# A Reverse Analytic Inequality for the Elementary Symmetric Function with Applications

Authors: Huan-Nan Shi; Jing Zhang
Journal: Journal of Applied Mathematics
 (2013)

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## 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 then-dimensional simplex are established.


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## Body

## 1. Introduction

Letx=(x1,x2,…,xn)∈ℝn=(-∞,+∞)n. The elementary symmetric functions are defined by
(1)Ek(x)=Ek(x1,…,xn)=∑1≤i1<⋯<ik≤n‍∏j=1k‍xij,hhhhhhhhhhhhhhhhhhhhhhhhk=1,…,n,E0(x)=1, and Ek(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.Letx∈(0,1/2)n, and E1(x)=1. Then
(2)En-1(1x-2)=En-1(1x1-2,…,1xn-2)≥n(n-2)n-1.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 Ek((1/x)-2) for k=1,2,…,n-1, where x∈ℝ+n.Theorem A.
Letx∈ℝ+n=(0,+∞)n, n≥2 and E1(x)=1. Then for k=1,2,…,n-1, one has
(3)Ek(1x-2)=Ek(1x1-2,…,1xn-2)≥(nk)(n-2)k,
where (nk)=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.
Letx∈ℝ+n, n, k∈ℕ={1,2,…}, μ>0, n≥2 and E1(x)=1. Then for n≥k+μ-1, one has
(4)Ek(1x-μ)=Ek(1x1-μ,…,1xn-μ)≥(nk)(n-μ)k.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 ofEk((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.
Letx∈ℝ+n, n, k∈ℕ, μ>0, n≥2. Then for n≥k+μ-1, Ek((1/x)-μ) is Schur harmonically concave on Ω={x:x∈ℝ+n,E1(x)≤1}. When E1(x)≤1, one has
(5)Ek(1x-μ)=Ek(1x1-μ,…,1xn-μ)≤(nk)(1H-μ)k,
where H=n(∑i=1n‍xi-1)-1 is the harmonic mean of x.Takingμ=2 in Theorem 1, we infer the following corollary.Corollary 2.
Letx∈ℝ+n, n, k∈ℕ, and E1(x)≤1. Then for n≥k+1, one has
(6)Ek(1x-2)=Ek(1x1-2,…,1xn-2)≤(nk)(1H-2)k,
where H=n(∑i=1n‍xi-1)-1 is the harmonic mean of x.Takingk=n-1 in Corollary 2, we obtain the following inequality.Corollary 3.
Letx∈ℝ+n, n∈ℕ, and E1(x)≤1. Then for n≥2, one has
(7)∑i=1n‍∏j=1,j≠in‍(1xj-2)≤n(1H-2)n-1,
where H=n(∑i=1n‍xi-1)-1 is the harmonic mean of x.


## 2. Definitions and Lemmas

Definition 4 (see [11, 12]).
Letx=(x1,…,xn) and y=(y1,…,yn)∈ℝn.(i)
x is said to be majorized by y (in symbols x≺y) if ∑i=1k‍x[i]≤∑i=1k‍y[i] for k=1,2,…,n-1 and ∑i=1n‍xi=∑i=1n‍yi, where x[1]≥⋯≥x[n] and y[1]≥⋯≥y[n] are rearrangements of x and y in a descending order.(ii)
Ω⊆ℝn is called a convex set if (αx1+βy1,…,αxn+βyn)∈Ω for every x and y∈Ω, where α and β∈[0,1] with α+β=1.(iii)
LetΩ⊆ℝn. 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]).
Letx=(x1,x2,…,xn)∈ℝn. Then
(8)(A,…,A)≺(x1,x2,…,xn),
where A=(1/n)∑i=1n‍xi is the arithmetic mean of x.Definition 6 (see [11, 12]).
LetΩ⊆ℝn.  (i)
Ω is called a symmetric set, if x∈Ω implies xP∈Ω for every n×n permutation matrix P.(ii)
The functionφ: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]).
LetΩ⊂ℝn 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
(9)(x1-x2)(∂φ(x)∂x1-∂φ(x)∂x2)≥0(≤0)
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]).
LetΩ⊂ℝ+n.  (i)
A setΩ is said to be harmonically convex if xy/(λx+(1-λ)y)∈Ω for every x,y∈Ω and λ∈[0,1], where xy=∑i=1n‍xiyi and (1/x)=(1/x1,…,1/xn).(ii)
A functionφ: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]).
LetΩ⊂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
(10)(x1-x2)(x12∂φ(x)∂x1-x22∂φ(x)∂x2)≥0(≤0),x∈Ω0.Lemma 10.
Letx∈ℝ+n, n, k∈ℕ, μ>0, n≥2, and E1(x)≤1. Then for n≥k+μ-1, one has
(11)Ek(1x-μ)≥0.Proof.
LetE1(x)=σ≤1. Then E1(x/σ)=(x1/σ)+⋯+(xn/σ)=1; and n≥k+μ-1≥k+σμ-1, from Theorem B, we have
(12)Ek(σx-σμ)=Ek(σx1-σμ,…,σxn-σμ)≥(nk)(n-σμ)k,
and then
(13)Ek(1x-μ)=Ek(1x1-μ,…,1xn-μ)≥(nk)(n-σμσ)k≥0.
Lemma 10 is proved.


## 3. Proof of Main Result

Proof of Theorem1.
Whenn≥2, k=1, obviously, inequality (5) holds. When n≥2, k≥2, noting that
(14)φ(x):=Ek(1x-μ)=(1x1-μ)(1x2-μ)Ek-2(1x3-μ,…,1xn-μ)+(1x1-μ+1x2-μ)Ek-1(1x3-μ,…,1xn-μ)+Ek(1x3-μ,…,1xn-μ),
we have
(15)∂φ(x)∂x1=-1x12(1x2-μ)Ek-2(1x3-μ,…,1xn-μ)-1x12Ek-1(1x3-μ,…,1xn-μ),(16)∂φ(x)∂x2=-1x22(1x1-μ)Ek-2(1x3-μ,…,1xn-μ)-1x22Ek-1(1x3-μ,…,1xn-μ).
And then
(17)Δ:=(x1-x2)(x12∂φ(x)∂x1-x22∂φ(x)∂x2)=-(x1-x2)2x1x2Ek-2(1x3-μ,…,1xn-μ).
Now we distinguish two cases to proveΔ≤0.
Case
1. If n≥2, k=2, then from the definition of elementary symmetric functions, it follows that Δ≤0.
Ifn=3, k=3, then from the condition n≥k+μ-1, it follows that μ≤n-k+1=1, and by xi≤E1(x)≤1, it follows that 1/xi≥1, i=3,…,n; therefore,
(18)Ek-2(1x3-μ,…,1xn-μ)=E1(1x3-μ,…,1xn-μ)=∑i=3n‍1xi-(n-2)μ≥0.
Case
2. If n≥4; then n-2≥2, from n≥k+μ-1, it follows that
(19)(n-2)≥(k-2)+μ-1
and ∑i=3n‍xi≤1. So by Lemma 10, it follows that Ek-2((1/x3)-μ,…,(1/xn)-μ)≥0.
Thus, for alln≥2, Δ≤0. By Lemma 9, we can derive that Ek((1/x)-μ) is Schur harmonically concave on Ω.
From Lemma5, it is seen that
(20)(1H,…,1H)≺(1x1,…,1xn).
According to Definition 8(ii), inequality (5) follows.
The proof of Theorem1 is completed.


## 4. Applications in Geometry

LetΩn be an n-dimensional simplex in the n-dimensional Euclidean space ℝn(n≥2) with vertices A0,A1,…,An whose volume is V. For i=0,1,…,n, let ri be the radius of ith escribed sphere of Ωn, and Fi the area of the ith face fi={A0,A1,…,Ai-1,Ai+1,…,An} of Ωn, F=∑i=0n‍Fi. R and r are circumradius and inradius of Ωn, respectively. Let P be an arbitrary interior point of the simplex Ωn, di the distance from the point P to the ith face fi of Ωn, and hi the altitude of Ωn from vertex Ai for i=0,1,…,n.In [9], by Theorem B, Ma and Pu obtained the following two theorems.Theorem C.
In the simplexΩn (n≥2), for k=1,2,…,n, one has
(21)Ek(h0r0,h1r1,…,hnrn)≥(n+1k)(n-1)k
with equality if and only if  Ωn is an orthocentric simplex.Theorem D.
LetP be an arbitrary point in the interior of n-dimensional simplex Ωn. Let Bi be the intersection of the line AiP with the hyperplane fi; further let Ri=|AiP|, di′=|PBi|, Mi=|AiBi|, i=0,1,…,n. Then for k=1,2,…,n+1, one has
(22)Ek(R0d0′,R1d1′,…,Rndn′)≥(n+1k)nk
with equality if and only if Ωn is an orthocentric simplex, and P is its orthocentre.Now, by Theorem1, we give the reverse of inequalities (21) and (22).Theorem 11.
In the simplexΩn, for k=1,2,…,n, one has
(23)Ek(h0r0,h1r1,…,hnrn)≤(n+1k)(1H-2)k,
where H=(n+1)(∑i=0n((hi/ri)+2))-1.Proof.
In then-dimensional simplex, for k=0,1,…,n, it is well known that
(24)ri=nV∑j=0n‍Fj-2Fi,hi=nVFi.
Letxi=Fi/∑j=0n‍Fj, i=0,1,…,n. Then, E1(x)=1. From equalities (24), we have
(25)1xi-2=∑j=0n‍Fj-2FiFi=hiri,i=0,1,…,n.
Thus from the inequality (6), it is deduced that the inequality (23) holds.
The proof of Theorem11 is completed.Theorem 12.
LetP be an arbitrary point in the interior of n-dimensional simplex Ωn. Let Bi be the intersection of the line AiP with the hyperplane fi; further let Ri=|AiP|, di′=|PBi|, Mi=|AiBi|, i=0,1,…,n. Then, for k=1,2,…,n+1,
(26)Ek(R0d0′,R1d1′,…,Rndn′)≤(n+1k)(1H-1)k,
where H=(n+1)(∑i=0n(Mi/di′))-1.Proof.
It is easy to see that(27)∑i=0n‍di′Mi=1,RiMi=1-di′Mi,i=0,1,…,n.
In the inequality (5), let μ=1, xi=di′/Mi. Then, 1≤k≤n+1, (1/xi)-1=Ri/di′, i=0,1,…,n. Thus, inequality (26) holds.
The proof of Theorem12 is completed.Whenk=n+1, inequality (26) is reduced to the following.Corollary 13.
Under the conditions of Theorem12, one has
(28)∏i=0n‍Ri≤(1H-1)n+1∏i=0n‍di′.Theorem 14.
In the simplexΩn, let
(29)λi=F-2FiFi,μi=F02+⋯+Fn2-2Fi2Fi2,hhhhhhhhhhhhhhhhhhhhh(i=0,1,…,n).
Then,
(30)∑i=0n‍∏j=0,j≠in‍λj≤(n+1)(1H-2)n,
where H=n(∑i=0n‍(λi+2))-1;
(31)∑i=0n∏j=0,j≠in‍μj≤(n+1)(1H-2)n,
where H=n(∑i=0n‍(μi+2))-1.Proof.
Takingxi=(Fi/F)(i=0,1,…,n) and applying Corollary 3 for these n+1 positive real numbers x0,x1,…,xn, inequality (30) follows.
Takingxi=Fi2/(F02+F12+⋯+Fn2) and applying Corollary 3, the desired inequality (31) follows.
The proof of Theorem14 is completed.Remark 15.
Inequalities (30) and (31) give the reverse of the inequalities (2.10) and (2.11) in [10], respectively.



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