Báo cáo hóa học: "THE OPTIMIZATION FOR THE INEQUALITIES OF POWER MEANS" docx

Inthis paper, we will state the important background and meaning of the inequality m j =1{M n[α j]a } λ j ≤≥M n[θ]a; a necessary and sufficient condition and another teresting sufficient con

OF POWER MEANS

JIAJIN WEN AND WAN-LAN WANG

Received 14 November 2005; Revised 5 May 2006; Accepted 14 July 2006

Let M n[t](a) be the tth power mean of a sequence a of positive real numbers, where

a =(a1,a2, , an),n ≥2, andα, λ ∈ R m

++,m ≥2,m

j =1λ j =1, min{ α } ≤ θ ≤max{ α } Inthis paper, we will state the important background and meaning of the inequality

m

j =1{M n[α j](a) } λ j ≤(≥)M n[θ](a); a necessary and sufficient condition and another teresting sufficient condition that the foregoing inequality holds are obtained; an openproblem posed by Wang et al in 2004 is solved and generalized; a rulable criterion of thesemipositivity of homogeneous symmetrical polynomial is also obtained Our methodsused are the procedure of descending dimension and theory of majorization; and applytechniques of mathematical analysis and permanents in algebra

in-Copyright © 2006 J Wen and W.-L Wang This is an open access article distributed der the Creative Commons Attribution License, which permits unrestricted use, distri-bution, and reproduction in any medium, provided the original work is properly cited

un-1 Symbols and introduction

We will use some symbols in the well-known monographs [1,5,13]:

Recall that the definitions of thetth power mean and Hardy mean of order r for a

sequencea =(a1, , an) (n ≥2) are, respectively,

M[t]

n (a) =

1

Journal of Inequalities and Applications

Volume 2006, Article ID 46782, Pages 1 25

DOI 10.1155/JIA/2006/46782

H n(a; r) =

1

is called Hardy function, wherei1, , i nis the total permutation of 1, , n.

Definition 1.1 Let α ∈ R n, letλ αbe a function ofα, λ α ∈ R,x ∈ R n

++ Then the function

f (x) =α ∈ B d λαhn(x; α) is called the generalized homogeneous symmetrical polynomial

ofn variables and degree d When Bd ⊂ Z n

+, f (x) is called the homogeneous symmetrical

polynomial ofn variables and degree d, simply, homogeneous symmetrical polynomial

(see [24, page 431])

Definition 1.2 Let ai j be the complex numbers,i, j =1, 2, , n, and let the matrix A =

(ai j)n × nbe ann × n matrix Then the permanent (of order n) of A is a function of matrix,

written perA, it is defined by

perA =

σ

where the summation extends over all one-to-one functions from 1, , n to 1, , n (See

[12].) It is often convenient in the proof ofLemma 2.2andCorollary 2.6that we will alsoapply a symbol similar to determinant as follows:

perA =

de-we have “the determinant changes sign if two adjacent rows are interchanged.” But the

affirmative proposition and its corollaries do not hold for permanents

As pointed out in [1], the theory of inequalities plays an important role in all thefields of mathematics And the power mean is the most important one in all the means.Many mathematicians wrote a great number of papers, and established the inequalitiesinvolving the power means and the related problems (see, e.g., [1,4,5,8,9,13,14,17–

19,21]) Recently, the authors studied the optimal real numberλ such that the following

20]) Although it is so, the true worth for inequalities is as follows: if an inequality cludes some parameters, we study that these parameters should satisfy some necessaryand sufficient conditions such that this inequality holds, then we call that the inequality

in-is optimized In thin-is paper, we want to din-iscuss the following optimal problems that aremore general inequalities than inequality (1.5)

y(t) θ dt

 1/θ (1.9)

By the above facts, inequality (1.8) can also be expressed as

prob-obtain not only a necessary and sufficient condition, but also an interesting sufficientcondition such that inequality (1.6) holds Note that the inequalities (1.6), (1.8), and(1.10) play some roles in the geometry of convex body (see, e.g., [3,7]) Our methodsare, of late years, the approach of descending dimension and theory of majorization; andapply some techniques of mathematical analysis and permanents [12] in algebra Notethat the way of descending dimension used in this paper is different from [15,23,25];and the majorization is an effective theory that “it can state the inwardness and the rela-tion between the quantities” (see [4,11,16]) It is very interesting that the mathematicalanalysis and permanent can skillfully be combined.

2 The background of inequality ( 1.6 )

The following theorem can display the background and meaning of inequality (1.6)

a11 a12 ··· a1n

a21 a22 ··· a2n

. . an1 an2 ··· ann

Proof We will prove the general case by the induction for m,

1

n! ·

All the elements ofn − m rows in the above permanent are 1.

Whenm =1, then the sign of equality is valid in (2.5) Assume thatm =2 below

We delete the element at ith row and jth column from the permanent per A, then we

construct a permanent of ordern −1, and it is called cofactor ofai j and is denoted by

Mi j Note the following identities and inequalities:

where we used ˇCebyˇsev’s inequality

Assume that the elements in the left-hand side of (2.5) are not all 1, and the count ofthese rows is equal tom −1 (m ≥3), inequality (2.5) holds We will prove that inequality(2.5) holds as follows

First we prove that inequalities (2.6) hold still

Note that the expansion of permanentM1jin terms of elements of the first column isgiven by

i =2(ai2 − ai1)M i1 ∗ ≥0,namely,

1(n −1)!M11≥ 1

Similarly,

1(n −1)!M12≥ 1

(n −1)!M13≥ ··· ≥ 1

(n −1)!M1n . (2.11)Thus, the first chain in (2.6) is proven; and the second chain of (2.6) is given

By inequality (2.6) and ˇCebyˇsev’s inequality, we obtain that

the left-hand side of (2.5) =1

.

(2.12)

It is noteworthy that the sign of equality of (2.12) is valid whena11= a12= ··· = a1n =1

If we change two rows (columns) in permanent, then permanent keeps invariable, then,from the assumption of the induction, we get

n! ·

From inequalities (2.12) and (2.13), we obtain inequality (2.5)

Lettingm = n in (2.5), we get inequality (2.4) So the proof is complete 

Proof Just as well assume that 0 < x1≤ x2≤ ··· ≤ xn, then

0< x α i

1 ≤ x α i

2 ≤ ··· ≤ x α i

n, i =1, 2, , n. (2.15)Thus, by the definition of permanent and byLemma 2.2, we obtain that

to the convex functions, where x ∈ R n

++, α, β ∈ R n; [24] generalizes the well-knownˇ

Cebyˇsev inequality to the generalized homogeneous symmetrical polynomial; [22] ied a necessary and sufficient condition such that

+, λα > 0 (for all

α ∈ Bd ), [1/ f (x)] δ is integrable on measurable set G, (0, 1] n ⊂ G ⊂ R n , then, for arbitrary

t d+n −1

δ dt

hn(x; α) = n!1 ·

n

(2.26)

InSection 1throughSection 2, these pioneer studies that the authors attempted woulddemonstrate that these results of this paper occupy some important positions in the the-ory of inequalities, as well as they are often used in several function spaces 

3 A necessary and sufficient condition that inequality ( 1.6 ) holds

We have known fromSection 2that investigation that inequalities (1.6) and (1.7) hold hasconsiderable meaning In this section, we will discuss how to transform inequality (1.6)into an inequality involving fewer variables so that there is a possibility that inequality(1.6) can be proven by means of mathematical software

Theorem 3.1 Let a ∈ R n

++, α, λ ∈ R m

++, n ≥ m ≥ 2,m

j =1λ j = 1, min { α } ≤ θ ≤max{ α } Then, a necessary and su fficient condition such that inequality ( 1.6 ) holds is that inequality

be a common polynomial of one variable Then u(t) has at most m zeroes onR 1

++, that is, the count of elements of the set Um = { t | u(t) =0,t > 0 } is | Um | , where | Um | ≤ m.

Proof We will prove by means of the induction for m.

Whenm =1, the conclusion is clear Assume that when 1≤ k ≤ m −1 (m ≥2), theinequality| U k | ≤ k holds We will prove that | U m | ≤ m holds as follows We can assume

++ Sincet r1 −1> 0, therefore u(t) has at most m −1 zeroes on

R 1

++,u(t) has at most m −1 extreme points onR 1

++ Let all the extreme points ofu(t) on

R 1

++be

t1,t2, , t p, t1< t2< ··· < t p, 0≤ p ≤ m −1. (3.4)

Ifp =0, thenu(t) is a monotonic function onR 1

++ We may assume thatu(t) is a

increas-ing function onR 1

++ We will prove thatu(t) is a strictly increasing function onR 1

++ asfollows

Let 0< t1< t2, thenu(t1)≤ u(t2) Ifu(t1)= u(t2), then for allt ∈[t1,t2],u(t1)≤ u(t) ≤

u(t2)= u(t1),u(t) ≡ u(t1),u(t) ≡0 Thus, for allt ∈[t1,t2],t is the zero of u(t) onR 1

++.This contradicts withu(t) which has m −1 zeroes onR 1

++ Therefore,u(t1)< u(t2),u(t)

is a strictly increasing function onR 1

++ Based on these facts, the count of the zeroes of

u(t) onR 1

++is| Um | ≤1≤ m.

Ifp ≥1, we can assert by using the above method thatu(t) is a strictly monotonic

func-tion on each of the followingp + 1 intervals: (0, t1], [t1,t2], , [tp, +∞) And the number

of zeroes ofu(t) is at most 1 on each of these intervals, then the amount of zeroes of u(t)

If Aq is a critical point of F(Aq,On − q ) (for all q : m ≤ q ≤ n) on Dq:= { Aq |q r =1ar =

q, Aq ∈ R q+}, then a1,a2, , aq satisfying r ≤ m, where m denotes largest number of the pair

(a i,a j ) with a i = a j , i < j, for i, j =1, 2, , q, that is, the amount of the elements in the set

{ a1,a2, , aq } is |{ a1,a2, , aq }| , and |{ a1,a2, , aq }| ≤ m.

Proof Make the Lagrange function L = F(Aq,On − q) +μ(q

r =1ar − q) Then Aqis a cal point ofF(Aq,On − q) on the domainDqif and only if∂L/∂ak = ∂F(Aq,On − q)/(∂ak) +

criti-μ =0,k =1, 2, , q, and Aq ∈ Dq Since lnF(Aq,On − q)=m

Proof of Theorem 3.1 Necessity If inequality (1.6) holds, in (1.6), we put that a =

(Am −1,Ik,On − m − k+1), (for allk : 0 ≤ k ≤ n − m + 1), then, (1.6) reduces to (3.1), thus (3.1)holds

Sufficiency Assume that (3.1) holds We will prove that inequality (1.6) holds.

Note that we will prove a more general conclusion, that is,

F

A q,O n − q ≤ q

n, ∀ q : m ≤ q ≤ n, ∀ A q ∈ D q, (3.9)where the definitions ofF(a) and Dqare inLemma 3.3 We can prove that (3.9) holds for

we just have to prove that for the critical pointA qofF(A q,O n − q) onD q, for the pointA q

on the boundary ofD q, (3.9) holds still

Case 1 If Aqis a critical point ofF(Aq,On − q) onDq, fromLemma 3.3, we know that theamount of unequal terms ofa1,a2, , a qis at mostm.

By the symmetry, we may assume thata m = a m+1 = ··· = a q > 0 Thus, taking k =

Case 2 Let Aqbe a point on the boundary ofDq Then there exists a term ina1,a2, , aq,this term must be zero We may assume thataq =0 FromAq ∈ Dq,a1+a2+···+aq −1=

Based on the principle of induction, (3.9) has been proven

Second, we will prove the general caseθ =1 Letting

Summarizing the above mentioned, inequality (3.8) has been proven Takingq = n in

inequality (3.8), we obtain inequality (1.6).Theorem 3.1is thus proved 

Proof ByTheorem 3.1, inequality (1.5) holds if and only if

t + k)/15) and − g(t, k) Thus our problem can be explained from the graphs, our

result is the following: ifa ∈ R15

+, thenλ ≤0.4160944179212302 if and only if



M15[1/2](a) 1− λ

M15[3/2](a)λ

Remark 3.6. Corollary 3.4is an open problem posed in [19]

In this section, we merely discuss the optimal problem of inequality (1.6) under theconditionn ≥ m ≥2 Whenm is sufficiently large, it is impossible that we applyTheorem3.1artificially Owing to this reason, we will discuss the general case of inequality (1.6) in

Section 4 In other words, we will search for the necessary and sufficient condition suchthatm ≥2,n ≥2 hold Our aim is to work artificially

4 The sufficient condition that inequality ( 1.6 ) holds

Recall the definition (see, e.g., [5, pages 41–42] and [9,19]) of generalized logarithmicmeansE(r, s; x, y),

Without loss of generality, we may assume thata1> a2, 0< α p −1< 1 < αp Note that

so, inequality (4.9) is equivalent to

Sufficiency Assume that (4.6) holds, we will prove that inequality (4.11) holds In fact,

we will follow every step in the following

Step 1 We will prove that

1 +a α p

2 )/2)1/α p, whenn > 2, we have v =(n

i =3a α p

i /(n −2))1/α p, whenn =2,

we may define an arbitrary value ofv Now we define that v = u When 1 ≤ j ≤ p −1,

we haveλ j(1− α j)> 0, 0 < α j < α p Therefore, by the inequality with power means, weobtain that

Thus, inequality (4.12) holds When j ≥ p, λj(1− αj)< 0, αj ≥ αp > 1, the reverse

inequality of (4.13) holds, therefore inequality (4.12) holds still

Step 2 We will prove that

therefore, byLemma 4.2, inequality (4.13) holds

Thus, we only have to prove that

It follows that inequality (4.17) and the assertion ofStep 2have been proven

Step 3 We will prove that inequality (4.11) holds

Since (1− αj)(αj − αp)< 0 ( j =1, , p −1,p + 1, , m), therefore, by the inequalities

It follows that inequality (4.11) holds

Necessity Assume that inequality (4.11) holds We will prove that inequality (4.6) holds

Proof of Theorem 4.1 We first prove a special case θ =1 By the hypothesis ofTheorem4.1andLemma 4.3, the function

Let A =(a1+a2+···+an)/n Then a =(A, A, , A) ≺ a From the definition of

Schur-convex function, we observe thatΦ(a) ≤ Φ(a) By reason of λ1+λ2+···+λn =1,

it is easy to see that inequality (1.6) is equivalent to the inequalityΦ(a) ≤ Φ(a) Thus

inequality (1.6) holds

Second, we prove the general caseθ =1 as follows

By the hypothesis ofTheorem 4.1, we obtain that

Replacinga by a θin (4.26), then inequality (4.26) reduces to inequality (1.6) This

Remark 4.4 Let m =2 inTheorem 4.1 Then we get [19, Theorem 1]

with equality holds if and only if x = y or a = b.

Proof InTheorem 4.1, lettingn =2,m =2 −1, p ≥2,λj =1/m, αj = a + ( j/m)(b −

a), b > a > 0, j =1, 2, , m, (a1,a2)=(x, y), θ =(1/m)m

j =1α j = a + (p/m)(b − a) = α p,then we haveα ∈ R m

therefore, byTheorem 4.1we have

m



j =1

x α j+y α j2

 1/mα j

≤

x θ+y θ2



· b − a

m ≤ln

x θ+y θ2

(4.29)

In other words, inequality (4.27) has been proven.Corollary 4.5is thus proved 

Example 4.6 Consider the condition such that the inequality

(4.31)

inequality (4.30) holds By means of Mathematica software, we can work out infg(t) =

0.297911 Namely, when λ ≤0.297911 , inequality (4.30) holds

5 The necessary and sufficient condition that inequality ( 1.7 ) holds

Theorem 5.1 Let a ∈ R n

++, n ≥ 2, α, λ ∈ R m

++, m ≥ 2,m

j =1λ j = 1, min { α } ≤ θ ≤max{ α } Then, a necessary and su fficient condition such that inequality ( 1.7 ) holds is that

Lemma 5.2 Let ajk > 0, q j > 0,m

j =1q j ≤ 1, 1 ≤ j ≤ m, 1 ≤ k ≤ n Then, an analogue of H¨older’s inequality is

In other words, inequality (1.7) holds

Necessity Assume that inequality (1.7) holds We will prove that inequality (5.1) holds

as follows: lettinga1=1,a2= a3= ··· = an →0 in inequality (1.7), (1.7) can be reducedto

Remark 5.3 Applying the approach of [19], we can establish some results that are similar

to [19, (33) and (37)] By using the definition of Riemann’ integral, we can obtain ananalogue of integral of (1.7) as follows

Corollary 5.4 Let the measurable function on the measurable sets E and E0,

| E |

 1/θ

where | E | and | E0|denote the measures of E and E0.

Inequality (5.6) has important background in the geometry of convex body (see, e.g.,[3,7])

19,21]) Recently, the authors studied the optimal real numberλ such that the