We extend the results in parts I–III on certain inequalities involving the weighted power means as well as the symmetric means. These inequalities can be largely viewed as concerning the bounds for ratios of differences of means and can be traced back to the work of Diananda.

1. Introduction

Let 𝑀𝑛,𝑟(𝐱;𝐪) be the weighted power means: 𝑀𝑛,𝑟(𝐱;𝐪)=(𝑛𝑖=1𝑞𝑖𝑥𝑟𝑖)1/𝑟, where 𝑀𝑛,0(𝐱;𝐪) denotes the limit of 𝑀𝑛,𝑟(𝐱;𝐪) as 𝑟0+, 𝐱=(𝑥1,,𝑥𝑛), 𝐪=(𝑞1,,𝑞𝑛), and 𝑞𝑖>0 (1𝑖𝑛) are positive real numbers with 𝑛𝑖=1𝑞𝑖=1. In this paper, we let 𝑞=min𝑞𝑖 and unless otherwise specified, we assume 0𝑥1<𝑥2<<𝑥𝑛.

For 𝑘{0,1,,𝑛}, the 𝑘th symmetric function 𝐸𝑛,𝑘 of 𝐱 and its mean 𝑃𝑛,𝑘 are defined by 𝐸𝑛,𝑟(𝐱)=1𝑖1<<𝑖𝑟𝑟𝑛𝑗=1𝑥𝑖𝑗,𝑃𝑟𝑛,𝑟𝐸(𝐱)=𝑛,𝑟(𝐱)(𝑛𝑟),1𝑟𝑛;𝐸𝑛,0=𝑃𝑛,0=1.(1.1)

We define 𝐴𝑛(𝐱;𝐪)=𝑀𝑛,1(𝐱;𝐪), 𝐺𝑛(𝐱)=𝑀𝑛,0(𝐱;𝐪), and 𝐻𝑛(𝐱;𝐪)=𝑀𝑛,1(𝐱;𝐪) and we shall write 𝑀𝑛,𝑟 for 𝑀𝑛,𝑟(𝐱;𝐪) and similarly for other means when there is no risk of confusion.

For a real number 𝛼 and mutually distinct numbers 𝑟, 𝑠, 𝑡, we define Δ𝑟,𝑠,𝑡,𝛼=||||𝑀𝛼𝑛,𝑟𝑀𝛼𝑛,𝑡𝑀𝛼𝑛,𝑟𝑀𝛼𝑛,𝑠||||,(1.2) where we interpret 𝑀0𝑛,𝑟𝑀0𝑛,𝑠 as ln𝑀𝑛,𝑟ln𝑀𝑛,𝑠. We also define Δ𝑟,𝑠,𝑡 to be Δ𝑟,𝑠,𝑡,1.

For 𝑟>𝑠>𝑡0, 𝛼>0, the author studied in [13] inequalities of the following two types: 𝐶𝑟,𝑠,𝑡((1𝑞)𝛼)Δ𝑟,𝑠,𝑡,𝛼,Δ(1.3)𝑟,𝑠,𝑡,𝛼𝐶𝑟,𝑠,𝑡(𝑞𝛼),(1.4) where for 0<𝑥<1, 𝐶𝑟,𝑠,𝑡(𝑥)=1𝑥1/𝑡1/𝑟1𝑥1/𝑠1/𝑟,𝑡>0;𝐶𝑟,𝑠,01(𝑥)=1𝑥1/𝑠1/𝑟.(1.5)

For any set {𝑎,𝑏,𝑐} with 𝑎,𝑏,𝑐 mutually distinct and nonnegative, we let 𝑟=max{𝑎,𝑏,𝑐}, 𝑡=min{𝑎,𝑏,𝑐}, 𝑠={𝑎,𝑏,𝑐}{𝑟,𝑡}. By saying that (1.3) (resp., (1.4)) holds for the set {𝑎,𝑏,𝑐}, 𝛼>0, we mean that (1.3) (resp., (1.4)) holds for 𝑟>𝑠>𝑡0, 𝛼>0.

A result of Diananda [4, 5] (see also [6, 7]) shows that (1.3) and (1.4) hold for {1,1/2,0}, 𝛼=1. When 𝛼=1, the sets {𝑎,𝑏,1}'s for which (1.3) or (1.4) holds have been completely determined by the author in [13]. Moreover, it is shown in [3] that (1.4) does not hold in general unless 0{𝑎,𝑏,𝑐}.

For general 𝛼's, the cases 𝑟>𝑠>𝑡0 for which inequality (1.3) or (1.4) holds are open (with 𝑡=0 in (1.4)). It is our main goal in this paper to study these cases. We first restrict our attention to the case {𝑎,𝑏,0}. This is partially because of the following result in [2] (note that there is a typo in the original statement though).

Theorem 1.1 (see [2, Theorem  3.2]). Let 𝑟>𝑠>0. If (1.3) holds for {𝑟,𝑠,0}, 𝛼>0, then it also holds for {𝑟,𝑠,0}, 𝑘𝛼 with 0<𝑘<1. If (1.4) holds for {𝑟,𝑠,0}, 𝛼>0, then it also holds for {𝑟,𝑠,0}, 𝑘𝛼 with 𝑘>1.

Moreover, for the unweighted case 𝑞1=𝑞2==𝑞𝑛=1/𝑛, the author [3, Theorem  3.5] has shown that (1.3) holds for {1,1/𝑟,0} with 𝛼=𝑛/𝑟 when 𝑟2 and (1.4) holds for {1,1/𝑟,0} with 𝛼=𝑛/((𝑛1)𝑟) when 1<𝑟2. The values of 𝛼's are best possible here; namely, no larger 𝛼's can make (1.3) hold for {1,1/𝑟,0} and similarly no smaller 𝛼's can make (1.4) hold for {1,1/𝑟,0}.

More generally, for arbitrary weights {𝑞𝑖}'s, by using similar arguments as in [3], one sees that the largest 𝛼 that can make (1.3) hold for {1,1/𝑟,0} is 1/(𝑞𝑟) and the smallest 𝛼 that can make (1.4) hold for {1,1/𝑟,0} is 1/((1𝑞)𝑟).

In Section 2, we will extend Theorem  3.5 of [3] to the case of arbitrary weights. Namely, we will prove the following.

Theorem 1.2. For 𝑟2, 0<𝑝1/𝑞, 𝐴𝑝𝑛(1𝑞)(𝑟1)𝑝/𝑟𝑀𝑝𝑛,𝑟+1(1𝑞)(𝑟1)𝑝/𝑟𝐺𝑝𝑛.(1.6) For 1<𝑟2, 𝑝1/(1𝑞), 𝐴𝑝𝑛𝑞(𝑟1)𝑝/𝑟𝑀𝑝𝑛,𝑟+1𝑞(𝑟1)𝑝/𝑟𝐺𝑝𝑛.(1.7)

We note here that by a change of variables 𝑥𝑖𝑥𝑖1/𝑟, one can easily rewrite (1.6) (resp., (1.7)) in the form of (1.3) (resp., (1.4)).

After studying (1.3) and (1.4) for the set {𝑎,𝑏,0}, we move on in Section 3 to the case {𝑎,𝑏,𝑐} with min(𝑎,𝑏,𝑐)>0. Our remark earlier allows us to focus on (1.3) only. In this case, we can recast (1.3), via a change of variables, as the following: 𝐴𝛼𝑛𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑛,𝑟+1𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑛,𝑠,(1.8) where 𝛼>0, 𝑟>1>𝑠>0 and 𝜆𝑟,𝑠1(𝑥)=1𝐶𝑟,1,𝑠.(𝑥)(1.9) We will show that (1.8) holds for all 𝑛 if and only if it holds for the case 𝑛=2. Based on this, we will then be able to prove (1.8) for certain 𝑟, 𝑠, 𝛼's satisfying a natural condition.

One certainly expects that analogues of (1.3) and (1.4) hold with weighted power means replaced by the symmetric means; one such example is given by the following result in [8].

Theorem 1.3. Let 𝑞𝑖=1/𝑛, then for any integer 2𝑘𝑛, 𝑛𝑖=1𝑥𝑖𝑘𝑛𝑘̃𝜆2,𝑘𝑛𝑘𝑀(𝑛)𝑘𝑛,2+̃𝜆2,𝑘(𝑛)𝐸𝑛,𝑘,(1.10) where for 2𝑟𝑘𝑛 (with (𝑛𝑛1)=0 here), ̃𝜆𝑟,𝑘𝑛(𝑛)=𝑘(11/𝑛)𝑘/𝑟(𝑛1)𝑘(𝑛𝑘)(11/𝑛)𝑘/𝑟𝑘𝑛1.(1.11)

As was pointed out in [3], the proof given in [8] for the above theorem is not quite correct. In Section 4, we will study inequalities involving the symmetric means and our results include a proof of Theorem 1.3.

2. Proof of Theorem 1.2

In view of Theorem 1.1, one only needs to prove (1.6) for 𝑝=1/𝑞 and similarly (1.7) for 𝑝=1/(1𝑞). In this proof we assume that 0<𝑥1𝑥𝑛. The case 𝑥1=0 will follow by taking the limit. We first prove (1.6) and we define 𝐴𝑓(𝐱)=𝑛1/𝑞(1𝑞)(𝑟1)/𝑞𝑟𝑀1/𝑞𝑛,𝑟𝐺𝑛1/𝑞.(2.1) If 𝑥1==𝑥𝑛, then 𝑓=0; otherwise we may assume 𝑛2 and 0<𝑥=𝑥1==𝑥𝑘<𝑥𝑘+1 for some 1𝑘<𝑛, then 𝜕𝑓=𝜕𝑥𝑘𝑖=1𝜕𝑓𝜕𝑥𝑖.(2.2) We want to show that the right-hand side above is nonnegative. It suffices to show that each single term in the sum is nonnegative. Without loss of generality, we now show that 𝜕𝑓𝜕𝑥10.(2.3) We have 𝑞𝑥1𝐺𝑛1/𝑞𝑞1𝜕𝑓𝜕𝑥1=𝐴𝑛1/𝑞1𝑥1𝐴𝑛(1𝑞)(𝑟1)/𝑞𝑟𝑀1/𝑞𝑟𝑛,𝑟𝑥𝑟1𝑀𝑟𝑛,𝑟.(2.4) Now we set 𝑦(𝑟)=(𝑞1𝑞)𝑥𝑟1+𝑛𝑖=2𝑞𝑖𝑥𝑟𝑖1𝑞1/𝑟,(2.5) so that 𝐴𝑛1/𝑞1𝑥1𝐴𝑛=(1𝑞)𝑞𝑥1+(1𝑞)𝑦(1)1/𝑞1𝑥1𝑦(1)(1𝑞)𝑞𝑥1+(1𝑞)𝑦(𝑟)1/𝑞1𝑥1.𝑦(𝑟)(2.6) Hence 𝑞𝐺𝑛1/𝑞𝑞1(1𝑞)1/𝑞𝑥11/𝑞1𝜕𝑓𝜕𝑥1𝑞+1𝑞𝑦(𝑟)𝑥11/𝑞11𝑦(𝑟)𝑥1𝑞+𝑦1𝑞𝑟(𝑟)𝑥𝑟11/𝑞𝑟1𝑦1𝑟(𝑟)𝑥𝑟1.(2.7) We want to show that the right-hand side expression above is nonnegative, and by setting 𝑧=𝑦(𝑟)/𝑥1, this is equivalent to show that 𝑔(𝑧,𝑞)=(𝑞/(1𝑞)+𝑧)1/𝑞1(𝑧1)(𝑞/(1𝑞)+𝑧𝑟)1/𝑞𝑟1(𝑧𝑟1)1(2.8) for 𝑧1, 0𝑞1/2 and calculation yields that (𝑞/(1𝑞)+𝑧𝑟)1/𝑞𝑟1(𝑧𝑟1)2(𝑞/(1𝑞)+𝑧)1/𝑞2(𝑞/(1𝑞)+𝑧𝑟)1/𝑞𝑟2𝜕𝑔=1𝜕𝑧𝑞𝑞𝑧1+1𝑞(𝑧𝑟𝑧1)𝑟+𝑞𝑞1𝑞𝑧+𝑧1𝑞𝑟𝑧𝑟1𝑧𝑟+𝑟𝑞.1𝑞1(2.9) We now set 𝑠=𝑞/(1𝑞) with 0𝑠1 and we consider 𝑎(𝑧,𝑠)=(𝑧1+𝑠)(𝑧𝑟1)(𝑧𝑟𝑧+𝑠)(𝑧+𝑠)𝑟𝑧𝑟1(𝑧𝑟+𝑟𝑠1).(2.10) By Cauchy's mean value theorem, 𝜕2𝑎𝜕𝑠2𝑧=2𝑟1𝑟𝑧𝑟1(𝑧1)0.(2.11) It follows that for 0𝑠1, 𝑎(𝑧,𝑠)min{𝑎(𝑧,0),𝑎(𝑧,1)}=min{0,𝑎(𝑧,1)}.(2.12) It follows from the discussion in [3] (see the function 𝑎(𝑦,1) defined in the proof of Theorem 3.5 there and note that we have 𝑟2 here) that 𝑎(𝑧,1)0. This implies that 𝑎(𝑧,𝑠)0 so that 𝑔(𝑧,𝑞) is an increasing function of 𝑧 and we then deduce that 𝑔(𝑧,𝑞)lim𝑧+𝑔(𝑧,𝑞)=1.(2.13) This shows that 𝜕𝑓/𝜕𝑥10 and hence 𝜕𝑓/𝜕𝑥0 and by letting 𝑥𝑥𝑘+1 and repeating the above argument, we conclude that 𝑓(𝐱)𝑓(𝑥𝑛,𝑥𝑛,,𝑥𝑛)=1(1𝑞)(𝑟1)/𝑞𝑟 which completes the proof for (1.6).

Now, to prove (1.7), we consider 𝐴(𝐱)=𝑛1/(1𝑞)𝑞(𝑟1)/(1𝑞)𝑟𝑀1/(1𝑞)𝑛,𝑟𝐺𝑛1/(1𝑞).(2.14) Similar to our discussion above, it suffices to show 𝜕/𝜕𝑥𝑛0. Now 𝑥𝑛𝐺𝑛1/(1𝑞)(1𝑞)𝑞𝑛𝜕𝜕𝑥𝑛=𝐴𝑛1/(1𝑞)1𝑥𝑛𝐴𝑛𝑞(𝑟1)/(1𝑞)𝑟𝑀1/(1𝑞)𝑟𝑛,𝑟𝑥𝑟𝑛𝑀𝑟𝑛,𝑟.(2.15) Now we set 𝑤(𝑟)=(𝑞𝑛𝑞)𝑥𝑟𝑛+𝑛1𝑖=1𝑞𝑖𝑥𝑟𝑖1𝑞1/𝑟,(2.16) so that 𝐴𝑛1/(1𝑞)1𝑥𝑛𝐴𝑛=(1𝑞)𝑞𝑥𝑛+(1𝑞)𝑤(1)1/(1𝑞)1𝑥𝑛𝑤(1)(1𝑞)𝑞𝑥𝑛+(1𝑞)𝑤(𝑟)1/(1𝑞)1𝑥𝑛,𝑤(𝑟)(2.17) where the inequality follows from the observation that the function 𝑧𝑞𝑥𝑛+(1𝑞)𝑧1/(1𝑞)1𝑥𝑛𝑧(2.18) is decreasing for 0<𝑧<𝑥𝑛.

We then deduce that 𝑥𝑛𝐺𝑛1/(1𝑞)𝑞𝑛𝑞1/(1𝑞)1𝑤1/(1𝑞)(𝑟)𝜕𝜕𝑥𝑛𝑥𝑛+𝑤(𝑟)1𝑞𝑞1/(1𝑞)1𝑥𝑛𝑥𝑤(𝑟)1𝑟𝑛𝑤𝑟+(𝑟)1𝑞𝑞1/(1𝑞)𝑟1𝑥𝑟𝑛𝑤𝑟.(𝑟)1(2.19) By proceeding similarly as in the proof of (1.6) above, one is then able to establish (1.7) and we shall omit all the details here.

3. A General Discussion on (1.8)

Theorem 3.1. For fixed 𝛼>0, 𝑟>1>𝑠>0, (1.8) holds for all 𝑛 if and only if it holds for 𝑛=2.

Proof. We consider the function 𝑓𝑛(𝐱;𝐪,𝑞)=𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑛,𝑟+1𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑛,𝑠𝐴𝛼𝑛.(3.1) The theorem asserts that in order to show 𝑓𝑛0, it suffices to check the case 𝑛=2. To see this, we may assume by homogeneity that 0𝑥1𝑥2𝑥𝑛1𝑥𝑛=1 and we let 𝐚=(𝑎1,,𝑎𝑛)[0,1]𝑛1 be the point in which the absolute minimum of 𝑓𝑛 is reached. We may assume that 0𝑎1<𝑎2<<𝑎𝑛1<𝑎𝑛=1 for otherwise if 𝑎𝑖=𝑎𝑖+1 for some 1𝑖𝑛1, by combining 𝑎𝑖 with 𝑎𝑖+1 and 𝑞𝑖 with 𝑞𝑖+1, and noticing that 𝜆𝑟,𝑠(𝑥) is an increasing function of 𝑥 by [2, Lemma  2.1], we have 𝑓𝑛(𝐚;𝐪,𝑞)𝑓𝑛1𝐚;𝐪,𝑞,(3.2) where 𝐚=(𝑎1,,𝑎𝑖1,𝑎𝑖+𝑎𝑖+1,𝑎𝑖+2,,𝑎𝑛), 𝐪=(𝑞1,,𝑞𝑖1,𝑞𝑖+𝑞𝑖+1,𝑞𝑖+2,,𝑞𝑛), and 𝑞=min(𝑞1,,𝑞𝑖1,𝑞𝑖+𝑞𝑖+1,𝑞𝑖+2,,𝑞𝑛). We can then reduce the determination of the absolute minimum of 𝑓𝑛 to that of 𝑓𝑛1 with different weights.
If 𝐚 is a boundary point of [0,1]𝑛1, then 𝑎1=0, and in this case we show that 𝑓𝑛(𝐚;𝐪𝑛,𝑞)0 follows from 𝑓𝑛1(𝐚;𝐪,𝑞)0, where 𝐚=(𝑎2,,𝑎𝑛) and 𝐪=(𝑞2,,𝑞𝑛)/(1𝑞1). On writing 𝑓𝑛1(𝐚;𝐪,𝑞)0 explicitly, we get 𝑛𝑖=2𝑞𝑖𝑎𝑖𝛼𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼𝑀𝛼𝑛1,𝑟𝐚;𝐪+1𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼𝑀𝛼𝑛1,𝑠𝐚;𝐪.(3.3) Meanwhile, 𝑓𝑛(𝐚;𝐪𝑛,𝑞)0 is equivalent to 𝑛𝑖=2𝑞𝑖𝑎𝑖𝛼𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼/𝑟𝑀𝛼𝑛1,𝑟𝐚;𝐪+1𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼/𝑠𝑀𝛼𝑛1,𝑠𝐚;𝐪.(3.4) Thus, it amounts to show that 𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼𝑀𝛼𝑛1,𝑟𝐚;𝐪+1𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼𝑀𝛼𝑛1,𝑠𝐚;𝐪𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼/𝑟𝑀𝛼𝑛1,𝑟𝐚;𝐪+1𝜆𝑟,𝑠((1𝑞)𝛼)1𝑞1𝛼/𝑠𝑀𝛼𝑛1,𝑠𝐚;𝐪,(3.5) which is equivalent to 1𝜆𝑟,𝑠((1𝑞)𝛼)𝜆𝑟,𝑠1𝑞1𝛼1𝜆𝑟,𝑠1𝑞1𝛼𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑛1,𝑠𝑀𝛼𝑛1,𝑟.(3.6) Now the above inequality follows from 𝑀𝑛1,𝑠𝑀𝑛1,𝑟 and 1𝜆𝑟,𝑠((1𝑞)𝛼)𝜆𝑟,𝑠1𝑞1𝛼1𝜆𝑟,𝑠1𝑞1𝛼𝜆𝑟,𝑠((1𝑞)𝛼),(3.7) since 𝜆𝑟,𝑠(𝑥) is an increasing function of 𝑥.
Thus 𝑓𝑛(𝐚;𝐪𝑛,𝑞)0 follows from 𝑓𝑛1(𝐚;𝐪,𝑞)0. Moreover, on writing 𝑞=min(𝑞2/(1𝑞1),,𝑞𝑛/(1𝑞1)) and noticing that 𝑞>𝑞, we deduce that 𝑓𝑛1(𝐚;𝐪,𝑞)𝑓𝑛1(𝐚;𝐪,𝑞). Hence the determination of 𝑓𝑛0 can be reduced to the determination of 𝑓𝑛10 with different weights.
It remains to show the case 𝑎1>0, so that 𝐚 is an interior point of [0,1]𝑛1. In this case we have 𝑓𝑛(𝐚;𝐪,𝑞)=0.(3.8) Thus 𝑎1,,𝑎𝑛1 solve the equation 𝑔(𝑥)=𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑟𝑛,𝑟𝑥𝑟1+1𝜆𝑟,𝑠((1𝑞)𝛼)𝑀𝛼𝑠𝑛,𝑠𝑥𝑠1𝐴𝑛𝛼1=0.(3.9) Note that 𝑓𝑛(𝐚;𝐪,𝑞)=𝑛𝑖=1𝑞𝑖𝑎𝑖𝑔𝑎𝑖=𝑞𝑛𝑎𝑛𝑔𝑎𝑛.(3.10) Thus if 𝑔(𝑎𝑛)0, then 𝑓𝑛0. If 𝑔(𝑎𝑛)<0, we note that 𝑔(𝑥)=0 can have at most two roots in (0,𝑎𝑛] since it is easy to see that 𝑔(𝑥)=0 can have at most one positive root. As lim𝑥0+𝑔(𝑥)=+, this implies that 𝑔(𝑥)=0 has only one root in (0,𝑎𝑛]. As 𝑎1,,𝑎𝑛1 are the distinct roots of 𝑔(𝑥)=0, this implies that we must have 𝑛1=1 so that we only need to show 𝑓𝑛0 for the case 𝑛=2 and this completes the proof.

In what follows, we will apply Theorem 3.1 to establish (1.8) for certain 𝑟,𝑠,𝛼's. Before we proceed, we note that there is a natural condition to be satisfied by 𝑟, 𝑠, 𝛼 in order for (1.8) to hold, namely, if we take 𝑛=2 and rewrite it as 𝑀𝛼2,𝑟𝑀𝛼2,𝑠𝑀𝛼2,𝑟𝐴𝛼211𝜆𝑟,𝑠((1𝑞)𝛼).(3.11) On taking 𝑥1𝑥2, we conclude that 𝑟𝑠1𝑟11𝜆𝑟,𝑠((1𝑞)𝛼).(3.12) Before we prove our next result, we need two lemmas.

Lemma 3.2. Fixing 𝑢<0, the function 𝑓(𝑡)=1𝑡1𝑡𝑢,0<𝑡1(3.13) is concave for 𝑢<1 and is convex for 1<𝑢<0.

Proof. Calculation yields that 𝑓(𝑡)=𝑢𝑡𝑢2(1𝑡𝑢)3𝑔(𝑡),(3.14) where 𝑔(𝑡)=(𝑢1)𝑡𝑢+1+(1+𝑢)𝑡𝑢(1+𝑢)𝑡+𝑢1.(3.15) Now 𝑔(𝑡)=𝑢(𝑢+1)(𝑢1)𝑡𝑢2(1𝑡).(3.16) Thus if 1<𝑢<0, then 𝑔(𝑡)>0 for 0<𝑡<1 and 𝑔(𝑡)<0 for 𝑡>1. Since 𝑔(1)=0, this implies that 𝑔(𝑡)<0 for 0<𝑡1. As 𝑔(1)=0, we then conclude that 𝑔(𝑡)>0 for 0<𝑡<1 and 𝑔(𝑡)<0 for 𝑡>1. It follows from this that 𝑓(𝑡) is convex for 1<𝑢<0 and the other assertion can be shown similarly, which completes the proof.

Lemma 3.3. For 𝑟>1, 0<𝑞1, 𝑞2<1, 𝑞1+𝑞2=1, the function 𝑞𝑓(𝑡)=1𝑡1/(𝑟1)+𝑞2𝛼1𝑞1𝑡𝑟/(𝑟1)+𝑞21𝛼/𝑟(3.17) is convex for 𝑡>0 when 𝑟2, 1>𝛼>0, or 𝛼=1, 𝑟>1.

Proof. Direct calculation shows that 𝑓𝑞(𝑡)=1𝑞𝑟11𝑡1/(𝑟1)+𝑞2𝛼3𝑞1𝑡𝑟/(𝑟1)+𝑞2𝛼/𝑟1𝑡1/(𝑟1)2𝑔(𝑡),(3.18) where 𝑔(𝑡)=(1𝛼)(𝑟2)𝑞𝑟121𝑞2𝑡2+2/(𝑟1)+(1𝛼)(𝑟𝛼)𝑞𝑟11𝑞22𝑡2+1/(𝑟1)+𝑟𝛼𝑞𝑟121𝑞2𝑡1+2/(𝑟1)(+211𝛼)2𝑞𝑟11𝑞22𝑡1+1/(𝑟1)+𝑟𝛼𝑞𝑟132(𝑡+1𝛼)(𝑟𝛼)𝑞𝑟11𝑞22𝑡1/(𝑟1)+(1𝛼)(𝑟2)𝑞𝑟132.(3.19) One then easily deduce the assertion of the lemma from the above expression of 𝑔(𝑡) and this completes the proof.

Theorem 3.4. Inequality (1.8) holds for the cases 𝑟2, 1>𝛼>0 or 𝛼=1, 𝑟>1, 𝑟+𝑠2, provided that (3.12) holds with strict inequality.

Proof. By Theorem 3.1, it suffices to prove the theorem for the case 𝑛=2. We write 𝜆=𝜆𝑟,𝑠((1𝑞)𝛼) for short in this proof. What we need to prove is the following: 𝑞1𝑥1+𝑞2𝑥2𝛼𝑞𝜆1𝑥𝑟1+𝑞2𝑥𝑟2𝛼/𝑟𝑞+(1𝜆)1𝑥𝑠1+𝑞2𝑥𝑠2𝛼/𝑠.(3.20) Without loss of generality, we may assume that 𝑞1=𝑞, 𝑞2=1𝑞 and define 𝑞𝑓(𝑡)=1𝑡𝑠+𝑞2𝛼/𝑠𝑞1𝑡+𝑞2𝛼𝑞𝜆1𝑡𝑟+𝑞2𝛼/𝑟,(3.21) so that we need to show that 𝑓(𝑡)𝑓(1) for 𝑡0. We have 𝑓(𝑡)=𝛼𝑞1𝑞2𝑞1𝑡𝑠+𝑞2𝛼/𝑠1𝑞1𝑡+𝑞2𝛼11𝑡𝑠1𝑞+𝜆1𝑡𝑟+𝑞2𝛼/𝑟1𝑡𝑠1𝑡𝑟1=𝛼𝑞1𝑞2𝑞1𝑡𝑠+𝑞2𝛼/𝑠1𝑞1𝑡𝑟+𝑞2𝛼/𝑟11𝑡𝑠1𝑔𝑡𝑟1,(3.22) where 𝑞𝑔(𝑡)=1𝑡1/(𝑟1)+𝑞2𝛼1𝑞1𝑡𝑟/(𝑟1)+𝑞21𝛼/𝑟+𝜆1𝑡1𝑡(𝑠1)/(𝑟1)𝜆.(3.23) By Lemmas 3.2 and 3.3, we see that 𝑔(𝑡) is a strictly convex function for 𝑟2, 1>𝛼>0 or 𝛼=1, 𝑟>1, 𝑟+𝑠2. Note that by our assumption (3.12) with strict inequality, lim𝑡1𝑔(𝑡)=1+𝜆𝑟𝑠𝑠1<0.(3.24) On the other hand, note that 𝜆 satisfies (1𝑞)𝛼=𝜆(1𝑞)𝛼/𝑟+(1𝜆)(1𝑞)𝛼/𝑠,(3.25) so that lim𝑡0+𝑔(𝑡)=(1𝑞)𝛼𝛼/𝑟𝜆=(1𝜆)(1𝑞)𝛼/𝑠𝛼/𝑟>0.(3.26) As 𝑔(𝑡) is strictly convex, this implies that there are exactly two roots 𝑡1, 𝑡2 of 𝑔(𝑡)=0 satisfying that 𝑡1(0,1) and 𝑡2>1. Note further that 𝑓(0)=𝑓(1) and lim𝑡0+𝑓(𝑡)<0, which implies that 𝑓(𝑡)𝑓(1) for 0𝑡1. Similarly, we note that 𝑓(𝑡)<0 for 𝑡(1,1+𝜖) with 𝜖>0 being small enough. This combined with the observation that lim𝑡+𝑓(𝑡)𝑓(1) as 𝜆𝑟,𝑠(𝑥) is an increasing function of 𝑥 implies that 𝑓(𝑡)𝑓(1) for 𝑡1 which completes the proof.

We remark here that if condition (3.12) is satisfied for some 𝑟, 𝑠, 𝛼, then it is also satisfied for 𝑟, 𝑠, 𝛼 with 0<𝛼<𝛼. Thus it is not surprising to expect a result like Theorem 3.4 to hold.

To end this section, we prove a variant of (1.3) which is motivated by the following inequality due to Mesihović [9] (with 𝑞𝑖=1/𝑛 here): 11𝑛𝐴𝑛+1𝑛𝐺𝑛1𝑛𝑛𝑖=1𝑥𝑖11/𝑛1𝑛𝑛𝑖=1𝑥𝑖1/𝑛.(3.27) We now generalize the above inequality to the arbitrary weight case.

Theorem 3.5. For 0𝑥1𝑥2𝑥𝑛, (1𝑞)𝐴𝑛+𝑞𝐺𝑛𝑛𝑖=1𝑞𝑖𝑥𝑖1𝑞𝑛𝑖=1𝑞𝑖𝑥𝑞𝑖(3.28) with equality holding if and only if 𝑥1==𝑥𝑛 or 𝑥1=0, 𝑞1=𝑞, 𝑥2==𝑥𝑛, or 𝑛=2, 𝑞=1/2.

Proof. Define 𝐷𝑛(𝐱)=(1𝑞)𝐴𝑛+𝑞𝐺𝑛𝑛𝑖=1𝑞𝑖𝑥𝑖1𝑞𝑛𝑖=1𝑞𝑖𝑥𝑞𝑖.(3.29) We need to show 𝐷𝑛0 and we have 1𝑞𝑛𝜕𝐷𝑛𝜕𝑥𝑛𝐺=1𝑞+𝑞𝑛𝑥𝑛𝑥(1𝑞)𝑛𝑛𝑞𝑖=1𝑞𝑖𝑥𝑞𝑖𝑥𝑞𝑛𝑛𝑞1𝑖=1𝑞𝑖𝑥𝑖1𝑞.(3.30) By a change of variables: 𝑥𝑖/𝑥𝑛𝑥𝑖, 1𝑖𝑛, we may assume 0𝑥1𝑥2𝑥𝑛=1 in (3.30) and rewrite it as 𝑔𝑛𝑥1,,𝑥𝑛1=1𝑞+𝑞𝐺𝑛(1𝑞)𝑛𝑖=1𝑞𝑖𝑥𝑞𝑖𝑞𝑛𝑖=1𝑞𝑖𝑥𝑖1𝑞.(3.31) We want to show 𝑔𝑛0. Let 𝐚=(𝑎1,,𝑎𝑛1)[0,1]𝑛1 be the point in which the absolute minimum of 𝑔𝑛 is reached. We may assume 𝑎1𝑎2𝑎𝑛1. If 𝑎1=0, then 𝐚 is a boundary point of [0,1]𝑛1, and in this case we have 𝑔𝑛𝑎1=0,,𝑎𝑛1=1𝑞(1𝑞)𝑛𝑖=2𝑞𝑖𝑎𝑞𝑖𝑞𝑛𝑖=2𝑞𝑖𝑎𝑖1𝑞1𝑞(1𝑞)1𝑞1𝑞1𝑞1=𝑞1𝑞0(3.32) with equality holding if and only if 𝑞1=𝑞, 𝑎2==𝑎𝑛=1. Now suppose 𝑎1>0 and 𝑎𝑚1<𝑎𝑚==𝑎𝑛=1 for some 1𝑚𝑛, then 𝑎1,,𝑎𝑚1 solve the equation 𝑔𝑛𝑎1,,𝑎𝑛1=0.(3.33) Equivalently, 𝑎1,,𝑎𝑚1 solve the equation 𝐺𝑛𝑥=(1𝑞)𝑞+𝑥1𝑞.(3.34) As the right-hand side expression above is an increasing function of 𝑥, the above equation has at most one root (regarding 𝐺𝑛 as constant). So we only need to show 𝑔𝑛0 for the case 𝑎1==𝑎𝑚1=𝑥<𝑎𝑚==𝑎𝑛=1 in (3.31) for some 1𝑚𝑛. In this case we regard 𝑔𝑛 as a function of 𝑥 and we recast it as (𝜔,𝑥)=1𝑞+𝑞𝑥𝜔(1𝑞)(𝜔𝑥𝑞+1𝜔)𝑞𝜔𝑥1𝑞+1𝜔=𝜔𝑞+𝑞𝑥𝜔(1𝑞)𝜔𝑥𝑞𝑞𝜔𝑥1𝑞.(3.35) Here 0<𝑥1 and 𝑞𝜔1𝑞. Note first that when 𝑞=1/2, (𝜔,𝑥)=0 so that we may now assume 0<𝑞<1/2. We have 𝜕|||𝜕𝜔𝜔=𝑞=1+𝑞𝑥𝑞ln𝑥(1𝑞)𝑥𝑞𝑞𝑥1𝑞=𝑑(𝑥).(3.36) Now 𝑑(𝑥)=𝑞𝑥𝑞1𝑒(𝑥),(3.37) where 𝑒(𝑥)=1+𝑞ln𝑥(1𝑞)(1𝑞)𝑥12𝑞.(3.38) Note that 𝑒(𝑥)=0 has one root (1𝑞)(12𝑞)𝑥012𝑞=𝑞 so that if 0<𝑥0<1, then at this point 𝑒𝑥0=𝑞ln𝑥0𝑞+𝑞12𝑞<0.(3.39) Note also that lim𝑥0+𝑒(𝑥)<0, 𝑒(1)=2𝑞1<0. This implies that 𝑒(𝑥)<0 for 0<𝑥1. Hence 𝑑(𝑥)𝑑(1)=0 for 0<𝑥1. As it is easy to see that (𝜔,𝑥) is a convex function of 𝜔 for fixed 𝑥, we conclude that (𝜔,𝑥) is an increasing function of 𝑞𝜔1𝑞 for fixed 𝑥. Thus for 0<𝑥𝑥, 𝑞𝜔1𝑞, (𝜔,𝑥)(𝑞,𝑥)=𝑞2𝑥𝑞𝑥1𝑞0(3.40) with equality holding if and only if 𝑥=1.
Thus we have shown 𝑔𝑛0; hence 𝜕𝐷𝑛/𝜕𝑥𝑛0 with equality holding if and only if 𝑥1==𝑥𝑛 or 𝑥1=0, 𝑞1=𝑞, 𝑥2==𝑥𝑛 or 𝑛=2, 𝑞1=1/2. By letting 𝑥𝑛 tend to 𝑥𝑛1, we have 𝐷𝑛𝐷𝑛1 (with weights 𝑞1,,𝑞𝑛2, 𝑞𝑛1+𝑞𝑛) with equality holding if and only if 𝑥𝑛=𝑥𝑛1 or 𝑛=2 and either 𝑞=1/2 or 𝑥1=0, 𝑞1=𝑞. It follows by induction that 𝐷𝑛0 with equality holding if and only if 𝑥1==𝑥𝑛 or 𝑥1=0, 𝑞1=𝑞, 𝑥2==𝑥𝑛 or 𝑛=2, 𝑞=1/2 and this completes the proof.

We remark here that if we define 𝑆(𝛽)=𝑛𝑖=1𝑞𝑖𝑥𝑖1𝛽𝑛𝑖=1𝑞𝑖𝑥𝛽𝑖,(3.41) then for 1𝛽1/2, 𝑑𝑆=𝑑𝛽1𝑖𝑗𝑛𝑞𝑖𝑞𝑗𝑥𝛽𝑖𝑥𝛽𝑗𝑥𝑗12𝛽𝑥𝑖12𝛽𝑥ln𝑖𝑥𝑗0.(3.42) Hence Theorem 3.5 improves (1.3) for the case {1,1/2,0}, 𝛼=1.

4. Inequalities Involving the Symmetric Means

In this section, we set 𝑞𝑖=1/𝑛, 1𝑖𝑛. As an analogue of (1.8) (or (1.3)), we first consider 𝐴𝛼𝑛𝜆𝛼,𝑟,𝑘(𝑛)𝑀𝛼𝑛,𝑟+1𝜆𝛼,𝑟,𝑘𝑃(𝑛)𝛼𝑛,𝑘,(4.1) where 𝛼>0, 𝑟>1, 𝑛𝑘2, and 𝜆𝛼,𝑟,𝑘(𝑛)=(11/𝑛)𝛼((𝑛𝑘)/𝑛)𝛼/𝑘(11/𝑛)𝛼/𝑟((𝑛𝑘)/𝑛)𝛼/𝑘.(4.2) The case 𝑟=2, 𝛼=𝑘 in (4.1) is just Theorem 1.3. In what follows, we will give a proof of Theorem 1.3 by combining the methods in [8, 10]. Before we prove our result, we would like to first recast (4.1) for the case 𝛼=𝑘 as 𝑛𝑖=1𝑥𝑖𝑘𝑛𝑘̃𝜆𝑟,𝑘𝑛𝑘𝑀(𝑛)𝑘𝑛,𝑟+̃𝜆𝑟,𝑘(𝑛)𝐸𝑛,𝑘,(4.3) where ̃𝜆𝑟,𝑘(𝑛) is defined as in the statement of Theorem 1.3. Now we need two lemmas.

Lemma 4.1. For 2𝑟𝑘𝑛1, ̃𝜆𝑟,𝑘̃𝜆(𝑛)𝑟,𝑘(𝑛1).(4.4)

Proof. We follow the method in the proof of Lemma 2 in [8]. We write ̃𝜆𝑟,𝑘(𝑛) as ̃𝜆𝑟,𝑘(𝑛𝑛)=𝑘!𝑘(11/𝑛)𝑘/𝑟(𝑛1)𝑘𝑘1𝑖=1(𝑛𝑖)𝑛(11/𝑛)𝑘/𝑟.𝑛+𝑘(4.5) From the above we see that in order for (4.4) to hold, it suffices to show that 𝑓(𝑡)=𝑘!(1𝑡)𝑘/𝑟(1𝑡)𝑘𝑘1𝑖=1(1𝑖𝑡)(1𝑡)𝑘/𝑟1+𝑘𝑡(4.6) is increasing on (0,1/𝑘). The logarithmic derivative of 𝑓(𝑡) is 𝑓(𝑡)𝑓=𝑘(𝑡)𝑟(1𝑡)1𝑟+𝑟11(1𝑡)𝑘(11/𝑟)𝑟1+(𝑘𝑟)𝑡(1𝑡)𝑘/𝑟+1+𝑘𝑡𝑘1𝑖=1𝑖.1𝑖𝑡(4.7) Note that for 0<𝑡<1, we have (1𝑡)𝑘/𝑟1+𝑘𝑡1(1𝑡)𝑘(11/𝑟)>0(4.8) by considering the Taylor expansions to the order of 𝑡2. It follows from this that 𝑓(𝑡)𝑓𝑘(𝑡)𝑟(1𝑡)1𝑟+𝑟11(1𝑡)𝑘(11/𝑟)𝑟1+(𝑘𝑟)𝑡1(1𝑡)𝑘(11/𝑟)+𝑘1𝑖=1𝑖=𝑘1𝑡1𝑡1𝑟𝑟(𝑘𝑟)𝑡/𝑟1(1𝑡)𝑘(11/𝑟)+𝑘12.(4.9) It is easy to see that the function 𝑡𝑡1(1𝑡)𝑘(11/𝑟)(4.10) is an increasing function for 0<𝑡1. Hence 1𝑟𝑟(𝑘𝑟)𝑡/𝑟1(1𝑡)𝑘(11/𝑟)+𝑘121𝑟𝑟𝑘𝑟𝑟+𝑘121=(𝑘1)21𝑟0.(4.11)

Lemma 4.2. Inequality (4.1) holds for all 𝐱 in the case 𝑛𝑘𝑟2, 𝛼=𝑘 if it holds for 𝐱=(𝑎,𝑏,,𝑏) with 0<𝑎𝑏.

Proof. In this proof we assume that 0𝑥1𝑥2𝑥𝑛. We prove the lemma by induction on 𝑛. When 𝑛=𝑘, the assertion holds as a special case of (1.6). Now assume that the assertion holds for 𝑛1, and we proceed to show it also holds for 𝑛. If 𝑥1=0, we use the equivalent form (4.3) of (4.1) for the case 𝛼=𝑘 to see that what we need to prove is 𝑛1𝑖=2𝑥𝑖𝑘𝑛𝑘̃𝜆𝑟,𝑘𝑛𝑘(𝑛)𝑛1𝑛𝑘/𝑟𝑀𝑘𝑛1,𝑟𝑥2,,𝑥𝑛+̃𝜆𝑟,𝑘(𝑛)𝐸𝑛1,𝑘𝑥2,,𝑥𝑛.(4.12) By the induction case 𝑛1, we have 𝑛1𝑖=2𝑥𝑖𝑘(𝑛1)𝑘̃𝜆𝑟,𝑘𝑘𝑀(𝑛1)𝑛1𝑘𝑛1,𝑟𝑥2,,𝑥𝑛+̃𝜆𝑟,𝑘(𝑛1)𝐸𝑛1,𝑘𝑥2,,𝑥𝑛.(4.13) Note that 𝑛𝑘̃𝜆𝑟,𝑘𝑛𝑘(𝑛)𝑛1𝑛𝑘/𝑟=(𝑛1)𝑘̃𝜆𝑟,𝑘𝑘(𝑛)𝑛1.(4.14) Using this with Lemma 4.1 together with the observation that 𝐸𝑛1,𝑘𝑘𝑛1=𝑃𝑘𝑛1,𝑘𝑀𝑘𝑛1,𝑟,(4.15) we see that inequality (4.12) follows from (4.13).
Thus from now on we may focus on the case 𝑥1>0. Since both sides of (4.1) are homogeneous functions, it suffices to show that min𝐱Δ𝜆𝑘,𝑟,𝑘(𝑛)𝑀𝑘𝑛,𝑟+1𝜆𝑘,𝑟,𝑘𝑃(𝑛)𝑘𝑛,𝑘1,(4.16) where 𝑥Δ=𝐱=1,,𝑥𝑛𝑥𝑖0,1𝑖𝑛,𝑛𝑖=1𝑥𝑖.=𝑛(4.17) Assume that 𝜆𝑘,𝑟,𝑘(𝑛)𝑀𝑘𝑛,𝑟+(1𝜆𝑘,𝑟,𝑘(𝑛))𝑃𝑘𝑛,𝑘 attains its minimum at some point (𝑎1,,𝑎𝑛) with 𝑎𝑖>0, 1𝑖𝑛. If 𝑎1=𝑎2==𝑎𝑛, then (4.16) holds. Furthermore, if 𝑛=2, then (4.16) also holds, being a special case of (1.6). Thus without loss of generality, we may assume 𝑛3 and 𝑎𝑛>𝑎𝑛1𝑎𝑛2 here. We may also assume that when 𝑟=2, 𝑘>2 since otherwise inequality (4.1) becomes an identity. Consider the function 𝑓(𝑥,𝑦)=𝜆𝑘,𝑟,𝑘(𝑛)𝑀𝑘𝑛,𝑟𝑎1,,𝑎𝑛2+,𝑥,𝑦1𝜆𝑘,𝑟,𝑘𝑃(𝑛)𝑘𝑛,𝑘𝑎1,,𝑎𝑛2,𝑥,𝑦(4.18) on the set (𝑥,𝑦)𝑥0,𝑦0,𝑥+𝑦=𝑛𝑛2𝑖=1𝑎𝑖.(4.19) It is minimized at (𝑎𝑛1,𝑎𝑛). It is easy to see that 𝑓 has the form 𝜆𝑘,𝑟,𝑘1(𝑛)𝑛(𝑥𝑟+𝑦𝑟)+𝐵𝑘/𝑟𝑎+𝐶1,,𝑎𝑛2𝑥𝑦+𝐷,(4.20) where 𝐵, 𝐶, and 𝐷 are nonnegative constants with 𝐶 depends on 𝑎1,,𝑎𝑛2. We now set 𝑥+𝑦=𝑐 and 𝑥𝑦=𝑧 with 0𝑧=𝑥𝑦(𝑥+𝑦)2/4=𝑐2/4. Note here that equality holds if and only if 𝑥𝑦=0 or 𝑥=𝑦. We regard the above function as a function of 𝑧=𝑥𝑦 and recast it as (𝑧)=𝜆𝑘,𝑟,𝑘(1𝑛)𝑛𝑐+𝑐24𝑧2𝑟+𝑐𝑐24𝑧2𝑟+𝐵𝑘/𝑟𝑎+𝐶1,,𝑎𝑛2𝑧+𝐷.(4.21) For 𝑦>𝑥>0, calculation yields 𝜆(𝑥𝑦)=𝑘,𝑟,𝑘(𝑛)𝑘𝑛1𝑛(𝑥𝑟+𝑦𝑟)+𝐵𝑘/𝑟1𝑦𝑟1𝑥𝑟1𝑎𝑦𝑥+𝐶1,,𝑎𝑛2.(4.22) Since 𝑎𝑛>𝑎𝑛1>0, we must have (𝑎𝑛1𝑎𝑛)=0 and we can further recast this as 𝐶𝑎1,,𝑎𝑛2=𝜆𝑘,𝑟,𝑘(𝑛)𝑘𝑛𝑀𝑘𝑟𝑛,𝑟𝑎1,,𝑎𝑛𝑎𝑛𝑟1𝑎𝑟1𝑛1𝑎𝑛𝑎𝑛1.(4.23) Now if 𝑎𝑛2>0, we can repeat the same argument for the pair (𝑎𝑛1,𝑎𝑛2). By a slightly abuse of notation, we obtain 𝑎𝑛2𝑎𝑛1𝑎=𝐶1,,𝑎𝑛3,𝑎𝑛𝜆𝑘,𝑟,𝑘(𝑛)𝑘𝑛𝑀𝑘𝑟𝑛,𝑟𝑎1,,𝑎𝑛𝑎𝑟1𝑛1𝑎𝑟1𝑛2𝑎𝑛1𝑎𝑛2.(4.24) It is easy to see that (since we assume 𝑎𝑖>0 and 𝑘3 when 𝑟=2) 𝐶𝑎1,,𝑎𝑛3,𝑎𝑛𝑎>𝐶1,,𝑎𝑛2.(4.25) Moreover, one checks easily that the function 𝑦(𝑥,𝑦)𝑟1𝑥𝑟1𝑦𝑥(4.26) is increasing with respect to each variable 𝑥,𝑦>0 when 𝑟2. It follows that when 𝑟2, 𝑎𝑛2𝑎𝑛1>𝑎𝑛1𝑎𝑛=0,(4.27) which implies that by decreasing the value of 𝑎𝑛2𝑎𝑛1 while keeping 𝑎𝑛2+𝑎𝑛1 fixed, one is able to get a smaller value for 𝜆𝑘,𝑟,𝑘(𝑛)𝑀𝑘𝑛,𝑟+(1𝜆𝑘,𝑟,𝑘(𝑛))𝑃𝑘𝑛,𝑘, contradicting the assumption that it attains its minimum at (𝑎1,,𝑎𝑛).
Hence we conclude that 𝜆𝑘,𝑟,𝑘(𝑛)𝑀𝑘𝑛,𝑟+(1𝜆𝑘,𝑟,𝑘(𝑛))𝑃𝑘𝑛,𝑘 is minimized at (𝑎,𝑏,,𝑏) with 0<𝑎𝑏 satisfying 𝑎+(𝑛1)𝑏=𝑛. In this case (4.16) holds by our assumption which completes the proof.

Now we are ready to prove a slightly generalization of Theorem 1.3.

Corollary 4.3. Inequality (4.1) holds in the cases 𝑛𝑘𝑟=2, 𝛼=𝑘 and 𝑛𝛼=𝑟=𝑘2.

Proof. The first case is just Theorem 1.3 and by Lemma 4.2, it suffices to show that inequality (4.1) holds for the case 𝐱=(𝑎,𝑏,,𝑏) with 0<𝑎𝑏 and this has been already treated in the proof of Theorem 2 in [8]. For the case 𝑛𝛼=𝑟=𝑘2, by Lemma 4.2 again, we only need to check the case 𝐱=(𝑎,𝑏,,𝑏) with 0<𝑎𝑏. In this case we define 𝑓(𝑎,𝑏)=𝜆𝑘,𝑘,𝑘𝑎(𝑛)𝑘𝑛+(𝑛1)𝑏𝑘𝑛+1𝜆𝑘,𝑘,𝑘(𝑛)𝑛𝑘𝑛𝑏𝑘+𝑘𝑛𝑎𝑏𝑘1𝑎𝑛+(𝑛1)𝑏𝑛𝑘.(4.28) As in the proof of Theorem 1.2, it suffices to show that 𝑛𝑘(𝑛1)𝑏𝑘1𝜕𝑓𝜕𝑏=𝜆𝑘,𝑘,𝑘(𝑛)+1𝜆𝑘,𝑘,𝑘(𝑛)𝑛𝑘+𝑛1𝑘1𝑎𝑛1𝑏𝑎+𝑛𝑏𝑛1𝑛𝑘10.(4.29) By a change of variables 𝑎/𝑏𝑎, we can recast inequality (4.29) as 𝑔(𝑎)=𝜆𝑘,𝑘,𝑘(𝑛)+1𝜆𝑘,𝑘,𝑘(𝑛)𝑛𝑘+𝑛1𝑘1𝑎𝑛1𝑛1𝑛+𝑎𝑛𝑘1.(4.30) As 𝑔(1)=0 and 𝑔(0)=0, we conclude that 𝑔(𝑎)0 for 0𝑎1 as 𝑔(𝑎) is a concave function for 0𝑎1 and this completes the proof.

We recall a result of Kuczma [10].

Theorem 4.4. For 𝑛3, 1𝑘𝑛1, 𝑃𝑛,𝑘𝑀𝑛,𝜂(𝑘) with 𝜂(𝑘)=𝑘(ln𝑛ln(𝑛1)),ln𝑛ln(𝑛𝑘)(4.31) and the result is best possible.

The above theorem combined with Corollary 4.3 immediately yields the following.

Corollary 4.5. Let 𝑞𝑖=1/𝑛 with 𝑛1 being an integer. Then for any integer 𝑛1𝑘𝑟=2 or 𝑛1𝑘=𝑟2, 𝐴𝑘𝑛𝜆𝑘,𝑟,𝑘(𝑛)𝑀𝑘𝑛,𝑟+1𝜆𝑘,𝑟,𝑘𝑀(𝑛)𝑘𝑛,𝜂(𝑘),(4.32) where 𝜆𝑘,𝑟,𝑘(𝑛) is defined as in (4.2) and 𝜂(𝑘) is defined as in Theorem 4.4.

Next, we consider the following inequality: 𝑃𝛼𝑛,𝑙𝜇𝛼,𝑘,𝑙(𝑛)𝐴𝛼𝑛+1𝜇𝛼,𝑘,𝑙𝑃(𝑛)𝛼𝑛,𝑘,(4.33) where 𝛼>0,𝑛𝑘>𝑙2 and 𝜇𝛼,𝑘,𝑙(𝑛)=((𝑛𝑙)/𝑛)𝛼/𝑙((𝑛𝑘)/𝑛)𝛼/𝑘((𝑛1)/𝑛)𝛼((𝑛𝑘)/𝑛)𝛼/𝑘.(4.34) We note here that it is easy to check that the function 𝑥𝑛𝑥𝑛1/𝑥(4.35) is a decreasing function for 1𝑥<𝑛 so that we have 0<𝜇𝛼,𝑘,𝑙(𝑛)<1.

As an analogue of Theorem 1.1, one can show similarly the following.

Proposition 4.6. Let 𝑛=𝑘>𝑙2; if (4.33) holds for 𝛼0>0, then it also holds for any 0<𝛼<𝛼0.

The case 𝑛=𝑘, 𝛼=1 of (4.33) was established in [11]. In the case of 𝑛=𝑘, one possible way of establishing (4.33) is to combine Theorems 4.4 and 1.2 together. However, this is not always applicable as one checks via certain change of variables that one needs to have 1/𝜂(𝑘)2 in order to apply Theorem 1.2, a condition which is not always satisfied. We now proceed directly to show the following.

Theorem 4.7. Inequality (4.33) holds for 𝑛=𝑘>𝑙2 and 0<𝛼𝑛.

Proof. In view of Proposition 4.6, it suffices to prove the theorem for the case 𝛼=𝑛. We write 𝜇(𝑛)=𝜇𝑛,𝑛,𝑙(𝑛) in this proof and note that since both sides of (4.33) are homogeneous functions, it suffices to show that max𝐱Δ𝑃𝑛𝑛,𝑙(1𝜇(𝑛))𝐺𝑛𝑛𝜇(𝑛),(4.36) where 𝑥Δ=𝐱=1,,𝑥𝑛𝑥𝑖0,1𝑖𝑛,𝑛𝑖=1𝑥𝑖.=𝑛(4.37) Assume that 𝑃𝑛𝑛,𝑙(1𝜇(𝑛))𝐺𝑛𝑛 attains its maximum at some point (𝑎1,,𝑎𝑛) with 𝑎𝑖0, 1𝑖𝑛. Consider the function 𝑓(𝑥,𝑦)=𝑃𝑛𝑛,𝑙𝑥,𝑦,𝑎3,,𝑎𝑛𝑥𝑦(1𝜇(𝑛))𝑛𝑖=3𝑎𝑖(4.38) on the set (𝑥,𝑦)𝑥0,𝑦0,𝑥+𝑦=𝑛𝑛𝑖=3𝑎𝑖.(4.39) It is maximized at (𝑎1,𝑎2). It is easy to see that 𝑓 has the form (𝐴𝑥𝑦+𝐵)𝑛/𝑙𝐶𝑥𝑦,(4.40) where 𝐴, 𝐵, and 𝐶 are nonnegative constants. The above function is certainly convex with respect to 𝑥𝑦. As 0𝑥𝑦(𝑥+𝑦)2/4 with equality holding if and only if 𝑥𝑦=0 or 𝑥=𝑦, 𝑓 is maximized at 𝑥𝑦=0 or 𝑥=𝑦. Repeating the same argument for other pairs (𝑎𝑖,𝑎𝑗), we conclude that in order to show (4.33) for 𝛼=𝑛=𝑘, it suffices to check that it holds for 𝐱 being of the following form (0,,0,𝑎,𝑎) or (𝑎,,𝑎) for some positive constant 𝑎. It is easy to see that (4.33) holds when 𝐱 is of the second form and when 𝐱 is of the first form, let 𝑚 denote the number of 𝑎's in 𝐱; if 𝑚<𝑙, then it is easy to see that (4.33) holds. So we may now assume that 𝑙𝑚𝑛=𝑘 and we need to show that (𝑚𝑙)𝑚1/𝑙𝑘𝜇𝑘,𝑘,𝑙((𝑛)𝑛𝑙)1/𝑙𝑛𝑘.(4.41) Certainly the left-hand side above increases with 𝑚; hence one only needs to verify the above inequality for the case 𝑚=𝑛1, which becomes an identity and this completes the proof.

We note here that Alzer [12] has shown that for 𝑛3, 𝑃𝑛1𝑛,𝑛1𝑛𝐴𝑛+1𝑛𝑛1+1𝐺𝑛+1𝑛𝑛1.(4.42) The case 𝛼=𝑙=𝑛1 of Theorem 4.7 now improves the above result, namely, for 𝑛3, 𝑃𝑛1𝑛,𝑛1𝑛𝑛2(𝑛1)𝑛1𝐴𝑛𝑛1+𝑛1𝑛2(𝑛1)𝑛1𝐺𝑛𝑛1,(4.43) as one checks easily that for 𝑛3, 𝑛𝑛2(𝑛1)𝑛1𝑛.𝑛+1(4.44)

Similar to Theorem 4.7, we have the following.

Theorem 4.8. For 1𝑘𝑛, 𝑞𝑖=1/𝑛, 𝑃𝑘𝑛,𝑘𝑛𝑘𝑀𝑛1𝑘𝑛,𝑘+𝑘1𝐺𝑛1𝑘𝑛.(4.45)

Proof. We define 𝑓(𝐱)=𝑛𝑘𝑀𝑛1𝑘𝑛,𝑘+𝑘1𝐺𝑛1𝑘𝑛𝑃𝑘𝑛,𝑘.(4.46) We now set 𝑃𝑛1,𝑘1=𝑃𝑛1,𝑘1𝑥1,,𝑥𝑛1,𝐺𝑛1=𝐺𝑛1𝑥1,,𝑥𝑛1,(4.47) so that 𝜕𝑓𝜕𝑥𝑛=𝑘𝑛𝑛𝑘𝑥𝑛1𝑛𝑘1+𝑘𝑛𝑘1𝐺𝑛1𝑘𝑛𝑥𝑛𝑘𝑛𝑃𝑘1𝑛1,𝑘1,(4.48) where we have also used the following relation: 𝑃𝑘𝑛,𝑘=𝑘𝑛𝑥𝑛