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volume 6, issue 1, article 24, 2005.

Received 30 September, 2003;

accepted 07 November, 2003.

Communicated by:T.M. Mills

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Journal of Inequalities in Pure and Applied Mathematics

ON SOME POLYNOMIAL–LIKE INEQUALITIES OF BRENNER AND ALZER

C.E.M. PEARCE AND J. PE ˇCARI ´C

School of Applied Mathematics The University of Adelaide Adelaide SA 5005 Australia

EMail:cpearce@maths.adelaide.edu.au

URL:http://www.maths.adelaide.edu.au/applied/staff/cpearce.html Faculty of Textile Technology

University of Zagreb Pierottijeva 6, 10000 Zagreb Croatia

EMail:pecaric@mahazu.hazu.hr

URL:http://mahazu.hazu.hr/DepMPCS/indexJP.html

c

2000Victoria University ISSN (electronic): 1443-5756 135-03

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On Some Polynomial–Like Inequalities of Brenner and

Alzer

C.E.M. Pearce and J. Peˇcari´c

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J. Ineq. Pure and Appl. Math. 6(1) Art. 24, 2005

Abstract

Refinements and extensions are presented for some inequalities of Brenner and Alzer for certain polynomial–like functions.

2000 Mathematics Subject Classification:Primary 26D15.

Key words: Polynomial inequalities, Switching inequalities, Jensen’s inequality

1. Introduction

Brenner [2] has given some interesting inequalities for certain polynomial–like functions. In particular he derived the following.

Theorem A. Supposem >1,0< p1, . . . , pk <1andPk=Pk

i=1pi ≤1. Then

(1.1)

k

X

i=1

(1−p^m_i )^m > k−1 + (1−P_k)^m.

Alzer [1] considered the sum A_k(x, s) =

k

X

i=0

s i

xⁱ(1−x)^s−i (0≤x≤1) and proved the following companion inequality to (1.1).

Theorem B. Let p, q, m and nbe positive real numbers and k a nonnegative integer. Ifp+q ≤1andm, n > k+ 1, then

(1.2) A_k(p^m, n) +A_k(qⁿ, m)>1 +A_k((p+q)^min(m,n),max(m, n)).

In the special casek = 0this provides

(1.3) (1−p^m)ⁿ+ (1−qⁿ)^m >1 + (1−(p+q)^min(m,n))^max(m,n) forp, q >0.

In Section 2 we use (1.3) to derive an improvement of Theorem A and a corresponding version of Theorem B. In Section 3 we give a related Jensen inequality and concavity result.

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Alzer

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2. Basic Results

Theorem 2.1. Under the conditions of TheoremAwe have

(2.1)

k

X

i=1

(1−p^m_i )^m > k−1 + (1−P_k^m)^m.

Proof. We proceed by mathematical induction, (1.3) with n = m providing a basis

(2.2) (1−p^m)^m+(1−q^m)^m >1+(1−(p+q)^m)^m forp, q >0andp+q ≤1 fork = 2. For the inductive step, suppose that (2.1) holds for somek ≥ 2, so that

k+1

X

i=1

(1−p^m_i )^m =

k

X

i=1

(1−p^m_i )^m+ (1−p^m_k+1)^m

> k−1 + (1−P_k^m)^m+ (1−p^m_k+1)^m. Applying (2.2) yields

k+1

X

i=1

(1−p^m_i )^m > k−1 + 1 + (1−(P_k+p_k+1)^m)^m

=k+ 1−P_k+1^m m

.

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Alzer

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For the remaining results in this paper it is convenient, for a fixed nonnegative integerkandm > k+ 1, to define

B(x) := A_k(x^m, m).

Theorem 2.2. Letp₁, . . . , p_` andmbe positive real numbers. If

P_`:=

`

X

i=1

p_i,

then

(2.3)

`

X

j=1

B(p_j)> `−1 +B(P_`).

Proof. We establish the result by induction, (1.2) withn=mproviding a basis

(2.4) B(p) +B(q)>1 +B(p+q) forp, q >0andp+q ≤1

for ` = 2. Suppose (2.3) to be true for some ` ≥ 2. Then by the inductive hypothesis

`+1

X

j=1

B(p_j) =

`

X

j=1

B(p_j) +B(p_`+1)

> `−1 +B(P_`) +B(p_`+1).

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Alzer

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Now applying (2.4) yields

`+1

X

j=1

B(p_j)> `−1 + 1 +B(P_`+p_`+1)

=`+B(P_`+1) (2.5)

as desired.

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Alzer

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3. Concavity of B

Inequality (2.3) is of the form

n

X

j=1

f(p_j)>(n−1)f(0) +f

n

X

j=1

p_i

! ,

that is, the Petrovi´c inequality for a concave functionf. A natural question is whetherBsatisfies the corresponding Jensen inequality

(3.1) B 1

n

X

j=1

p_j

!

≥ 1 n

n

X

j=1

B(p_j) for positivep₁, p₂, . . . , p_nsatisfyingPn

j=1p_j ≤1and indeed whetherBis concave. We now address these questions. It is convenient to first deal separately with the casen= 2.

Theorem 3.1. Supposep,qare positive and distinct withp+q≤1. Then

(3.2) B

p+q 2

> 1

2[B(p) +B(q)]. Proof. Letu∈[0,1). Forp∈[0,1−u]we define

G(p) = B(p) +B(1−u−p).

By an argument of Alzer [1] we have (3.3) G⁰(p) =m

k

(m−k)mp^m−1(1−p^m)^m−1

p^m 1−p^m

k

[g(p)−1],

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Alzer

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where

(3.4) g(p) =

1−u−p 1−p^m

m−1

1−(1−u−p)^m p

m−1

×

(1−u−p)^m 1−(1−u−p)^m

k

1−p^m p^m

k

is a strictly decreasing function.

It was shown in [1] that there existsp₀ ∈(0,1−u)such thatG(p)is strictly increasing on[0, p₀]and strictly decreasing on[p₀,1−u], so that

G(p)< G(p0) for p∈[0,1−u], p6=p0.

On the other hand, we have by (3.4) that g((1−u)/2) = 1 and so from (3.3) G⁰((1−u)/2) = 0. Hencep₀ = (1−u)/2and therefore

G(p)< G

1−u 2

for p6= (1−u)/2.

Setu= 1−(p+q). Sincep6=q, we must havep6= (1−u)/2. Therefore G(p)< G

p+q 2

, which is simply (3.2).

Corollary 3.2. The mapB is concave on(0,1).

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Alzer

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Proof. Theorem 3.1 gives that B is Jensen concave, so that −B is Jensen–

convex. Since B is continuous, we have by a classical result [3, Chapter 3]

that−Bmust also be convex and soB is concave.

The following result funishes additional information about strictness.

Theorem 3.3. Let p1, . . . , pn, be positive numbers with Pn

j=1pj ≤ 1. Then (3.1) applies. If not all thep_j are equal, then the inequality is strict.

Proof. The result is trivial with equality if the p_j all share a common value, so we assume at least two different values.

We proceed by induction, Theorem3.1providing a basis forn = 2. For the inductive step, suppose that (3.1) holds for somen ≥ 2and thatPn+1

j=1 p_j ≤1.

Without loss of generality we may assume thatp_n+1is the greatest of the values p_j. Since not all the valuesp_j are equal, we therefore have

pn+1 > 1 n

n

X

j=1

pj.

This rearranges to give 1 n

n

X

j=1

pj < 1 n

"

pn+1+n−1 n+ 1

n+1

X

j=1

pj

# . Both sides of this inequality take values in(0,1).

Also we have 1 n+ 1

n+1

X

j=1

p_j = 1 2

"

1 n

n

X

j=1

p_j + 1 n

(

p_n+1+ n−1 n+ 1

n+1

X

j=1

p_j )#

.

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Alzer

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Hence applying (3.2) provides

B 1

n+ 1

n+1

X

j=1

pj

!

> 1 2

"

B 1

n

X

j=1

p_j

!

+B 1 n

(

p_n+1+n−1 n+ 1

n+1

X

j=1

p_j )!#

.

By the inductive hypothesis

B 1

n

X

j=1

pj

!

≥ 1 n

n

X

j=1

B(pj)

and

B 1

n (

p_n+1+n−1 n+ 1

n+1

X

j=1

p_j )!

≥ 1 n

"

B(p_n+1) + (n−1)B 1 n+ 1

n+1

X

j=1

p_j

!#

.

Hence

B 1

n+ 1

n+1

X

j=1

p_j

!

> 1 2n

"_n+1 X

j=1

B(p_j) + (n−1)B 1 n+ 1

n+1

X

j=1

p_j

!#

.

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Alzer

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Rearrangement of this inequality yields

B 1

n+ 1

n+1

X

j=1

p_j

!

> 1 n+ 1

n+1

X

j=1

B(p_j), the desired result.

Remark 1. Taken together, relations (2.5) and (3.1) give (3.5) n−1 +B

n

X

j=1

p_j

!

<

n

X

j=1

B(p_j)≤nB 1 n

n

X

j=1

p_j

! ,

the second inequality being strict unless all the valuesp_jare equal. IfPn

j=1p_j = 1, this simplifies to

(3.6) n−1<

n

X

j=1

B(p_j)≤nB(n⁻¹), sinceB(1) = 0.

Fork = 0, (3.5) and (3.6) become (form >1) respectively n−1 + 1−

n

X

j=1

p_j

!m!m

<

n

X

j=1

(1−p^m_j )^m ≤n 1− 1 n

n

X

j=1

p_j

!m!m

and

n−1<

n

X

j=1

(1−p^m_j )^m ≤n(1−n^−m)^m.

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Alzer

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References

[1] H. ALZER, On an inequality of J.L. Brenner, J. Math. Anal. Appl., 183 (1994), 547–550.

[2] J.L. BRENNER, Analytical inequalities with applications to special func- tions, J. Math. Anal. Appl., 106 (1985), 427–442.

[3] G.H. HARDY, J. E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cam- bridge University Press, Cambridge (1934).

[4] J.E. PE ˇCARI ´C, F. PROSCHAN AND Y.L. TONG, Convex Functions, Par- tial Orderings and Statistical Applications, Academic Press, New York (1992).