On Satnoianu-Wu’s Inequality

Abstract

By applying techniques in the theory of convex functions and Schur-geometrically convex functions, the author investigates a conjecture of Satnoianu on an algebraic inequality and generalizes some known results in recent years.

1. Introduction

For 𝑛∈ℕ={1,2,3,…} and 𝑥_𝑖∈ℝ₊=(0,∞) with 𝑖=1,2,…,𝑛, let 𝐴_𝑛(𝑥)=1𝑛^𝑛_𝑖=1𝑥_𝑖,𝐺_𝑛(𝑥)=^𝑛_𝑖=1𝑥_𝑖^1/𝑛,𝐻_𝑛(𝑥)=𝑛∑^𝑛_𝑖=11/𝑥_𝑖. (1) These quantities are, respectively, called the arithmetic, geometric, and harmonic means of a positive sequence 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛). For more information on the theory of means, see the monograph [1] or the papers [2–5] and plenty of references therein.

For convenience, in what follows, we use the notation 𝑥^𝑞=(𝑥^𝑞₁,𝑥^𝑞₂,…,𝑥^𝑞_𝑛) for 0<𝑚≤𝑥^𝑞₁,𝑥^𝑞₂,…,𝑥^𝑞_𝑛≤𝑀, (2) and 𝑞∈ℝ=(−∞,∞).

In [6], Satnoianu posed the following conjecture.

Conjecture 1. For 𝑛≥2, 𝑥_𝑖>0, and 𝜆≥𝑛^𝑛−1−1, it is valid that ^𝑛_𝑖=1𝑥_𝑖^𝑛−1𝑥_𝑖^𝑛−1+𝜆∏^𝑛_{𝑘=1,𝑘≠𝑖}𝑥_𝑘^1/(𝑛−1)≥𝑛(1+𝜆)^1/(𝑛−1). (3)

This conjecture has been solved and researched in [7–10]. Among them, Wu obtained in [10] the following result.

Theorem 2 (see [10, Theorem  1]). Let 𝑛 be a positive integer and let 𝛼, 𝛽, and 𝑥_𝑖 for 𝑖=1,2,…,𝑛 be positive numbers. If 𝑝≠0 and 𝛽≥𝑛^max{𝑝,1}−1𝛼, (4) then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺(𝑥^𝑞)^1/p≥𝑛𝛼+𝛽^1/𝑝. (5)

It is clear that inequality (5) generalizes (3). Therefore, we would like to call inequality (5) Satnoianu-Wu’s inequality.

Now we naturally pose the following questions: (1)Can one improve the condition (4)? (2)Does the reversed inequality of (5) exist? (3)Are there other types of inequalities of Satnoianu-Wu type?

The goal of this paper is to answer these questions.

2. Definitions and Lemmas

We need the following definitions and lemmas.

Definition 3 (see [11, page 8]). Let 𝑥=(𝑥₁,𝑥₂,…,𝑥_𝑛)∈ℝ^𝑛 and 𝑦=(𝑦₁,𝑦₂,…,𝑦_𝑛)∈ℝ^𝑛. We say that 𝑥 is majorized by 𝑦 (in symbols 𝑥≺𝑦) if ^𝑘_𝑖=1𝑥_[𝑖]≤^𝑘_𝑖=1𝑦_[𝑖],^𝑛_𝑖=1𝑥_𝑖=^𝑛_𝑖=1𝑦_𝑖, (6) for 𝑘=1,2,…,𝑛−1, where 𝑥_[1]≥⋯≥𝑥_[𝑛] and 𝑦_[1]≥⋯≥𝑦_[𝑛] are rearrangements of 𝑥 and 𝑦 in a descending order.

Definition 4 (see [12]). Let Ω⊆ℝ^𝑛₊.(1)The set Ω is said to be geometrically convex if (𝑥^𝜆₁𝑦₁^1−𝜆,…,𝑥^𝜆_𝑛𝑦_𝑛^1−𝜆)∈Ω for every 𝑥,𝑦∈Ω and 𝜆∈[0,1].(2)A function 𝜑∶Ω→ℝ₊ is said to be Schur-geometrically convex on Ω if ln𝑥=(ln𝑥₁,…,ln𝑥_𝑛)≺ln𝑦=(ln𝑦₁,…,ln𝑦_𝑛) implies 𝜑(𝑥)≤𝜑(𝑦) for every 𝑥,𝑦∈Ω.(3)A function 𝜑∶Ω→ℝ₊ is said to be Schur-geometrically concave on Ω if ln𝑥=(ln𝑥₁,…,ln𝑥_𝑛)≺ln𝑦=(ln𝑦₁,…,ln𝑦_𝑛) implies 𝜑(𝑥)≥𝜑(𝑦) for every 𝑥,𝑦∈Ω.

Lemma 5 (see [12]). Let Ω⊆ℝ^𝑛₊ be a symmetric and geometrically convex set with inner points and 𝜑∶Ω→ℝ₊ a symmetric and differentiable function in Ω^∘. Then 𝜑 is a Schur-geometrically convex (or Schur-geometrically concave, resp.) function on Ω if and only if ln𝑥₁−ln𝑥₂𝑥₁𝜕𝜑(𝑥)𝜕𝑥₁−𝑥₂𝜕𝜑(𝑥)𝜕𝑥₂≥0𝑜𝑟≤0,𝑟𝑒𝑠𝑝.,𝑥∈Ω^∘. (7)

Lemma 6 (see [1, page 4, Bernoulli's inequality]). The inequality (1+𝑡)^𝑟≥1+𝑟𝑡 (8) holds for 𝑟≥1 and 𝑡≥−1 or for 𝑟≤0 and 𝑡>−1. If  0<𝑟<1 and 𝑡≥−1, inequality (8) is reversed.

3. Main Results

Now we start off to demonstrate our main results.

Theorem 7. Let 𝑚,𝑀 be defined as in (2) and 𝑛≥2, let 𝛼,𝛽,𝑥_𝑖∈ℝ₊ for 𝑖=1,2,…,𝑛, and let 𝑝,𝑞∈ℝ with 𝑝≠0. If (1−𝑝)𝑚𝛽≥2𝑝𝑀𝛼, then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝. (9) If (1−𝑝)𝑀𝛽≤2𝑝𝑚𝛼, inequality (9) is reversed.

Proof. For 𝑀^′>𝑚^′>0 and 𝑟∈ℝ with 𝑟≠0, since 𝑡1+𝑡^𝑟′′=𝑟(𝑟−1−2𝑡)𝑡^𝑟−1(1+𝑡)^𝑟+2,𝑡∈𝑚^′,𝑀^′, (10) then(1)for 𝑟<0, or for 𝑟>1 and 𝑟−1−2𝑀^′>0, the function (𝑡/(1+𝑡))^𝑟 is convex on [𝑚^′,𝑀^′];(2)for 0<𝑟≤1, or for 𝑟>1 and 𝑟−1−2𝑚^′<0, the function (𝑡/(1+𝑡))^𝑟 is concave on [𝑚^′,𝑀^′].By Jensen’s inequality, when 𝑟<0, or when 𝑟>1 and 𝑟−1−2𝑀^′>0, we have ^𝑛_𝑖=1𝑢_𝑖1+𝑢_𝑖^𝑟𝐴≥𝑛^𝑛(𝑢)1+𝐴_𝑛(𝑢)^𝑟, (11) where 𝑢=(𝑢₁,𝑢₂,…,𝑢_𝑛)∈[𝑚^′,𝑀^′]^𝑛. When 0<𝑟≤1, or when 𝑟>1 and 𝑟−1−2𝑚^′<0, inequality (11) is reversed.
Let 𝑢_𝑖=𝛼𝑥^𝑞_𝑖/𝛽𝐴_𝑛(𝑥^𝑞) for 𝑖=1,2,…,𝑛, 𝑟=1/𝑝, 𝑚^′=𝛼𝑚/𝛽𝑀, and 𝑀^′=𝛼𝑀/𝛽𝑚. Then it is easy to show that 𝐴_𝑛(𝑢)=𝛼/𝛽. Making use of (11), we obtain inequality (9). The proof of Theorem 7 is complete.

Corollary 8. Under the conditions of Theorem 7 and when 𝑀=(𝑛−1)𝑚, if (1−𝑝)𝛽>2𝑝(𝑛−1)𝛼, then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (12) if (1−𝑝)(𝑛−1)𝛽<2𝑝𝛼, inequality (12) is reversed.

Corollary 9. Let 𝑛≥2, 𝛼,𝛽,𝑥_𝑖∈ℝ₊ for 𝑖=1,2,…,𝑛, and 𝑝,𝑞∈ℝ. If 𝑝<0, then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (13) if 𝑝≥1, inequality (13) is reversed.

Proof. This follows from Theorem 7 by considering that if 𝑝<0, we have (1−𝑝)𝑚𝛽>2𝑝𝑀𝛼 and that if 𝑝≥1, we have (1−𝑝)𝑀𝛽<2𝑝𝑚𝛼.

Theorem 10. Let 𝑛≥2, 𝛼,𝛽,𝑥_𝑖∈ℝ₊ for  𝑖=1,2,…,𝑛, 𝑝,𝑞∈ℝ with 𝑝≠0, and 𝑚,𝑀 defined by (2). If  𝑚𝛽≥𝑝𝑀𝛼, then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (14) if 𝑀𝛽≤𝑚𝑝𝛼, inequality (14) is reversed.

Proof. For 𝑀^′>𝑚^′>0, let 𝑓(𝑢)=^𝑛_𝑖=11+𝑢_𝑖^𝑟,𝑢∈𝑚^′,𝑀^′𝑛. (15) Then 𝜕𝑓(𝑢)𝜕𝑢_𝑖=𝑟1+𝑢_𝑖^𝑟−1,𝜕𝜕𝑢_𝑖𝑢_𝑖𝜕𝑓(𝑢)𝜕𝑢_𝑖=𝑟1+𝑢_𝑖^𝑟−21+𝑟𝑢_𝑖, (16) for 1≤𝑖≤𝑛. When 𝑟>0, or when 𝑟<0 and 1+𝑟𝑚^′<0, we have ln𝑢₁−ln𝑢₂𝑢₁𝜕𝑓(𝑢)𝜕𝑢₁−𝑢₂𝜕𝑓(𝑢)𝜕𝑢₂≥0,𝑢∈𝑚^′,𝑀^′𝑛; (17) when 𝑟<0 and 1+𝑟𝑀^′>0, inequality (17) is reversed.
Using Lemma 5, we have the following conclusions:(1)if 𝑟>0, or if 𝑟<0 and 1+𝑟𝑚^′<0, the function 𝑓(𝑢) is Schur-geometrically convex on [𝑚^′,𝑀^′]^𝑛;(2)if 𝑟<0 and 1+𝑟𝑀^′>0, the function 𝑓(𝑢) is Schur-geometrically concave on [𝑚^′,𝑀^′]^𝑛.By the fact that ln𝐺_𝑛(𝑢),ln𝐺_𝑛(𝑢),…,ln𝐺_𝑛(𝑢)≺ln𝑢₁,ln𝑢₂,…,ln𝑢_𝑛,𝑢∈ℝ^𝑛₊, (18) and by Definition 4, if 𝑟>0, or if 𝑟<0 and 1+𝑟𝑚^′<0, we have ^𝑛_𝑖=11+𝑢_𝑖^𝑟≥𝑛1+𝐺_𝑛(𝑢)^𝑟,𝑢∈𝑚^′,𝑀^′𝑛; (19) if 𝑟<0 and 1+𝑟𝑀^′>0, inequality (19) is reversed.
Letting 𝑢_𝑖=𝛽𝐺_𝑛(𝑥^𝑞)/𝛼𝑥^𝑞_𝑖 for 𝑖=1,2,…,𝑛 leads to 𝐺_𝑛(𝑢)=𝛽/𝛼. Putting 𝑟=−1/𝑝, 𝑚^′=𝛽𝑚/𝛼𝑀 and 𝑀^′=𝛽𝑀/𝛼𝑚 in inequality (19) results in inequality (14). The proof of Theorem 10 is complete.

Corollary 11. Under the conditions of Theorem 10 and when 𝑀=(𝑛−1)𝑚, if 𝛽≥𝑝(𝑛−1)𝛼, then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (20) if (𝑛−1)𝛽≤𝑝𝛼, inequality (20) is reversed.

Remark 12. When 𝑛≥2 and 𝑝>1, from Lemma 6 and (4), it follows that 𝛽≥𝑛^max{𝑝,1}−1𝛼>𝑝(𝑛−1)𝛼. (21)

Theorem 13. Let 𝑛≥2, 𝛼,𝛽,𝑥_𝑖∈ℝ₊ for 𝑖=1,2,…,𝑛, and 𝑝,𝑞∈ℝ with 𝑝≠0. If 𝑝≥−1, then ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐻_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (22) if 𝑝≤−1, inequality (22) is reversed.

Proof. Since (1+𝑡)^𝑟 is a convex (or concave, resp.) function on ℝ₊ for 𝑟≥1 or 𝑟<0 (or for 0<𝑟≤1, resp.), by Jensen's inequality, if 𝑟≥1 or 𝑟<0, we have ^𝑛_𝑖=11+𝑢_𝑖^𝑟≥𝑛1+𝐴_𝑛(𝑢)^𝑟,𝑢∈ℝ^𝑛₊; (23) if 0<𝑟≤1, inequality (23) is reversed.
Letting 𝑢_𝑖=𝛽𝐻_𝑛(𝑥^𝑞)/𝛼𝑥^𝑞_𝑖 and 𝑟=−1/𝑝 shows 𝐴_𝑛(𝑢)=𝛽/𝛼. Further from (23), we obtain inequality (22). The proof of Theorem 13 is complete.

Remark 14. It is clear that inequalities (9) and (22) both generalize inequality (3).

Corollary 15. Let 𝑛≥2, 𝛼,𝛽,𝑥_𝑖∈ℝ₊ for 𝑖=1,2,…,𝑛, 𝑝,𝑞∈ℝ with 𝑝≠0, and 𝑚,𝑀 defined as in (2).(1)When  −1≤𝑝<0, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐻_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝. (24)(2)When 0<𝑝<1 and (1−𝑝)𝑚𝛽>2𝑝𝑀𝛼, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐻_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝. (25)(3)When 𝑝>0 and 𝑚𝛽>𝑝𝑀𝛼, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐻_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝. (26)(4)When 𝑝>0 and 𝑀𝛽<𝑚𝑝𝛼, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≤^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≤𝑛𝛼+𝛽^1/𝑝. (27)

Proof. This follows from utilizing the well-known harmonic-geometric-arithmetic mean inequality 𝐻_𝑛(𝑥)≤𝐺_𝑛(𝑥)≤𝐴_𝑛(𝑥) (28) and Theorems 7 to 13.

Corollary 16. Under the conditions of Corollary 15 and when 𝑀=(𝑛−1)𝑚,(1)if  0<𝑝<1 and (1−𝑝)𝛽>2𝑝(𝑛−1)𝛼, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐻_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (29)(2)if 𝑝>0 and 𝛽>𝑝(𝑛−1)𝛼, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐻_𝑛(𝑥^𝑞)^1/𝑝≥^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≥𝑛𝛼+𝛽^1/𝑝; (30)(3)if 𝑝>0 and (𝑛−1)𝛽<𝑝𝛼, one has ^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐴_𝑛(𝑥^𝑞)^1/𝑝≤^𝑛_𝑖=1𝑥^𝑞_𝑖𝛼𝑥^𝑞_𝑖+𝛽𝐺_𝑛(𝑥^𝑞)^1/𝑝≤𝑛𝛼+𝛽^1/𝑝. (31)

Acknowledgments

The author thanks four anonymous referees for their careful corrections to and valuable comments on the original version of this paper. This work was partially supported by the Foundation of the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region under Grant no. NJZY13159, China.