# A New Method to Prove and Find Analytic Inequalities

Authors: Xiao-Ming Zhang; Bo-Yan Xi; Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2010)

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## 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.


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## Body

## 1. Monotonicity Theorem

Throughout the paperℝ 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,…,xn)∈ℝn; the arithmetic mean A(x) and the power mean Mr(x) of order r with respect to the positive real numbers x1,x2,…,xn are defined by A(x)=(1/n)∑i=1nxi, Mr(x)=((1/n)∑i=1nxir)1/r for r≠0, and M0(x)=(∏i=1nxi)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 thatm<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 ξl∈(0,l) such that
(1.1)l(-∂f(x1-ξl,x2+ξl)∂x1+∂f(x1-ξl,x2+ξl)∂x2)≤0,-∂f(x1-ξl,x2+ξl)∂x1+∂f(x1-ξl,x2+ξl)∂x2≤0.
Letting l→0+, we get
(1.2)∂f(x1,x2)∂x1≥∂f(x1,x2)∂x2.
According to the continuity of partial derivatives, we know that
(1.3)∂f(x1,x1)∂x1≥∂f(x1,x1)∂x2
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 ξl∈(0,l) such that
(1.4)f(a,b)-f(a-l,b+l)=-(f(a-l,b+l)-f(a,b))=-l(-∂f(a-ξl,b+ξl)∂x1+∂f(a-ξl,b+ξl)∂x2)=l(∂f(a-ξl,b+ξl)∂x1-∂f(a-ξl,b+ξl)∂x2)≥0.
Therefore the proof of Lemma1.1 is completed.Theorem 1.2.
Suppose thatD⊂ℝn is a symmetric convex set with nonempty interior, f:D→ℝ has continuous partial derivatives, and
(1.5)D⌣i={x∈D∣xi=max1≤j≤n{xj}}-{x∈D∣x1=x2=⋯=xn},D⌢i={x∈D∣xi=min1≤j≤n{xj}}-{x∈D∣x1=x2=⋯=xn},i=1,2,…,n. If for all i,j=1,2,…,n with i≠j,
(1.6)∂f∂xi>(<)∂f∂xj
holds in D⌣i∩D⌢j, then
(1.7)f(a1,a2,…,an)≥(≤)f(A(a),A(a),…,A(a))
for all a=(a1,a2,…,an)∈D, with equality if only if a1=a2=⋯=an.Proof.
Ifn=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/∂xi>∂f/∂xj with i≠j.
Ifa1=a2=⋯=an, then inequality (1.7) is clearly true. If max1≤j≤n{aj}≠min1≤j≤n{aj}, then without loss of generality we assume that a1≥a2≥⋯≥an-1≥an.
(1)   Ifa1>max2≤j≤n{aj} and an<min1≤j≤n-1{aj}, then (a1,a2,…,an)∈D⌣1∩D⌢n. From Lemma 1.1 and the conditions in Theorem 1.2 we know that there exist a1(1) and an(1) such that l=a1-a1(1)=an(1)-an>0, a1(1)=a2 or an(1)=an-1, and
(1.8)f(a1,a2,a3,…,an)≥f(a1(1),a2,a3,…,an(1)).
For the sake of convenience, we denote ai(1)=ai,2≤i≤n-1. Consequently,
(1.9)f(a1,a2,a3,…,an)≥f(a1(1),a2(1),a3(1),…,an(1)).
If a1(1)=a2(1)=⋯=an(1), then Theorem 1.2 holds. Otherwise, for the case of a1(1)=a2(1)>an(1), (a1(1),a2(1),a3(1),…,an(1))∈D⌣1∩D⌢n and
(1.10)∂f(x)∂x1|x=(a1(1),a2(1),a3(1),…,an(1))>∂f(x)∂xn|x=(a1(1),a2(1),a3(1),…,an(1)).
From the continuity of partial derivatives we know that there exists ɛ>0 such that
(1.11)∂f(x)∂x1|x=(s,a2(1),a3(1),…,t)>∂f(x)∂xn|x=(s,a2(1),a3(1),…,t),
where s∈[a1(1)-ɛ,a1(1)] and t∈[an(1),an(1)+ɛ]. Denote a1(2)=a1(1)-ɛ, an(2)=an(1)+ɛ, ai(2)=ai(1)(2≤i≤n-1). By Lemma 1.1, we get
(1.12)f(a1(1),a2(1),a3(1),…,an(1))≥f(a1(2),a2(2),a3(2),…,an(2)),
and a2(2)=max1≤i≤n{ai(2)}. For the case of a1(1)>an-1(1)=an(1), after a similar argument, we get inequality (1.12) with an-1(2)=min1≤i≤n{ai(2)}.
Repeating the above steps, we get{a1(i),a2(i),…,an(i)}(i=1,2,…) such that ∑j=1naj(i) is a constant and {aj(i)}(i=1,2,…) are monotone increasing (decreasing) sequences if aj≥(≤)A(a),j=1,2,3,…,n, and
(1.13)f(a1(1),a2(1),a3(1),…,an(1))≥f(a1(i),a2(i),a3(i),…,an(i)).
If there exists i∈ℕ such that a1(i)=a2(i)=⋯=an(i), then the proof of Theorem 1.2 is completed. Otherwise, we denote α=infi∈ℕ{max{a1(i),a2(i),…,an(i)}}; without loss of generality, we assume that
(1.14)max{a1(ij),a2(ij),…,an(ij)}=a1(ij)→α,limj→+∞(a1(ij),a2(ij),…,an(ij))=(α,b2,b3,…,bn),
where {ij}j=1+∞ is a subsequence of ℕ. Then from the continuity of function f, we get
(1.15)f(a1,a2,a3,…,an)≥f(α,b2,b3,…,bn).
If α≠min{b2,b3,…,bn}, then we repeat the above arguments and get a contradiction to the definition of α. Hence α=b2=b3=⋯=bn. From α+∑i=2nbi=∑i=1nai we get α=b2=b3=⋯=bn=A(a); the proof of Theorem 1.2 is completed.
(2) The proof for the case of a1=max2≤j≤n{aj} or an=min1≤j≤n-1{aj} is implied in the proof of (1).In particular, according to Theorem1.2 the following corollary holds.Corollary 1.3.
Suppose thatD⊂ℝn is a symmetric convex set with nonempty interior, f:D→ℝ is a symmetric function with continuous partial derivatives, and
(1.16)D⌣1={x∈D∣x1=max1≤j≤n{xj}}-{x∈D∣x1=x2=⋯=xn},D⌢2={x∈D∣x2=min1≤j≤n{xj}}-{x∈D∣x1=x2=⋯=xn},D*=D⌣1∩D⌢2.
If ∂f/∂x1>(<)∂f/∂x2 holds in D*, then
(1.17)f(a1,a2,…,an)≥(≤)f(A(a),A(a),…,A(a))
for all a=(a1,a2,…,an)∈D, and equality holds if and only if a1=a2=⋯=an.


## 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.
Foru=(u1,u2,…un),v=(v1,v2,…vn)∈ℝn, without loss of generality one assumes that u1≥u2≥⋯≥un and v1≥v2≥⋯≥vn. Then u is said to be majorized by v (in symbols u≺v) if ∑i=1kui≤∑i=1kvi for k=1,2,…,n-1 and ∑i=1nui=∑i=1nvi.Definition 2.2.
Suppose thatΩ⊂ℝn. 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 thatΩ⊂ℝn 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
(2.1)(u1-u2)(∂φ∂u1-∂φ∂u2)≥(≤)0
holds for any u=(u1,u2,…,un)∈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, Theorem1.2 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 Theorem1.2 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(3.1)(x1,x2,…,xn),(y1,y2,…,yn)∈ℝ+n.
If p,q>1 and 1/p+1/q=1, then
(3.2)(∑k=1nxkp)1/p(∑k=1nykq)1/q≥∑k=1nxkyk.Proof.
Let(a1,a2,…,an)∈ℝ+n and
(3.3)f:b→(∑k=1n‍ak)1/p(∑k=1nakbk)1/q-∑k=1nakbk1/q,b∈ℝ+n.
Then
(3.4)∂f∂bi=1q·(∑k=1nak)1/p(∑k=1nakbk)1/q-1ai-1q·aibi1/q-1,∂f∂bi-∂f∂bj=1q(∑k=1nak∑k=1nbkak)1/p(ai-aj)-1q(aibi-1/p-ajbj-1/p).
Let b∈D⌣i∩D⌢j (see (1.5)).
(1) Ifai≥aj, then
(3.5)∂f∂bi-∂f∂bj≥1q(∑k=1nakbi∑k=1nak)1/p(ai-aj)-1q(aibi-1/p-ajbj-1/p)=1qaj(bj-1/p-bi-1/p)>0.
(2) If ai≤aj, then
(3.6)∂f∂bi-∂f∂bj≥1q(∑k=1nakbj∑k=1nak)1/p(ai-aj)-1q(aibi-1/p-ajbj-1/p)=1qai(bj-1/p-bi-1/p)>0.
From Theorem 1.2 we get
(3.7)f(b)≥f(A(b),A(b),…,A(b)),
that is,
(3.8)(∑k=1nak)1/p(∑k=1nakbk)1/q-∑k=1nakbk1/q≥0.
Therefore, the Hölder inequality follows from (3.8) with ak=xkp and bk=ykq/xkp.


## 4. Improvement of the Sierpiński Inequality

In the section, we give some improvements of the well-known Sierpiński inequality:(4.1)[M-1(a)](n-1)/n[A(a)]1/n≤M0(a)≤[M-1(a)]1/n[A(a)](n-1)/n.Theorem 4.1.
Suppose thatn≥3, a=(a1,a2,…,an)∈ℝ+n, β>0>α. If λ=-2α/n(β-α) for β+α>0 and λ=1/n for β+α≤0, then
(4.2)[Mα(a)]1-λ·[Mβ(a)]λ≤M0(a).Proof.
Letf(x)=(1/nβ)ln(∏i=1nxi)-((1-λ)/α)ln((1/n)∑i=1nxiα/β), x∈ℝ+n. Then
(4.3)∂f(x)∂xj=1nβxj-1-λβxjα/β-1∑i=1nxiα/β,j=1,2,∂f(x)∂x1-∂f(x)∂x2=x2-x1nβx1x2-1-λβx1α/β-1-x2α/β-1∑i=1nxiα/β.
Case 1.
α+β>0. Let
(4.4)g(t)=β+αβ-αtβ-α-tβ+t-α-β+αβ-α,t∈(1,+∞).
Then
(4.5)tα+1g′(t)=(β+α)tβ-βtβ+α-α,[tα+1g′(t)]′=(β+α)βtβ+α-1(t-α-1)>0.
Therefore, tα+1g′(t) is monotone increasing in (1,+∞). From
(4.6)limt→1+tα+1g′(t)=limt→1+[(β+α)tβ-βtβ+α-α]=0,
we know that tα+1g′(t)>0, g′(t)>0. Then limt→1+g(t)=0 leads to g(t)>0 and
(4.7)β+αβ-αtβ-α-tβ+t-α-β+αβ-α>0,β+αβ-αtβ-(1+2αβ-α)tα-tα+β+1>0,β+αβ-αtβ-(n-1+2αβ-α)tα-tα+β+(n-1)>0,(1-nλ)tβ-(n-1-nλ)tα-tα+β+(n-1)>0,(4.8)(1-λ)1-tα-βtα+(n-1)>tβ-1ntβ.
We assume that x∈D* (see (1.16)). Let t=(x1/x2)1/β. Then inequality (4.8) becomes
(4.9)(1-λ)x2α/β-1-x1α/β-1x1α/β+(n-1)x2α/β>x1-x2nx1x2,1-λβx2α/β-1-x1α/β-1∑i=1nxiα/β>x1-x2nβx1x2.
Combining inequalities (4.3) and (4.9) yields that ∂f(x)/∂x1-∂f(x)/∂x2>0. Using Corollary 1.3 we have
(4.10)f(x1,x2,…,xn)≥f(A(x),A(x),…,A(x)),1nβln(∏i=1nxi)-1-λαln(1n∑i=1nxiα/β)≥λβln(1n∑i=1nxi).
Letting ai=xi1/β,i=1,2,…,n, we get
(4.11)[Mα(a)]1-λ·[Mβ(a)]λ≤M0(a).Case 2.
α+β<0. Let t>1. Then from α<0 and α+β<0, one has
(4.12)(n-1)>(n-2)tα+tα+β.
Hence inequality (4.8) holds. The rest is similar to above, so we omit it.
The proof of Theorem4.2 is similar to the proof of Theorem 4.1, and so we omit it.Theorem 4.2.
Suppose thatn≥3, a=(a1,a2,…,an)∈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
(4.13)M0(a)≤[Mα(a)]1-θ·[Mβ(a)]θ.Theorem 4.3.
Suppose thatn≥3, a=(a1,a2,…,an)∈ℝ+n. If r=-lnn/(n-1)[lnn-ln(n-1)], then r<-1 and
(4.14)[M1/r(a)](n-1)/n[A(a)]1/n≤M0(a)≤[Mr(a)]1/n[A(a)](n-1)/n.Proof.
Letn≥3 and
(4.15)f:x∈(0,+∞)n→∑k=1nxk1/rn(∑k=1nxkn)1/r(n-1)-(∏k=1nxk)1/r(n-1).
Then
(4.16)r=-lnnln(1+1/(n-1))n-1<-lnnlne=-lnn<-1,∂f∂x1=x1(1-r)/rrn(∑k=1nxkn)1/r(n-1)+∑k=1nxk1/rrn2(n-1)(∑k=1nxkn)(1/r(n-1))-1-1r(n-1)x1(∏k=1nxk)1/r(n-1).
Therefore, we get
(4.17)∂f∂x1-∂f∂x2=1rn(x1(1-r)/r-x2(1-r)/r)(∑k=1nxkn)1/r(n-1)-1r(n-1)(∏k=1nxk)1/r(n-1)(1x1-1x2),(∏k=1nxk)1/(-r(n-1))(∂f∂x1-∂f∂x2)=x1(1-r)/(-r)-x2(1-r)/(-r)-rnx1(1-r)/(-r)x2(1-r)/(-r)(n∏k=1nxk∑k=1nxk)1/(-r)(n-1)+x1-x2r(n-1)x1x2=x1(1-r)/(-r)-x2(1-r)/(-r)-rnx1(1-r)/(-r)x2(1-r)/(-r)(n∑k=1n∏i=1,≠knxi-1)1/(-r)(n-1)+x1-x2r(n-1)x1x2.
We assume that x∈D* (see (1.16)). Then we have
(4.18)(∏k=1nxk)1/-r(n-1)(∂f∂x1-∂f∂x2)≥x1(1-r)/(-r)-x2(1-r)/(-r)-rnx1(1-r)/(-r)x2(1-r)/(-r)(nx2-(n-1)+(n-1)x1-1x2-(n-2))1/(-r)(n-1)+x1-x2r(n-1)x1x2.
Letting x1/x2=t>1, from n1+1/r(n-1)=n1-(lnn-ln(n-1))/lnn=n-1, we get
(4.19)(∏k=1nxk)1/-r(n-1)(∂f∂x1-∂f∂x2)≥1n-1·x1(1-r)/(-r)-x2(1-r)/(-r)-rx1(1-r)/(-r)x2(1-r)/(-r)(x1x2n-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)(tt+n-1)(lnn-ln(n-1))/lnn-t-1t]=1-r(n-1)x2[t(1-r)/(-r)-1t(1-r)/(-r)·(1+n-1t)-(lnn-ln(n-1))/lnn-t-1t].
According to Bernoulli's inequality(1+x)α<1+αx with x≥-1,x≠0, and 0<α<1, one has
(4.20)(∏k=1nxk)1/-r(n-1)(∂f∂x1-∂f∂x2)>1-r(n-1)x2[t(1-r)/(-r)-1t(1-r)/(-r)·11+(lnn-ln(n-1))/lnn·(n-1)/t-t-1t]=1-r(n-1)x2[t(1-r)/(-r)-1t(1-r)/(-r)-t1/(-r)/r-t-1t]=1-r(n-1)x2·(1+1/r)t1/(-r)-1/r·t(1+r)/(-r)-1t(1-r)/(-r)-t1/(-r)/r.
For 0<s<1 and t>1, it is not difficult to verify that (1-s)ts+sts-1-1>0. Letting s=-1/r, we have
(4.21)(1+1r)t1/(-r)-1r·t(1+r)/(-r)-1>0,∂f∂x1-∂f∂x2>0.
Using Corollary 1.3, we know that
(4.22)f(x1,x2,…,xn)≥f(A(x),A(x),…,A(x)),∑k=1nxk1/rn(∑k=1nxkn)1/r(n-1)≥(∏k=1nxk)1/r(n-1).
Letting ai=xi1/r(i=1,2,…,n), we get
(4.23)∑k=1nakn(∑k=1nakrn)1/r(n-1)≥(∏k=1nak)1/(n-1),(4.24)[A(a)](n-1)/n[Mr(a)]1/n≥M0(a).
From (4.23), we get
(4.25)(∏k=1nak)r≥(∑k=1nakn)(n-1)r·∑k=1nakrn.
Letting ai→ai1/r(i=1,2,…,n), we have
(4.26)∏k=1nak≥(∑k=1nak1/rn)(n-1)r·∑k=1nakn,M0(a)≥[M1/r(a)](n-1)/n[A(a)]1/n.
Inequality (4.14) is proved.


## 5. Five New Inequalities

Letn≥3 and a=(a1,a2,…,an)∈ℝ+n. Then(5.1)∏nk(a)=(∏1≤i1<⋯<ik≤n1k∑j=1kaij)1/(nk)References [5, 6] is the third symmetric mean of a.Theorem 5.1.
If2≤k≤n-1, p=(k-1)/(n-1), then
(5.2)∏nk(a)≥[A(a)]p[M0(a)]1-p
with the best possible constant p=(k-1)/(n-1).Proof.
Leta=(a1,a2,…,an)∈ℝ+n and
(5.3)f(a)=[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤n1k∑j=1kaij.
Then
(5.4)∂f∂a1=-(n-k)·(nk)n(n-1)a1[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤n1k∑j=1kaij+[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤n1k∑j=1kaij·(∑2≤i1<⋯ik-1≤n1a1+∑i=1k-1aij),(5.5)∂f∂a1-∂f∂a2=-(n-k)·(nk)n(n-1)[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤n1k∑j=1kaij(1a1-1a2)+[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤n1k∑j=1kaij·(∑3≤i1<⋯ik-1≤n(1a1+∑i=1k-1aij-1a2+∑i=1k-1aij))=(a1-a2)·[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤n1k∑j=1kaij·[(n-k)·(nk)n(n-1)a1a2-∑3≤i1<⋯ik-1≤n1(a1+∑i=1k-1aij)(a2+∑i=1k-1aij)].
If a∈D* (see (1.16)), then
(5.6)a1+(k-1)a2>a1,(n-k)·(nk)n(n-1)a1a2>(n-2k-1)ka2(a1+(k-1)a2),(n-k)·(nk)n(n-1)a1a2>∑3≤i1<⋯ik-1≤n1(a1+(k-1)a2)ka2,(n-k)·(nk)n(n-1)a1a2>∑3≤i1<⋯ik-1≤n1(a1+∑i=1k-1aij)(a2+∑i=1k-1aij).
Combining inequalities (5.5) and (5.6) yields that ∂f/∂a1-∂f/∂a2>0. Then from Corollary 1.3 we have
(5.7)f(a1,a2,…,an)≥f(A(a),A(a),…,A(a))
for all a=(a1,a2,…,an)∈ℝ+n, which implies that
(5.8)[∏i=1nai]-(n-k)·(nk)/n(n-1)·∏1≤i1<⋯<ik≤nk-1∑j=1kaij≥[A(a)](k-1)·(nk)/(n-1).
Therefore, inequality (5.2) is proved.Takinga1=a2=⋯=an-1=1 and an=x in inequality (5.2), we get(5.9)(x+k-1k)(n-1k-1)/(nk)≥(x+n-1n)p(xn)1-p,p≤(k/n)ln((x+k-1)/k)-(1/n)lnxln(x+n-1)-ln(nxn).
Letting x→+∞, we get(5.10)p≤limx→+∞k/n·1/(x+k-1)-1/nx1/(x+n-1)-1/nx=limx→+∞kx/(x+k-1)-1nx/(x+n-1)-1=k-1n-1.
So p=(k-1)/(n-1) is the best possible constant.Forn≥2, a=(a1,a2,…,an)∈ℝ+n, Alzer [7] established the following inequality:(5.11)n-1nA(a)+1nM-1(a)≥M0(a).Theorems5.2 and 5.3 are the improvements of Alzer's inequality.Theorem 5.2.
Ifp=n2/(n2+4n-4), then
(5.12)pA(a)+(1-p)M-1(a)≥M0(a).Proof.
Firstly, letp>n2/(n2+4n-4), and
(5.13)f(x)=p/n·∑i=1nexi+(1-p)n·(∑i=1ne-xi)-1,x=(x1,x2,…,xn)∈ℝn.
Then
(5.14)∂f∂x1=pnex1+(1-p)n(∑i=1ne-xi)2e-x1,∂f∂x1-∂f∂x2=pn(ex1-ex2)-(1-p)n(∑i=1ne-xi)2(e-x2-e-x1).
If x1=max1≤i≤n{xi}>x2=min1≤i≤n{xi}, t=ex1-x2>1, then
(5.15)∂f∂x1-∂f∂x2≥pn(ex1-ex2)-(1-p)n((n-1)e-x1+e-x2)2(e-x2-e-x1)=(ex1-ex2)n((n-1)ex2+ex1)2[p((n-1)ex2+ex1)2-(1-p)n2ex1ex2]=e3x2(t-1)n((n-1)ex2+ex1)2[p(n-1+t)2-n2t+pn2t]>e3x2(t-1)n((n-1)ex2+ex1)2[n2n2+4n-4(n-1+t)2-n2t+n2n2+4n-4n2t]=ne3x2(t-1)(t-n+1)2(n2+4n-4)((n-1)ex2+ex1)2≥0.
Then from Corollary 1.3, we get
(5.16)f(x)≥f(A(x),A(x),…,A(x)),pn∑i=1nexi+(1-p)n∑i=1ne-xi≥eA(x)=∏i=1nexin.
Let exi=ai 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.
Ifp=(1-n-5n2-6n+1)/(2n), then
(5.17)n-1nA(a)+1nMp(a)≥M0(a).Proof.
Let(5.18)f(a)=∏i=1nai1/pn-(n-1)n2·∑i=1nai1/p,a∈ℝ+n.
Then
(5.19)∂f∂a1=1npa1∏i=1nai1/pn-n-1n2pa11/p-1,∂f∂a1-∂f∂a2=-a1-a2npa1a2∏i=1nai1/np-n-1n2p(a11/p-1-a21/p-1).
If a1=max1≤i≤n{ai}>a2=min1≤i≤n{ai}>0 and a1/a2=t>1, then from p<0 and -(a1-a2)/npa1a2>0 we get
(5.20)∂f∂a1-∂f∂a2≤-a1-a2npa1a2a11/(np)a2(n-1)/(np)-n-1n2p(a11/p-1-a21/p-1)=a11/p-1n2p[-nt-1tt1-(n-1)/(np)-(n-1)(1-t1-1/p)].
Letg(t)=-nt1-(n-1)/(np)+nt-(n-1)/(np)+(n-1)t1-1/p-(n-1),t>1. Then
(5.21)g′(t)=(-n+n-1p)t-(n-1)/(np)-n-1pt-1-(n-1)/(np)+(n-1)(1-1p)t-1/p,t1+(n-1)/(np)g′(t)=(-n+n-1p)t-n-1p+(n-1)(1-1p)t1-1/(np),(t1+(n-1)/(np)g′(t))′=(-n+n-1p)+(n-1)(1-1p)(1-1np)t-1/(np)>(-n+n-1p)+(n-1)(1-1p)(1-1np)=-1p2[p2+(1-1n)p-1+1n]=0.
Thus t1+(n-1)/(np)g′(t) is a monotone increasing function. This monotonicity and
(5.22)limt→1+t1+(n-1)/(np)g′(t)=limt→1+[(-n+n-1p)t-n-1p+(n-1)(1-1p)t1-(1/np)]=-1-n-1p≥0
lead to t1+(n-1)/(np)g′(t)>0. Therefore g′(t)>0 and g(t) is a monotone increasing function. From limt→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
(5.23)f(a)≤f(A(a),A(a),…,A(a)),∏i=1nai1/pn-n-1n2·∑i=1nai1/p≤1n·A1/p(a).
Finally, let ai→aip(i=1,2,…,n) in the above inequality. Then we know that Theorem 5.3 holds.
Ifn≥2 and 0<a1≤a2≤⋯≤an, then the following inequalities can be found in [8–10]:
(5.24)∑1≤i<j≤n(ai-aj)22n2an≤A(a)-M0(a)≤∑1≤i<j≤n(ai-aj)22n2a1,a132n2an4∑1≤i<j≤n(ai-aj)2≤M0(a)-M-1(a)≤an32n2a14∑1≤i<j≤n(ai-aj)2.
Theorems 5.4 and 5.5 are the improvements of inequalities (5.24).Theorem 5.4.
Ifn≥2 and 0<m≤a1,a2,…,an≤M, then
(5.25)∑1≤i<j≤n(ai-aj)22n2·M(n-1)/n·A1/n(a)≤A(a)-M0(a)≤∑1≤i<j≤n(ai-aj)22n2·m(n-1)/n·A1/n(a).Proof.
Let(5.26)f:a∈[m,M]n→(1n∑k=1nak)1/n(1n∑k=1nak-∏k=1nakn)-∑1≤i<j≤n(ai-aj)22n2M(n-1)/n.
Then
(5.27)∂f∂a1=1n2(1n∑k=1nak)1/n-1(1n∑k=1nak-∏k=1nakn)+(1n∑k=1nak)1/n(1n-1na1∏k=1nakn)-∑2≤i≤n(a1-ai)n2M(n-1)/n,∂f∂a1-∂f∂a2=a1-a2na1a2∏k=1nakn·(1n∑k=1nak)1/n-a1-a2nM(n-1)/n=a1-a2na1a2M(n-1)/n[M(n-1)/n∏k=1nakn·(1n∑k=1nak)1/n-a1a2].
We assume that a∈D* (see (1.16)). Then
(5.28)∂f∂a1-∂f∂a2>a1-a2na1a2M(n-1)/n[M(n-1)/na11/na2(n-1)/n·a21/n-a1a2]≥a1-a2na1a2M(n-1)/n[a1(n-1)/n·a11/na2(n-1)/n·a21/n-a1a2]=0.
According to Corollary 1.3, we get
(5.29)f(a)≥f(A(a),A(a),…,A(a)),A(a)-M0(a)≥∑1≤i<j≤n(ai-aj)22n2M(n-1)/nA1/n(a).
Let(5.30)g:a∈[m,M]→∑1≤i<j≤n(ai-aj)22n2m(n-1)/n-[1n∑k=1nak]1/n[1n∑k=1nak-∏k=1nakn].
A similar argument as above leads to
(5.31)A(a)-M0(a)≤∑1≤i<j≤n(ai-aj)22n2m(n-1)/nA1/n(a).
The proof of Theorem5.4 is completed.Let(5.32)f:x∈[1M,1m]→12n2·M(n-3)/nm(2n-3)/n∑1≤i<j≤n(1xi-1xj)2-1M0(x),g:x∈[1M,1m]→1M0(x)-m(n-1)/n2n2M(2n-1)/n∑1≤i<j≤n(1xi-1xj)2.The proof of Theorem5.5 is similar to the proof of Theorem 5.4, and so we omit it.Theorem 5.5.
Letn≥2, 0<m≤a1,a2,…,an≤M. Then
(5.33)m(n-1)/n2n2M(2n-1)/n∑1≤i<j≤n(ai-aj)2≤M0(a)-M-1(a)≤M(n-3)/n2n2m(2n-3)/n∑1≤i<j≤n(ai-aj)2.Remark 5.6.
More applications for Theorem1.2 and Corollary 1.3 were shown in [11].



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*Source: 128934-2010-04-07.xml*