβ‰€2, we have π‘Ÿ+𝑠<2. When π‘Ÿ+𝑠≀0, it suffices to check the case π‘ž2=1 and in this case one uses the relation π·π‘Ÿ,𝑠(𝑑;0,1)=π‘‘π‘Ÿ+π‘ π·βˆ’π‘ ,βˆ’π‘Ÿ(𝑑;1,0) to convert this to the previous case that has been discussed.

Theorem 2.2. Let π‘Ÿ>𝑠. The right-hand side inequality of (1.2) holds for 𝛼=0 when 0≀𝑠≀1,π‘Ÿ+𝑠≀3 or 𝑠<0,βˆ’1β‰€π‘Ÿβ‰€0 and βˆ’3β‰€π‘Ÿ+𝑠≀0. The left-hand side inequality of (1.2) holds for 𝛼=0 when βˆ’2≀𝑠≀0,βˆ’3β‰€π‘Ÿ+𝑠≀0.

Proof. To prove the first assertion of the theorem, we may assume π‘Ÿ>2 or βˆ’1β‰€π‘Ÿβ‰€0 in view of our discussion in the last paragraph of Section 1 and for the case π‘Ÿ>2, we define 𝑔𝑛(πͺ,𝐱)=ln𝑀𝑛,π‘Ÿβˆ’ln𝑀𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ 2π‘₯2π‘›πœŽπ‘›.(2.3) Similar to the proof of Theorem 5.1 [2], it suffices to show that πœ•π‘”π‘›/πœ•π‘₯1≀0. Calculation shows that 1π‘ž1πœ•π‘”π‘›πœ•π‘₯1=π‘₯1π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯1π‘ βˆ’1𝑀𝑠𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯1βˆ’π΄π‘›ξ€ΈβˆΆ=𝑓𝑛(πͺ,𝐱).(2.4) We now show by induction on 𝑛 that 𝑓𝑛(πͺ,𝐱)≀0. When 𝑛=1, there is nothing to prove. When 𝑛=2, this becomes 1π‘ž2𝑓2π‘₯(πͺ,𝐱)=2π‘Ÿ+π‘ βˆ’1π·π‘Ÿ,𝑠π‘₯1/π‘₯2;π‘ž2,π‘ž1ξ€Έπ‘€π‘Ÿ2,π‘Ÿπ‘€π‘ 2,𝑠≀0,(2.5) by Lemma 2.1.
Suppose now 𝑛β‰₯3; in order to show 𝑓𝑛(πͺ,𝐱)≀0, we may assume that 0<π‘₯1<π‘₯𝑛 are being fixed and it suffices to show that the maximum value of 𝑓𝑛(πͺ,𝐱) is non-positive on the region π‘…π‘›Γ—π‘†π‘›βˆ’2, where 𝑅𝑛={(π‘ž1,π‘ž2,…,π‘žπ‘›)∢0β‰€π‘žπ‘–βˆ‘β‰€1,1≀𝑖≀𝑛,𝑛𝑖=1π‘žπ‘–=1} and π‘†π‘›βˆ’2={(π‘₯2,…,π‘₯π‘›βˆ’1)∢π‘₯π‘–βˆˆ[π‘₯1,π‘₯𝑛],2β‰€π‘–β‰€π‘›βˆ’1}.
Let (πͺξ…ž,π±ξ…ž) be a point of π‘…π‘›Γ—π‘†π‘›βˆ’2 in which the absolute maximum of 𝑓𝑛 is reached. If π‘₯ξ…žπ‘–=π‘₯ξ…žπ‘–+1 for some 1β‰€π‘–β‰€π‘›βˆ’1, by combining π‘₯ξ…žπ‘– with π‘₯ξ…žπ‘–+1 and π‘žξ…žπ‘– with π‘žξ…žπ‘–+1, we are back to the case of π‘›βˆ’1 variables with different weights. If π‘žξ…žπ‘–=1 for some 𝑖, then we have π‘₯1π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯1π‘ βˆ’1𝑀𝑠𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯1βˆ’π΄π‘›ξ€Έ=π‘₯1π‘Ÿβˆ’1π‘₯π‘Ÿπ‘–βˆ’π‘₯1π‘ βˆ’1π‘₯π‘ π‘–βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯1βˆ’π‘₯𝑖≀π‘₯1π‘Ÿβˆ’1π‘₯π‘Ÿπ‘–βˆ’π‘₯1π‘ βˆ’1π‘₯π‘ π‘–βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑖π‘₯1βˆ’π‘₯𝑖=1π‘₯π‘–π·π‘Ÿ,𝑠π‘₯1π‘₯𝑖;1,0≀0,(2.6) by Lemma 2.1. If π‘žξ…žπ‘–=0 for some 1<𝑖<𝑛, we are back to the case of π‘›βˆ’1 variables. If π‘žξ…žπ‘›=0, then we may assume that π‘žξ…žπ‘›βˆ’1β‰ 0 and note that we have 𝑀𝑛,π‘Ÿ=π‘€π‘›βˆ’1,π‘Ÿ,𝑀𝑛,𝑠=π‘€π‘›βˆ’1,𝑠,𝐴𝑛=π΄π‘›βˆ’1 and that π‘₯1π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯1π‘ βˆ’1𝑀𝑠𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯1βˆ’π΄π‘›ξ€Έβ‰€π‘₯1π‘Ÿβˆ’1π‘€π‘Ÿπ‘›βˆ’1,π‘Ÿβˆ’π‘₯1π‘ βˆ’1π‘€π‘ π‘›βˆ’1,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2π‘›βˆ’1ξ€·π‘₯1βˆ’π΄π‘›βˆ’1ξ€Έ,(2.7) and we are again back to the case π‘›βˆ’1. If π‘žξ…ž1=0, then similarly we may assume that π‘žξ…ž2β‰ 0 and if we can show that (again with 𝑀𝑛,π‘Ÿ=π‘€π‘›βˆ’1,π‘Ÿ,𝑀𝑛,𝑠=π‘€π‘›βˆ’1,𝑠 and 𝐴𝑛=π΄π‘›βˆ’1 here) π‘₯1π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯1π‘ βˆ’1𝑀𝑠𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯1βˆ’π΄π‘›ξ€Έβ‰€π‘₯2π‘Ÿβˆ’1π‘€π‘Ÿπ‘›βˆ’1,π‘Ÿβˆ’π‘₯2π‘ βˆ’1π‘€π‘ π‘›βˆ’1,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯2βˆ’π΄π‘›βˆ’1ξ€Έ,(2.8) then we are back to the case of π‘›βˆ’1 variables. Note that the above inequality will follow if the function π‘₯π‘₯βŸΌπ‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯π‘ βˆ’1𝑀𝑠𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛π‘₯(2.9) is an increasing function for 0<π‘₯≀𝑀𝑛,π‘Ÿ (in fact, one only needs this for 0<π‘₯≀π‘₯2) and its derivative is (π‘Ÿβˆ’1)π‘₯π‘Ÿβˆ’2π‘€π‘Ÿπ‘›,π‘Ÿ+(1βˆ’π‘ )π‘₯π‘ βˆ’2𝑀𝑠𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ π‘₯2𝑛β‰₯(π‘Ÿβˆ’1)π‘₯π‘Ÿβˆ’2π‘₯π‘Ÿπ‘›+(1βˆ’π‘ )π‘₯π‘ βˆ’2π‘₯π‘ π‘›βˆ’π‘Ÿβˆ’π‘ π‘₯2π‘›βˆΆ=β„Ž(π‘₯),(2.10) with the inequality holding for the case π‘Ÿ>2 (note that together with π‘Ÿ+𝑠≀3, this implies that 𝑠<1). It also follows from π‘Ÿ+𝑠≀3 that β„Žξ…ž(π‘₯)=0 has no root in (0,π‘₯𝑛). One then deduces from β„Ž(π‘₯𝑛)=0 and limπ‘₯β†’0+β„Ž(π‘₯)=+∞ that β„Ž(π‘₯)β‰₯0 for 0<π‘₯≀π‘₯𝑛.
So from now on it remains to consider the case π‘žξ…žπ‘–β‰ 0,1,π‘₯ξ…žπ‘–β‰ π‘₯ξ…žπ‘— for 1≀𝑖,𝑗≀𝑛,𝑖≠𝑗 and this implies that (πͺξ…ž,π±ξ…ž) is an interior point of π‘…π‘›Γ—π‘†π‘›βˆ’2. We will now show that this cannot happen.
We define π‘₯𝑝(π‘₯)=βˆ’1π‘Ÿβˆ’1π‘₯π‘Ÿπ‘€2π‘Ÿπ‘›,π‘Ÿ+π‘₯1π‘ βˆ’1π‘₯𝑠𝑀2𝑠𝑛,𝑠+(π‘Ÿβˆ’π‘ )π‘₯π‘₯2π‘›βˆ’πœ†.(2.11) Note here in the definition of 𝑝(π‘₯) that 𝑀𝑛,π‘Ÿand𝑀𝑛,𝑠 are not functions of π‘₯, they take values at some point (πͺ,𝐱) to be specified, and πœ† is also a constant to be specified.
As (πͺξ…ž,π±ξ…ž) is an interior point of π‘…π‘›Γ—π‘†π‘›βˆ’2, we may use the Lagrange multiplier method to obtain a real number πœ† so that at (πͺξ…ž,π±ξ…ž), πœ•π‘“π‘›πœ•π‘žπ‘–πœ•=πœ†πœ•π‘žπ‘–ξƒ©π‘›ξ“π‘–=1π‘žπ‘–ξƒͺ,1βˆ’1π‘žπ‘—πœ•π‘“π‘›πœ•π‘₯𝑗=0(2.12) for all 1≀𝑖≀𝑛 and 2β‰€π‘—β‰€π‘›βˆ’1.
By (2.12), a computation shows that each π‘₯′𝑖 (1≀𝑖≀𝑛) is a root of 𝑝(π‘₯)=0 (where 𝑀𝑛,π‘Ÿ,𝑀𝑛,𝑠 take their values at (πͺξ…ž,π±ξ…ž)) and each π‘₯′𝑖 (2β‰€π‘–β‰€π‘›βˆ’1) is a root of π‘ξ…ž(π‘₯)=0. Now 𝑛β‰₯3 implies 𝑝(π‘₯2)=0. As 𝑝(π‘₯1)=𝑝(π‘₯2)=𝑝(π‘₯𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers π‘₯1<π‘Ž<π‘₯2<𝑏<π‘₯𝑛 such that π‘ξ…ž(π‘Ž)=π‘ξ…ž(π‘₯2)=π‘ξ…ž(𝑏)=0. However, it is easy to see that π‘ξ…ž(π‘₯)=0 has at most two positive roots and this contradiction implies the first assertion of the theorem for the case 0≀𝑠≀1.
Now to show the right-hand side inequality hold of (1.2) for the case 𝑠<0,βˆ’1β‰€π‘Ÿβ‰€0 and βˆ’3β‰€π‘Ÿ+𝑠≀0, once again it suffices to show that the function 𝑔𝑛(πͺ,𝐱) defined above is nonnegative for any integer 𝑛β‰₯1. We note that when 𝑛=1, this is obvious and when 𝑛=2, this follows again from πœ•π‘”2/πœ•π‘₯1≀0 by Lemma 2.1.
Suppose now 𝑛β‰₯3; in order to show 𝑔𝑛(πͺ,𝐱)β‰₯0, we may assume that 0<π‘₯1<π‘₯𝑛 are being fixed and it suffices to show the minimum value of 𝑔𝑛(πͺ,𝐱) is nonnegative on the region π‘…π‘›Γ—π‘†π‘›βˆ’2, where 𝑅𝑛 and π‘†π‘›βˆ’2 are defined as above.
Let (πͺξ…ž,π±ξ…ž) be a point of π‘…π‘›Γ—π‘†π‘›βˆ’2 in which the absolute minimum of 𝑔𝑛 is reached. Note that πœŽπ‘›=𝑀2𝑛,2βˆ’π΄2𝑛; thus if π‘₯ξ…žπ‘–=π‘₯ξ…žπ‘–+1 for some 1β‰€π‘–β‰€π‘›βˆ’1, by combining π‘₯ξ…žπ‘– with π‘₯ξ…žπ‘–+1 and π‘žξ…žπ‘– with π‘žξ…žπ‘–+1, we are back to the case of π‘›βˆ’1 variables with different weights. Similarly, if π‘žξ…žπ‘–=1 for some 𝑖, then we are back to the case 𝑛=1. If π‘žξ…žπ‘–=0 for some 1≀𝑖<𝑛, we are back to the case of π‘›βˆ’1 variables. If π‘žξ…žπ‘›=0, then we may assume that π‘žξ…žπ‘›βˆ’1β‰ 0 and note that we have 𝑀𝑛,π‘Ÿ=π‘€π‘›βˆ’1,π‘Ÿ,𝑀𝑛,𝑠=π‘€π‘›βˆ’1,𝑠,𝑀𝑛,2=π‘€π‘›βˆ’1,2,𝐴𝑛=π΄π‘›βˆ’1 and that ln𝑀𝑛,π‘Ÿβˆ’ln𝑀𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ 2π‘₯2π‘›πœŽπ‘›=ln𝑀𝑛,π‘Ÿβˆ’ln𝑀𝑛,π‘ βˆ’π‘Ÿβˆ’π‘ 2π‘₯2𝑛𝑀2𝑛,2βˆ’π΄2𝑛β‰₯lnπ‘€π‘›βˆ’1,π‘Ÿβˆ’lnπ‘€π‘›βˆ’1,π‘ βˆ’π‘Ÿβˆ’π‘ 2π‘₯2π‘›βˆ’1𝑀2π‘›βˆ’1,2βˆ’π΄2π‘›βˆ’1=lnπ‘€π‘›βˆ’1,π‘Ÿβˆ’lnπ‘€π‘›βˆ’1,π‘ βˆ’π‘Ÿβˆ’π‘ 2π‘₯2π‘›βˆ’1πœŽπ‘›βˆ’1,(2.13) and we are again back to the case of π‘›βˆ’1 variables.
So from now on it remains to consider the case π‘žξ…žπ‘–β‰ 0,1,π‘₯ξ…žπ‘–β‰ π‘₯ξ…žπ‘— for 1≀𝑖,𝑗≀𝑛,𝑖≠𝑗, and this implies that (πͺξ…ž,π±ξ…ž) is an interior point of π‘…π‘›Γ—π‘†π‘›βˆ’2. We will now show that this cannot happen.
We define π‘Žπ‘₯(π‘₯)=π‘Ÿπ‘Ÿπ‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯𝑠𝑠𝑀𝑠𝑛,π‘ βˆ’ξ€·π‘₯(π‘Ÿβˆ’π‘ )2βˆ’2𝐴𝑛π‘₯ξ€Έ2π‘₯2π‘›βˆ’πœ†.(2.14) Here we define π‘₯0/0=lnπ‘₯. Also note here in the definition of π‘Ž(π‘₯), that 𝑀𝑛,π‘Ÿ,𝑀𝑛,𝑠,and𝐴𝑛 are not functions of π‘₯, they take values at some point (πͺ,𝐱) to be specified, and πœ† is also a constant to be specified.
As (πͺξ…ž,π±ξ…ž) is an interior point of π‘…π‘›Γ—π‘†π‘›βˆ’2, we may use the Lagrange multiplier method to obtain a real number πœ† so that at (πͺξ…ž,π±ξ…ž), πœ•π‘”π‘›πœ•π‘žπ‘–πœ•=πœ†πœ•π‘žπ‘–ξƒ©π‘›ξ“π‘–=1π‘žπ‘–ξƒͺ,1βˆ’1π‘žπ‘—πœ•π‘”π‘›πœ•π‘₯𝑗=0(2.15) for all 1≀𝑖≀𝑛 and 2β‰€π‘—β‰€π‘›βˆ’1.
By (2.15), a computation shows that each π‘₯ξ…žπ‘– (1≀𝑖≀𝑛) is a root of π‘Ž(π‘₯)=0 (where 𝑀𝑛,π‘Ÿ,𝑀𝑛,𝑠,and𝐴𝑛 take their values at (πͺξ…ž,π±ξ…ž)) and each π‘₯ξ…žπ‘– (2β‰€π‘–β‰€π‘›βˆ’1) is a root of π‘Žξ…ž(π‘₯)=0. Now 𝑛β‰₯3 implies π‘Ž(π‘₯2)=0. As π‘Ž(π‘₯1)=π‘Ž(π‘₯2)=π‘Ž(π‘₯𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers π‘₯1<𝑐<π‘₯2<𝑑<π‘₯𝑛 such that π‘Žξ…ž(𝑐)=π‘Žξ…ž(π‘₯2)=π‘Žξ…ž(𝑑)=0. However, we have π‘Žξ…žπ‘₯(π‘₯)=π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘₯π‘ βˆ’1𝑀𝑠𝑛,π‘ βˆ’ξ€·(π‘Ÿβˆ’π‘ )π‘₯βˆ’π΄π‘›ξ€Έπ‘₯2𝑛.(2.16) It is easy to see that π‘Žξ…žξ…žξ…ž(π‘₯)=0 has at most one positive root, which implies that π‘Žξ…ž(π‘₯)=0 has at most three positive roots. As π‘Ÿβ‰€0, it follows from limπ‘₯β†’0+π‘Žξ…ž(π‘₯)=βˆ’βˆž and limπ‘₯β†’+βˆžπ‘Žξ…ž(π‘₯)=βˆ’βˆž that π‘Žξ…ž(π‘₯)=0 has even numbers of roots so that π‘Žξ…ž(π‘₯)=0 can have at most two positive roots. This contradiction now establishes the right-hand side inequality of (1.2) for the case 𝑠<0,βˆ’1β‰€π‘Ÿβ‰€0, and βˆ’3β‰€π‘Ÿ+𝑠≀0.
One can show the second assertion of the theorem using an argument similar to the above and we shall leave this to the reader.

3. Further Discussions

As we have pointed out in Section 1 that if either one of the inequalities (1.2)–(1.4) holds for some π‘Ÿ,𝑠,𝛼≀2, then one often expects inequality (1.5) to hold as well for the same π‘Ÿ,𝑠,𝛼. In view of this, one may ask whether it is feasible to prove so for those pairs π‘Ÿ,𝑠,𝛼=0 satisfying Theorem 2.2. We now prove a special case here.

Theorem 3.1. Let βˆ’3β‰€π‘Ÿβ‰€3,π‘Ÿβ‰ 0,𝑑β‰₯0, then the following inequality holds: π‘₯2𝑛||lnπΊπ‘›βˆ’ln𝑀𝑛,π‘Ÿ||β‰₯ξ€·π‘₯𝑛+𝑑2||ln𝐺𝑛,π‘‘βˆ’ln𝑀𝑛,π‘Ÿ,𝑑||.(3.1)

Proof. We first prove the theorem for the case βˆ’3β‰€π‘Ÿ<0. For this, we may assume that 𝑑>0 is fixed and replace π‘Ÿ with βˆ’π‘Ÿ so that 0<π‘Ÿβ‰€3 in what follows. We define 𝑓𝑛(πͺ,𝐱)=π‘₯2𝑛lnπΊπ‘›βˆ’ln𝑀𝑛,βˆ’π‘Ÿξ€Έβˆ’ξ€·π‘₯𝑛+𝑑2ξ€·ln𝐺𝑛,π‘‘βˆ’ln𝑀𝑛,βˆ’π‘Ÿ,𝑑.(3.2) As in the proof of Theorem 2.2, it suffices to show that πœ•π‘“π‘›/πœ•π‘₯1≀0 and calculation shows βˆ’1π‘ž1πœ•π‘“π‘›πœ•π‘₯1=𝑔𝑛(πͺ,𝐱)βˆ’π‘”π‘›ξ€·πͺ,𝐱𝑑,(3.3) where 𝑔𝑛(πͺ,𝐱)=π‘₯2π‘›ξƒ©βˆ‘π‘›π‘–=1π‘žπ‘–ξ€·π‘₯π‘Ÿπ‘–βˆ’π‘₯π‘Ÿ1ξ€Έ/π‘₯π‘Ÿπ‘–π‘₯1βˆ‘π‘›π‘–=1π‘žπ‘–ξ€·π‘₯1/π‘₯π‘–ξ€Έπ‘Ÿξƒͺ.(3.4) It is easy to check that π‘₯𝑛π‘₯𝑖β‰₯π‘₯𝑛+𝑑π‘₯𝑖,π‘₯+𝑑1π‘₯𝑖≀π‘₯1+𝑑π‘₯𝑖.+𝑑(3.5) In view of (3.5), the inequality πœ•π‘“π‘›/πœ•π‘₯1≀0 will follow from 𝑑1ξ€·π‘₯𝑖=π‘₯2𝑛π‘₯π‘Ÿπ‘–βˆ’π‘₯π‘Ÿ1ξ€Έπ‘₯1π‘₯π‘Ÿπ‘–βˆ’ξ€·π‘₯𝑛+𝑑2π‘₯𝑖+π‘‘π‘Ÿβˆ’ξ€·π‘₯1ξ€Έ+π‘‘π‘Ÿξ€Έξ€·π‘₯1π‘₯+𝑑𝑖+π‘‘π‘Ÿβ‰₯0(3.6) for π‘₯1≀π‘₯𝑖≀π‘₯𝑛. We may assume that π‘₯𝑛>π‘₯1 here and it is easy to see that π‘‘ξ…ž1(π‘₯)=0 can have at most one root π‘₯0 in between π‘₯1 and π‘₯𝑛. This combined with the observation that 𝑑1(π‘₯1)=0,π‘‘ξ…ž1(π‘₯1)>0 implies that 𝑑1(π‘₯) reaches its local maximum at π‘₯0 if it exists. Hence we are left to check that 𝑑1(π‘₯𝑛)β‰₯0. In this case we note that π‘₯π‘›βˆ’π‘₯1=(π‘₯𝑛+𝑑)βˆ’(π‘₯1+𝑑) and we rewrite 𝑑1(π‘₯𝑛) as 𝑑1ξ€·π‘₯𝑛=π‘₯2𝑛π‘₯π‘Ÿπ‘›βˆ’π‘₯π‘Ÿ1ξ€Έπ‘₯1π‘₯π‘Ÿπ‘›ξ€·π‘₯π‘›βˆ’π‘₯1ξ€Έβˆ’ξ€·π‘₯𝑛+𝑑2π‘₯𝑛+π‘‘π‘Ÿβˆ’ξ€·π‘₯𝑛+π‘‘π‘Ÿξ€Έξ€·π‘₯1π‘₯+𝑑𝑛+π‘‘π‘Ÿπ‘₯ξ€·ξ€·π‘›ξ€Έβˆ’ξ€·π‘₯+𝑑1ξ‚΅π‘₯+𝑑=𝑒𝑛π‘₯1ξ‚Άξ‚΅π‘₯βˆ’π‘’π‘›+𝑑π‘₯1ξ‚Ά,+𝑑(3.7) where π‘₯𝑒(π‘₯)=π‘Ÿβˆ’1π‘₯π‘Ÿβˆ’2.(π‘₯βˆ’1)(3.8) In view of (3.5) again, we just need to show that 𝑒(π‘₯) is an increasing function for π‘₯>1. Note that π‘’ξ…žπ‘₯(π‘₯)=π‘Ÿβˆ’3ξ€·π‘₯π‘Ÿ+1βˆ’2π‘₯π‘Ÿξ€Έ+(π‘Ÿβˆ’1)π‘₯βˆ’(π‘Ÿβˆ’2)ξ€·π‘₯π‘Ÿβˆ’2ξ€Έ(π‘₯βˆ’1)2,(3.9) and it is easy to see that the function π‘₯π‘Ÿ+1βˆ’2π‘₯π‘Ÿ+(π‘Ÿβˆ’1)π‘₯βˆ’(π‘Ÿβˆ’2) is non-negative for π‘₯β‰₯1 when 0<π‘Ÿβ‰€3 by considering its Taylor expansion at π‘₯=1 and this completes the proof for the assertion of the theorem for the case βˆ’3β‰€π‘Ÿ<0.
To prove the theorem for the case 0<π‘Ÿβ‰€3, we may again assume that 𝑑>0 is fixed and define 𝑒𝑛(πͺ,𝐱)=π‘₯2𝑛ln𝑀𝑛,π‘Ÿβˆ’lnπΊπ‘›ξ€Έβˆ’ξ€·π‘₯𝑛+𝑑2ξ€·ln𝑀𝑛,π‘Ÿ,π‘‘βˆ’ln𝐺𝑛,𝑑.(3.10) Again it suffices to show that πœ•π‘’π‘›/πœ•π‘₯1≀0 and calculation shows βˆ’1π‘ž1πœ•π‘’π‘›πœ•π‘₯1=𝑣𝑛(πͺ,𝐱)βˆ’π‘£π‘›ξ€·πͺ,𝐱𝑑,(3.11) where 𝑣𝑛(πͺ,𝐱)=π‘₯2π‘›ξƒ©βˆ‘π‘›π‘–=1π‘žπ‘–ξ€·π‘₯π‘Ÿπ‘–βˆ’π‘₯π‘Ÿ1ξ€Έ/π‘₯π‘Ÿπ‘›π‘₯1βˆ‘π‘›π‘–=1π‘žπ‘–ξ€·π‘₯𝑖/π‘₯π‘›ξ€Έπ‘Ÿξƒͺ.(3.12) In view of (3.5), the inequality πœ•π‘’π‘›/πœ•π‘₯1≀0 will follow from 𝑑2ξ€·π‘₯𝑖=π‘₯2𝑛π‘₯π‘Ÿπ‘–βˆ’π‘₯π‘Ÿ1ξ€Έπ‘₯1π‘₯π‘Ÿπ‘›βˆ’ξ€·π‘₯𝑛+𝑑2π‘₯𝑖+π‘‘π‘Ÿβˆ’ξ€·π‘₯1ξ€Έ+π‘‘π‘Ÿξ€Έξ€·π‘₯1π‘₯+𝑑𝑛+π‘‘π‘Ÿβ‰₯0(3.13) for π‘₯1≀π‘₯𝑖≀π‘₯𝑛. We may assume that π‘₯𝑛>π‘₯1 here and it is easy to see that π‘‘ξ…ž2(π‘₯)=0 can have at most one root π‘₯0 in between π‘₯1 and π‘₯𝑛. This combined with the observation that 𝑑2(π‘₯1)=0,π‘‘ξ…ž2(π‘₯1)>0 implies that 𝑑1(π‘₯) reaches its local maximum at π‘₯0 if it exists. Hence we are left to check that 𝑑2(π‘₯𝑛)β‰₯0. As 𝑑2(π‘₯𝑛)=𝑑1(π‘₯𝑛), this completes the proof for the remaining case 0<π‘Ÿβ‰€3 of the theorem.

Now we show that, in general, it is not true that for βˆ’3β‰€π‘Ÿβ‰€3,π‘Ÿβ‰ 0,𝑑β‰₯0,

π‘₯21||lnπΊπ‘›βˆ’ln𝑀𝑛,π‘Ÿ||≀π‘₯1ξ€Έ+𝑑2||ln𝐺𝑛,π‘‘βˆ’ln𝑀𝑛,π‘Ÿ,𝑑||.(3.14) To proceed, we first look at the following related inequalities (with π‘Ÿ>𝑠 here):

ln𝑀𝑛,π‘Ÿβˆ’ln𝑀𝑛,π‘ βˆ’(π‘Ÿβˆ’π‘ )πœŽπ‘›2π‘₯21≀ln𝑀𝑛,π‘Ÿ,π‘‘βˆ’ln𝑀𝑛,𝑠,π‘‘βˆ’(π‘Ÿβˆ’π‘ )πœŽπ‘›2ξ€·π‘₯1ξ€Έ+𝑑2,(3.15)ln𝑀𝑛,π‘Ÿβˆ’ln𝑀𝑛,π‘ βˆ’(π‘Ÿβˆ’π‘ )πœŽπ‘›2π‘₯2𝑛β‰₯ln𝑀𝑛,π‘Ÿ,π‘‘βˆ’ln𝑀𝑛,𝑠,π‘‘βˆ’(π‘Ÿβˆ’π‘ )πœŽπ‘›2ξ€·π‘₯𝑛+𝑑2.(3.16) Let 𝑓𝑛(πͺ,𝐱,𝑑) denote the right-hand side expression of (3.15); then (3.15) holds if and only if πœ•π‘“π‘›(πͺ,𝐱,0)/πœ•π‘‘β‰₯0. As 𝐱 is arbitrary, we can recast this condition as

π‘€π‘Ÿβˆ’1𝑛,π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘€π‘ βˆ’1𝑛,π‘ βˆ’1𝑀𝑠𝑛,𝑠+(π‘Ÿβˆ’π‘ )πœŽπ‘›π‘₯31β‰₯0.(3.17) Similarly, (3.16) holds if and only if the following inequality holds:

π‘€π‘Ÿβˆ’1𝑛,π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’π‘€π‘ βˆ’1𝑛,π‘ βˆ’1𝑀𝑠𝑛,𝑠+(π‘Ÿβˆ’π‘ )πœŽπ‘›π‘₯3𝑛≀0.(3.18) As a first step towards establishing (3.17), we consider the case 𝑛=2 here; in this case we let π‘₯1=1≀𝑑=π‘₯2 and rewrite the left-hand side of (3.17) as

π‘ž1+π‘ž2π‘‘π‘Ÿβˆ’1π‘ž1+π‘ž2π‘‘π‘Ÿβˆ’π‘ž1+π‘ž2π‘‘π‘ βˆ’1π‘ž1+π‘ž2𝑑𝑠+(π‘Ÿβˆ’π‘ )π‘ž1π‘ž2(π‘‘βˆ’1)2=π‘ž1π‘ž2(1βˆ’π‘‘)ξ€·π‘ž1+π‘ž2π‘‘π‘Ÿπ‘žξ€Έξ€·1+π‘ž2π‘‘π‘ ξ€Έπ·π‘Ÿ,𝑠𝑑;π‘ž1,π‘ž2ξ€Έ(3.19) with π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2) being defined as in Lemma 2.1. Using the same notations as in Lemma 2.1, we see that in order for (3.17) to hold for 𝑛=2, we need to have π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 for (𝑑,π‘ž1,π‘ž2)∈𝐼2×𝐸. Similar treatment of (3.18) shows that in order for it to hold in the case 𝑛=2, one needs to have π·π‘Ÿ,𝑠(𝑑;π‘ž1,π‘ž2)≀0 for (𝑑,π‘ž1,π‘ž2)∈𝐼1×𝐸.

It follows from the proof of Lemma 2.1 that π·π‘Ÿ,0(𝑑;1,0)≀0 fails to hold for all 𝑑β‰₯1 when π‘Ÿ>2. In another words, there exists 𝐱,πͺ such that when π‘Ÿ>2,

π‘€π‘Ÿβˆ’1𝑛,π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’1𝐻𝑛+(π‘Ÿβˆ’π‘ )πœŽπ‘›π‘₯31<0(3.20) holds. Now we return to the inequality (3.14) and we take π‘Ÿ>0 there. Just as in the discussion above, one sees that (3.14) is equivalent to

2ξ€·ln𝑀𝑛,π‘Ÿβˆ’ln𝐺𝑛π‘₯1+π‘€π‘Ÿβˆ’1𝑛,π‘Ÿβˆ’1π‘€π‘Ÿπ‘›,π‘Ÿβˆ’1𝐻𝑛β‰₯0.(3.21) This combined with (3.20) now implies that for π‘Ÿ>2,

ln𝑀𝑛,π‘Ÿβˆ’ln𝐺𝑛>(π‘Ÿβˆ’π‘ )πœŽπ‘›2π‘₯21.(3.22)

However, on taking 𝑑→+∞ on (3.14), we get the above inequality reversed (with > replaced by ≀) and this leads to a contradiction; hence (3.14) does not hold for π‘Ÿ>2 in general.

To end this paper, we note that it is an open problem to determine all the triples (π‘Ÿ,𝑠,𝛼) so that inequality (1.2) holds. However, when 𝛼=0 with π‘Ÿ=0 or 𝑠=0, the result given in Theorem 2.2 is best possible. In this case Theorem 2.2 implies that for |π‘Ÿ|≀3,π‘Ÿβ‰ 0,

||ln𝑀𝑛,π‘Ÿβˆ’ln𝐺𝑛||β‰₯|π‘Ÿ|2π‘₯2π‘›πœŽπ‘›.(3.23)

We point out here that inequality (3.23) does not hold in general when |π‘Ÿ|>3. To see this, it suffices to consider the case 𝑛=2 and in this case we can set 0<π‘₯1=𝑑≀π‘₯2=1 and consider more generally for π‘Ÿ>𝑠, the function 𝑔2(πͺ,𝐱) defined in the proof of Theorem 2.2, regarding it as a function 𝑓(𝑑) of 𝑑. It is easy to check that 𝑓(1)=π‘“ξ…ž(1)=π‘“ξ…žξ…ž(1)=0; hence by the Taylor expansion of 𝑓(𝑑) around 𝑑=1, we need 𝑓(3)(1)≀0 in order for 𝑓(𝑑)β‰₯0 to hold for any 0<𝑑≀1. Calculation shows that

𝑓(3)(1)=π‘ž1(ξ€·π‘Ÿβˆ’π‘ )π‘Ÿ+π‘ βˆ’3βˆ’3(π‘Ÿ+π‘ βˆ’1)π‘ž1+2(π‘Ÿ+𝑠)π‘ž21ξ€Έ.(3.24) On taking π‘ž1β†’0+, one sees immediately that we must have π‘Ÿ+𝑠≀3 here in order for 𝑓(𝑑)β‰₯0 for all 0<𝑑≀1. On taking 𝑠=0, we see that one needs π‘Ÿβ‰€3 in order for (3.23) to hold for positive π‘Ÿ. Similarly, one checks easily that in the case 𝑛=2, if inequality (3.23) holds for some π‘Ÿ, then it also holds for βˆ’π‘Ÿ by a change of variables π‘₯𝑖→1/π‘₯2βˆ’π‘–+1. Hence one needs π‘Ÿβ‰₯βˆ’3 in order for (3.23) to hold for any negative π‘Ÿ.

References

  1. P. Gao, β€œKy Fan inequality and bounds for differences of means,” International Journal of Mathematics and Mathematical Sciences, vol. 2003, no. 16, pp. 995–1002, 2003. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  2. P. Gao, β€œA new approach to Ky Fan-type inequalities,” International Journal of Mathematics and Mathematical Sciences, vol. 2005, no. 22, pp. 3551–3574, 2005. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  3. D. I. Cartwright and M. J. Field, β€œA refinement of the arithmetic mean-geometric mean inequality,” Proceedings of the American Mathematical Society, vol. 71, no. 1, pp. 36–38, 1978. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  4. H. Alzer, β€œA new refinement of the arithmetic meanβ€”geometric mean inequality,” The Rocky Mountain Journal of Mathematics, vol. 27, no. 3, pp. 663–667, 1997. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  5. P. Gao, β€œCertain bounds for the difference of means,” Journal of Inequalities in Pure and Applied Mathematics, vol. 4, no. 4, article 76, p. 10, 2003. View at Google Scholar Β· View at MathSciNet
  6. A. McD. Mercer, β€œBounds for A-G, A-H, G-H, and a family of inequalities of Ky Fan's type, using a general method,” Journal of Mathematical Analysis and Applications, vol. 243, no. 1, pp. 163–173, 2000. View at Publisher Β· View at Google Scholar Β· View at MathSciNet
  7. A. McD. Mercer, β€œImproved upper and lower bounds for the difference Anβˆ’Gn,” The Rocky Mountain Journal of Mathematics, vol. 31, no. 2, pp. 553–560, 2001. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  8. P. R. Mercer, β€œRefined arithmetic, geometric and harmonic mean inequalities,” The Rocky Mountain Journal of Mathematics, vol. 33, no. 4, pp. 1459–1464, 2003. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  9. H. Alzer, β€œThe inequality of Ky Fan and related results,” Acta Applicandae Mathematicae, vol. 38, no. 3, pp. 305–354, 1995. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  10. E. F. Beckenbach and R. Bellman, Inequalities, Ergebnisse der Mathematik und ihrer Grenzgebiete. Neue Folge, Band 30, Springer, New York, NY, USA, 2nd edition, 1965. View at MathSciNet
  11. Z. Wang and J. Chen, β€œGeneralization of Ky Fan inequality,” Mathematica Balkanica, vol. 5, no. 4, pp. 373–380, 1991. View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet
  12. P. Gao, β€œA generalization of Ky Fan's inequality,” International Journal of Mathematics and Mathematical Sciences, vol. 28, no. 7, pp. 419–425, 2001. View at Publisher Β· View at Google Scholar Β· View at Zentralblatt MATH Β· View at MathSciNet