A Reverse Analytic Inequality for the Elementary Symmetric Function with Applications

Abstract

We give a reverse inequality involving the elementary symmetric function by use of the Schur harmonic convexity theory. As applications, several new analytic inequalities for the 𝑛-dimensional simplex are established.

1. Introduction

Let 𝐱=(𝑥₁,𝑥₂,…,𝑥_𝑛)∈ℝ^𝑛=(−∞,+∞)^𝑛. The elementary symmetric functions are defined by 𝐸_𝑘(𝐱)=𝐸_𝑘𝑥₁,…,𝑥_𝑛=_{1≤𝑖1<⋯<𝑖𝑘}^𝑘_{≤𝑛𝑗=1}𝑥_𝑖𝑗,𝑘=1,…,𝑛, (1)𝐸₀(𝐱)=1, and 𝐸_𝑘(𝐱)=0 for 𝑘<0 or 𝑘>𝑛.

In 1997, Leng and Tang [1] in order to achieve higher dimensional generalization of the famous Pedoe inequality [2–5] presented the following analytic inequality.

Let 𝐱∈(0,1/2)^𝑛, and 𝐸₁(𝐱)=1. Then 𝐸_𝑛−11𝐱−2=𝐸_𝑛−11𝑥₁1−2,…,𝑥_𝑛−2≥𝑛(𝑛−2)^𝑛−1. (2)

Chen et al. [6] gave an analytic proof which is skillful. By the method of successive adjustment, Fan [7] gave a simple elementary proof for inequality under conditions 𝐱∈(0,1)^𝑛, 𝑛≥2, and 𝐸₁(𝐱)=1. Using the same method, Shi [8] extended inequality (2) to the cases of 𝐸_𝑘((1/𝐱)−2) for 𝑘=1,2,…,𝑛−1, where 𝐱∈ℝ^𝑛₊.

Theorem A. Let 𝐱∈ℝ^𝑛₊=(0,+∞)^𝑛, 𝑛≥2 and 𝐸₁(𝐱)=1. Then for 𝑘=1,2,…,𝑛−1, one has 𝐸_𝑘1𝐱−2=𝐸_𝑘1𝑥₁1−2,…,𝑥_𝑛≥𝑛𝑘−2(𝑛−2)^𝑘, (3) where (^𝑛_𝑘)=𝑛!/𝑘!(𝑛−𝑘)!.

And Shi also pointed out that inequality (3) does not hold for 𝑘=𝑛.

In 2002, using analytical methods, Ma and Pu [9] gave parameter extension of the inequality (3).

Theorem B. Let 𝐱∈ℝ^𝑛₊, 𝑛, 𝑘∈ℕ={1,2,…}, 𝜇>0, 𝑛≥2 and 𝐸₁(𝐱)=1. Then for 𝑛≥𝑘+𝜇−1, one has 𝐸_𝑘1𝐱−𝜇=𝐸_𝑘1𝑥₁1−𝜇,…,𝑥_𝑛≥𝑛𝑘−𝜇𝑛−𝜇^𝑘. (4)

In 2005, Li et al. [10] gave an elementary proof for Theorem B when 𝑘≤𝑛−1.

In this paper, by studying the Schur harmonic convexity of 𝐸_𝑘((1/𝐱)−𝜇), we obtain a reverse of inequality (4) and accordingly establish the reverse of the corresponding simplex inequalities.

Our main result is the following theorem.

Theorem 1. Let 𝐱∈ℝ^𝑛₊, 𝑛, 𝑘∈ℕ, 𝜇>0, 𝑛≥2. Then for 𝑛≥𝑘+𝜇−1, 𝐸_𝑘((1/𝐱)−𝜇) is Schur harmonically concave on Ω={𝐱∶𝐱∈ℝ^𝑛₊,𝐸₁(𝐱)≤1}. When 𝐸₁(𝐱)≤1, one has 𝐸_𝑘1𝐱−𝜇=𝐸_𝑘1𝑥₁1−𝜇,…,𝑥_𝑛≤𝑛𝑘1−𝜇𝐻−𝜇^𝑘, (5) where 𝐻=𝑛(∑^𝑛_𝑖=1𝑥_𝑖⁻¹)⁻¹ is the harmonic mean of 𝐱.

Taking 𝜇=2 in Theorem 1, we infer the following corollary.

Corollary 2. Let 𝐱∈ℝ^𝑛₊, 𝑛, 𝑘∈ℕ, and 𝐸₁(𝐱)≤1. Then for 𝑛≥𝑘+1, one has 𝐸_𝑘1𝐱−2=𝐸_𝑘1𝑥₁1−2,…,𝑥_𝑛≤𝑛𝑘1−2𝐻−2^𝑘, (6) where 𝐻=𝑛(∑^𝑛_𝑖=1𝑥_𝑖⁻¹)⁻¹ is the harmonic mean of 𝐱.

Taking 𝑘=𝑛−1 in Corollary 2, we obtain the following inequality.

Corollary 3. Let 𝐱∈ℝ^𝑛₊, 𝑛∈ℕ, and 𝐸₁(𝐱)≤1. Then for 𝑛≥2, one has ^𝑛𝑛_{𝑖=1𝑗=1,𝑗≠𝑖}1𝑥_𝑗1−2≤𝑛𝐻−2^𝑛−1, (7) where 𝐻=𝑛(∑^𝑛_𝑖=1𝑥_𝑖⁻¹)⁻¹ is the harmonic mean of 𝐱.

2. Definitions and Lemmas

Definition 4 (see [11, 12]). Let 𝐱=(𝑥₁,…,𝑥_𝑛) and 𝐲=(𝑦₁,…,𝑦_𝑛)∈ℝ^𝑛.(i)𝐱 is said to be majorized by 𝐲 (in symbols 𝐱≺𝐲) if ∑^𝑘_𝑖=1𝑥_[𝑖]≤∑^𝑘_𝑖=1𝑦_[𝑖] for 𝑘=1,2,…,𝑛−1 and ∑^𝑛_𝑖=1𝑥_𝑖=∑^𝑛_𝑖=1𝑦_𝑖, where 𝑥_[1]≥⋯≥𝑥_[𝑛] and 𝑦_[1]≥⋯≥𝑦_[𝑛] are rearrangements of 𝐱 and 𝐲 in a descending order.(ii)Ω⊆ℝ^𝑛 is called a convex set if (𝛼𝑥₁+𝛽𝑦₁,…,𝛼𝑥_𝑛+𝛽𝑦_𝑛)∈Ω for every 𝐱 and 𝐲∈Ω, where 𝛼 and 𝛽∈[0,1] with 𝛼+𝛽=1.(iii)Let Ω⊆ℝ^𝑛. The function 𝜑: Ω→ℝ is said to be a Schur convex function on Ω if 𝐱≺𝐲 on Ω implies 𝜑(𝐱)≤𝜑(𝐲). 𝜑 is said to be a Schur concave function on Ω if and only if −𝜑 is Schur convex.

Lemma 5 (see [11, 12]). Let 𝐱=(𝑥₁,𝑥₂,…,𝑥_𝑛)∈ℝ^𝑛. Then (𝐴,…,𝐴)≺𝑥₁,𝑥₂,…,𝑥_𝑛, (8) where 𝐴=(1/𝑛)∑^𝑛_𝑖=1𝑥_𝑖 is the arithmetic mean of 𝐱.

Definition 6 (see [11, 12]). Let Ω⊆ℝ^𝑛. (i)Ω is called a symmetric set, if 𝐱∈Ω implies 𝐱𝑃∈Ω for every 𝑛×𝑛 permutation matrix 𝑃.(ii)The function 𝜑∶Ω→ℝ is called symmetric if for every permutation matrix 𝑃, 𝜑(𝐱𝑃)=𝜑(𝐱) for all 𝐱∈Ω.

Lemma 7 (see [11, p. 84]). Let Ω⊂ℝ^𝑛 be symmetric and have a nonempty interior convex set Ω⁰, and let 𝜑∶Ω→ℝ be continuous on Ω and differentiable in Ω⁰. Then 𝜑 is the Schur-convex (Schur-concave) function if and only if 𝜑 is symmetric on Ω and 𝑥₁−𝑥₂𝜕𝜑(𝐱)𝜕𝑥₁−𝜕𝜑(𝐱)𝜕𝑥₂≥0(≤0) (9) holds for any 𝐱∈Ω⁰.

The Schur-convexity described the ordering of majorization; the order-preserving functions were first comprehensively studied by Issai Schur in 1923. It has important applications in analytic inequalities, combinatorial optimization, quantum physics, information theory, and so on. See [11].

In 2009, Chu and Lv [13] introduced the notion of harmonically Schur convex function and some interesting inequalities were obtained. The Schur harmonic convexity involving some special functions has been investigated; see, for example, [14–20].

Definition 8 (see [13]). Let Ω⊂ℝ^𝑛₊. (i)A set Ω is said to be harmonically convex if 𝐱𝐲/(𝜆𝐱+(1−𝜆)𝐲)∈Ω for every 𝐱,𝐲∈Ω and 𝜆∈[0,1], where 𝐱𝐲=∑^𝑛_𝑖=1𝑥_𝑖𝑦_𝑖 and (1/𝐱)=(1/𝑥₁,…,1/𝑥_𝑛).(ii)A function 𝜑∶Ω→ℝ₊ is said to be Schur harmonically convex on Ω if 1/𝐱≺1/𝐲 implies 𝜑(𝐱)≤𝜑(𝐲). A function 𝜑 is said to be a Schur harmonically concave function on Ω if and only if −𝜑 is a Schur harmonically convex function.

Lemma 9 (see [13]). Let Ω⊂ℝ^𝑛₊ be a symmetric and harmonically convex set with inner points and let 𝜑∶Ω→ℝ₊ be a continuously symmetric function which is differentiable on Ω⁰. Then, 𝜑 is Schur harmonically convex (Schur harmonically concave) on Ω if and only if 𝑥₁−𝑥₂𝑥²₁𝜕𝜑(𝐱)𝜕𝑥₁−𝑥²₂𝜕𝜑(𝐱)𝜕𝑥₂≥0(≤0),𝐱∈Ω⁰. (10)

Lemma 10. Let 𝐱∈ℝ^𝑛₊, 𝑛, 𝑘∈ℕ, 𝜇>0, 𝑛≥2, and 𝐸₁(𝐱)≤1. Then for 𝑛≥𝑘+𝜇−1, one has 𝐸_𝑘1𝐱−𝜇≥0. (11)

Proof. Let 𝐸₁(𝐱)=𝜎≤1. Then 𝐸₁(𝐱/𝜎)=(𝑥₁/𝜎)+⋯+(𝑥_𝑛/𝜎)=1; and 𝑛≥𝑘+𝜇−1≥𝑘+𝜎𝜇−1, from Theorem B, we have 𝐸_𝑘𝜎𝐱−𝜎𝜇=𝐸_𝑘𝜎𝑥₁𝜎−𝜎𝜇,…,𝑥_𝑛≥𝑛𝑘−𝜎𝜇𝑛−𝜎𝜇^𝑘, (12) and then 𝐸_𝑘1𝐱−𝜇=𝐸_𝑘1𝑥₁1−𝜇,…,𝑥_𝑛≥𝑛𝑘−𝜇𝑛−𝜎𝜇𝜎^𝑘≥0. (13) Lemma 10 is proved.

3. Proof of Main Result

Proof of Theorem 1. When 𝑛≥2, 𝑘=1, obviously, inequality (5) holds. When 𝑛≥2, 𝑘≥2, noting that 𝜑(𝐱)∶=𝐸_𝑘1𝐱=1−𝜇𝑥₁1−𝜇𝑥₂−𝜇𝐸_𝑘−21𝑥₃1−𝜇,…,𝑥_𝑛+1−𝜇𝑥₁1−𝜇+𝑥₂−𝜇𝐸_𝑘−11𝑥₃1−𝜇,…,𝑥_𝑛−𝜇+𝐸_𝑘1𝑥₃1−𝜇,…,𝑥_𝑛−𝜇, (14) we have 𝜕𝜑(𝐱)𝜕𝑥₁1=−𝑥²₁1𝑥₂−𝜇𝐸_𝑘−21𝑥₃1−𝜇,…,𝑥_𝑛−1−𝜇𝑥²₁𝐸_𝑘−11𝑥₃1−𝜇,…,𝑥_𝑛,(𝐱)−𝜇𝜕𝜑𝜕𝑥₂1=−𝑥²₂1𝑥₁−𝜇𝐸_𝑘−21𝑥₃1−𝜇,…,𝑥_𝑛−1−𝜇𝑥²₂𝐸_𝑘−11𝑥₃1−𝜇,…,𝑥_𝑛−𝜇. (15)(16) And then Δ∶=𝑥₁−𝑥₂𝑥²₁𝜕𝜑(𝐱)𝜕𝑥₁−𝑥²₂𝜕𝜑(𝐱)𝜕𝑥₂𝑥=−¹−𝑥₂²𝑥₁𝑥₂𝐸_𝑘−21𝑥₃1−𝜇,…,𝑥_𝑛−𝜇. (17)
Now we distinguish two cases to prove Δ≤0.
Case 1. If 𝑛≥2, 𝑘=2, then from the definition of elementary symmetric functions, it follows that Δ≤0.
If 𝑛=3, 𝑘=3, then from the condition 𝑛≥𝑘+𝜇−1, it follows that 𝜇≤𝑛−𝑘+1=1, and by 𝑥_𝑖≤𝐸₁(𝐱)≤1, it follows that 1/𝑥_𝑖≥1, 𝑖=3,…,𝑛; therefore, 𝐸_𝑘−21𝑥₃1−𝜇,…,𝑥_𝑛−𝜇=𝐸₁1𝑥₃1−𝜇,…,𝑥_𝑛=−𝜇_𝑛𝑖=31𝑥_𝑖−(𝑛−2)𝜇≥0. (18)
Case 2. If 𝑛≥4; then 𝑛−2≥2, from 𝑛≥𝑘+𝜇−1, it follows that (𝑛−2)≥(𝑘−2)+𝜇−1 (19) and ∑^𝑛_𝑖=3𝑥_𝑖≤1. So by Lemma 10, it follows that 𝐸_𝑘−2((1/𝑥₃)−𝜇,…,(1/𝑥_𝑛)−𝜇)≥0.
Thus, for all 𝑛≥2, Δ≤0. By Lemma 9, we can derive that 𝐸_𝑘((1/𝐱)−𝜇) is Schur harmonically concave on Ω.
From Lemma 5, it is seen that 1𝐻1,…,𝐻≺1𝑥₁1,…,𝑥_𝑛. (20) According to Definition 8(ii), inequality (5) follows.
The proof of Theorem 1 is completed.

4. Applications in Geometry

Let Ω_𝑛 be an 𝑛-dimensional simplex in the 𝑛-dimensional Euclidean space ℝ^𝑛(𝑛≥2) with vertices 𝐴₀,𝐴₁,…,𝐴_𝑛 whose volume is 𝑉. For 𝑖=0,1,…,𝑛, let 𝑟_𝑖 be the radius of 𝑖th escribed sphere of Ω_𝑛, and 𝐹_𝑖 the area of the 𝑖th face 𝑓_𝑖={𝐴₀,𝐴₁,…,𝐴_𝑖−1,𝐴_𝑖+1,…,𝐴_𝑛} of Ω_𝑛, 𝐹=∑^𝑛_𝑖=0𝐹_𝑖. 𝑅 and 𝑟 are circumradius and inradius of Ω_𝑛, respectively. Let 𝑃 be an arbitrary interior point of the simplex Ω_𝑛, 𝑑_𝑖 the distance from the point 𝑃 to the 𝑖th face 𝑓_𝑖 of Ω_𝑛, and ℎ_𝑖 the altitude of Ω_𝑛 from vertex 𝐴_𝑖 for 𝑖=0,1,…,𝑛.

In [9], by Theorem B, Ma and Pu obtained the following two theorems.

Theorem C. In the simplex Ω_𝑛 (𝑛≥2), for 𝑘=1,2,…,𝑛, one has 𝐸_𝑘ℎ₀𝑟₀,ℎ₁𝑟₁ℎ,…,^𝑛𝑟_𝑛≥𝑘()𝑛+1𝑛−1^𝑘 (21) with equality if and only if Ω_𝑛 is an orthocentric simplex.

Theorem D. Let 𝑃 be an arbitrary point in the interior of 𝑛-dimensional simplex Ω_𝑛. Let 𝐵_𝑖 be the intersection of the line 𝐴_𝑖𝑃 with the hyperplane 𝑓_𝑖; further let 𝑅_𝑖=|𝐴_𝑖𝑃|, 𝑑^′_𝑖=|𝑃𝐵_𝑖|, 𝑀_𝑖=|𝐴_𝑖𝐵_𝑖|, 𝑖=0,1,…,𝑛. Then for 𝑘=1,2,…,𝑛+1, one has 𝐸_𝑘𝑅₀𝑑^′₀,𝑅₁𝑑^′₁𝑅,…,^𝑛𝑑^′_𝑛≥𝑘𝑛𝑛+1_𝑘 (22) with equality if and only if Ω_𝑛 is an orthocentric simplex, and 𝑃 is its orthocentre.

Now, by Theorem 1, we give the reverse of inequalities (21) and (22).

Theorem 11. In the simplex Ω_𝑛, for 𝑘=1,2,…,𝑛, one has 𝐸_𝑘ℎ₀𝑟₀,ℎ₁𝑟₁ℎ,…,^𝑛𝑟_𝑛≤𝑘𝑛+11𝐻−2^𝑘, (23) where 𝐻=(𝑛+1)(∑^𝑛_𝑖=0((ℎ_𝑖/𝑟_𝑖)+2))⁻¹.

Proof. In the 𝑛-dimensional simplex, for 𝑘=0,1,…,𝑛, it is well known that 𝑟_𝑖=𝑛𝑉∑^𝑛_𝑗=0𝐹_𝑗−2𝐹_𝑖,ℎ_𝑖=𝑛𝑉𝐹_𝑖. (24)
Let 𝑥_𝑖=𝐹_𝑖/∑^𝑛_𝑗=0𝐹_𝑗, 𝑖=0,1,…,𝑛. Then, 𝐸₁(𝐱)=1. From equalities (24), we have 1𝑥_𝑖∑−2=^𝑛𝑗=0𝐹_𝑗−2𝐹_𝑖𝐹_𝑖=ℎ_𝑖𝑟_𝑖,𝑖=0,1,…,𝑛. (25)
Thus from the inequality (6), it is deduced that the inequality (23) holds.
The proof of Theorem 11 is completed.

Theorem 12. Let 𝑃 be an arbitrary point in the interior of 𝑛-dimensional simplex Ω_𝑛. Let 𝐵_𝑖 be the intersection of the line 𝐴_𝑖𝑃 with the hyperplane 𝑓_𝑖; further let 𝑅_𝑖=|𝐴_𝑖𝑃|, 𝑑^′_𝑖=|𝑃𝐵_𝑖|, 𝑀_𝑖=|𝐴_𝑖𝐵_𝑖|, 𝑖=0,1,…,𝑛. Then, for 𝑘=1,2,…,𝑛+1, 𝐸_𝑘𝑅₀𝑑^′₀,𝑅₁𝑑^′₁𝑅,…,^𝑛𝑑^′_𝑛≤𝑘𝑛+11𝐻−1^𝑘, (26) where 𝐻=(𝑛+1)(∑^𝑛_𝑖=0(𝑀_𝑖/𝑑^′_𝑖))⁻¹.

Proof. It is easy to see that ^𝑛_𝑖=0𝑑^′_𝑖𝑀_𝑖𝑅=1,^𝑖𝑀_𝑖𝑑=1−^′𝑖𝑀_𝑖,𝑖=0,1,…,𝑛. (27)
In the inequality (5), let 𝜇=1, 𝑥_𝑖=𝑑^′_𝑖/𝑀_𝑖. Then, 1≤𝑘≤𝑛+1, (1/𝑥_𝑖)−1=𝑅_𝑖/𝑑^′_𝑖, 𝑖=0,1,…,𝑛. Thus, inequality (26) holds.
The proof of Theorem 12 is completed.

When 𝑘=𝑛+1, inequality (26) is reduced to the following.

Corollary 13. Under the conditions of Theorem 12, one has ^𝑛_𝑖=0𝑅_𝑖≤1𝐻−1^𝑛𝑛+1_𝑖=0𝑑^′_𝑖. (28)

Theorem 14. In the simplex Ω_𝑛, let 𝜆_𝑖=𝐹−2𝐹_𝑖𝐹_𝑖,𝜇_𝑖=𝐹²₀+⋯+𝐹²_𝑛−2𝐹²_𝑖𝐹²_𝑖,(𝑖=0,1,…,𝑛). (29) Then, ^𝑛𝑛_{𝑖=0𝑗=0,𝑗≠𝑖}𝜆_𝑗≤(1𝑛+1)𝐻−2^𝑛, (30) where 𝐻=𝑛(∑^𝑛_𝑖=0(𝜆_𝑖+2))⁻¹; ^𝑛𝑛_{𝑖=0𝑗=0,𝑗≠𝑖}𝜇_𝑗≤(1𝑛+1)𝐻−2^𝑛, (31) where 𝐻=𝑛(∑^𝑛_𝑖=0(𝜇_𝑖+2))⁻¹.

Proof. Taking 𝑥_𝑖=(𝐹_𝑖/𝐹)(𝑖=0,1,…,𝑛) and applying Corollary 3 for these 𝑛+1 positive real numbers 𝑥₀,𝑥₁,…,𝑥_𝑛, inequality (30) follows.
Taking 𝑥_𝑖=𝐹²_𝑖/(𝐹²₀+𝐹²₁+⋯+𝐹²_𝑛) and applying Corollary 3, the desired inequality (31) follows.
The proof of Theorem 14 is completed.

Remark 15. Inequalities (30) and (31) give the reverse of the inequalities (2.10) and (2.11) in [10], respectively.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work was supported by the Importation and Development of High-Caliber Talents Project of Beijing Municipal Institutions (IDHT201304089) and the Funding Project for Academic Human Resources Development in Institutions of Higher Learning under the Jurisdiction of Beijing Municipality (PHR (IHLB)) (PHR201108407). The authors are grateful to Professor Fernando Simões and the four anonymous reviewers for helpful pieces of advice that improved its presentation.