On a Class of Ky Fan-Type Inequalities

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 𝑀_𝑛,𝑟(𝐱;𝐪) be the generalized weighted means: 𝑀_𝑛,𝑟(𝐱;𝐪)=(∑^𝑛𝑖=1𝑞_𝑖𝑥^𝑟𝑖)^1/𝑟, where 𝑀_𝑛,0(𝐱;𝐪) denotes the limit of 𝑀_𝑛,𝑟(𝐱;𝐪) as 𝑟→0⁺. Here 𝐱=(𝑥₁,…,𝑥_𝑛), 𝐪=(𝑞₁,…,𝑞_𝑛) with 𝑞_𝑖>0(1≤𝑖≤𝑛) satisfying ∑^𝑛𝑖=1𝑞_𝑖=1. In this paper, we always assume 0<𝑥₁≤𝑥₂≤⋯≤𝑥_𝑛. To any given 𝐱 and 𝑡≥0, we set 𝐱^=(1−𝑥₁,…,1−𝑥_𝑛),𝐱_𝑡=(𝑥₁+𝑡,…,𝑥_𝑛+𝑡).

We define 𝐴_𝑛(𝐱;𝐪)=𝑀_𝑛,1(𝐱;𝐪),𝐺_𝑛(𝐱;𝐪)=𝑀_𝑛,0(𝐱;𝐪),𝐻_𝑛(𝐱;𝐪)=𝑀_𝑛,−1(𝐱;𝐪), and we shall write 𝑀_𝑛,𝑟 for 𝑀_𝑛,𝑟(𝐱;𝐪), 𝑀_{𝑛,𝑟,𝑡} for 𝑀_𝑛,𝑟(𝐱_𝑡;𝐪), and 𝑀^𝑛,𝑟 for 𝑀_𝑛,𝑟(𝐱^;𝐪) if 𝑥_𝑛<1 and similarly for other means when there is no risk of confusion. We further denote 𝜎_𝑛=∑^𝑛𝑖=1𝑞_𝑖(𝑥_𝑖−𝐴_𝑛)².

When 𝑥_𝑛<1, we define

Δ_{𝑟,𝑠,𝛼,𝑡}=𝑀^{𝛼𝑛,𝑟,𝑡}−𝑀^{𝛼𝑛,𝑠,𝑡}/𝛼𝑀^{𝛼𝑛,𝑟,𝑡}−𝑀^{𝛼𝑛,𝑠,𝑡}/𝛼,(1.1) where we set 𝑀^0𝑛,𝑟/0=ln𝑀_𝑛,𝑟 and we shall write Δ_{𝑟,𝑠,𝛼} for Δ_{𝑟,𝑠,𝛼,0} and Δ_𝑟,𝑠 for Δ_𝑟,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 𝑟>𝑠 and 𝛼≤2, the following inequalities are equivalent: 𝑟−𝑠2𝑥^12−𝛼𝜎_𝑛≥𝑀^{𝛼𝑛,𝑟}−𝑀^{𝛼𝑛,𝑠}𝛼≥2𝑟𝑥−𝑠^𝑛2−𝛼𝑥𝜎_𝑛,(1.2)_𝑛1−𝑥_𝑛^2−𝛼≥Δ_{𝑟,𝑠,𝛼}≥𝑥¹1−𝑥₁^2−𝛼,(1.3) where in (1.3) one requires 𝑥_𝑛<1.

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

𝑡+𝑥^𝑛𝑥₁^2−𝛼≥Δ_{𝑟,𝑠,𝛼,𝑡}≥𝑥¹𝑡+𝑥_𝑛^2−𝛼(1.4) being valid for all 𝑡≥0. 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,

𝑡+𝑥^𝑛𝑥_𝑛^2−𝛼≥Δ_{𝑟,𝑠,𝛼,𝑡}≥𝑥¹𝑡+𝑥₁^2−𝛼.(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, 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 𝑥_𝑖∈(0,1/2], Δ_1,0,0≤1, with equality holding if and only if 𝑥₁=⋅⋅⋅=𝑥_𝑛.

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

Theorem 1.3. For 𝑟>𝑠,𝑥_𝑖∈(0,1/2], Δ_𝑟,𝑠,0≤1 holds if and only if |𝑟+𝑠|≤3,2^𝑠/𝑠≥2^𝑟/𝑟 when 𝑠>0, 𝑠2^𝑠≤𝑟2^𝑟 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,𝛼≤2, then the same side inequality of (1.2) also holds for 𝑟,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 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 −1≤𝑠≤1,𝑠≤𝑟≤1+𝑠,𝛼=0. On the other hand, [5, Corollary 3.2] combined with Theorem 1.4 implies that (1.3) holds for 𝛼=0,0≤𝑠≤𝑟≤1 and 𝑟−1≤𝑠≤1,0≤𝑟≤2. 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,0≤𝑟≤2,−1≤𝑠≤1.

2. The Main Theorem

Lemma 2.1. Let 𝑟>𝑠, 𝐼₁=(0,1],𝐼₂=[1,+∞) and let 𝐸 denote the region 𝐸={(𝑞₁,𝑞₂)∣𝑞₁≥0,𝑞₂≥0,𝑞₁+𝑞₂=1}. Define 𝐷_𝑟,𝑠𝑡;𝑞₁,𝑞₂=𝑡^𝑟−1−𝑡^𝑠−1+(𝑟−𝑠)(1−𝑡)𝑞₁+𝑞₂𝑡^𝑟𝑞₁+𝑞₂𝑡^𝑠.(2.1) Then for 𝑠≥0, 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 holds for all (𝑡,𝑞₁,𝑞₂)∈𝐼₁×𝐸 if and only if 𝑠≤1 and 𝑟+𝑠≤3 and 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 holds for all (𝑡,𝑞₁,𝑞₂)∈𝐼₂×𝐸 if and only if 𝑟≤2 and 𝑟+𝑠≤3.
For 𝑠<0, if 𝑟≤0, then 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 holds for all (𝑡,𝑞₁,𝑞₂)∈𝐼₁×𝐸 if and only if −1≤𝑟≤0 and −3≤𝑟+𝑠≤0 and 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 holds for all (𝑡,𝑞₁,𝑞₂)∈𝐼₂×𝐸 if and only if 𝑠≥−2 and −3≤𝑟+𝑠≤0.
For 𝑠<0<𝑟, 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 holds for all (𝑡,𝑞₁,𝑞₂)∈𝐼₂×𝐸 if and only if 𝑟≤2 and 𝑟+𝑠≥0 or 𝑠≥−2 and 𝑟+𝑠≤0.

Proof. When 𝑠≥0, in order for 𝐷_𝑟,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 𝐷_𝑟,𝑠(𝑡;1,0) as 𝑓(𝑡)=𝑡^𝑟−1−𝑡^𝑠−1+(𝑟−𝑠)(1−𝑡).(2.2) Note that 𝑓(1)=𝑓^(1)=0; hence in order for 𝑓(𝑡)≤0 to hold for all 0<𝑡≤1, it is necessary that 𝑓^(1)≤0. Note that 𝑓^(𝑡)=(𝑟−1)(𝑟−2)𝑡^𝑟−3−(𝑠−1)(𝑠−2)𝑡^𝑠−3 and from this one checks easily that 𝑓^(1)≤0 is equivalent to 𝑟+𝑠≤3. On the other hand, on taking 𝑡→0⁺, we see that one needs to have 𝑠≤1 in order for 𝑓(𝑡)≤0 to hold for all 0<𝑡≤1. Now, it also follows from 𝑠≤1 that 𝑡^3−𝑠𝑓^(𝑡)=(𝑟−1)(𝑟−2)𝑡^𝑟−𝑠−(𝑠−1)(𝑠−2)≤max{(𝑟−1)(𝑟−2)−(𝑠−1)(𝑠−2),−(𝑠−1)(𝑠−2)}≤0. Hence one deduces via Taylor expansion of 𝑓(𝑡) at 1 that 𝑓(𝑡)≤0 for all 0<𝑡≤1.
Similarly, when 𝑠≥0, in order for 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 to hold for all (𝑡,𝑞₁,𝑞₂)∈𝐼₂×𝐸, one just needs to check 𝑓(𝑡)≤0 for 𝑡≥1. As 𝑓(1)=𝑓^(1)=0, certainly it is necessary to have 𝑓^(1)≤0 and lim_𝑡→+∞𝑓(𝑡)≤0. These imply that 𝑟≤2 and 𝑟+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 𝑟≤0 by noting that 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)=𝑡^𝑟+𝑠𝐷_{−𝑠,−𝑟}(𝑡;𝑞₂,𝑞₁).
It remains to treat the case 𝑟>0>𝑠. We let 𝑔(𝑞)=(1−𝑞+𝑞𝑡^𝑟)(1−𝑞+𝑞𝑡^𝑠), and note that 𝑔^(𝑞)=2(𝑡^𝑟−1)(𝑡^𝑠−1)≤0. It follows from this that in order for 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 to hold for (𝑡,𝑞₁,𝑞₂)∈𝐼₂×𝐸, it suffices to check the cases 𝑞₂=0,1. When 𝑟+𝑠≥0, we only need to check the case 𝑞₂=0 and in this case one can discuss similarly to the case 𝑠≥0 above to conclude the assertion of the lemma. We just point out here that as 𝑠<0<𝑟≤2, we have 𝑟+𝑠<2. When 𝑟+𝑠≤0, it suffices to check the case 𝑞₂=1 and in this case one uses the relation 𝐷_𝑟,𝑠(𝑡;0,1)=𝑡^𝑟+𝑠𝐷_{−𝑠,−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 0≤𝑠≤1,𝑟+𝑠≤3 or 𝑠<0,−1≤𝑟≤0 and −3≤𝑟+𝑠≤0. The left-hand side inequality of (1.2) holds for 𝛼=0 when −2≤𝑠≤0,−3≤𝑟+𝑠≤0.

Proof. To prove the first assertion of the theorem, we may assume 𝑟>2 or −1≤𝑟≤0 in view of our discussion in the last paragraph of Section 1 and for the case 𝑟>2, we define 𝑔_𝑛(𝐪,𝐱)=ln𝑀_𝑛,𝑟−ln𝑀_𝑛,𝑠−𝑟−𝑠2𝑥^2𝑛𝜎_𝑛.(2.3) Similar to the proof of Theorem 5.1 [2], it suffices to show that 𝜕𝑔_𝑛/𝜕𝑥₁≤0. Calculation shows that 1𝑞₁𝜕𝑔_𝑛𝜕𝑥₁=𝑥^1𝑟−1𝑀^{𝑟𝑛,𝑟}−𝑥^1𝑠−1𝑀^{𝑠𝑛,𝑠}−𝑟−𝑠𝑥^2𝑛𝑥₁−𝐴_𝑛∶=𝑓_𝑛(𝐪,𝐱).(2.4) We now show by induction on 𝑛 that 𝑓_𝑛(𝐪,𝐱)≤0. When 𝑛=1, there is nothing to prove. When 𝑛=2, this becomes 1𝑞₂𝑓₂(𝐪,𝐱)=𝑥^{2𝑟+𝑠−1}𝐷_𝑟,𝑠𝑥₁/𝑥₂;𝑞₂,𝑞₁𝑀^𝑟2,𝑟𝑀^𝑠2,𝑠≤0,(2.5) by Lemma 2.1.
Suppose now 𝑛≥3; in order to show 𝑓_𝑛(𝐪,𝐱)≤0, we may assume that 0<𝑥₁<𝑥_𝑛 are being fixed and it suffices to show that the maximum value of 𝑓_𝑛(𝐪,𝐱) is non-positive on the region 𝑅_𝑛×𝑆_𝑛−2, where 𝑅_𝑛={(𝑞₁,𝑞₂,…,𝑞_𝑛)∶0≤𝑞_𝑖≤1,1≤𝑖≤𝑛,∑^𝑛𝑖=1𝑞_𝑖=1} and 𝑆_𝑛−2={(𝑥₂,…,𝑥_𝑛−1)∶𝑥_𝑖∈[𝑥₁,𝑥_𝑛],2≤𝑖≤𝑛−1}.
Let (𝐪^,𝐱^) be a point of 𝑅_𝑛×𝑆_𝑛−2 in which the absolute maximum of 𝑓_𝑛 is reached. If 𝑥^𝑖=𝑥^𝑖+1 for some 1≤𝑖≤𝑛−1, by combining 𝑥^𝑖 with 𝑥^𝑖+1 and 𝑞^𝑖 with 𝑞^𝑖+1, we are back to the case of 𝑛−1 variables with different weights. If 𝑞^𝑖=1 for some 𝑖, then we have 𝑥^1𝑟−1𝑀^{𝑟𝑛,𝑟}−𝑥^1𝑠−1𝑀^{𝑠𝑛,𝑠}−𝑟−𝑠𝑥^2𝑛𝑥₁−𝐴_𝑛=𝑥𝑥^{1𝑟−1𝑟𝑖}−𝑥𝑥^{1𝑠−1𝑠𝑖}−𝑟−𝑠𝑥^2𝑛≤𝑥_1𝑟−1𝑥₁−𝑥_𝑖𝑥^𝑟𝑖−𝑥^1𝑠−1𝑥^𝑠𝑖−𝑟−𝑠𝑥^2𝑖=1𝑥₁−𝑥_𝑖𝑥_𝑖𝐷_𝑟,𝑠𝑥𝑥¹_𝑖;1,0≤0,(2.6) by Lemma 2.1. If 𝑞^𝑖=0 for some 1<𝑖<𝑛, we are back to the case of 𝑛−1 variables. If 𝑞^𝑛=0, then we may assume that 𝑞^𝑛−1≠0 and note that we have 𝑀_𝑛,𝑟=𝑀_{𝑛−1,𝑟},𝑀_𝑛,𝑠=𝑀_{𝑛−1,𝑠},𝐴_𝑛=𝐴_𝑛−1 and that 𝑥^1𝑟−1𝑀^{𝑟𝑛,𝑟}−𝑥^1𝑠−1𝑀^{𝑠𝑛,𝑠}−𝑟−𝑠𝑥^2𝑛𝑥₁−𝐴_𝑛≤𝑀𝑥^1𝑟−1_{𝑟𝑛−1,𝑟}−𝑀𝑥^1𝑠−1_{𝑠𝑛−1,𝑠}−𝑟−𝑠𝑥^2𝑛−1𝑥₁−𝐴_𝑛−1,(2.7) and we are again back to the case 𝑛−1. If 𝑞^1=0, then similarly we may assume that 𝑞^2≠0 and if we can show that (again with 𝑀_𝑛,𝑟=𝑀_{𝑛−1,𝑟},𝑀_𝑛,𝑠=𝑀_{𝑛−1,𝑠} and 𝐴_𝑛=𝐴_𝑛−1 here) 𝑥^1𝑟−1𝑀^{𝑟𝑛,𝑟}−𝑥^1𝑠−1𝑀^{𝑠𝑛,𝑠}−𝑟−𝑠𝑥^2𝑛𝑥₁−𝐴_𝑛≤𝑀𝑥^2𝑟−1_{𝑟𝑛−1,𝑟}−𝑀𝑥^2𝑠−1_{𝑠𝑛−1,𝑠}−𝑟−𝑠𝑥^2𝑛𝑥₂−𝐴_𝑛−1,(2.8) then we are back to the case of 𝑛−1 variables. Note that the above inequality will follow if the function 𝑥⟼𝑥^𝑟−1𝑀^{𝑟𝑛,𝑟}−𝑥^𝑠−1𝑀^{𝑠𝑛,𝑠}−𝑟−𝑠𝑥^2𝑛𝑥(2.9) is an increasing function for 0<𝑥≤𝑀_𝑛,𝑟 (in fact, one only needs this for 0<𝑥≤𝑥₂) and its derivative is (𝑟−1)𝑥^𝑟−2𝑀^{𝑟𝑛,𝑟}+(1−𝑠)𝑥^𝑠−2𝑀^{𝑠𝑛,𝑠}−𝑟−𝑠𝑥^2𝑛≥(𝑟−1)𝑥^𝑟−2𝑥^𝑟𝑛+(1−𝑠)𝑥^𝑠−2𝑥^𝑠𝑛−𝑟−𝑠𝑥^2𝑛∶=ℎ(𝑥),(2.10) with the inequality holding for the case 𝑟>2 (note that together with 𝑟+𝑠≤3, this implies that 𝑠<1). It also follows from 𝑟+𝑠≤3 that ℎ^(𝑥)=0 has no root in (0,𝑥_𝑛). One then deduces from ℎ(𝑥_𝑛)=0 and lim_𝑥→0⁺ℎ(𝑥)=+∞ that ℎ(𝑥)≥0 for 0<𝑥≤𝑥_𝑛.
So from now on it remains to consider the case 𝑞^𝑖≠0,1,𝑥^𝑖≠𝑥^𝑗 for 1≤𝑖,𝑗≤𝑛,𝑖≠𝑗 and this implies that (𝐪^,𝐱^) is an interior point of 𝑅_𝑛×𝑆_𝑛−2. We will now show that this cannot happen.
We define 𝑝(𝑥)=−𝑥^1𝑟−1𝑥^𝑟𝑀^{2𝑟𝑛,𝑟}+𝑥^1𝑠−1𝑥^𝑠𝑀^{2𝑠𝑛,𝑠}+(𝑟−𝑠)𝑥𝑥^2𝑛−𝜆.(2.11) Note here in the definition of 𝑝(𝑥) that 𝑀_𝑛,𝑟and𝑀_𝑛,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 𝑅_𝑛×𝑆_𝑛−2, we may use the Lagrange multiplier method to obtain a real number 𝜆 so that at (𝐪^,𝐱^), 𝜕𝑓_𝑛𝜕𝑞_𝑖=𝜆𝜕𝜕𝑞_𝑖^𝑛_𝑖=1𝑞_𝑖−1,1𝑞_𝑗𝜕𝑓_𝑛𝜕𝑥_𝑗=0(2.12) for all 1≤𝑖≤𝑛 and 2≤𝑗≤𝑛−1.
By (2.12), a computation shows that each 𝑥′_𝑖 (1≤𝑖≤𝑛) is a root of 𝑝(𝑥)=0 (where 𝑀_𝑛,𝑟,𝑀_𝑛,𝑠 take their values at (𝐪^,𝐱^)) and each 𝑥′_𝑖 (2≤𝑖≤𝑛−1) is a root of 𝑝^(𝑥)=0. Now 𝑛≥3 implies 𝑝(𝑥₂)=0. As 𝑝(𝑥₁)=𝑝(𝑥₂)=𝑝(𝑥_𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers 𝑥₁<𝑎<𝑥₂<𝑏<𝑥_𝑛 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 0≤𝑠≤1.
Now to show the right-hand side inequality hold of (1.2) for the case 𝑠<0,−1≤𝑟≤0 and −3≤𝑟+𝑠≤0, once again it suffices to show that the function 𝑔_𝑛(𝐪,𝐱) defined above is nonnegative for any integer 𝑛≥1. We note that when 𝑛=1, this is obvious and when 𝑛=2, this follows again from 𝜕𝑔₂/𝜕𝑥₁≤0 by Lemma 2.1.
Suppose now 𝑛≥3; in order to show 𝑔_𝑛(𝐪,𝐱)≥0, we may assume that 0<𝑥₁<𝑥_𝑛 are being fixed and it suffices to show the minimum value of 𝑔_𝑛(𝐪,𝐱) is nonnegative on the region 𝑅_𝑛×𝑆_𝑛−2, where 𝑅_𝑛 and 𝑆_𝑛−2 are defined as above.
Let (𝐪^,𝐱^) be a point of 𝑅_𝑛×𝑆_𝑛−2 in which the absolute minimum of 𝑔_𝑛 is reached. Note that 𝜎_𝑛=𝑀^2𝑛,2−𝐴^2𝑛; thus if 𝑥^𝑖=𝑥^𝑖+1 for some 1≤𝑖≤𝑛−1, by combining 𝑥^𝑖 with 𝑥^𝑖+1 and 𝑞^𝑖 with 𝑞^𝑖+1, we are back to the case of 𝑛−1 variables with different weights. Similarly, if 𝑞^𝑖=1 for some 𝑖, then we are back to the case 𝑛=1. If 𝑞^𝑖=0 for some 1≤𝑖<𝑛, we are back to the case of 𝑛−1 variables. If 𝑞^𝑛=0, then we may assume that 𝑞^𝑛−1≠0 and note that we have 𝑀_𝑛,𝑟=𝑀_{𝑛−1,𝑟},𝑀_𝑛,𝑠=𝑀_{𝑛−1,𝑠},𝑀_𝑛,2=𝑀_𝑛−1,2,𝐴_𝑛=𝐴_𝑛−1 and that ln𝑀_𝑛,𝑟−ln𝑀_𝑛,𝑠−𝑟−𝑠2𝑥^2𝑛𝜎_𝑛=ln𝑀_𝑛,𝑟−ln𝑀_𝑛,𝑠−𝑟−𝑠2𝑥^2𝑛𝑀^2𝑛,2≥ln𝑀_{𝑛−1,𝑟}−𝐴^2𝑛−ln𝑀_{𝑛−1,𝑠}−𝑟−𝑠2𝑥^2𝑛−1𝑀^2𝑛−1,2=ln𝑀_{𝑛−1,𝑟}−𝐴^2𝑛−1−ln𝑀_{𝑛−1,𝑠}−𝑟−𝑠2𝑥^2𝑛−1𝜎_𝑛−1,(2.13) and we are again back to the case of 𝑛−1 variables.
So from now on it remains to consider the case 𝑞^𝑖≠0,1,𝑥^𝑖≠𝑥^𝑗 for 1≤𝑖,𝑗≤𝑛,𝑖≠𝑗, and this implies that (𝐪^,𝐱^) is an interior point of 𝑅_𝑛×𝑆_𝑛−2. We will now show that this cannot happen.
We define 𝑎(𝑥)=𝑥^𝑟𝑟𝑀^{𝑟𝑛,𝑟}−𝑥^𝑠𝑠𝑀^{𝑠𝑛,𝑠}−(𝑟−𝑠)𝑥²−2𝐴_𝑛𝑥2𝑥^2𝑛−𝜆.(2.14) Here we define 𝑥⁰/0=ln𝑥. Also note here in the definition of 𝑎(𝑥), that 𝑀_𝑛,𝑟,𝑀_𝑛,𝑠,and𝐴_𝑛 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 𝑅_𝑛×𝑆_𝑛−2, we may use the Lagrange multiplier method to obtain a real number 𝜆 so that at (𝐪^,𝐱^), 𝜕𝑔_𝑛𝜕𝑞_𝑖=𝜆𝜕𝜕𝑞_𝑖^𝑛_𝑖=1𝑞_𝑖−1,1𝑞_𝑗𝜕𝑔_𝑛𝜕𝑥_𝑗=0(2.15) for all 1≤𝑖≤𝑛 and 2≤𝑗≤𝑛−1.
By (2.15), a computation shows that each 𝑥^𝑖 (1≤𝑖≤𝑛) is a root of 𝑎(𝑥)=0 (where 𝑀_𝑛,𝑟,𝑀_𝑛,𝑠,and𝐴_𝑛 take their values at (𝐪^,𝐱^)) and each 𝑥^𝑖 (2≤𝑖≤𝑛−1) is a root of 𝑎^(𝑥)=0. Now 𝑛≥3 implies 𝑎(𝑥₂)=0. As 𝑎(𝑥₁)=𝑎(𝑥₂)=𝑎(𝑥_𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers 𝑥₁<𝑐<𝑥₂<𝑑<𝑥_𝑛 such that 𝑎^(𝑐)=𝑎^(𝑥₂)=𝑎^(𝑑)=0. However, we have 𝑎^(𝑥)=𝑥^𝑟−1𝑀^{𝑟𝑛,𝑟}−𝑥^𝑠−1𝑀^{𝑠𝑛,𝑠}−(𝑟−𝑠)𝑥−𝐴^𝑛𝑥^2𝑛.(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 𝑟≤0, it follows from lim_𝑥→0⁺𝑎^(𝑥)=−∞ and lim_𝑥→+∞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,−1≤𝑟≤0, and −3≤𝑟+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 𝑟,𝑠,𝛼≤2, 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 −3≤𝑟≤3,𝑟≠0,𝑡≥0, then the following inequality holds: 𝑥^2𝑛||ln𝐺_𝑛−ln𝑀_𝑛,𝑟||≥𝑥_𝑛+𝑡²||ln𝐺_𝑛,𝑡−ln𝑀_{𝑛,𝑟,𝑡}||.(3.1)

Proof. We first prove the theorem for the case −3≤𝑟<0. For this, we may assume that 𝑡>0 is fixed and replace 𝑟 with −𝑟 so that 0<𝑟≤3 in what follows. We define 𝑓_𝑛(𝐪,𝐱)=𝑥^2𝑛ln𝐺_𝑛−ln𝑀_𝑛,−𝑟−𝑥_𝑛+𝑡²ln𝐺_𝑛,𝑡−ln𝑀_{𝑛,−𝑟,𝑡}.(3.2) As in the proof of Theorem 2.2, it suffices to show that 𝜕𝑓_𝑛/𝜕𝑥₁≤0 and calculation shows −1𝑞₁𝜕𝑓_𝑛𝜕𝑥₁=𝑔_𝑛(𝐪,𝐱)−𝑔_𝑛𝐪,𝐱_𝑡,(3.3) where 𝑔_𝑛(𝐪,𝐱)=𝑥^2𝑛∑^𝑛𝑖=1𝑞_𝑖𝑥^𝑟𝑖−𝑥^𝑟1/𝑥^𝑟𝑖𝑥₁∑^𝑛𝑖=1𝑞_𝑖𝑥₁/𝑥_𝑖^𝑟.(3.4) It is easy to check that 𝑥_𝑛𝑥_𝑖≥𝑥^𝑛+𝑡𝑥_𝑖+𝑡,𝑥𝑥¹_𝑖≤𝑥¹+𝑡𝑥_𝑖+𝑡.(3.5) In view of (3.5), the inequality 𝜕𝑓_𝑛/𝜕𝑥₁≤0 will follow from 𝑑₁𝑥_𝑖=𝑥^2𝑛𝑥^𝑟𝑖−𝑥^𝑟1𝑥₁𝑥^𝑟𝑖−𝑥^𝑛+𝑡²𝑥_𝑖+𝑡^𝑟−𝑥₁+𝑡^𝑟𝑥₁+𝑡𝑥_𝑖+𝑡^𝑟≥0(3.6) for 𝑥₁≤𝑥_𝑖≤𝑥_𝑛. We may assume that 𝑥_𝑛>𝑥₁ here and it is easy to see that 𝑑^1(𝑥)=0 can have at most one root 𝑥₀ in between 𝑥₁ and 𝑥_𝑛. This combined with the observation that 𝑑₁(𝑥₁)=0,𝑑^1(𝑥₁)>0 implies that 𝑑₁(𝑥) reaches its local maximum at 𝑥₀ if it exists. Hence we are left to check that 𝑑₁(𝑥_𝑛)≥0. In this case we note that 𝑥_𝑛−𝑥₁=(𝑥_𝑛+𝑡)−(𝑥₁+𝑡) and we rewrite 𝑑₁(𝑥_𝑛) as 𝑑₁𝑥_𝑛=𝑥^2𝑛𝑥^𝑟𝑛−𝑥^𝑟1𝑥₁𝑥^𝑟𝑛𝑥_𝑛−𝑥₁−𝑥^𝑛+𝑡²𝑥_𝑛+𝑡^𝑟−𝑥_𝑛+𝑡^𝑟=𝑒𝑥_𝑛𝑥₁+𝑡𝑥_𝑛+𝑡^𝑟𝑥_𝑛+𝑡−𝑥₁+𝑡𝑥₁−𝑒𝑥^𝑛+𝑡𝑥₁+𝑡,(3.7) where 𝑒(𝑥)=𝑥^𝑟−1𝑥^𝑟−2(𝑥−1).(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 𝑒^(𝑥)=𝑥^𝑟−3𝑥^𝑟+1−2𝑥^𝑟+(𝑟−1)𝑥−(𝑟−2)𝑥^𝑟−2(𝑥−1)²,(3.9) and it is easy to see that the function 𝑥^𝑟+1−2𝑥^𝑟+(𝑟−1)𝑥−(𝑟−2) is non-negative for 𝑥≥1 when 0<𝑟≤3 by considering its Taylor expansion at 𝑥=1 and this completes the proof for the assertion of the theorem for the case −3≤𝑟<0.
To prove the theorem for the case 0<𝑟≤3, we may again assume that 𝑡>0 is fixed and define 𝑢_𝑛(𝐪,𝐱)=𝑥^2𝑛ln𝑀_𝑛,𝑟−ln𝐺_𝑛−𝑥_𝑛+𝑡²ln𝑀_{𝑛,𝑟,𝑡}−ln𝐺_𝑛,𝑡.(3.10) Again it suffices to show that 𝜕𝑢_𝑛/𝜕𝑥₁≤0 and calculation shows −1𝑞₁𝜕𝑢_𝑛𝜕𝑥₁=𝑣_𝑛(𝐪,𝐱)−𝑣_𝑛𝐪,𝐱_𝑡,(3.11) where 𝑣_𝑛(𝐪,𝐱)=𝑥^2𝑛∑^𝑛𝑖=1𝑞_𝑖𝑥^𝑟𝑖−𝑥^𝑟1/𝑥^𝑟𝑛𝑥₁∑^𝑛𝑖=1𝑞_𝑖𝑥_𝑖/𝑥_𝑛^𝑟.(3.12) In view of (3.5), the inequality 𝜕𝑢_𝑛/𝜕𝑥₁≤0 will follow from 𝑑₂𝑥_𝑖=𝑥^2𝑛𝑥^𝑟𝑖−𝑥^𝑟1𝑥₁𝑥^𝑟𝑛−𝑥^𝑛+𝑡²𝑥_𝑖+𝑡^𝑟−𝑥₁+𝑡^𝑟𝑥₁+𝑡𝑥_𝑛+𝑡^𝑟≥0(3.13) for 𝑥₁≤𝑥_𝑖≤𝑥_𝑛. We may assume that 𝑥_𝑛>𝑥₁ here and it is easy to see that 𝑑^2(𝑥)=0 can have at most one root 𝑥₀ in between 𝑥₁ and 𝑥_𝑛. This combined with the observation that 𝑑₂(𝑥₁)=0,𝑑^2(𝑥₁)>0 implies that 𝑑₁(𝑥) reaches its local maximum at 𝑥₀ if it exists. Hence we are left to check that 𝑑₂(𝑥_𝑛)≥0. As 𝑑₂(𝑥_𝑛)=𝑑₁(𝑥_𝑛), this completes the proof for the remaining case 0<𝑟≤3 of the theorem.

Now we show that, in general, it is not true that for −3≤𝑟≤3,𝑟≠0,𝑡≥0,

𝑥²¹||ln𝐺_𝑛−ln𝑀_𝑛,𝑟||≤𝑥₁+𝑡²||ln𝐺_𝑛,𝑡−ln𝑀_{𝑛,𝑟,𝑡}||.(3.14) To proceed, we first look at the following related inequalities (with 𝑟>𝑠 here):

ln𝑀_𝑛,𝑟−ln𝑀_𝑛,𝑠−(𝑟−𝑠)𝜎^𝑛2𝑥²¹≤ln𝑀_{𝑛,𝑟,𝑡}−ln𝑀_{𝑛,𝑠,𝑡}−(𝑟−𝑠)𝜎^𝑛2𝑥₁+𝑡²ln𝑀_𝑛,𝑟,(3.15)−ln𝑀_𝑛,𝑠−(𝑟−𝑠)𝜎^𝑛2𝑥^2𝑛≥ln𝑀_{𝑛,𝑟,𝑡}−ln𝑀_{𝑛,𝑠,𝑡}−(𝑟−𝑠)𝜎^𝑛2𝑥_𝑛+𝑡².(3.16) Let 𝑓_𝑛(𝐪,𝐱,𝑡) denote the right-hand side expression of (3.15); then (3.15) holds if and only if 𝜕𝑓_𝑛(𝐪,𝐱,0)/𝜕𝑡≥0. As 𝐱 is arbitrary, we can recast this condition as

𝑀^{𝑟−1𝑛,𝑟−1}𝑀^{𝑟𝑛,𝑟}−𝑀^{𝑠−1𝑛,𝑠−1}𝑀^{𝑠𝑛,𝑠}+(𝑟−𝑠)𝜎^𝑛𝑥³¹≥0.(3.17) Similarly, (3.16) holds if and only if the following inequality holds:

𝑀^{𝑟−1𝑛,𝑟−1}𝑀^{𝑟𝑛,𝑟}−𝑀^{𝑠−1𝑛,𝑠−1}𝑀^{𝑠𝑛,𝑠}+(𝑟−𝑠)𝜎^𝑛𝑥^3𝑛≤0.(3.18) As a first step towards establishing (3.17), we consider the case 𝑛=2 here; in this case we let 𝑥₁=1≤𝑡=𝑥₂ and rewrite the left-hand side of (3.17) as

𝑞₁+𝑞₂𝑡^𝑟−1𝑞₁+𝑞₂𝑡^𝑟−𝑞¹+𝑞₂𝑡^𝑠−1𝑞₁+𝑞₂𝑡^𝑠+(𝑟−𝑠)𝑞₁𝑞₂(𝑡−1)²=𝑞¹𝑞₂(1−𝑡)𝑞₁+𝑞₂𝑡^𝑟𝑞₁+𝑞₂𝑡^𝑠𝐷^𝑟,𝑠𝑡;𝑞₁,𝑞₂(3.19) with 𝐷_𝑟,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 𝐷_𝑟,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 𝐷_𝑟,𝑠(𝑡;𝑞₁,𝑞₂)≤0 for (𝑡,𝑞₁,𝑞₂)∈𝐼₁×𝐸.

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

𝑀^{𝑟−1𝑛,𝑟−1}𝑀^{𝑟𝑛,𝑟}−1𝐻_𝑛+(𝑟−𝑠)𝜎^𝑛𝑥³¹<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

2ln𝑀_𝑛,𝑟−ln𝐺_𝑛𝑥₁+𝑀^{𝑟−1𝑛,𝑟−1}𝑀^{𝑟𝑛,𝑟}−1𝐻_𝑛≥0.(3.21) This combined with (3.20) now implies that for 𝑟>2,

ln𝑀_𝑛,𝑟−ln𝐺_𝑛>(𝑟−𝑠)𝜎^𝑛2𝑥²¹.(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,𝑟≠0,

||ln𝑀_𝑛,𝑟−ln𝐺_𝑛||≥|𝑟|2𝑥^2𝑛𝜎_𝑛.(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=𝑡≤𝑥₂=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<𝑡≤1. Calculation shows that

𝑓⁽³⁾(1)=𝑞₁(𝑟−𝑠)𝑟+𝑠−3−3(𝑟+𝑠−1)𝑞₁+2(𝑟+𝑠)𝑞²¹.(3.24) On taking 𝑞₁→0⁺, one sees immediately that we must have 𝑟+𝑠≤3 here in order for 𝑓(𝑡)≥0 for all 0<𝑡≤1. On taking 𝑠=0, we see that one needs 𝑟≤3 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 𝑥_𝑖→1/𝑥_2−𝑖+1. Hence one needs 𝑟≥−3 in order for (3.23) to hold for any negative 𝑟.