A New Method to Prove and Find Analytic Inequalities

Abstract

We present a new method to study analytic inequalities. As for its applications, we prove the well-known Hölder inequality and establish several new analytic inequalities.

1. Monotonicity Theorem

Throughout the paper ℝ denotes the set of real numbers and ℝ₊ denotes the set of strictly positive real numbers. Let 𝑛≥2, 𝑛∈ℕ, and 𝐱=(𝑥₁,𝑥₂,…,𝑥_𝑛)∈ℝ^𝑛; the arithmetic mean 𝐴(𝐱) and the power mean 𝑀_𝑟(𝐱) of order 𝑟 with respect to the positive real numbers 𝑥₁,𝑥₂,…,𝑥_𝑛 are defined by 𝐴(𝐱)=(1/𝑛)∑^𝑛𝑖=1𝑥_𝑖, 𝑀_𝑟(𝐱)=((1/𝑛)∑^𝑛𝑖=1𝑥^𝑟𝑖)^1/𝑟 for 𝑟≠0, and 𝑀₀(𝐱)=(∏^𝑛𝑖=1𝑥_𝑖)^1/𝑛, respectively.

In [1], Pachpatte gave many basic methods and tools for researchers working in inequalities. In this section, we present a monotonicity theorem which can be used as powerful tool to prove and find analytic inequalities.

Lemma 1.1. Suppose that 𝑚<𝑀, 𝐷={(𝑥₁,𝑥₂)∣𝑚≤𝑥₂≤𝑥₁≤𝑀}. If 𝑓∶𝐷→ℝ has continuous partial derivatives, then 𝜕𝑓/𝜕𝑥₁≥(≤)𝜕𝑓/𝜕𝑥₂ holds in 𝐷 if and only if 𝑓(𝑎,𝑏)≥(≤)𝑓(𝑎−𝑙,𝑏+𝑙) holds for all (𝑎,𝑏)∈𝐷 and 𝑙>0 with 𝑏<𝑏+𝑙≤𝑎−𝑙<𝑎.

Proof. We only prove the case of 𝜕𝑓/𝜕𝑥₁≥𝜕𝑓/𝜕𝑥₂.
Necessity. For all (𝑥₁,𝑥₂)∈𝐷 and 𝑙∈ℝ₊ with 𝑚≤𝑥₂<𝑥₂+𝑙≤𝑥₁−𝑙<𝑥₁≤𝑀, by the assumption we have 𝑓(𝑥₁−𝑙,𝑥₂+𝑙)−𝑓(𝑥₁,𝑥₂)≤0. Then from the Langrange's mean value theorem we know that there exists 𝜉_𝑙∈(0,𝑙) such that 𝑙−𝜕𝑓𝑥¹−𝜉_𝑙,𝑥₂+𝜉_𝑙𝜕𝑥₁+𝜕𝑓𝑥¹−𝜉_𝑙,𝑥₂+𝜉_𝑙−𝜕𝑓𝑥₁𝜕𝑥₂≤0,−𝜉_𝑙,𝑥₂+𝜉_𝑙𝜕𝑥₁+𝜕𝑓𝑥¹−𝜉_𝑙,𝑥₂+𝜉_𝑙𝜕𝑥₂≤0.(1.1) Letting 𝑙→0+, we get 𝜕𝑓𝑥₁,𝑥₂𝜕𝑥₁≥𝜕𝑓𝑥¹,𝑥₂𝜕𝑥₂.(1.2) According to the continuity of partial derivatives, we know that 𝜕𝑓𝑥₁,𝑥₁𝜕𝑥₁≥𝜕𝑓𝑥¹,𝑥₁𝜕𝑥₂(1.3) holds also.
Sufficiency. For all (𝑎,𝑏)∈𝐷 and 𝑙>0 with 𝑏<𝑏+𝑙≤𝑎−𝑙<𝑎, from the assumption and the Langrange's mean value theorem we know that there exists 𝜉_𝑙∈(0,𝑙) such that 𝑓(𝑎,𝑏)−𝑓(𝑎−𝑙,𝑏+𝑙)==−−𝑙(𝑓−(𝑎𝜕−𝑓𝑙,𝑎𝑏−+𝜉𝑙)_𝑙−𝑓(𝑎,𝑏)),𝑏+𝜉_𝑙𝜕𝑥₁+𝜕𝑓𝑎−𝜉^𝑙,𝑏+𝜉_𝑙=𝑙𝜕𝑓𝑎−𝜉_𝑙𝜕𝑥₂,𝑏+𝜉_𝑙𝜕𝑥₁−𝜕𝑓𝑎−𝜉^𝑙,𝑏+𝜉_𝑙≥0.𝜕𝑥₂(1.4)
Therefore the proof of Lemma 1.1 is completed.

Theorem 1.2. Suppose that 𝐷⊂ℝ^𝑛 is a symmetric convex set with nonempty interior, 𝑓∶𝐷→ℝ has continuous partial derivatives, and 𝐷^⌣_𝑖=𝐱∈𝐷∣𝑥_𝑖=max_{1≤𝑗≤𝑛}𝑥_𝑗−𝐱∈𝐷∣𝑥₁=𝑥₂=⋯=𝑥_𝑛,_⌢𝐷_𝑖=𝐱∈𝐷∣𝑥_𝑖=min_{1≤𝑗≤𝑛}𝑥_𝑗−𝑥∈𝐷∣𝑥₁=𝑥₂=⋯=𝑥_𝑛,(1.5)𝑖=1,2,…,𝑛. If for all 𝑖,𝑗=1,2,…,𝑛 with 𝑖≠𝑗, 𝜕𝑓𝜕𝑥_𝑖>(<)𝜕𝑓𝜕𝑥_𝑗(1.6) holds in 𝐷^⌣_𝑖∩𝐷^⌢_𝑗, then 𝑓𝑎₁,𝑎₂,…,𝑎_𝑛≥(≤)𝑓(𝐴(𝐚),𝐴(𝐚),…,𝐴(𝐚))(1.7) for all 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈𝐷, with equality if only if 𝑎₁=𝑎₂=⋯=𝑎_𝑛.

Proof. If 𝑛=2, then Theorem 1.2 follows from Lemma 1.1 and 𝑙=|𝑎₁−𝑎₂|/2. We assume that 𝑛≥3 in the next discussion. Without loss of generality, we only prove the case of 𝜕𝑓/𝜕𝑥_𝑖>𝜕𝑓/𝜕𝑥_𝑗 with 𝑖≠𝑗.
If 𝑎₁=𝑎₂=⋯=𝑎_𝑛, then inequality (1.7) is clearly true. If max_{1≤𝑗≤𝑛}{𝑎_𝑗}≠min_{1≤𝑗≤𝑛}{𝑎_𝑗}, then without loss of generality we assume that 𝑎₁≥𝑎₂≥⋯≥𝑎_𝑛−1≥𝑎_𝑛.
(1)   If 𝑎₁>max_{2≤𝑗≤𝑛}{𝑎_𝑗} and 𝑎_𝑛<min_{1≤𝑗≤𝑛−1}{𝑎_𝑗}, then (𝑎₁,𝑎₂,…,𝑎_𝑛)∈𝐷^⌣₁∩𝐷^⌢_𝑛. From Lemma 1.1 and the conditions in Theorem 1.2 we know that there exist 𝑎¹⁽¹⁾ and 𝑎^𝑛(1) such that 𝑙=𝑎₁−𝑎¹⁽¹⁾=𝑎^𝑛(1)−𝑎_𝑛>0, 𝑎¹⁽¹⁾=𝑎₂ or 𝑎^𝑛(1)=𝑎_𝑛−1, and 𝑓𝑎₁,𝑎₂,𝑎₃,…,𝑎_𝑛≥𝑓𝑎¹⁽¹⁾,𝑎₂,𝑎₃,…,𝑎^𝑛(1).(1.8) For the sake of convenience, we denote 𝑎^𝑖(1)=𝑎_𝑖,2≤𝑖≤𝑛−1. Consequently, 𝑓𝑎₁,𝑎₂,𝑎₃,…,𝑎_𝑛≥𝑓𝑎¹⁽¹⁾,𝑎²⁽¹⁾,𝑎³⁽¹⁾,…,𝑎^𝑛(1).(1.9) If 𝑎¹⁽¹⁾=𝑎²⁽¹⁾=⋯=𝑎^𝑛(1), then Theorem 1.2 holds. Otherwise, for the case of 𝑎¹⁽¹⁾=𝑎²⁽¹⁾>𝑎^𝑛(1), (𝑎¹⁽¹⁾,𝑎²⁽¹⁾,𝑎³⁽¹⁾,…,𝑎^𝑛(1))∈𝐷^⌣₁∩𝐷^⌢_𝑛 and 𝜕𝑓(𝐱)𝜕𝑥₁||||_{𝐱=(𝑎1(1),𝑎2(1),𝑎3(1),…,𝑎𝑛(1))}>𝜕𝑓(𝐱)𝜕𝑥_𝑛||||_{𝐱=(𝑎1(1),𝑎2(1),𝑎3(1),…,𝑎𝑛(1))}.(1.10) From the continuity of partial derivatives we know that there exists 𝜀>0 such that 𝜕𝑓(𝐱)𝜕𝑥₁||||_{𝐱=(𝑠,𝑎2(1),𝑎3(1),…,𝑡)}>𝜕𝑓(𝐱)𝜕𝑥_{𝑛|||||𝐱=(𝑠,𝑎2(1),𝑎3(1),…,𝑡)},(1.11) where 𝑠∈[𝑎¹⁽¹⁾−𝜀,𝑎¹⁽¹⁾] and 𝑡∈[𝑎^𝑛(1),𝑎^𝑛(1)+𝜀]. Denote 𝑎¹⁽²⁾=𝑎¹⁽¹⁾−𝜀, 𝑎^𝑛(2)=𝑎^𝑛(1)+𝜀, 𝑎^𝑖(2)=𝑎^𝑖(1)(2≤𝑖≤𝑛−1). By Lemma 1.1, we get 𝑓𝑎¹⁽¹⁾,𝑎²⁽¹⁾,𝑎³⁽¹⁾,…,𝑎^𝑛(1)≥𝑓𝑎¹⁽²⁾,𝑎²⁽²⁾,𝑎³⁽²⁾,…,𝑎^𝑛(2),(1.12) and 𝑎²⁽²⁾=max_{1≤𝑖≤𝑛}{𝑎^𝑖(2)}. For the case of 𝑎¹⁽¹⁾>𝑎^(1)𝑛−1=𝑎^𝑛(1), after a similar argument, we get inequality (1.12) with 𝑎^(2)𝑛−1=min_{1≤𝑖≤𝑛}{𝑎^𝑖(2)}.
Repeating the above steps, we get {𝑎^1(𝑖),𝑎^2(𝑖),…,𝑎^𝑛(𝑖)}(𝑖=1,2,…) such that ∑^𝑛𝑗=1𝑎^𝑗(𝑖) is a constant and {𝑎^𝑗(𝑖)}(𝑖=1,2,…) are monotone increasing (decreasing) sequences if 𝑎_𝑗≥(≤)𝐴(𝐚),𝑗=1,2,3,…,𝑛, and 𝑓𝑎¹⁽¹⁾,𝑎²⁽¹⁾,𝑎³⁽¹⁾,…,𝑎^𝑛(1)≥𝑓𝑎^1(𝑖),𝑎^2(𝑖),𝑎^3(𝑖),…,𝑎^𝑛(𝑖).(1.13) If there exists 𝑖∈ℕ such that 𝑎^1(𝑖)=𝑎^2(𝑖)=⋯=𝑎^𝑛(𝑖), then the proof of Theorem 1.2 is completed. Otherwise, we denote 𝛼=inf_𝑖∈ℕ{max{𝑎^1(𝑖),𝑎^2(𝑖),…,𝑎^𝑛(𝑖)}}; without loss of generality, we assume that limm_𝑗→+∞ax𝑎^(𝑖𝑗)1,𝑎^(𝑖𝑗)2,…,𝑎^{(𝑖𝑗)𝑛}=𝑎^(𝑖𝑗)1⟶𝛼,𝑎^(𝑖𝑗)1,𝑎^(𝑖𝑗)2,…,𝑎^{(𝑖𝑗)𝑛}=𝛼,𝑏²,𝑏₃,…,𝑏_𝑛,(1.14) where {𝑖_𝑗}^+∞𝑗=1 is a subsequence of ℕ. Then from the continuity of function 𝑓, we get 𝑓𝑎₁,𝑎₂,𝑎₃,…,𝑎_𝑛≥𝑓𝛼,𝑏₂,𝑏₃,…,𝑏_𝑛.(1.15) If 𝛼≠min{𝑏₂,𝑏₃,…,𝑏_𝑛}, then we repeat the above arguments and get a contradiction to the definition of 𝛼. Hence 𝛼=𝑏₂=𝑏₃=⋯=𝑏_𝑛. From 𝛼+∑^𝑛𝑖=2𝑏_𝑖=∑^𝑛𝑖=1𝑎_𝑖 we get 𝛼=𝑏₂=𝑏₃=⋯=𝑏_𝑛=𝐴(𝐚); the proof of Theorem 1.2 is completed.
(2) The proof for the case of 𝑎₁=max_2≤j≤𝑛{𝑎_𝑗} or 𝑎_𝑛=min_{1≤𝑗≤𝑛−1}{𝑎_𝑗} is implied in the proof of (1).

In particular, according to Theorem 1.2 the following corollary holds.

Corollary 1.3. Suppose that 𝐷⊂ℝ^𝑛 is a symmetric convex set with nonempty interior, 𝑓∶𝐷→ℝ is a symmetric function with continuous partial derivatives, and 𝐷^⌣₁=𝐱∈𝐷∣𝑥₁=max_{1≤𝑗≤𝑛}𝑥_𝑗−𝐱∈𝐷∣𝑥₁=𝑥₂=⋯=𝑥_𝑛,_⌢𝐷₂=𝐱∈𝐷∣𝑥₂=min_{1≤𝑗≤𝑛}𝐷_∗𝑥_𝑗−𝐱∈𝐷∣𝑥₁=𝑥₂=⋯=𝑥_𝑛,=𝐷^⌣₁∩𝐷^⌢₂.(1.16) If 𝜕𝑓/𝜕𝑥₁>(<)𝜕𝑓/𝜕𝑥₂ holds in 𝐷^∗, then 𝑓𝑎₁,𝑎₂,…,𝑎_𝑛≥(≤)𝑓(𝐴(𝐚),𝐴(𝐚),…,𝐴(𝐚))(1.17) for all 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈𝐷, and equality holds if and only if 𝑎₁=𝑎₂=⋯=𝑎_𝑛.

2. Comparing with Schur's Condition

The Schur convexity was introduced by I. Schur [2] in 1923; the following Definitions 2.1 and 2.2 can be found in [2, 3].

Definition 2.1. For 𝐮=(𝑢₁,𝑢₂,…𝑢_𝑛),𝐯=(𝑣₁,𝑣₂,…𝑣_𝑛)∈ℝ^𝑛, without loss of generality one assumes that 𝑢₁≥𝑢₂≥⋯≥𝑢_𝑛 and 𝑣₁≥𝑣₂≥⋯≥𝑣_𝑛. Then 𝐮 is said to be majorized by 𝐯 (in symbols 𝐮≺𝐯) if ∑^𝑘𝑖=1𝑢_𝑖≤∑^𝑘𝑖=1𝑣_𝑖 for 𝑘=1,2,…,𝑛−1 and ∑^𝑛𝑖=1𝑢_𝑖=∑^𝑛𝑖=1𝑣_𝑖.

Definition 2.2. Suppose that Ω⊂ℝ^𝑛. A real-valued function 𝜑∶Ω→ℝ is said to be Schur convex (Schur concave) if 𝐮≺𝐯 implies that 𝜑(𝐮)≤(≥)𝜑(𝐯).
Recall that the following so-called Schur's condition is very useful for determining whether or not a given function is Schur convex or concave.

Theorem 2.3 (see [2, page 57]). Suppose that Ω⊂ℝ^𝑛 is a symmetric convex set with nonempty interior intΩ. If 𝜑∶Ω→ℝ is continuous on Ω and differentiable in intΩ, then 𝜑 is Schur convex (Schur concave) on Ω if and only if it is symmetric and 𝑢₁−𝑢₂𝜕𝜕𝑢𝜑₁−𝜕𝜕𝑢𝜑₂≥(≤)0(2.1) holds for any 𝑢=(𝑢₁,𝑢₂,…,𝑢_𝑛)∈intΩ.

It is well known that a convex function is not necessarily a Schur convex function, and a Schur convex function need notbe convex in the ordinary sense either. However, under the assumption of ordinary convexity, 𝑓 is Schur convex if and only if it is symmetric [4].

Although the Schur convexity is an important tool in researching analytic inequalities, but the restriction of symmetry cannot be used in dealing with nonsymmetric functions. Obviously, Theorem 1.2 is the generalization and development of Theorem 2.3; the following results in Sections 3–5 show that a large number of inequalities can be proved, improved, and found by Theorem 1.2.

3. A Proof for the Hölder Inequality

Using Theorem 1.2 and Corollary 1.3, we can prove some well-known inequalities, for example, power mean inequality, Hölder inequality, and Minkowski inequality. In this section, we only prove the Hölder inequality.

Proposition 3.1 (Hölder inequality). Suppose that 𝑥₁,𝑥₂,…,𝑥_𝑛,𝑦₁,𝑦₂,…,𝑦_𝑛∈ℝ^𝑛+.(3.1) If 𝑝,𝑞>1 and 1/𝑝+1/𝑞=1, then ^𝑛_𝑘=1𝑥^𝑝𝑘^1/𝑝^𝑛_𝑘=1𝑦^𝑞𝑘^1/𝑞≥^𝑛_𝑘=1𝑥_𝑘𝑦_𝑘.(3.2)

Proof. Let (𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+ and 𝑓∶𝐛⟶^𝑛_𝑘=10𝑥0200𝑑𝑎_𝑘^1/𝑝^𝑛_𝑘=1𝑎_𝑘𝑏_𝑘^1/𝑞−^𝑛_𝑘=1𝑎_𝑘𝑏^𝑘1/𝑞,𝑏∈ℝ^𝑛+.(3.3) Then 𝜕𝑓𝜕𝑏_𝑖=1𝑞⋅^𝑛_𝑘=1𝑎_𝑘^1/𝑝^𝑛_𝑘=1𝑎_𝑘𝑏_𝑘^1/𝑞−1𝑎_𝑖−1𝑞⋅𝑎^𝑖𝜕𝑓𝑏^{𝑖1/𝑞−1},𝜕𝑏_𝑖−𝜕𝑓𝜕𝑏_𝑗=1𝑞∑^𝑛𝑘=1𝑎_𝑘∑^𝑛𝑘=1𝑏_𝑘𝑎_𝑘^1/𝑝𝑎_𝑖−𝑎_𝑗−1𝑞𝑎^𝑖𝑏^{𝑖−1/𝑝}−𝑎_𝑗𝑏^{𝑗−1/𝑝}.(3.4) Let 𝐛∈𝐷^⌣_𝑖∩𝐷^⌢_𝑗 (see (1.5)).
(1) If 𝑎_𝑖≥𝑎_𝑗, then 𝜕𝑓𝜕𝑏_𝑖−𝜕𝑓𝜕𝑏_𝑗≥1𝑞∑^𝑛𝑘=1𝑎_𝑘𝑏_𝑖∑^𝑛𝑘=1𝑎_𝑘^1/𝑝=1𝑎_𝑖−𝑎_𝑗−1𝑞𝑎^𝑖𝑏^{𝑖−1/𝑝}−𝑎_𝑗𝑏^{𝑗−1/𝑝}𝑞𝑎^𝑗𝑏^{𝑗−1/𝑝}>0.−𝑏^{𝑖−1/𝑝}(3.5)
(2) If 𝑎_𝑖≤𝑎_𝑗, then 𝜕𝑓𝜕𝑏_𝑖−𝜕𝑓𝜕𝑏_𝑗≥1𝑞∑^𝑛𝑘=1𝑎_𝑘𝑏_𝑗∑^𝑛𝑘=1𝑎_𝑘^1/𝑝=1𝑎_𝑖−𝑎_𝑗−1𝑞𝑎^𝑖𝑏^{𝑖−1/𝑝}−𝑎_𝑗𝑏^{𝑗−1/𝑝}𝑞𝑎^𝑖𝑏^{𝑗−1/𝑝}>0.−𝑏^{𝑖−1/𝑝}(3.6) From Theorem 1.2 we get 𝑓(𝐛)≥𝑓(𝐴(𝐛),𝐴(𝐛),…,𝐴(𝐛)),(3.7) that is, ^𝑛_𝑘=1𝑎_𝑘^1/𝑝^𝑛_𝑘=1𝑎_𝑘𝑏_𝑘^1/𝑞−^𝑛_𝑘=1𝑎_𝑘𝑏^𝑘1/𝑞≥0.(3.8) Therefore, the Hölder inequality follows from (3.8) with 𝑎_𝑘=𝑥^𝑝𝑘 and 𝑏_𝑘=𝑦^𝑞𝑘/𝑥^𝑝𝑘.

4. Improvement of the Sierpiński Inequality

In the section, we give some improvements of the well-known Sierpiński inequality:

𝑀₋₁(𝐚)^{(𝑛−1)/𝑛}[𝐴(𝐚)]^1/𝑛≤𝑀₀(𝐚)≤𝑀₋₁(𝐚)^1/𝑛[𝐴(𝐚)]^{(𝑛−1)/𝑛}.(4.1)

Theorem 4.1. Suppose that 𝑛≥3, 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+, 𝛽>0>𝛼. If 𝜆=−2𝛼/𝑛(𝛽−𝛼) for 𝛽+𝛼>0 and 𝜆=1/𝑛 for 𝛽+𝛼≤0, then 𝑀_𝛼(𝐚)^1−𝜆⋅𝑀_𝛽(𝐚)^𝜆≤𝑀₀(𝐚).(4.2)

Proof. Let 𝑓(𝐱)=(1/𝑛𝛽)ln(∏^𝑛𝑖=1𝑥_𝑖)−((1−𝜆)/𝛼)ln((1/𝑛)∑^𝑛𝑖=1𝑥^{𝑖𝛼/𝛽}), 𝐱∈ℝ^𝑛+. Then 𝜕𝑓(𝐱)𝜕𝑥_𝑗=1𝑛𝛽𝑥_𝑗−1−𝜆𝛽∑𝑥^𝑛𝑖=1𝑥^{𝑗𝛼/𝛽−1𝑖𝛼/𝛽}𝜕𝑓(𝐱),𝑗=1,2,𝜕𝑥₁−𝜕𝑓(𝐱)𝜕𝑥₂=𝑥²−𝑥₁𝑛𝛽𝑥₁𝑥₂−1−𝜆𝛽𝑥^{1𝛼/𝛽−1}−𝑥^{2𝛼/𝛽−1}∑^𝑛𝑖=1𝑥^{𝑖𝛼/𝛽}.(4.3)
Case 1. 𝛼+𝛽>0. Let 𝑔(𝑡)=𝛽+𝛼𝛽−𝛼𝑡^𝛽−𝛼−𝑡^𝛽+𝑡^−𝛼−𝛽+𝛼𝛽−𝛼,𝑡∈(1,+∞).(4.4) Then 𝑡𝑡^𝛼+1𝑔^(𝑡)=(𝛽+𝛼)𝑡^𝛽−𝛽𝑡^𝛽+𝛼−𝛼,_𝛼+1𝑔^(𝑡)^=(𝛽+𝛼)𝛽𝑡^{𝛽+𝛼−1}(𝑡^−𝛼−1)>0.(4.5) Therefore, 𝑡^𝛼+1𝑔^(𝑡) is monotone increasing in (1,+∞). From lim_𝑡→1+𝑡^𝛼+1𝑔^(𝑡)=lim_𝑡→1+(𝛽+𝛼)𝑡^𝛽−𝛽𝑡^𝛽+𝛼−𝛼=0,(4.6) we know that 𝑡^𝛼+1𝑔^(𝑡)>0, 𝑔^(𝑡)>0. Then lim_𝑡→1+𝑔(𝑡)=0 leads to 𝑔(𝑡)>0 and 𝛽+𝛼𝛽−𝛼𝑡^𝛽−𝛼−𝑡^𝛽+𝑡^−𝛼−𝛽+𝛼𝛽+𝛼𝛽−𝛼>0,𝛽−𝛼𝑡^𝛽−1+2𝛼𝛽−𝛼𝑡^𝛼𝛽+𝛼−𝑡^𝛼+𝛽+1>0,𝛽−𝛼𝑡^𝛽−𝑛−1+2𝛼𝛽−𝛼𝑡^𝛼(1−𝑛𝜆)𝑡_𝛽−𝑡^𝛼+𝛽+(𝑛−1)>0,(1−−(𝑛𝜆−)11−−𝑛𝑡𝜆)𝑡^𝛼_𝛼−𝛽−𝑡^𝛼+𝛽+(𝑛−1)>0,(4.7)𝑡^𝛼+(𝑛−1)>𝑡^𝛽−1𝑛𝑡^𝛽.(4.8) We assume that 𝐱∈𝐷^∗ (see (1.16)). Let 𝑡=(𝑥₁/𝑥₂)^1/𝛽. Then inequality (4.8) becomes (1−𝜆)𝑥^{2𝛼/𝛽−1}−𝑥^{1𝛼/𝛽−1}𝑥^1𝛼/𝛽+(𝑛−1)𝑥^2𝛼/𝛽>𝑥¹−𝑥₂𝑛𝑥₁𝑥₂1−𝜆,𝛽𝑥^{2𝛼/𝛽−1}−𝑥^{1𝛼/𝛽−1}∑^𝑛𝑖=1𝑥^{𝑖𝛼/𝛽}>𝑥¹−𝑥₂𝑛𝛽𝑥₁𝑥₂.(4.9) Combining inequalities (4.3) and (4.9) yields that 𝜕𝑓(𝐱)/𝜕𝑥₁−𝜕𝑓(𝐱)/𝜕𝑥₂>0. Using Corollary 1.3 we have 1𝑓𝑥₁,𝑥₂,…,𝑥_𝑛≥𝑓(𝐴(𝐱),𝐴(𝐱),…,𝐴(𝐱)),𝑛𝛽ln^𝑛_𝑖=1𝑥_𝑖−1−𝛼𝜆ln1𝑛^𝑛_𝑖=1𝑥^{𝑖𝛼/𝛽}≥𝛽𝜆ln1𝑛^𝑛_𝑖=1𝑥_𝑖.(4.10) Letting 𝑎_𝑖=𝑥^𝑖1/𝛽,𝑖=1,2,…,𝑛, we get 𝑀_𝛼(𝐚)^1−𝜆⋅𝑀_𝛽(𝐚)^𝜆≤𝑀₀(𝐚).(4.11)Case 2. 𝛼+𝛽<0. Let 𝑡>1. Then from 𝛼<0 and 𝛼+𝛽<0, one has (𝑛−1)>(𝑛−2)𝑡^𝛼+𝑡^𝛼+𝛽.(4.12) Hence inequality (4.8) holds. The rest is similar to above, so we omit it.
The proof of Theorem 4.2 is similar to the proof of Theorem 4.1, and so we omit it.

Theorem 4.2. Suppose that 𝑛≥3, 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+, 𝛽>0>𝛼. If 𝜃=(𝑛−1)/𝑛 for 𝛽+𝛼>0 and 𝜃=1−2𝛽/𝑛(𝛽−𝛼) for 𝛽+𝛼≤0, then 𝑀₀(𝐚)≤𝑀_𝛼(𝐚)^1−𝜃⋅𝑀_𝛽(𝐚)^𝜃.(4.13)

Theorem 4.3. Suppose that 𝑛≥3, 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+. If 𝑟=−ln𝑛/(𝑛−1)[ln𝑛−ln(𝑛−1)], then 𝑟<−1 and 𝑀_1/𝑟(𝐚)^{(𝑛−1)/𝑛}[𝐴(𝐚)]^1/𝑛≤𝑀₀(𝐚)≤𝑀_𝑟(𝐚)^1/𝑛[𝐴(𝐚)]^{(𝑛−1)/𝑛}.(4.14)

Proof. Let 𝑛≥3 and 𝑓∶𝐱∈(0,+∞)^𝑛⟶∑^𝑛𝑘=1𝑥^𝑘1/𝑟𝑛∑^𝑛𝑘=1𝑥_𝑘𝑛^{1/𝑟(𝑛−1)}−^𝑛_𝑘=1𝑥_𝑘^{1/𝑟(𝑛−1)}.(4.15) Then 𝑟=−ln𝑛ln(1+1/(𝑛−1))^𝑛−1<−ln𝑛𝜕𝑓ln𝑒=−ln𝑛<−1,𝜕𝑥₁=𝑥^{1(1−𝑟)/𝑟}𝑟𝑛∑^𝑛𝑘=1𝑥_𝑘𝑛^{1/𝑟(𝑛−1)}+∑^𝑛𝑘=1𝑥^𝑘1/𝑟𝑟𝑛²(𝑛−1)∑^𝑛𝑘=1𝑥_𝑘𝑛^{(1/𝑟(𝑛−1))−1}−1𝑟(𝑛−1)𝑥₁^𝑛_𝑘=1𝑥_𝑘^{1/𝑟(𝑛−1)}.(4.16) Therefore, we get 𝜕𝑓𝜕𝑥₁−𝜕𝑓𝜕𝑥₂=1𝑟𝑛𝑥^{1(1−𝑟)/𝑟}−𝑥^{2(1−𝑟)/𝑟}∑^𝑛𝑘=1𝑥_𝑘𝑛^{1/𝑟(𝑛−1)}−1𝑟(𝑛−1)^𝑛_𝑘=1𝑥_𝑘^{1/𝑟(𝑛−1)}𝑥1₁−1_𝑛𝑥₂,_𝑘=1𝑥_𝑘^{1/(−𝑟(𝑛−1))}𝜕𝜕𝑥𝑓₁−𝜕𝑓𝜕𝑥₂=𝑥^{1(1−𝑟)/(−𝑟)}−𝑥^{2(1−𝑟)/(−𝑟)}−𝑟𝑛𝑥^{1(1−𝑟)/(−𝑟)}𝑥^{2(1−𝑟)/(−𝑟)}𝑛∏^𝑛𝑘=1𝑥_𝑘∑^𝑛𝑘=1𝑥_𝑘^{1/(−𝑟)(𝑛−1)}+𝑥¹−𝑥₂𝑟(𝑛−1)𝑥₁𝑥₂=𝑥^{1(1−𝑟)/(−𝑟)}−𝑥^{2(1−𝑟)/(−𝑟)}−𝑟𝑛𝑥^{1(1−𝑟)/(−𝑟)}𝑥^{2(1−𝑟)/(−𝑟)}∑^𝑛𝑘=1∏𝑛^{𝑛𝑖=1,≠𝑘}𝑥^𝑖−1^{1/(−𝑟)(𝑛−1)}+𝑥¹−𝑥₂𝑟(𝑛−1)𝑥₁𝑥₂.(4.17) We assume that 𝐱∈𝐷^∗ (see (1.16)). Then we have ^𝑛_𝑘=1𝑥_𝑘^{1/−𝑟(𝑛−1)}𝜕𝜕𝑥𝑓₁−𝜕𝑓≥𝑥_{1(1−𝑟)/(−𝑟)}𝜕𝑥₂−𝑥^{2(1−𝑟)/(−𝑟)}−𝑟𝑛𝑥^{1(1−𝑟)/(−𝑟)}𝑥^{2(1−𝑟)/(−𝑟)}𝑥^{2−(𝑛−1)}+(𝑛−𝑛1)𝑥¹⁻¹𝑥^{2−(𝑛−2)}^{1/(−𝑟)(𝑛−1)}+𝑥¹−𝑥₂𝑟(𝑛−1)𝑥₁𝑥₂.(4.18) Letting 𝑥₁/𝑥₂=𝑡>1, from 𝑛^{1+1/𝑟(𝑛−1)}=𝑛^{1−(ln𝑛−ln(𝑛−1))/ln𝑛}=𝑛−1, we get ^𝑛_𝑘=1𝑥_𝑘^{1/−𝑟(𝑛−1)}𝜕𝜕𝑥𝑓₁−𝜕𝑓≥1𝜕𝑥₂𝑛−1⋅𝑥^{1(1−𝑟)/(−𝑟)}−𝑥^{2(1−𝑟)/(−𝑟)}−𝑟𝑥^{1(1−𝑟)/(−𝑟)}𝑥^{2(1−𝑟)/(−𝑟)}𝑥¹𝑥^2𝑛−1𝑥₁+(𝑛−1)𝑥₂^{(ln𝑛−ln(𝑛−1))/ln𝑛}+𝑥¹−𝑥₂𝑟(𝑛−1)𝑥₁𝑥₂=1−𝑟(𝑛−1)𝑥₂𝑡^{(1−𝑟)/(−𝑟)}−1𝑡^{(1−𝑟)/(−𝑟)}𝑡+𝑛𝑡−1^{(ln𝑛−ln(𝑛−1))/ln𝑛}−𝑡−1=1𝑡−𝑟(𝑛−1)𝑥₂𝑡^{(1−𝑟)/(−𝑟)}−1𝑡^{(1−𝑟)/(−𝑟)}⋅1+𝑛−1𝑡^{−(ln𝑛−ln(𝑛−1))/ln𝑛}−𝑡−1𝑡.(4.19)
According to Bernoulli's inequality (1+𝑥)^𝛼<1+𝛼𝑥 with 𝑥≥−1,𝑥≠0, and 0<𝛼<1, one has ^𝑛_𝑘=1𝑥_𝑘^{1/−𝑟(𝑛−1)}𝜕𝜕𝑥𝑓₁−𝜕𝑓>1𝜕𝑥₂−𝑟(𝑛−1)𝑥₂𝑡^{(1−𝑟)/(−𝑟)}−1𝑡^{(1−𝑟)/(−𝑟)}⋅1=11+(ln𝑛−ln(𝑛−1))/ln𝑛⋅(𝑛−1)/𝑡−𝑡−𝑡1−𝑟(𝑛−1)𝑥₂𝑡^{(1−𝑟)/(−𝑟)}−1=1𝑡^{(1−𝑟)/(−𝑟)}−𝑡^1/(−𝑟)/𝑟−𝑡−𝑡1−𝑟(𝑛−1)𝑥₂⋅(1+1/𝑟)𝑡^1/(−𝑟)−1/𝑟⋅𝑡^{(1+𝑟)/(−𝑟)}−1𝑡^{(1−𝑟)/(−𝑟)}−𝑡^1/(−𝑟)/𝑟.(4.20) For 0<𝑠<1 and 𝑡>1, it is not difficult to verify that (1−𝑠)𝑡^𝑠+𝑠𝑡^𝑠−1−1>0. Letting 𝑠=−1/𝑟, we have 1+1𝑟𝑡𝜕𝑓^1/(−𝑟)−1𝑟⋅𝑡^{(1+𝑟)/(−𝑟)}−1>0,𝜕𝑥₁−𝜕𝑓𝜕𝑥₂>0.(4.21) Using Corollary 1.3, we know that 𝑓∑𝑥₁,𝑥₂,…,𝑥_𝑛≥𝑓(𝐴(𝐱),𝐴(𝐱),…,𝐴(𝐱)),_𝑛𝑘=1𝑥^𝑘1/𝑟𝑛∑^𝑛𝑘=1𝑥_𝑘𝑛^{1/𝑟(𝑛−1)}≥^𝑛_𝑘=1𝑥_𝑘^{1/𝑟(𝑛−1)}.(4.22) Letting 𝑎_𝑖=𝑥^𝑖1/𝑟(𝑖=1,2,…,𝑛), we get ∑^𝑛𝑘=1𝑎_𝑘𝑛∑^𝑛𝑘=1𝑎^𝑟𝑘𝑛^{1/𝑟(𝑛−1)}≥^𝑛_𝑘=1𝑎_𝑘^1/(𝑛−1)[𝐴(𝐚)]_{(𝑛−1)/𝑛},(4.23)𝑀_𝑟(𝐚)^1/𝑛≥𝑀₀(𝐚).(4.24)
From (4.23), we get ^𝑛_𝑘=1𝑎_𝑘^𝑟≥∑^𝑛𝑘=1𝑎_𝑘𝑛^{(𝑛−1)𝑟}⋅∑^𝑛𝑘=1𝑎^𝑟𝑘𝑛.(4.25) Letting 𝑎_𝑖→𝑎^𝑖1/𝑟(𝑖=1,2,…,𝑛), we have ^𝑛_𝑘=1𝑎_𝑘≥∑^𝑛𝑘=1𝑎^𝑘1/𝑟𝑛^{(𝑛−1)𝑟}⋅∑^𝑛𝑘=1𝑎_𝑘𝑀₀𝑛,(𝐚)≥𝑀_1/𝑟(𝐚)^{(𝑛−1)/𝑛}[𝐴(𝐚)]^1/𝑛.(4.26) Inequality (4.14) is proved.

5. Five New Inequalities

Let 𝑛≥3 and 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+. Then

^𝑘_𝑛(𝐚)=⎛⎜⎜⎝_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗⎞⎟⎟⎠^𝑛𝑘)1/((5.1)

References [5, 6] is the third symmetric mean of 𝐚.

Theorem 5.1. If 2≤𝑘≤𝑛−1, 𝑝=(𝑘−1)/(𝑛−1), then ^𝑘_𝑛(𝐚)≥[𝐴(𝐚)]^𝑝𝑀₀(𝐚)^1−𝑝(5.2) with the best possible constant 𝑝=(𝑘−1)/(𝑛−1).

Proof. Let 𝑎=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+ and 𝑓(𝐚)=^𝑛_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗.(5.3) Then 𝜕𝑓𝜕𝑎₁=−(𝑛−𝑘)⋅(^𝑛𝑘)𝑛(𝑛−1)𝑎₁^𝑛_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗+^𝑛_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗⋅⎛⎜⎜⎝_{2≤𝑖1<⋯𝑖𝑘−1≤𝑛}1𝑎₁+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗⎞⎟⎟⎠,(5.4)𝜕𝑓𝜕𝑎₁−𝜕𝑓𝜕𝑎₂=−(𝑛−𝑘)⋅(^𝑛𝑘)𝑛(𝑛−1)^𝑛_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗𝑎1₁−1+_𝑛𝑎₂_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗⋅⎛⎜⎜⎝_{3≤𝑖1<⋯𝑖𝑘−1≤𝑛}⎛⎜⎜⎝1𝑎₁+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗−1𝑎₂+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗=𝑎₁⎞⎟⎟⎠⎞⎟⎟⎠−𝑎₂⋅^𝑛_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}1𝑘^𝑘_𝑗=1𝑎_𝑖𝑗⋅⎡⎢⎢⎣(𝑛−𝑘)⋅(^𝑛𝑘)𝑛(𝑛−1)𝑎₁𝑎₂−_{3≤𝑖1<⋯𝑖𝑘−1≤𝑛}1𝑎¹+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗𝑎²+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗⎤⎥⎥⎦.(5.5) If 𝐚∈𝐷^∗ (see (1.16)), then (𝑛−𝑘)⋅𝑎(_1𝑛𝑘+(𝑘−1)𝑎₂>𝑎₁,)𝑛(𝑛−1)𝑎₁𝑎₂>^{𝑛−2𝑘−1}(𝑛−𝑘)⋅(_𝑛𝑘𝑘𝑎₂𝑎₁+(𝑘−1)𝑎₂,)𝑛(𝑛−1)𝑎₁𝑎₂>_{3≤𝑖1<⋯𝑖𝑘−1≤𝑛}1𝑎₁+(𝑘−1)𝑎₂𝑘𝑎₂(𝑛−𝑘)⋅(_𝑛𝑘,)𝑛(𝑛−1)𝑎₁𝑎₂>_{3≤𝑖1<⋯𝑖𝑘−1≤𝑛}1𝑎¹+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗𝑎²+∑^{𝑘−1𝑖=1}𝑎_𝑖𝑗.(5.6)
Combining inequalities (5.5) and (5.6) yields that 𝜕𝑓/𝜕𝑎₁−𝜕𝑓/𝜕𝑎₂>0. Then from Corollary 1.3 we have 𝑓𝑎₁,𝑎₂,…,𝑎_𝑛≥𝑓(𝐴(𝐚),𝐴(𝐚),…,𝐴(𝐚))(5.7) for all 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+, which implies that ^𝑛_𝑖=1𝑎_𝑖^{𝑛𝑘−(𝑛−𝑘)⋅()/𝑛(𝑛−1)}⋅_{1≤𝑖1<⋯<𝑖𝑘≤𝑛}𝑘^𝑘−1_𝑗=1𝑎_𝑖𝑗≥[𝐴(𝐚)]^{𝑛𝑘(𝑘−1)⋅()/(𝑛−1)}.(5.8) Therefore, inequality (5.2) is proved.

Taking 𝑎₁=𝑎₂=⋯=𝑎_𝑛−1=1 and 𝑎_𝑛=𝑥 in inequality (5.2), we get

𝑥+𝑝≤𝑘𝑘−(𝑘1/𝑛)ln((𝑥^{(𝑛𝑘)𝑛−1𝑘−1)/(}+≥𝑘−𝑥1)+/𝑘𝑛𝑛)−−1(1/^𝑝𝑛)√l^𝑛n𝑥𝑥^1−𝑝,ln(𝑥+𝑛−1)−ln𝑛√^𝑛𝑥.(5.9) Letting 𝑥→+∞, we get

𝑝≤lim_𝑥→+∞𝑘/𝑛⋅1/(𝑥+𝑘−1)−1/𝑛𝑥1/(𝑥+𝑛−1)−1/𝑛𝑥=lim^𝑥→+∞𝑘𝑥/(𝑥+𝑘−1)−1𝑛𝑥/(𝑥+𝑛−1)−1=𝑘𝑛−−11.(5.10) So 𝑝=(𝑘−1)/(𝑛−1) is the best possible constant.

For 𝑛≥2, 𝐚=(𝑎₁,𝑎₂,…,𝑎_𝑛)∈ℝ^𝑛+, Alzer [7] established the following inequality:

𝑛−1𝑛𝐴(𝐚)+1𝑛𝑀⁻¹(𝐚)≥𝑀₀(𝐚).(5.11)

Theorems 5.2 and 5.3 are the improvements of Alzer's inequality.

Theorem 5.2. If 𝑝=𝑛²/(𝑛²+4𝑛−4), then 𝑝𝐴(𝐚)+(1−𝑝)𝑀₋₁(𝐚)≥𝑀₀(𝐚).(5.12)

Proof. Firstly, let 𝑝>𝑛²/(𝑛²+4𝑛−4), and 𝑓(𝐱)=𝑝/𝑛⋅^𝑛_𝑖=1𝑒^𝑥𝑖+(1−𝑝)𝑛⋅^𝑛_𝑖=1𝑒^−𝑥𝑖⁻¹,𝑥=𝑥₁,𝑥₂,…,𝑥_𝑛∈ℝ^𝑛.(5.13) Then 𝜕𝑓𝜕𝑥₁=𝑝𝑛𝑒^𝑥1+(1−𝑝)𝑛∑^𝑛𝑖=1𝑒^−𝑥𝑖²𝜕𝑓𝑒^−𝑥1,𝜕𝑥₁−𝜕𝑓𝜕𝑥₂=𝑝𝑛(𝑒^𝑥1−𝑒^𝑥2)−(1−𝑝)𝑛∑^𝑛𝑖=1𝑒^−𝑥𝑖²(𝑒^−𝑥2−𝑒^−𝑥1).(5.14) If 𝑥₁=max_{1≤𝑖≤𝑛}{𝑥_𝑖}>𝑥₂=min_{1≤𝑖≤𝑛}{𝑥_𝑖}, 𝑡=𝑒^{𝑥1−𝑥2}>1, then 𝜕𝑓𝜕𝑥₁−𝜕𝑓𝜕𝑥₂≥𝑝𝑛(𝑒^𝑥1−𝑒^𝑥2)−(1−𝑝)𝑛((𝑛−1)𝑒^−𝑥1+𝑒^−𝑥2)²=(𝑒_𝑥1(𝑒^−𝑥2−𝑒^−𝑥1)−𝑒^𝑥2)𝑛((𝑛−1)𝑒^𝑥2+𝑒^𝑥1)²=𝑒_3𝑥2𝑝((𝑛−1)𝑒^𝑥2+𝑒^𝑥1)²−(1−𝑝)𝑛²𝑒^𝑥1𝑒^𝑥2(𝑡−1)𝑛((𝑛−1)𝑒^𝑥2+𝑒^𝑥1)²>𝑒_3𝑥2𝑝(𝑛−1+𝑡)²−𝑛²𝑡+𝑝𝑛²𝑡(𝑡−1)𝑛((𝑛−1)𝑒^𝑥2+𝑒^𝑥1)²𝑛²+𝑛4²𝑛−4(𝑛−1+𝑡)²−𝑛²𝑡+𝑛²𝑛²+4𝑛−4𝑛²=𝑛𝑒_3𝑥2𝑡(𝑡−1)(𝑡−𝑛+1)²𝑛²+4𝑛−4((𝑛−1)𝑒^𝑥2+𝑒^𝑥1)²≥0.(5.15) Then from Corollary 1.3, we get 𝑝𝑓(𝐱)≥𝑓(𝐴(𝐱),𝐴(𝐱),…,𝐴(𝐱)),𝑛^𝑛_𝑖=1𝑒^𝑥𝑖+(1−𝑝)𝑛∑^𝑛𝑖=1𝑒^−𝑥𝑖≥𝑒^𝐴(𝐱)=^𝑛⎷_𝑛_𝑖=1𝑒^𝑥𝑖.(5.16) Let 𝑒^𝑥𝑖=𝑎_𝑖 in above inequality. Then we know that inequality (5.12) holds. From continuity we know that inequality (5.12) holds also for 𝑝=𝑛²/(𝑛²+4𝑛−4).

Theorem 5.3. If 𝑝=(1−𝑛−√5𝑛²−6𝑛+1)/(2𝑛), then 𝑛−1𝑛𝐴(𝐚)+1𝑛𝑀^𝑝(𝐚)≥𝑀₀(𝐚).(5.17)

Proof. Let 𝑓(𝐚)=^𝑛⎷_𝑛_𝑖=1𝑎^𝑖1/𝑝−(𝑛−1)𝑛²⋅^𝑛_𝑖=1𝑎^𝑖1/𝑝,𝐚∈ℝ^𝑛+.(5.18) Then 𝜕𝑓𝜕𝑎₁=1𝑛𝑝𝑎₁^𝑛⎷_𝑛_𝑖=1𝑎^𝑖1/𝑝−𝑛−1𝑛²𝑝𝑎^11/𝑝−1𝜕𝑓,𝜕𝑎₁−𝜕𝑓𝜕𝑎₂=−𝑎¹−𝑎₂𝑛𝑝𝑎₁𝑎_2𝑛𝑖=1𝑎^{𝑖1/𝑛𝑝}−𝑛−1𝑛²𝑝𝑎^11/𝑝−1−𝑎^21/𝑝−1.(5.19) If 𝑎₁=max_{1≤𝑖≤𝑛}{𝑎_𝑖}>𝑎₂=min_{1≤𝑖≤𝑛}{𝑎_𝑖}>0 and 𝑎₁/𝑎₂=𝑡>1, then from 𝑝<0 and −(𝑎₁−𝑎₂)/𝑛𝑝𝑎₁𝑎₂>0 we get 𝜕𝑓𝜕𝑎₁−𝜕𝑓𝜕𝑎₂≤−𝑎¹−𝑎₂𝑛𝑝𝑎₁𝑎₂𝑎^{11/(𝑛𝑝)}𝑎^{2(𝑛−1)/(𝑛𝑝)}−𝑛−1𝑛²𝑝𝑎^11/𝑝−1=𝑎_11/𝑝−1−𝑎^21/𝑝−1𝑛²𝑝−𝑛𝑡−𝑡1𝑡^{1−(𝑛−1)/(𝑛𝑝)}−(𝑛−1)1−𝑡^1−1/𝑝.(5.20)
Let 𝑔(𝑡)=−𝑛𝑡^{1−(𝑛−1)/(𝑛𝑝)}+𝑛𝑡^{−(𝑛−1)/(𝑛𝑝)}+(𝑛−1)𝑡^1−1/𝑝−(𝑛−1),𝑡>1. Then 𝑔^(𝑡)=−𝑛+𝑛−1𝑝𝑡^{−(𝑛−1)/(𝑛𝑝)}−𝑛−1𝑝𝑡^{−1−(𝑛−1)/(𝑛𝑝)}+(𝑛−1)1−1𝑝𝑡^−1/𝑝𝑡_{1+(𝑛−1)/(𝑛𝑝)},𝑔^(𝑡)=−𝑛+𝑛−1𝑝𝑡−𝑛−𝑝1+(𝑛−1)1−1𝑝𝑡^{1−1/(𝑛𝑝)}𝑡_{1+(𝑛−1)/(𝑛𝑝)},𝑔^(𝑡)^=−𝑛+𝑛−1𝑝+(𝑛−1)1−1𝑝1−𝑛1𝑝𝑡^{−1/(𝑛𝑝)}>−𝑛+𝑛−1=−1𝑝+(𝑛−1)1−1𝑝1−𝑛1𝑝𝑝²𝑝²+1−1=0.𝑛𝑝−1+1𝑛(5.21) Thus 𝑡^{1+(𝑛−1)/(𝑛𝑝)}𝑔^(𝑡) is a monotone increasing function. This monotonicity and lim_𝑡→1+𝑡^{1+(𝑛−1)/(𝑛𝑝)}𝑔^(𝑡)=lim_𝑡→1+=−1−𝑛−−1𝑛+𝑛−𝑝1𝑡−𝑛−𝑝1+(𝑛−1)1−1𝑝𝑡^{1−(1/𝑛𝑝)}≥0𝑝(5.22) lead to 𝑡^{1+(𝑛−1)/(𝑛𝑝)}𝑔^(𝑡)>0. Therefore 𝑔^(𝑡)>0 and 𝑔(𝑡) is a monotone increasing function. From lim_𝑡→1+𝑔(𝑡)=0 and the monotonicity of 𝑔(𝑡) we know that 𝑔(𝑡)>0. By (5.20), we know that 𝜕𝑓/𝜕𝑎₁−𝜕𝑓/𝜕𝑎₂<0. According to Corollary 1.3 we get 𝑓(𝐚)≤𝑓(𝐴(𝐚),𝐴(𝐚),…,𝐴(𝐚)),_𝑛⎷_𝑛_𝑖=1𝑎^𝑖1/𝑝−𝑛−1𝑛²⋅^𝑛_𝑖=1𝑎^𝑖1/𝑝≤1𝑛⋅𝐴^1/𝑝(𝐚).(5.23) Finally, let 𝑎_𝑖→𝑎^𝑝𝑖(𝑖=1,2,…,𝑛) in the above inequality. Then we know that Theorem 5.3 holds.
If 𝑛≥2 and 0<𝑎₁≤𝑎₂≤⋯≤𝑎_𝑛, then the following inequalities can be found in [8–10]: ∑_{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²𝑎_𝑛≤𝐴(𝐚)−𝑀₀(𝐚)≤∑^{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²𝑎₁𝑎₃₁,2𝑛²𝑎^4𝑛_{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²≤𝑀₀(𝐚)−𝑀₋₁(𝐚)≤𝑎^3𝑛2𝑛²𝑎⁴¹_{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗².(5.24) Theorems 5.4 and 5.5 are the improvements of inequalities (5.24).

Theorem 5.4. If 𝑛≥2 and 0<𝑚≤𝑎₁,𝑎₂,…,𝑎_𝑛≤𝑀, then ∑_{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²⋅𝑀^{(𝑛−1)/𝑛}⋅𝐴^1/𝑛(𝐚)≤𝐴(𝐚)−𝑀⁰(𝐚)≤∑^{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²⋅𝑚^{(𝑛−1)/𝑛}⋅𝐴^1/𝑛(𝐚).(5.25)

Proof. Let 𝑓∶𝐚∈[𝑚,𝑀]^𝑛⟶1𝑛^𝑛_𝑘=1𝑎_𝑘^1/𝑛⎛⎜⎜⎝1𝑛^𝑛_𝑘=1𝑎_𝑘−^𝑛⎷_𝑛_𝑘=1𝑎_𝑘−∑_{1≤𝑖<𝑗≤𝑛}⎞⎟⎟⎠𝑎_𝑖−𝑎_𝑗²2𝑛²𝑀^{(𝑛−1)/𝑛}.(5.26) Then 𝜕𝑓𝜕𝑎₁=1𝑛²1𝑛^𝑛_𝑘=1𝑎_𝑘^1/𝑛−1⎛⎜⎜⎝1𝑛^𝑛_𝑘=1𝑎_𝑘−^𝑛⎷_𝑛_𝑘=1𝑎_𝑘+1⎞⎟⎟⎠𝑛^𝑛_𝑘=1𝑎_𝑘^1/𝑛⎛⎜⎜⎝1𝑛−𝑛1𝑎₁^𝑛⎷_𝑛_𝑘=1𝑎_𝑘⎞⎟⎟⎠−∑_{2≤𝑖≤𝑛}𝑎₁−𝑎_𝑖𝑛²𝑀^{(𝑛−1)/𝑛}𝜕𝑓,𝜕𝑎₁−𝜕𝑓𝜕𝑎₂=𝑎¹−𝑎₂𝑛𝑎₁𝑎₂^𝑛⎷_𝑛_𝑘=1𝑎_𝑘⋅1𝑛^𝑛_𝑘=1𝑎_𝑘^1/𝑛−𝑎¹−𝑎₂𝑛𝑀^{(𝑛−1)/𝑛}=𝑎¹−𝑎₂𝑛𝑎₁𝑎₂𝑀^{(𝑛−1)/𝑛}⎡⎢⎢⎣𝑀⎷_{(𝑛−1)/𝑛𝑛𝑛}_𝑘=1𝑎_𝑘⋅1𝑛^𝑛_𝑘=1𝑎_𝑘^1/𝑛−𝑎₁𝑎₂⎤⎥⎥⎦.(5.27) We assume that 𝐚∈𝐷^∗ (see (1.16)). Then 𝜕𝑓𝜕𝑎₁−𝜕𝑓𝜕𝑎₂>𝑎¹−𝑎₂𝑛𝑎₁𝑎₂𝑀^{(𝑛−1)/𝑛}𝑀^{(𝑛−1)/𝑛}≥𝑎₁𝑎^11/𝑛𝑎^{2(𝑛−1)/𝑛}⋅𝑎^21/𝑛−𝑎₁𝑎₂−𝑎₂𝑛𝑎₁𝑎₂𝑀^{(𝑛−1)/𝑛}𝑎^{1(𝑛−1)/𝑛}=0.⋅𝑎^11/𝑛𝑎^{2(𝑛−1)/𝑛}⋅𝑎^21/𝑛−𝑎₁𝑎₂(5.28) According to Corollary 1.3, we get 𝐴𝑓(𝐚()𝐚−)≥𝑀𝑓₀(𝐴(𝐚),𝐴(𝐚),…,𝐴(a)),(𝐚)≥∑^{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²𝑀^{(𝑛−1)/𝑛}𝐴^1/𝑛(𝐚).(5.29)
Let 𝑔∶𝐚∈[𝑚,𝑀]⟶∑^{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²𝑚^{(𝑛−1)/𝑛}−1𝑛^𝑛_𝑘=1𝑎_𝑘^1/𝑛⎡⎢⎢⎣1𝑛^𝑛_𝑘=1𝑎_𝑘−^𝑛⎷_𝑛_𝑘=1𝑎_𝑘⎤⎥⎥⎦.(5.30) A similar argument as above leads to 𝐴(𝐚)−𝑀₀(𝐚)≤∑^{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²2𝑛²𝑚^{(𝑛−1)/𝑛}𝐴^1/𝑛(𝐚).(5.31)
The proof of Theorem 5.4 is completed.

Let

𝑓∶𝐱∈1𝑀,𝑚1⟶21𝑛²⋅𝑀^{(𝑛−3)/𝑛}𝑚^{(2𝑛−3)/𝑛}_{1≤𝑖<𝑗≤𝑛}𝑥1_𝑖−1𝑥_𝑗²−1𝑔∶𝐱∈1𝑀₀(𝐱),𝑀,𝑚1⟶𝑀₀1(𝐱)−2𝑛²𝑚𝑀^{(𝑛−1)/𝑛(2𝑛−1)/𝑛}_{1≤𝑖<𝑗≤𝑛}𝑥1_𝑖−1𝑥_𝑗².(5.32)

The proof of Theorem 5.5 is similar to the proof of Theorem 5.4, and so we omit it.

Theorem 5.5. Let 𝑛≥2, 0<𝑚≤𝑎₁,𝑎₂,…,𝑎_n≤𝑀. Then 𝑚^{(𝑛−1)/𝑛}2𝑛²𝑀^{(2𝑛−1)/𝑛}_{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗²≤≤𝑀𝑀₀(𝐚)−𝑀₋₁(𝐚)_{(𝑛−3)/𝑛}2𝑛²𝑚^{(2𝑛−3)/𝑛}_{1≤𝑖<𝑗≤𝑛}𝑎_𝑖−𝑎_𝑗².(5.33)

Remark 5.6. More applications for Theorem 1.2 and Corollary 1.3 were shown in [11].

Acknowledgments

The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions. This research is partly supported by the N. S. Foundation of China under Grant no. 60850005, N. S. Foundation of Zhejiang Province under Grants nos. D7080080 and Y607128, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant no. T200924.