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=1ğ‘žğ‘–ğ‘¥ğ‘Ÿğ‘–)1/𝑟, where 𝑀𝑛,0(𝐱;𝐪) denotes the limit of 𝑀𝑛,𝑟(𝐱;𝐪) as 𝑟→0+. Here 𝐱=(𝑥1,…,𝑥𝑛), 𝐪=(ğ‘ž1,…,ğ‘žğ‘›) with ğ‘žğ‘–>0(1≤𝑖≤𝑛) satisfying ∑𝑛𝑖=1ğ‘žğ‘–=1. In this paper, we always assume 0<𝑥1≤𝑥2≤⋯≤𝑥𝑛. To any given 𝐱 and 𝑡≥0, we set ğ±î…ž=(1−𝑥1,…,1−𝑥𝑛),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 ğ‘€î…žğ‘›,𝑟 for 𝑀𝑛,𝑟(ğ±î…ž;𝐪) if 𝑥𝑛<1 and similarly for other means when there is no risk of confusion. We further denote ğœŽğ‘›=∑𝑛𝑖=1ğ‘žğ‘–(𝑥𝑖−𝐴𝑛)2.

When 𝑥𝑛<1, we define

Δ𝑟,𝑠,𝛼,𝑡=î€·ğ‘€î…žğ›¼ğ‘›,𝑟,ğ‘¡âˆ’ğ‘€î…žğ›¼ğ‘›,𝑠,𝑡/𝛼𝑀𝛼𝑛,𝑟,𝑡−𝑀𝛼𝑛,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 𝛼≤2 following the method there).

Theorem 1.1. For 𝑟>𝑠 and 𝛼≤2, the following inequalities are equivalent: 𝑟−𝑠2𝑥12âˆ’ğ›¼ğœŽğ‘›â‰¥î€·ğ‘€ğ›¼ğ‘›,𝑟−𝑀𝛼𝑛,𝑠𝛼≥𝑟−𝑠2𝑥𝑛2âˆ’ğ›¼ğœŽğ‘›,𝑥(1.2)𝑛1−𝑥𝑛2−𝛼≥Δ𝑟,𝑠,𝛼≥𝑥11−𝑥12−𝛼,(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+𝑥12−𝛼≥Δ𝑟,𝑠,𝛼,𝑡≥𝑥1𝑡+𝑥𝑛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,

𝑥𝑛𝑡+𝑥𝑛2−𝛼≥Δ𝑟,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 𝑥𝑖∈(0,1/2], Δ1,0,0≤1, 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,0≤1 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,0≤1 holds if and only if |𝑟+𝑠|≤3,2𝑠/𝑠≥2𝑟/𝑟 when 𝑠>0, 𝑠2𝑠≤𝑟2𝑟 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>ğ‘ î…ž, then it also holds for 𝑟>ğ‘ î…ž. 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={(ğ‘ž1,ğ‘ž2)âˆ£ğ‘ž1≥0,ğ‘ž2≥0,ğ‘ž1+ğ‘ž2=1}. Define 𝐷𝑟,𝑠𝑡;ğ‘ž1,ğ‘ž2=𝑡𝑟−1−𝑡𝑠−1î€·ğ‘ž+(𝑟−𝑠)(1−𝑡)1+ğ‘ž2ğ‘¡ğ‘Ÿğ‘žî€¸î€·1+ğ‘ž2𝑡𝑠.(2.1) Then for 𝑠≥0, 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 holds for all (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼1×𝐸 if and only if 𝑠≤1 and 𝑟+𝑠≤3 and 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 holds for all (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼2×𝐸 if and only if 𝑟≤2 and 𝑟+𝑠≤3.
For 𝑠<0, if 𝑟≤0, then 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 holds for all (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼1×𝐸 if and only if −1≤𝑟≤0 and −3≤𝑟+𝑠≤0 and 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 holds for all (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼2×𝐸 if and only if 𝑠≥−2 and −3≤𝑟+𝑠≤0.
For 𝑠<0<𝑟, 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 holds for all (𝑡,ğ‘ž1,ğ‘ž2)∈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 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 to hold for all (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼1×𝐸, one just needs to check the case ğ‘ž1=1,ğ‘ž2=0. In this case we can rewrite 𝐷𝑟,𝑠(𝑡;1,0) as 𝑓(𝑡)=𝑡𝑟−1−𝑡𝑠−1+(𝑟−𝑠)(1−𝑡).(2.2) Note that 𝑓(1)=ğ‘“î…ž(1)=0; hence in order for 𝑓(𝑡)≤0 to hold for all 0<𝑡≤1, it is necessary that ğ‘“î…žî…ž(1)≤0. Note that ğ‘“î…žî…ž(𝑡)=(𝑟−1)(𝑟−2)𝑡𝑟−3−(𝑠−1)(𝑠−2)𝑡𝑠−3 and from this one checks easily that ğ‘“î…žî…ž(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 𝑡3âˆ’ğ‘ ğ‘“î…žî…ž(𝑡)=(𝑟−1)(𝑟−2)𝑡𝑟−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 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 to hold for all (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼2×𝐸, one just needs to check 𝑓(𝑡)≤0 for 𝑡≥1. As 𝑓(1)=ğ‘“î…ž(1)=0, certainly it is necessary to have ğ‘“î…žî…ž(1)≤0 and lim𝑡→+âˆžğ‘“(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 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)=𝑡𝑟+𝑠𝐷−𝑠,−𝑟(𝑡;ğ‘ž2,ğ‘ž1).
It remains to treat the case 𝑟>0>𝑠. We let 𝑔(ğ‘ž)=(1âˆ’ğ‘ž+ğ‘žğ‘¡ğ‘Ÿ)(1âˆ’ğ‘ž+ğ‘žğ‘¡ğ‘ ), and note that ğ‘”î…žî…ž(ğ‘ž)=2(𝑡𝑟−1)(𝑡𝑠−1)≤0. It follows from this that in order for 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 to hold for (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼2×𝐸, it suffices to check the cases ğ‘ž2=0,1. When 𝑟+𝑠≥0, we only need to check the case ğ‘ž2=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 ğ‘ž2=1 and in this case one uses the relation 𝐷𝑟,𝑠(𝑡;0,1)=𝑡𝑟+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 𝑔𝑛(𝐪,𝐱)=ln𝑀𝑛,𝑟−ln𝑀𝑛,𝑠−𝑟−𝑠2𝑥2ğ‘›ğœŽğ‘›.(2.3) Similar to the proof of Theorem 5.1 [2], it suffices to show that 𝜕𝑔𝑛/𝜕𝑥1≤0. Calculation shows that 1ğ‘ž1𝜕𝑔𝑛𝜕𝑥1=𝑥1𝑟−1𝑀𝑟𝑛,𝑟−𝑥1𝑠−1𝑀𝑠𝑛,𝑠−𝑟−𝑠𝑥2𝑛𝑥1−𝐴𝑛∶=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 1ğ‘ž2𝑓2𝑥(𝐪,𝐱)=2𝑟+𝑠−1𝐷𝑟,𝑠𝑥1/𝑥2;ğ‘ž2,ğ‘ž1𝑀𝑟2,𝑟𝑀𝑠2,𝑠≤0,(2.5) by Lemma 2.1.
Suppose now 𝑛≥3; 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×𝑆𝑛−2, where 𝑅𝑛={(ğ‘ž1,ğ‘ž2,…,ğ‘žğ‘›)∶0â‰¤ğ‘žğ‘–âˆ‘â‰¤1,1≤𝑖≤𝑛,𝑛𝑖=1ğ‘žğ‘–=1} and 𝑆𝑛−2={(𝑥2,…,𝑥𝑛−1)∶𝑥𝑖∈[𝑥1,𝑥𝑛],2≤𝑖≤𝑛−1}.
Let (ğªî…ž,ğ±î…ž) be a point of 𝑅𝑛×𝑆𝑛−2 in which the absolute maximum of 𝑓𝑛 is reached. If ğ‘¥î…žğ‘–=ğ‘¥î…žğ‘–+1 for some 1≤𝑖≤𝑛−1, by combining ğ‘¥î…žğ‘– with ğ‘¥î…žğ‘–+1 and ğ‘žî…žğ‘– with ğ‘žî…žğ‘–+1, we are back to the case of 𝑛−1 variables with different weights. If ğ‘žî…žğ‘–=1 for some 𝑖, then we have 𝑥1𝑟−1𝑀𝑟𝑛,𝑟−𝑥1𝑠−1𝑀𝑠𝑛,𝑠−𝑟−𝑠𝑥2𝑛𝑥1−𝐴𝑛=𝑥1𝑟−1𝑥𝑟𝑖−𝑥1𝑠−1𝑥𝑠𝑖−𝑟−𝑠𝑥2𝑛𝑥1−𝑥𝑖≤𝑥1𝑟−1𝑥𝑟𝑖−𝑥1𝑠−1𝑥𝑠𝑖−𝑟−𝑠𝑥2𝑖𝑥1−𝑥𝑖=1𝑥𝑖𝐷𝑟,𝑠𝑥1𝑥𝑖;1,0≤0,(2.6) by Lemma 2.1. If ğ‘žî…žğ‘–=0 for some 1<𝑖<𝑛, we are back to the case of 𝑛−1 variables. If ğ‘žî…žğ‘›=0, then we may assume that ğ‘žî…žğ‘›âˆ’1≠0 and note that we have 𝑀𝑛,𝑟=𝑀𝑛−1,𝑟,𝑀𝑛,𝑠=𝑀𝑛−1,𝑠,𝐴𝑛=𝐴𝑛−1 and that 𝑥1𝑟−1𝑀𝑟𝑛,𝑟−𝑥1𝑠−1𝑀𝑠𝑛,𝑠−𝑟−𝑠𝑥2𝑛𝑥1−𝐴𝑛≤𝑥1𝑟−1𝑀𝑟𝑛−1,𝑟−𝑥1𝑠−1𝑀𝑠𝑛−1,𝑠−𝑟−𝑠𝑥2𝑛−1𝑥1−𝐴𝑛−1,(2.7) and we are again back to the case 𝑛−1. If ğ‘žî…ž1=0, then similarly we may assume that ğ‘žî…ž2≠0 and if we can show that (again with 𝑀𝑛,𝑟=𝑀𝑛−1,𝑟,𝑀𝑛,𝑠=𝑀𝑛−1,𝑠 and 𝐴𝑛=𝐴𝑛−1 here) 𝑥1𝑟−1𝑀𝑟𝑛,𝑟−𝑥1𝑠−1𝑀𝑠𝑛,𝑠−𝑟−𝑠𝑥2𝑛𝑥1−𝐴𝑛≤𝑥2𝑟−1𝑀𝑟𝑛−1,𝑟−𝑥2𝑠−1𝑀𝑠𝑛−1,𝑠−𝑟−𝑠𝑥2𝑛𝑥2−𝐴𝑛−1,(2.8) then we are back to the case of 𝑛−1 variables. Note that the above inequality will follow if the function 𝑥𝑥⟼𝑟−1𝑀𝑟𝑛,𝑟−𝑥𝑠−1𝑀𝑠𝑛,𝑠−𝑟−𝑠𝑥2𝑛𝑥(2.9) is an increasing function for 0<𝑥≤𝑀𝑛,𝑟 (in fact, one only needs this for 0<𝑥≤𝑥2) and its derivative is (𝑟−1)𝑥𝑟−2𝑀𝑟𝑛,𝑟+(1−𝑠)𝑥𝑠−2𝑀𝑠𝑛,𝑠−𝑟−𝑠𝑥2𝑛≥(𝑟−1)𝑥𝑟−2𝑥𝑟𝑛+(1−𝑠)𝑥𝑠−2𝑥𝑠𝑛−𝑟−𝑠𝑥2𝑛∶=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,𝑥𝑛). One then deduces from ℎ(𝑥𝑛)=0 and lim𝑥→0+ℎ(𝑥)=+∞ that ℎ(𝑥)≥0 for 0<𝑥≤𝑥𝑛.
So from now on it remains to consider the case ğ‘žî…žğ‘–â‰ 0,1,ğ‘¥î…žğ‘–â‰ ğ‘¥î…žğ‘— for 1≤𝑖,𝑗≤𝑛,𝑖≠𝑗 and this implies that (ğªî…ž,ğ±î…ž) is an interior point of 𝑅𝑛×𝑆𝑛−2. We will now show that this cannot happen.
We define 𝑥𝑝(𝑥)=−1𝑟−1𝑥𝑟𝑀2𝑟𝑛,𝑟+𝑥1𝑠−1𝑥𝑠𝑀2𝑠𝑛,𝑠+(𝑟−𝑠)𝑥𝑥2𝑛−𝜆.(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 (ğªî…ž,ğ±î…ž) is an interior point of 𝑅𝑛×𝑆𝑛−2, we may use the Lagrange multiplier method to obtain a real number 𝜆 so that at (ğªî…ž,ğ±î…ž), ğœ•ğ‘“ğ‘›ğœ•ğ‘žğ‘–ğœ•=ğœ†ğœ•ğ‘žğ‘–îƒ©ğ‘›î“ğ‘–=1ğ‘žğ‘–îƒª,1−1ğ‘žğ‘—ğœ•ğ‘“ğ‘›ğœ•ğ‘¥ğ‘—=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 𝑀𝑛,𝑟,𝑀𝑛,𝑠 take their values at (ğªî…ž,ğ±î…ž)) and each 𝑥′𝑖 (2≤𝑖≤𝑛−1) is a root of ğ‘î…ž(𝑥)=0. Now 𝑛≥3 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<ğ‘Ž<𝑥2<𝑏<𝑥𝑛 such that ğ‘î…ž(ğ‘Ž)=ğ‘î…ž(𝑥2)=ğ‘î…ž(𝑏)=0. However, it is easy to see that ğ‘î…ž(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 𝑔𝑛(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 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×𝑆𝑛−2, where 𝑅𝑛 and 𝑆𝑛−2 are defined as above.
Let (ğªî…ž,ğ±î…ž) be a point of 𝑅𝑛×𝑆𝑛−2 in which the absolute minimum of 𝑔𝑛 is reached. Note that ğœŽğ‘›=𝑀2𝑛,2−𝐴2𝑛; thus if ğ‘¥î…žğ‘–=ğ‘¥î…žğ‘–+1 for some 1≤𝑖≤𝑛−1, by combining ğ‘¥î…žğ‘– with ğ‘¥î…žğ‘–+1 and ğ‘žî…žğ‘– with ğ‘žî…žğ‘–+1, we are back to the case of 𝑛−1 variables with different weights. Similarly, if ğ‘žî…žğ‘–=1 for some 𝑖, then we are back to the case 𝑛=1. If ğ‘žî…žğ‘–=0 for some 1≤𝑖<𝑛, we are back to the case of 𝑛−1 variables. If ğ‘žî…žğ‘›=0, then we may assume that ğ‘žî…žğ‘›âˆ’1≠0 and note that we have 𝑀𝑛,𝑟=𝑀𝑛−1,𝑟,𝑀𝑛,𝑠=𝑀𝑛−1,𝑠,𝑀𝑛,2=𝑀𝑛−1,2,𝐴𝑛=𝐴𝑛−1 and that ln𝑀𝑛,𝑟−ln𝑀𝑛,𝑠−𝑟−𝑠2𝑥2ğ‘›ğœŽğ‘›=ln𝑀𝑛,𝑟−ln𝑀𝑛,𝑠−𝑟−𝑠2𝑥2𝑛𝑀2𝑛,2−𝐴2𝑛≥ln𝑀𝑛−1,𝑟−ln𝑀𝑛−1,𝑠−𝑟−𝑠2𝑥2𝑛−1𝑀2𝑛−1,2−𝐴2𝑛−1=ln𝑀𝑛−1,𝑟−ln𝑀𝑛−1,𝑠−𝑟−𝑠2𝑥2𝑛−1ğœŽğ‘›âˆ’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 ğ‘žî…žğ‘–â‰ 0,1,ğ‘¥î…žğ‘–â‰ ğ‘¥î…žğ‘— for 1≤𝑖,𝑗≤𝑛,𝑖≠𝑗, and this implies that (ğªî…ž,ğ±î…ž) is an interior point of 𝑅𝑛×𝑆𝑛−2. We will now show that this cannot happen.
We define ğ‘Žğ‘¥(𝑥)=𝑟𝑟𝑀𝑟𝑛,𝑟−𝑥𝑠𝑠𝑀𝑠𝑛,𝑠−𝑥(𝑟−𝑠)2−2𝐴𝑛𝑥2𝑥2𝑛−𝜆.(2.14) Here we define 𝑥0/0=ln𝑥. Also note here in the definition of ğ‘Ž(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 (ğªî…ž,ğ±î…ž) is an interior point of 𝑅𝑛×𝑆𝑛−2, we may use the Lagrange multiplier method to obtain a real number 𝜆 so that at (ğªî…ž,ğ±î…ž), ğœ•ğ‘”ğ‘›ğœ•ğ‘žğ‘–ğœ•=ğœ†ğœ•ğ‘žğ‘–îƒ©ğ‘›î“ğ‘–=1ğ‘žğ‘–îƒª,1−1ğ‘žğ‘—ğœ•ğ‘”ğ‘›ğœ•ğ‘¥ğ‘—=0(2.15) for all 1≤𝑖≤𝑛 and 2≤𝑗≤𝑛−1.
By (2.15), a computation shows that each ğ‘¥î…žğ‘– (1≤𝑖≤𝑛) is a root of ğ‘Ž(𝑥)=0 (where 𝑀𝑛,𝑟,𝑀𝑛,𝑠,and𝐴𝑛 take their values at (ğªî…ž,ğ±î…ž)) and each ğ‘¥î…žğ‘– (2≤𝑖≤𝑛−1) is a root of ğ‘Žî…ž(𝑥)=0. Now 𝑛≥3 implies ğ‘Ž(𝑥2)=0. As ğ‘Ž(𝑥1)=ğ‘Ž(𝑥2)=ğ‘Ž(𝑥𝑛)=0, it follows from Rolle’s Theorem that there must be two numbers 𝑥1<𝑐<𝑥2<𝑑<𝑥𝑛 such that ğ‘Žî…ž(𝑐)=ğ‘Žî…ž(𝑥2)=ğ‘Žî…ž(𝑑)=0. However, we have ğ‘Žî…žğ‘¥(𝑥)=𝑟−1𝑀𝑟𝑛,𝑟−𝑥𝑠−1𝑀𝑠𝑛,𝑠−(𝑟−𝑠)𝑥−𝐴𝑛𝑥2𝑛.(2.16) It is easy to see that ğ‘Žî…žî…žî…ž(𝑥)=0 has at most one positive root, which implies that ğ‘Žî…ž(𝑥)=0 has at most three positive roots. As 𝑟≤0, it follows from lim𝑥→0+ğ‘Žî…ž(𝑥)=−∞ and lim𝑥→+âˆžğ‘Žî…ž(𝑥)=−∞ that ğ‘Žî…ž(𝑥)=0 has even numbers of roots so that ğ‘Žî…ž(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: 𝑥2𝑛||ln𝐺𝑛−ln𝑀𝑛,𝑟||≥𝑥𝑛+𝑡2||ln𝐺𝑛,𝑡−ln𝑀𝑛,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 𝑓𝑛(𝐪,𝐱)=𝑥2𝑛ln𝐺𝑛−ln𝑀𝑛,−𝑟−𝑥𝑛+𝑡2ln𝐺𝑛,𝑡−ln𝑀𝑛,−𝑟,𝑡.(3.2) As in the proof of Theorem 2.2, it suffices to show that 𝜕𝑓𝑛/𝜕𝑥1≤0 and calculation shows −1ğ‘ž1𝜕𝑓𝑛𝜕𝑥1=𝑔𝑛(𝐪,𝐱)−𝑔𝑛𝐪,𝐱𝑡,(3.3) where 𝑔𝑛(𝐪,𝐱)=𝑥2𝑛∑𝑛𝑖=1ğ‘žğ‘–î€·ğ‘¥ğ‘Ÿğ‘–âˆ’ğ‘¥ğ‘Ÿ1/𝑥𝑟𝑖𝑥1∑𝑛𝑖=1ğ‘žğ‘–î€·ğ‘¥1/𝑥𝑖𝑟.(3.4) It is easy to check that 𝑥𝑛𝑥𝑖≥𝑥𝑛+𝑡𝑥𝑖,𝑥+𝑡1𝑥𝑖≤𝑥1+𝑡𝑥𝑖.+𝑡(3.5) In view of (3.5), the inequality 𝜕𝑓𝑛/𝜕𝑥1≤0 will follow from 𝑑1𝑥𝑖=𝑥2𝑛𝑥𝑟𝑖−𝑥𝑟1𝑥1𝑥𝑟𝑖−𝑥𝑛+𝑡2𝑥𝑖+𝑡𝑟−𝑥1+𝑡𝑟𝑥1𝑥+𝑡𝑖+𝑡𝑟≥0(3.6) for 𝑥1≤𝑥𝑖≤𝑥𝑛. We may assume that 𝑥𝑛>𝑥1 here and it is easy to see that ğ‘‘î…ž1(𝑥)=0 can have at most one root 𝑥0 in between 𝑥1 and 𝑥𝑛. This combined with the observation that 𝑑1(𝑥1)=0,ğ‘‘î…ž1(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 𝑥𝑛−𝑥1=(𝑥𝑛+𝑡)−(𝑥1+𝑡) and we rewrite 𝑑1(𝑥𝑛) as 𝑑1𝑥𝑛=𝑥2𝑛𝑥𝑟𝑛−𝑥𝑟1𝑥1𝑥𝑟𝑛𝑥𝑛−𝑥1−𝑥𝑛+𝑡2𝑥𝑛+𝑡𝑟−𝑥𝑛+𝑡𝑟𝑥1𝑥+𝑡𝑛+𝑡𝑟𝑥𝑛−𝑥+𝑡1𝑥+𝑡=𝑒𝑛𝑥1𝑥−𝑒𝑛+𝑡𝑥1,+𝑡(3.7) where 𝑥𝑒(𝑥)=𝑟−1𝑥𝑟−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 ğ‘’î…žğ‘¥(𝑥)=𝑟−3𝑥𝑟+1−2𝑥𝑟+(𝑟−1)𝑥−(𝑟−2)𝑥𝑟−2(𝑥−1)2,(3.9) and it is easy to see that the function 𝑥𝑟+1−2𝑥𝑟+(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 𝑢𝑛(𝐪,𝐱)=𝑥2𝑛ln𝑀𝑛,𝑟−ln𝐺𝑛−𝑥𝑛+𝑡2ln𝑀𝑛,𝑟,𝑡−ln𝐺𝑛,𝑡.(3.10) Again it suffices to show that 𝜕𝑢𝑛/𝜕𝑥1≤0 and calculation shows −1ğ‘ž1𝜕𝑢𝑛𝜕𝑥1=𝑣𝑛(𝐪,𝐱)−𝑣𝑛𝐪,𝐱𝑡,(3.11) where 𝑣𝑛(𝐪,𝐱)=𝑥2𝑛∑𝑛𝑖=1ğ‘žğ‘–î€·ğ‘¥ğ‘Ÿğ‘–âˆ’ğ‘¥ğ‘Ÿ1/𝑥𝑟𝑛𝑥1∑𝑛𝑖=1ğ‘žğ‘–î€·ğ‘¥ğ‘–/𝑥𝑛𝑟.(3.12) In view of (3.5), the inequality 𝜕𝑢𝑛/𝜕𝑥1≤0 will follow from 𝑑2𝑥𝑖=𝑥2𝑛𝑥𝑟𝑖−𝑥𝑟1𝑥1𝑥𝑟𝑛−𝑥𝑛+𝑡2𝑥𝑖+𝑡𝑟−𝑥1+𝑡𝑟𝑥1𝑥+𝑡𝑛+𝑡𝑟≥0(3.13) for 𝑥1≤𝑥𝑖≤𝑥𝑛. We may assume that 𝑥𝑛>𝑥1 here and it is easy to see that ğ‘‘î…ž2(𝑥)=0 can have at most one root 𝑥0 in between 𝑥1 and 𝑥𝑛. This combined with the observation that 𝑑2(𝑥1)=0,ğ‘‘î…ž2(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<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||ln𝐺𝑛−ln𝑀𝑛,𝑟||≤𝑥1+𝑡2||ln𝐺𝑛,𝑡−ln𝑀𝑛,𝑟,𝑡||.(3.14) To proceed, we first look at the following related inequalities (with 𝑟>𝑠 here):

ln𝑀𝑛,𝑟−ln𝑀𝑛,𝑠−(𝑟−𝑠)ğœŽğ‘›2𝑥21≤ln𝑀𝑛,𝑟,𝑡−ln𝑀𝑛,𝑠,𝑡−(𝑟−𝑠)ğœŽğ‘›2𝑥1+𝑡2,(3.15)ln𝑀𝑛,𝑟−ln𝑀𝑛,𝑠−(𝑟−𝑠)ğœŽğ‘›2𝑥2𝑛≥ln𝑀𝑛,𝑟,𝑡−ln𝑀𝑛,𝑠,𝑡−(𝑟−𝑠)ğœŽğ‘›2𝑥𝑛+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)/𝜕𝑡≥0. As 𝐱 is arbitrary, we can recast this condition as

𝑀𝑟−1𝑛,𝑟−1𝑀𝑟𝑛,𝑟−𝑀𝑠−1𝑛,𝑠−1𝑀𝑠𝑛,𝑠+(𝑟−𝑠)ğœŽğ‘›ğ‘¥31≥0.(3.17) Similarly, (3.16) holds if and only if the following inequality holds:

𝑀𝑟−1𝑛,𝑟−1𝑀𝑟𝑛,𝑟−𝑀𝑠−1𝑛,𝑠−1𝑀𝑠𝑛,𝑠+(𝑟−𝑠)ğœŽğ‘›ğ‘¥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

ğ‘ž1+ğ‘ž2𝑡𝑟−1ğ‘ž1+ğ‘ž2ğ‘¡ğ‘Ÿâˆ’ğ‘ž1+ğ‘ž2𝑡𝑠−1ğ‘ž1+ğ‘ž2𝑡𝑠+(𝑟−𝑠)ğ‘ž1ğ‘ž2(𝑡−1)2=ğ‘ž1ğ‘ž2(1−𝑡)î€·ğ‘ž1+ğ‘ž2ğ‘¡ğ‘Ÿğ‘žî€¸î€·1+ğ‘ž2𝑡𝑠𝐷𝑟,𝑠𝑡;ğ‘ž1,ğ‘ž2(3.19) with 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘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 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 for (𝑡,ğ‘ž1,ğ‘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 𝐷𝑟,𝑠(𝑡;ğ‘ž1,ğ‘ž2)≤0 for (𝑡,ğ‘ž1,ğ‘ž2)∈𝐼1×𝐸.

It follows from the proof of Lemma 2.1 that 𝐷𝑟,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,

𝑀𝑟−1𝑛,𝑟−1𝑀𝑟𝑛,𝑟−1𝐻𝑛+(𝑟−𝑠)ğœŽğ‘›ğ‘¥31<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ln𝑀𝑛,𝑟−ln𝐺𝑛𝑥1+𝑀𝑟−1𝑛,𝑟−1𝑀𝑟𝑛,𝑟−1𝐻𝑛≥0.(3.21) This combined with (3.20) now implies that for 𝑟>2,

ln𝑀𝑛,𝑟−ln𝐺𝑛>(𝑟−𝑠)ğœŽğ‘›2𝑥21.(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,

||ln𝑀𝑛,𝑟−ln𝐺𝑛||≥|𝑟|2𝑥2ğ‘›ğœŽğ‘›.(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)=ğ‘“î…ž(1)=ğ‘“î…žî…ž(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)=ğ‘ž1(𝑟−𝑠)𝑟+𝑠−3−3(𝑟+𝑠−1)ğ‘ž1+2(𝑟+𝑠)ğ‘ž21.(3.24) On taking ğ‘ž1→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 𝑥𝑖→1/𝑥2−𝑖+1. Hence one needs 𝑟≥−3 in order for (3.23) to hold for any negative 𝑟.