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International Journal of Mathematics and Mathematical Sciences
Volume 2009 (2009), Article ID 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 Silvestru Dragomir

Copyright © 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,𝑟(𝐱;𝐪) be the generalized weighted means: 𝑀𝑛,𝑟(𝐱;𝐪)=(𝑛𝑖=1𝑞𝑖𝑥𝑟𝑖)1/𝑟, where 𝑀𝑛,0(𝐱;𝐪) denotes the limit of 𝑀𝑛,𝑟(𝐱;𝐪) as 𝑟0+. Here 𝐱=(𝑥1,,𝑥𝑛), 𝐪=(𝑞1,,𝑞𝑛) with 𝑞𝑖>0(1𝑖𝑛) satisfying 𝑛𝑖=1𝑞𝑖=1. In this paper, we always assume 0<𝑥1𝑥2𝑥𝑛. To any given 𝐱 and 𝑡0, we set 𝐱=(1𝑥1,,1𝑥𝑛),𝐱𝑡=(𝑥1+𝑡,,𝑥𝑛+𝑡).

We define 𝐴𝑛(𝐱;𝐪)=𝑀𝑛,1(𝐱;𝐪),𝐺𝑛(𝐱;𝐪)=𝑀𝑛,0(𝐱;𝐪),𝐻𝑛(𝐱;𝐪)=𝑀𝑛,1(𝐱;𝐪), and we shall write 𝑀𝑛,𝑟 for 𝑀𝑛,𝑟(𝐱;𝐪), 𝑀𝑛,𝑟,𝑡 for 𝑀𝑛,𝑟(𝐱𝑡;𝐪), and 𝑀𝑛,𝑟 for 𝑀𝑛,𝑟(𝐱;𝐪) if 𝑥𝑛<1 and similarly for other means when there is no risk of confusion. We further denote 𝜎𝑛=𝑛𝑖=1𝑞𝑖(𝑥𝑖𝐴𝑛)2.

When 𝑥𝑛<1, we define

Δ𝑟,𝑠,𝛼,𝑡=𝑀𝛼𝑛,𝑟,𝑡𝑀𝛼𝑛,𝑠,𝑡/𝛼𝑀𝛼𝑛,𝑟,𝑡𝑀𝛼𝑛,𝑠,𝑡,/𝛼(1.1) where we set 𝑀0𝑛,𝑟/0=ln𝑀𝑛,𝑟 and we shall write Δ𝑟,𝑠,𝛼 for Δ𝑟,𝑠,𝛼,0 and Δ𝑟,𝑠 for Δ𝑟,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 𝛼2 following the method there).

Theorem 1.1. For 𝑟>𝑠 and 𝛼2, the following inequalities are equivalent: 𝑟𝑠2𝑥12𝛼𝜎𝑛𝑀𝛼𝑛,𝑟𝑀𝛼𝑛,𝑠𝛼𝑟𝑠2𝑥𝑛2𝛼𝜎𝑛,𝑥(1.2)𝑛1𝑥𝑛2𝛼Δ𝑟,𝑠,𝛼𝑥11𝑥12𝛼,(1.3) where in (1.3) one requires 𝑥𝑛<1.

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

𝑥𝑛𝑡+𝑥12𝛼Δ𝑟,𝑠,𝛼,𝑡𝑥1𝑡+𝑥𝑛2𝛼(1.4) being valid for all 𝑡0. We point out here that when inequality (1.2) holds for some 𝑟,𝑠, one can often expect for a better result than (1.4), namely,

𝑥𝑛𝑡+𝑥𝑛2𝛼Δ𝑟,𝑠,𝛼,𝑡𝑥1𝑡+𝑥12𝛼.(1.5)

We note that inequality (1.2) does not hold for all pairs 𝑟,𝑠 (see [1]). Cartwright and Field [3] first proved the validity of (1.2) for 𝑟=1,𝑠=0,𝛼=1. For other extensions and refinements of (1.2), see [2, 48]. 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 𝑥𝑖(0,1/2], Δ1,0,01, with equality holding if and only if 𝑥1==𝑥𝑛.

A nice result of Wang and Chen [11] determines all the pairs 𝑟,𝑠 with 𝑟>𝑠 such that Δ𝑟,𝑠,01 is satisfied when 𝑥𝑖(0,1/2]. Their result is contained in the following.

Theorem 1.3. For 𝑟>𝑠,𝑥𝑖(0,1/2], Δ𝑟,𝑠,01 holds if and only if |𝑟+𝑠|3,2𝑠/𝑠2𝑟/𝑟 when 𝑠>0, 𝑠2𝑠𝑟2𝑟 when 𝑟<0.

We note here that Theorem 1.2 follows from the left-hand side inequality of (1.3) for the case 𝑟=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 𝑟,𝑠,𝛼2, then the same side inequality of (1.2) also holds for 𝑟,𝑠 and any 𝛽𝛼. 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 𝑟=1,𝑠=𝛼=0 and consequently (1.3) holds for 𝑟=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 𝑟,𝑠 with 𝑟>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,𝑡=1 in [5, Proposition 2.3], we deduce with the help of Theorem 1.4 that (1.2) holds for 1𝑠1,𝑠𝑟1+𝑠,𝛼=0. On the other hand, [5, Corollary 3.2] combined with Theorem 1.4 implies that (1.3) holds for 𝛼=0,0𝑠𝑟1 and 𝑟1𝑠1,0𝑟2. We also observe that if (1.2) holds for 𝑟>𝑠 and 𝑠>𝑠, then it also holds for 𝑟>𝑠. As (1.2) and (1.3) are equivalent, we conclude that when 𝛼=0, (1.2) holds for any 𝑟>𝑠,0𝑟2,1𝑠1.

2. The Main Theorem

Lemma 2.1. Let 𝑟>𝑠, 𝐼1=(0,1],𝐼2=[1,+) and let 𝐸 denote the region 𝐸={(𝑞1,𝑞2)𝑞10,𝑞20,𝑞1+𝑞2=1}. Define 𝐷𝑟,𝑠𝑡;𝑞1,𝑞2=𝑡𝑟1𝑡𝑠1𝑞+(𝑟𝑠)(1𝑡)1+𝑞2𝑡𝑟𝑞1+𝑞2𝑡𝑠.(2.1) Then for 𝑠0, 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 holds for all (𝑡,𝑞1,𝑞2)𝐼1×𝐸 if and only if 𝑠1 and 𝑟+𝑠3 and 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 holds for all (𝑡,𝑞1,𝑞2)𝐼2×𝐸 if and only if 𝑟2 and 𝑟+𝑠3.
For 𝑠<0, if 𝑟0, then 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 holds for all (𝑡,𝑞1,𝑞2)𝐼1×𝐸 if and only if 1𝑟0 and 3𝑟+𝑠0 and 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 holds for all (𝑡,𝑞1,𝑞2)𝐼2×𝐸 if and only if 𝑠2 and 3𝑟+𝑠0.
For 𝑠<0<𝑟, 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 holds for all (𝑡,𝑞1,𝑞2)𝐼2×𝐸 if and only if 𝑟2 and 𝑟+𝑠0 or 𝑠2 and 𝑟+𝑠0.

Proof. When 𝑠0, in order for 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 to hold for all (𝑡,𝑞1,𝑞2)𝐼1×𝐸, one just needs to check the case 𝑞1=1,𝑞2=0. In this case we can rewrite 𝐷𝑟,𝑠(𝑡;1,0) as 𝑓(𝑡)=𝑡𝑟1𝑡𝑠1+(𝑟𝑠)(1𝑡).(2.2) Note that 𝑓(1)=𝑓(1)=0; hence in order for 𝑓(𝑡)0 to hold for all 0<𝑡1, it is necessary that 𝑓(1)0. Note that 𝑓(𝑡)=(𝑟1)(𝑟2)𝑡𝑟3(𝑠1)(𝑠2)𝑡𝑠3 and from this one checks easily that 𝑓(1)0 is equivalent to 𝑟+𝑠3. On the other hand, on taking 𝑡0+, we see that one needs to have 𝑠1 in order for 𝑓(𝑡)0 to hold for all 0<𝑡1. Now, it also follows from 𝑠1 that 𝑡3𝑠𝑓(𝑡)=(𝑟1)(𝑟2)𝑡𝑟𝑠(𝑠1)(𝑠2)max{(𝑟1)(𝑟2)(𝑠1)(𝑠2),(𝑠1)(𝑠2)}0. Hence one deduces via Taylor expansion of 𝑓(𝑡) at 1 that 𝑓(𝑡)0 for all 0<𝑡1.
Similarly, when 𝑠0, in order for 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 to hold for all (𝑡,𝑞1,𝑞2)𝐼2×𝐸, one just needs to check 𝑓(𝑡)0 for 𝑡1. As 𝑓(1)=𝑓(1)=0, certainly it is necessary to have 𝑓(1)0 and lim𝑡+𝑓(𝑡)0. These imply that 𝑟2 and 𝑟+𝑠3 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 𝑟0 by noting that 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)=𝑡𝑟+𝑠𝐷𝑠,𝑟(𝑡;𝑞2,𝑞1).
It remains to treat the case 𝑟>0>𝑠. We let 𝑔(𝑞)=(1𝑞+𝑞𝑡𝑟)(1𝑞+𝑞𝑡𝑠), and note that 𝑔(𝑞)=2(𝑡𝑟1)(𝑡𝑠1)0. It follows from this that in order for 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 to hold for (𝑡,𝑞1,𝑞2)𝐼2×𝐸, it suffices to check the cases 𝑞2=0,1. When 𝑟+𝑠0, we only need to check the case 𝑞2=0 and in this case one can discuss similarly to the case 𝑠0 above to conclude the assertion of the lemma. We just point out here that as 𝑠<0<𝑟2, we have 𝑟+𝑠<2. When 𝑟+𝑠0, it suffices to check the case 𝑞2=1 and in this case one uses the relation 𝐷𝑟,𝑠(𝑡;0,1)=𝑡𝑟+𝑠𝐷𝑠,𝑟(𝑡;1,0) to convert this to the previous case that has been discussed.

Theorem 2.2. Let 𝑟>𝑠. The right-hand side inequality of (1.2) holds for 𝛼=0 when 0𝑠1,𝑟+𝑠3 or 𝑠<0,1𝑟0 and 3𝑟+𝑠0. The left-hand side inequality of (1.2) holds for 𝛼=0 when 2𝑠0,3𝑟+𝑠0.

Proof. To prove the first assertion of the theorem, we may assume 𝑟>2 or 1𝑟0 in view of our discussion in the last paragraph of Section 1 and for the case 𝑟>2, we define 𝑔𝑛(𝐪,𝐱)=ln𝑀𝑛,𝑟ln𝑀𝑛,𝑠𝑟𝑠2𝑥2𝑛𝜎𝑛.(2.3) Similar to the proof of Theorem 5.1 [2], it suffices to show that 𝜕𝑔𝑛/𝜕𝑥10. Calculation shows that 1𝑞1𝜕𝑔𝑛𝜕𝑥1=𝑥1𝑟1𝑀𝑟𝑛,𝑟𝑥1𝑠1𝑀𝑠𝑛,𝑠𝑟𝑠𝑥2𝑛𝑥1𝐴𝑛=𝑓𝑛(𝐪,𝐱).(2.4) We now show by induction on 𝑛 that 𝑓𝑛(𝐪,𝐱)0. When 𝑛=1, there is nothing to prove. When 𝑛=2, this becomes 1𝑞2𝑓2𝑥(𝐪,𝐱)=2𝑟+𝑠1𝐷𝑟,𝑠𝑥1/𝑥2;𝑞2,𝑞1𝑀𝑟2,𝑟𝑀𝑠2,𝑠0,(2.5) by Lemma 2.1.
Suppose now 𝑛3; in order to show 𝑓𝑛(𝐪,𝐱)0, we may assume that 0<𝑥1<𝑥𝑛 are being fixed and it suffices to show that the maximum value of 𝑓𝑛(𝐪,𝐱) is non-positive on the region 𝑅𝑛×𝑆𝑛2, where 𝑅𝑛={(𝑞1,𝑞2,,𝑞𝑛)0𝑞𝑖1,1𝑖𝑛,𝑛𝑖=1𝑞𝑖=1} and 𝑆𝑛2={(𝑥2,,𝑥𝑛1)𝑥𝑖[𝑥1,𝑥𝑛],2𝑖𝑛1}.
Let (𝐪,𝐱) be a point of 𝑅𝑛×𝑆𝑛2 in which the absolute maximum of 𝑓𝑛 is reached. If 𝑥𝑖=𝑥𝑖+1 for some 1𝑖𝑛1, by combining 𝑥𝑖 with 𝑥𝑖+1 and 𝑞𝑖 with 𝑞𝑖+1, we are back to the case of 𝑛1 variables with different weights. If 𝑞𝑖=1 for some 𝑖, then we have 𝑥1𝑟1𝑀𝑟𝑛,𝑟𝑥1𝑠1𝑀𝑠𝑛,𝑠𝑟𝑠𝑥2𝑛𝑥1𝐴𝑛=𝑥1𝑟1𝑥𝑟𝑖𝑥1𝑠1𝑥𝑠𝑖𝑟𝑠𝑥2𝑛𝑥1𝑥𝑖𝑥1𝑟1𝑥𝑟𝑖𝑥1𝑠1𝑥𝑠𝑖𝑟𝑠𝑥2𝑖𝑥1𝑥𝑖=1𝑥𝑖𝐷𝑟,𝑠𝑥1𝑥𝑖;1,00,(2.6) by Lemma 2.1. If 𝑞𝑖=0 for some 1<𝑖<𝑛, we are back to the case of 𝑛1 variables. If 𝑞𝑛=0, then we may assume that 𝑞𝑛10 and note that we have 𝑀𝑛,𝑟=𝑀𝑛1,𝑟,𝑀𝑛,𝑠=𝑀𝑛1,𝑠,𝐴𝑛=𝐴𝑛1 and that 𝑥1𝑟1𝑀𝑟𝑛,𝑟𝑥1𝑠1𝑀𝑠𝑛,𝑠𝑟𝑠𝑥2𝑛𝑥1𝐴𝑛𝑥1𝑟1𝑀𝑟𝑛1,𝑟𝑥1𝑠1𝑀𝑠𝑛1,𝑠𝑟𝑠𝑥2𝑛1𝑥1𝐴𝑛1,(2.7) and we are again back to the case 𝑛1. If 𝑞1=0, then similarly we may assume that 𝑞20 and if we can show that (again with 𝑀𝑛,𝑟=𝑀𝑛1,𝑟,𝑀𝑛,𝑠=𝑀𝑛1,𝑠 and 𝐴𝑛=𝐴𝑛1 here) 𝑥1𝑟1𝑀𝑟𝑛,𝑟𝑥1𝑠1𝑀𝑠𝑛,𝑠𝑟𝑠𝑥2𝑛𝑥1𝐴𝑛𝑥2𝑟1𝑀𝑟𝑛1,𝑟𝑥2𝑠1𝑀𝑠𝑛1,𝑠𝑟𝑠𝑥2𝑛𝑥2𝐴𝑛1,(2.8) then we are back to the case of 𝑛1 variables. Note that the above inequality will follow if the function 𝑥𝑥𝑟1𝑀𝑟𝑛,𝑟𝑥𝑠1𝑀𝑠𝑛,𝑠𝑟𝑠𝑥2𝑛𝑥(2.9) is an increasing function for 0<𝑥𝑀𝑛,𝑟 (in fact, one only needs this for 0<𝑥𝑥2) and its derivative is (𝑟1)𝑥𝑟2𝑀𝑟𝑛,𝑟+(1𝑠)𝑥𝑠2𝑀𝑠𝑛,𝑠𝑟𝑠𝑥2𝑛(𝑟1)𝑥𝑟2𝑥𝑟𝑛+(1𝑠)𝑥𝑠2𝑥𝑠𝑛𝑟𝑠𝑥2𝑛=(𝑥),(2.10) with the inequality holding for the case 𝑟>2 (note that together with 𝑟+𝑠3, this implies that 𝑠<1). It also follows from 𝑟+𝑠3 that (𝑥)=0 has no root in (0,𝑥𝑛). One then deduces from (𝑥𝑛)=0 and lim𝑥0+(𝑥)=+ that (𝑥)0 for 0<𝑥𝑥𝑛.
So from now on it remains to consider the case 𝑞𝑖0,1,𝑥𝑖𝑥𝑗 for 1𝑖,𝑗𝑛,𝑖𝑗 and this implies that (𝐪,𝐱) is an interior point of 𝑅𝑛×𝑆𝑛2. We will now show that this cannot happen.
We define 𝑥𝑝(𝑥)=1𝑟1𝑥𝑟𝑀2𝑟𝑛,𝑟+𝑥1𝑠1𝑥𝑠𝑀2𝑠𝑛,𝑠+(𝑟𝑠)𝑥𝑥2𝑛𝜆.(2.11) Note here in the definition of 𝑝(𝑥) that 𝑀𝑛,𝑟and𝑀𝑛,𝑠 are not functions of 𝑥, they take values at some point (𝐪,𝐱) to be specified, and 𝜆 is also a constant to be specified.
As (𝐪,𝐱) is an interior point of 𝑅𝑛×𝑆𝑛2, we may use the Lagrange multiplier method to obtain a real number 𝜆 so that at (𝐪,𝐱), 𝜕𝑓𝑛𝜕𝑞𝑖𝜕=𝜆𝜕𝑞𝑖𝑛𝑖=1𝑞𝑖,11𝑞𝑗𝜕𝑓𝑛𝜕𝑥𝑗=0(2.12) for all 1𝑖𝑛 and 2𝑗𝑛1.
By (2.12), a computation shows that each 𝑥𝑖 (1𝑖𝑛) is a root of 𝑝(𝑥)=0 (where 𝑀𝑛,𝑟,𝑀𝑛,𝑠 take their values at (𝐪,𝐱)) and each 𝑥𝑖 (2𝑖𝑛1) is a root of 𝑝(𝑥)=0. Now 𝑛3 implies 𝑝(𝑥2)=0. As 𝑝(𝑥1)=𝑝(𝑥2)=𝑝(𝑥𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers 𝑥1<𝑎<𝑥2<𝑏<𝑥𝑛 such that 𝑝(𝑎)=𝑝(𝑥2)=𝑝(𝑏)=0. However, it is easy to see that 𝑝(𝑥)=0 has at most two positive roots and this contradiction implies the first assertion of the theorem for the case 0𝑠1.
Now to show the right-hand side inequality hold of (1.2) for the case 𝑠<0,1𝑟0 and 3𝑟+𝑠0, once again it suffices to show that the function 𝑔𝑛(𝐪,𝐱) defined above is nonnegative for any integer 𝑛1. We note that when 𝑛=1, this is obvious and when 𝑛=2, this follows again from 𝜕𝑔2/𝜕𝑥10 by Lemma 2.1.
Suppose now 𝑛3; in order to show 𝑔𝑛(𝐪,𝐱)0, we may assume that 0<𝑥1<𝑥𝑛 are being fixed and it suffices to show the minimum value of 𝑔𝑛(𝐪,𝐱) is nonnegative on the region 𝑅𝑛×𝑆𝑛2, where 𝑅𝑛 and 𝑆𝑛2 are defined as above.
Let (𝐪,𝐱) be a point of 𝑅𝑛×𝑆𝑛2 in which the absolute minimum of 𝑔𝑛 is reached. Note that 𝜎𝑛=𝑀2𝑛,2𝐴2𝑛; thus if 𝑥𝑖=𝑥𝑖+1 for some 1𝑖𝑛1, by combining 𝑥𝑖 with 𝑥𝑖+1 and 𝑞𝑖 with 𝑞𝑖+1, we are back to the case of 𝑛1 variables with different weights. Similarly, if 𝑞𝑖=1 for some 𝑖, then we are back to the case 𝑛=1. If 𝑞𝑖=0 for some 1𝑖<𝑛, we are back to the case of 𝑛1 variables. If 𝑞𝑛=0, then we may assume that 𝑞𝑛10 and note that we have 𝑀𝑛,𝑟=𝑀𝑛1,𝑟,𝑀𝑛,𝑠=𝑀𝑛1,𝑠,𝑀𝑛,2=𝑀𝑛1,2,𝐴𝑛=𝐴𝑛1 and that ln𝑀𝑛,𝑟ln𝑀𝑛,𝑠𝑟𝑠2𝑥2𝑛𝜎𝑛=ln𝑀𝑛,𝑟ln𝑀𝑛,𝑠𝑟𝑠2𝑥2𝑛𝑀2𝑛,2𝐴2𝑛ln𝑀𝑛1,𝑟ln𝑀𝑛1,𝑠𝑟𝑠2𝑥2𝑛1𝑀2𝑛1,2𝐴2𝑛1=ln𝑀𝑛1,𝑟ln𝑀𝑛1,𝑠𝑟𝑠2𝑥2𝑛1𝜎𝑛1,(2.13) and we are again back to the case of 𝑛1 variables.
So from now on it remains to consider the case 𝑞𝑖0,1,𝑥𝑖𝑥𝑗 for 1𝑖,𝑗𝑛,𝑖𝑗, and this implies that (𝐪,𝐱) is an interior point of 𝑅𝑛×𝑆𝑛2. We will now show that this cannot happen.
We define 𝑎𝑥(𝑥)=𝑟𝑟𝑀𝑟𝑛,𝑟𝑥𝑠𝑠𝑀𝑠𝑛,𝑠𝑥(𝑟𝑠)22𝐴𝑛𝑥2𝑥2𝑛𝜆.(2.14) Here we define 𝑥0/0=ln𝑥. Also note here in the definition of 𝑎(𝑥), that 𝑀𝑛,𝑟,𝑀𝑛,𝑠,and𝐴𝑛 are not functions of 𝑥, they take values at some point (𝐪,𝐱) to be specified, and 𝜆 is also a constant to be specified.
As (𝐪,𝐱) is an interior point of 𝑅𝑛×𝑆𝑛2, we may use the Lagrange multiplier method to obtain a real number 𝜆 so that at (𝐪,𝐱), 𝜕𝑔𝑛𝜕𝑞𝑖𝜕=𝜆𝜕𝑞𝑖𝑛𝑖=1𝑞𝑖,11𝑞𝑗𝜕𝑔𝑛𝜕𝑥𝑗=0(2.15) for all 1𝑖𝑛 and 2𝑗𝑛1.
By (2.15), a computation shows that each 𝑥𝑖 (1𝑖𝑛) is a root of 𝑎(𝑥)=0 (where 𝑀𝑛,𝑟,𝑀𝑛,𝑠,and𝐴𝑛 take their values at (𝐪,𝐱)) and each 𝑥𝑖 (2𝑖𝑛1) is a root of 𝑎(𝑥)=0. Now 𝑛3 implies 𝑎(𝑥2)=0. As 𝑎(𝑥1)=𝑎(𝑥2)=𝑎(𝑥𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers 𝑥1<𝑐<𝑥2<𝑑<𝑥𝑛 such that 𝑎(𝑐)=𝑎(𝑥2)=𝑎(𝑑)=0. However, we have 𝑎𝑥(𝑥)=𝑟1𝑀𝑟𝑛,𝑟𝑥𝑠1𝑀𝑠𝑛,𝑠(𝑟𝑠)𝑥𝐴𝑛𝑥2𝑛.(2.16) It is easy to see that 𝑎(𝑥)=0 has at most one positive root, which implies that 𝑎(𝑥)=0 has at most three positive roots. As 𝑟0, it follows from lim𝑥0+𝑎(𝑥)= and lim𝑥+𝑎(𝑥)= that 𝑎(𝑥)=0 has even numbers of roots so that 𝑎(𝑥)=0 can have at most two positive roots. This contradiction now establishes the right-hand side inequality of (1.2) for the case 𝑠<0,1𝑟0, and 3𝑟+𝑠0.
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 𝑟,𝑠,𝛼2, then one often expects inequality (1.5) to hold as well for the same 𝑟,𝑠,𝛼. In view of this, one may ask whether it is feasible to prove so for those pairs 𝑟,𝑠,𝛼=0 satisfying Theorem 2.2. We now prove a special case here.

Theorem 3.1. Let 3𝑟3,𝑟0,𝑡0, then the following inequality holds: 𝑥2𝑛||ln𝐺𝑛ln𝑀𝑛,𝑟||𝑥𝑛+𝑡2||ln𝐺𝑛,𝑡ln𝑀𝑛,𝑟,𝑡||.(3.1)

Proof. We first prove the theorem for the case 3𝑟<0. For this, we may assume that 𝑡>0 is fixed and replace 𝑟 with 𝑟 so that 0<𝑟3 in what follows. We define 𝑓𝑛(𝐪,𝐱)=𝑥2𝑛ln𝐺𝑛ln𝑀𝑛,𝑟𝑥𝑛+𝑡2ln𝐺𝑛,𝑡ln𝑀𝑛,𝑟,𝑡.(3.2) As in the proof of Theorem 2.2, it suffices to show that 𝜕𝑓𝑛/𝜕𝑥10 and calculation shows 1𝑞1𝜕𝑓𝑛𝜕𝑥1=𝑔𝑛(𝐪,𝐱)𝑔𝑛𝐪,𝐱𝑡,(3.3) where 𝑔𝑛(𝐪,𝐱)=𝑥2𝑛𝑛𝑖=1𝑞𝑖𝑥𝑟𝑖𝑥𝑟1/𝑥𝑟𝑖𝑥1𝑛𝑖=1𝑞𝑖𝑥1/𝑥𝑖𝑟.(3.4) It is easy to check that 𝑥𝑛𝑥𝑖𝑥𝑛+𝑡𝑥𝑖,𝑥+𝑡1𝑥𝑖𝑥1+𝑡𝑥𝑖.+𝑡(3.5) In view of (3.5), the inequality 𝜕𝑓𝑛/𝜕𝑥10 will follow from 𝑑1𝑥𝑖=𝑥2𝑛𝑥𝑟𝑖𝑥𝑟1𝑥1𝑥𝑟𝑖𝑥𝑛+𝑡2𝑥𝑖+𝑡𝑟𝑥1+𝑡𝑟𝑥1𝑥+𝑡𝑖+𝑡𝑟0(3.6) for 𝑥1𝑥𝑖𝑥𝑛. We may assume that 𝑥𝑛>𝑥1 here and it is easy to see that 𝑑1(𝑥)=0 can have at most one root 𝑥0 in between 𝑥1 and 𝑥𝑛. This combined with the observation that 𝑑1(𝑥1)=0,𝑑1(𝑥1)>0 implies that 𝑑1(𝑥) reaches its local maximum at 𝑥0 if it exists. Hence we are left to check that 𝑑1(𝑥𝑛)0. In this case we note that 𝑥𝑛𝑥1=(𝑥𝑛+𝑡)(𝑥1+𝑡) and we rewrite 𝑑1(𝑥𝑛) as 𝑑1𝑥𝑛=𝑥2𝑛𝑥𝑟𝑛𝑥𝑟1𝑥1𝑥𝑟𝑛𝑥𝑛𝑥1𝑥𝑛+𝑡2𝑥𝑛+𝑡𝑟𝑥𝑛+𝑡𝑟𝑥1𝑥+𝑡𝑛+𝑡𝑟𝑥𝑛𝑥+𝑡1𝑥+𝑡=𝑒𝑛𝑥1𝑥𝑒𝑛+𝑡𝑥1,+𝑡(3.7) where 𝑥𝑒(𝑥)=𝑟1𝑥𝑟2.(𝑥1)(3.8) In view of (3.5) again, we just need to show that 𝑒(𝑥) is an increasing function for 𝑥>1. Note that 𝑒𝑥(𝑥)=𝑟3𝑥𝑟+12𝑥𝑟+(𝑟1)𝑥(𝑟2)𝑥𝑟2(𝑥1)2,(3.9) and it is easy to see that the function 𝑥𝑟+12𝑥𝑟+(𝑟1)𝑥(𝑟2) is non-negative for 𝑥1 when 0<𝑟3 by considering its Taylor expansion at 𝑥=1 and this completes the proof for the assertion of the theorem for the case 3𝑟<0.
To prove the theorem for the case 0<𝑟3, we may again assume that 𝑡>0 is fixed and define 𝑢𝑛(𝐪,𝐱)=𝑥2𝑛ln𝑀𝑛,𝑟ln𝐺𝑛𝑥𝑛+𝑡2ln𝑀𝑛,𝑟,𝑡ln𝐺𝑛,𝑡.(3.10) Again it suffices to show that 𝜕𝑢𝑛/𝜕𝑥10 and calculation shows 1𝑞1𝜕𝑢𝑛𝜕𝑥1=𝑣𝑛(𝐪,𝐱)𝑣𝑛𝐪,𝐱𝑡,(3.11) where 𝑣𝑛(𝐪,𝐱)=𝑥2𝑛𝑛𝑖=1𝑞𝑖𝑥𝑟𝑖𝑥𝑟1/𝑥𝑟𝑛𝑥1𝑛𝑖=1𝑞𝑖𝑥𝑖/𝑥𝑛𝑟.(3.12) In view of (3.5), the inequality 𝜕𝑢𝑛/𝜕𝑥10 will follow from 𝑑2𝑥𝑖=𝑥2𝑛𝑥𝑟𝑖𝑥𝑟1𝑥1𝑥𝑟𝑛𝑥𝑛+𝑡2𝑥𝑖+𝑡𝑟𝑥1+𝑡𝑟𝑥1𝑥+𝑡𝑛+𝑡𝑟0(3.13) for 𝑥1𝑥𝑖𝑥𝑛. We may assume that 𝑥𝑛>𝑥1 here and it is easy to see that 𝑑2(𝑥)=0 can have at most one root 𝑥0 in between 𝑥1 and 𝑥𝑛. This combined with the observation that 𝑑2(𝑥1)=0,𝑑2(𝑥1)>0 implies that 𝑑1(𝑥) reaches its local maximum at 𝑥0 if it exists. Hence we are left to check that 𝑑2(𝑥𝑛)0. As 𝑑2(𝑥𝑛)=𝑑1(𝑥𝑛), this completes the proof for the remaining case 0<𝑟3 of the theorem.

Now we show that, in general, it is not true that for 3𝑟3,𝑟0,𝑡0,

𝑥21||ln𝐺𝑛ln𝑀𝑛,𝑟||𝑥1+𝑡2||ln𝐺𝑛,𝑡ln𝑀𝑛,𝑟,𝑡||.(3.14) To proceed, we first look at the following related inequalities (with 𝑟>𝑠 here):

ln𝑀𝑛,𝑟ln𝑀𝑛,𝑠(𝑟𝑠)𝜎𝑛2𝑥21ln𝑀𝑛,𝑟,𝑡ln𝑀𝑛,𝑠,𝑡(𝑟𝑠)𝜎𝑛2𝑥1+𝑡2,(3.15)ln𝑀𝑛,𝑟ln𝑀𝑛,𝑠(𝑟𝑠)𝜎𝑛2𝑥2𝑛ln𝑀𝑛,𝑟,𝑡ln𝑀𝑛,𝑠,𝑡(𝑟𝑠)𝜎𝑛2𝑥𝑛+𝑡2.(3.16) Let 𝑓𝑛(𝐪,𝐱,𝑡) denote the right-hand side expression of (3.15); then (3.15) holds if and only if 𝜕𝑓𝑛(𝐪,𝐱,0)/𝜕𝑡0. As 𝐱 is arbitrary, we can recast this condition as

𝑀𝑟1𝑛,𝑟1𝑀𝑟𝑛,𝑟𝑀𝑠1𝑛,𝑠1𝑀𝑠𝑛,𝑠+(𝑟𝑠)𝜎𝑛𝑥310.(3.17) Similarly, (3.16) holds if and only if the following inequality holds:

𝑀𝑟1𝑛,𝑟1𝑀𝑟𝑛,𝑟𝑀𝑠1𝑛,𝑠1𝑀𝑠𝑛,𝑠+(𝑟𝑠)𝜎𝑛𝑥3𝑛0.(3.18) As a first step towards establishing (3.17), we consider the case 𝑛=2 here; in this case we let 𝑥1=1𝑡=𝑥2 and rewrite the left-hand side of (3.17) as

𝑞1+𝑞2𝑡𝑟1𝑞1+𝑞2𝑡𝑟𝑞1+𝑞2𝑡𝑠1𝑞1+𝑞2𝑡𝑠+(𝑟𝑠)𝑞1𝑞2(𝑡1)2=𝑞1𝑞2(1𝑡)𝑞1+𝑞2𝑡𝑟𝑞1+𝑞2𝑡𝑠𝐷𝑟,𝑠𝑡;𝑞1,𝑞2(3.19) with 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2) 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 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 for (𝑡,𝑞1,𝑞2)𝐼2×𝐸. Similar treatment of (3.18) shows that in order for it to hold in the case 𝑛=2, one needs to have 𝐷𝑟,𝑠(𝑡;𝑞1,𝑞2)0 for (𝑡,𝑞1,𝑞2)𝐼1×𝐸.

It follows from the proof of Lemma 2.1 that 𝐷𝑟,0(𝑡;1,0)0 fails to hold for all 𝑡1 when 𝑟>2. In another words, there exists 𝐱,𝐪 such that when 𝑟>2,

𝑀𝑟1𝑛,𝑟1𝑀𝑟𝑛,𝑟1𝐻𝑛+(𝑟𝑠)𝜎𝑛𝑥31<0(3.20) holds. Now we return to the inequality (3.14) and we take 𝑟>0 there. Just as in the discussion above, one sees that (3.14) is equivalent to

2ln𝑀𝑛,𝑟ln𝐺𝑛𝑥1+𝑀𝑟1𝑛,𝑟1𝑀𝑟𝑛,𝑟1𝐻𝑛0.(3.21) This combined with (3.20) now implies that for 𝑟>2,

ln𝑀𝑛,𝑟ln𝐺𝑛>(𝑟𝑠)𝜎𝑛2𝑥21.(3.22)

However, on taking 𝑡+ 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 𝑟>2 in general.

To end this paper, we note that it is an open problem to determine all the triples (𝑟,𝑠,𝛼) so that inequality (1.2) holds. However, when 𝛼=0 with 𝑟=0 or 𝑠=0, the result given in Theorem 2.2 is best possible. In this case Theorem 2.2 implies that for |𝑟|3,𝑟0,

||ln𝑀𝑛,𝑟ln𝐺𝑛|||𝑟|2𝑥2𝑛𝜎𝑛.(3.23)

We point out here that inequality (3.23) does not hold in general when |𝑟|>3. To see this, it suffices to consider the case 𝑛=2 and in this case we can set 0<𝑥1=𝑡𝑥2=1 and consider more generally for 𝑟>𝑠, the function 𝑔2(𝐪,𝐱) defined in the proof of Theorem 2.2, regarding it as a function 𝑓(𝑡) of 𝑡. It is easy to check that 𝑓(1)=𝑓(1)=𝑓(1)=0; hence by the Taylor expansion of 𝑓(𝑡) around 𝑡=1, we need 𝑓(3)(1)0 in order for 𝑓(𝑡)0 to hold for any 0<𝑡1. Calculation shows that

𝑓(3)(1)=𝑞1(𝑟𝑠)𝑟+𝑠33(𝑟+𝑠1)𝑞1+2(𝑟+𝑠)𝑞21.(3.24) On taking 𝑞10+, one sees immediately that we must have 𝑟+𝑠3 here in order for 𝑓(𝑡)0 for all 0<𝑡1. On taking 𝑠=0, we see that one needs 𝑟3 in order for (3.23) to hold for positive 𝑟. Similarly, one checks easily that in the case 𝑛=2, if inequality (3.23) holds for some 𝑟, then it also holds for 𝑟 by a change of variables 𝑥𝑖1/𝑥2𝑖+1. Hence one needs 𝑟3 in order for (3.23) to hold for any negative 𝑟.

References

  1. P. Gao, “Ky 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 · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  2. P. Gao, “A 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 · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  3. D. I. Cartwright and M. J. Field, “A 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 · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  4. H. Alzer, “A new refinement of the arithmetic mean—geometric mean inequality,” The Rocky Mountain Journal of Mathematics, vol. 27, no. 3, pp. 663–667, 1997. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  5. P. Gao, “Certain 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 · View at MathSciNet
  6. A. McD. Mercer, “Bounds 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 · View at Google Scholar · View at MathSciNet
  7. A. McD. Mercer, “Improved upper and lower bounds for the difference AnGn,” The Rocky Mountain Journal of Mathematics, vol. 31, no. 2, pp. 553–560, 2001. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  8. P. R. Mercer, “Refined arithmetic, geometric and harmonic mean inequalities,” The Rocky Mountain Journal of Mathematics, vol. 33, no. 4, pp. 1459–1464, 2003. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  9. H. Alzer, “The inequality of Ky Fan and related results,” Acta Applicandae Mathematicae, vol. 38, no. 3, pp. 305–354, 1995. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · 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, “Generalization of Ky Fan inequality,” Mathematica Balkanica, vol. 5, no. 4, pp. 373–380, 1991. View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet
  12. P. Gao, “A generalization of Ky Fan's inequality,” International Journal of Mathematics and Mathematical Sciences, vol. 28, no. 7, pp. 419–425, 2001. View at Publisher · View at Google Scholar · View at Zentralblatt MATH · View at MathSciNet