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Refinements of Inequalities Yu Miao, Li-Min Liu

and Feng Qi vol. 9, iss. 2, art. 53, 2008

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REFINEMENTS OF INEQUALITIES BETWEEN THE SUM OF SQUARES AND THE EXPONENTIAL OF

SUM OF A NONNEGATIVE SEQUENCE

YU MIAO, LI-MIN LIU

College of Mathematics and Information Science Henan Normal University

Henan Province, 453007, P.R. China

EMail:yumiao728@yahoo.com.cn llim2004@163.com

FENG QI

Research Institute of Mathematical Inequality Theory Henan Polytechnic University

Jiaozuo City, Henan Province 454010, P.R. China

EMail:qifeng618@gmail.com

Received: 12 November, 2007

Accepted: 07 March, 2008

Communicated by: A. Sofo 2000 AMS Sub. Class.: 26D15, 60E15.

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Refinements of Inequalities Yu Miao, Li-Min Liu

and Feng Qi vol. 9, iss. 2, art. 53, 2008

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Close Key words: Inequality, Exponential of sum, Nonnegative sequence, Normal random

variable.

Abstract: Using probability theory methods, the following sharp inequality is es- tablished:

ek kk

n

X

i=1

xi

!k

exp

n

X

i=1

xi

! ,

wherek N,n Nandxi 0for1 i n. Upon takingk = 2 in the above inequality, the inequalities obtained in [F. Qi, Inequalities between the sum of squares and the exponential of sum of a nonnegative sequence, J. Inequal. Pure Appl. Math. 8(3) (2007), Art.78] are refined.

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Refinements of Inequalities Yu Miao, Li-Min Liu

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Contents

1 Introduction 4

2 Proofs of Theorem 1.1 and Theorem 1.2 6

3 Further Discussion 8

4 Remarks 10

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Refinements of Inequalities Yu Miao, Li-Min Liu

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1. Introduction

In [1], the following two inequalities were found.

Theorem A. For(x1, x2, . . . , xn)∈[0,∞)nandn ≥2, the inequality

(1.1) e2

4

n

X

i=1

x2i ≤exp

n

X

i=1

xi

!

is valid. Equality in (1.1) holds ifxi = 2for some given1 ≤ i≤ n andxj = 0for all1≤j ≤nwithj 6=i. Thus, the constant e42 in (1.1) is the best possible.

Theorem B. Let{xi}i=1 be a nonnegative sequence such thatP

i=1xi <∞. Then

(1.2) e2

4

X

i=1

x2i ≤exp

X

i=1

xi

! .

Equality in (1.2) holds ifxi = 2for some giveni∈Nandxj = 0for allj ∈Nwith j 6=i. Thus, the constant e42 in (1.2) is the best possible.

In this note, by using some inequalities of normal random variables in probability theory, we will establish the following two inequalities whose special cases refine inequalities (1.1) and (1.2).

Theorem 1.1. For(x1, x2, . . . , xn)∈[0,∞)nandn≥1, the inequality

(1.3) ek

kk

n

X

i=1

xi

!k

≤exp

n

X

i=1

xi

!

holds for allk ∈N. Equality in (1.3) holds ifPn

i=1xi =k.

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Theorem 1.2. Let{xi}i=1be a nonnegative sequence such thatP

i=1xi <∞. Then the inequality

(1.4) ek

kk

X

i=1

xi

!k

≤exp

X

i=1

xi

!

is valid for allk ∈N. Equality in (1.4) holds ifP

i=1xi =k.

Our original ideas stem from probability theory, so we will prove the above the- orems by using some normal inequalities. In fact, from the proofs of the above theorems in the next section, one will notice that there may be simpler proofs of them, by which we will obtain more the general results of Section3.

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2. Proofs of Theorem 1.1 and Theorem 1.2

In order to prove Theorem 1.1 and Theorem 1.2, the following two lemmas are necessary.

Lemma 2.1. LetSbe a normal random variable with meanµand varianceσ2. Then

(2.1) E exp{tS}

= exp

µt+t2σ2 2

, t∈R

and

(2.2) E(S−µ)2k2k(2k−1)!!, k ∈N. Proof. The proof is straightforward.

Lemma 2.2. LetSbe a normal random variable with mean0and varianceσ2. Then

(2.3) ek

(2k)k(2k−1)!!E(S2k)≤E eS

, k ∈N. Proof. Putting

f(x) =klogx−x

2, x∈(0,∞),

it is easy to check thatf(x)takes the maximum value atx= 2k. By Lemma2.1, we know that

E(S2k) = σ2k(2k−1)!!, and E eS

=eσ2/2. Therefore, for the function

g(σ2) = ek

(2k)k(2k−1)!!σ2k(2k−1)!!−eσ2/2 = ek

(2k)k(2k−1)!!E(S2k)−E eS , it is easy to check thatg(σ2)≤0and at the pointσ2 = 2k, the equality holds.

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Now we are in a position to prove Theorem1.1and Theorem1.2.

Proof of Theorem1.1. Leti}1≤i≤n be a sequence of independent normal random variables with mean zero and varianceσi2 = 2xi for alli = 1, . . . , n. Furthermore, let Sn = Pn

i=1ξi, and it is well known that Sn is a normal random variable with mean zero and varianceσ2 = 2Pn

i=1xi. Therefore, we have EeSn =eσ2/2 = exp

n

X

i=1

xi

!

and

E(Sn2) = σ2 = 2

n

X

i=1

xi.

From Lemma2.2, we have ek

(2k)k(2k−1)!!E(Sn2k)≤E eSn ,

that is,

(2.4) ek

kk

n

X

i=1

xi

!k

≤exp

n

X

i=1

xi

! ,

where the equality holds in (2.4) ifPn

i=1xi =k.

Proof of Theorem1.2. This can be concluded by lettingn→ ∞in Theorem1.1.

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3. Further Discussion

In this section, we will give the general results of Theorem1.1and Theorem 1.2by a simpler proof.

Theorem 3.1. For(x1, x2, . . . , xn)∈[0,∞)nandn≥1, the inequality

(3.1) ek

kk

n

X

i=1

xi

!k

≤exp

n

X

i=1

xi

!

holds for allk∈(0,∞). Equality in (3.1) holds ifPn

i=1xi =k. For(x1, x2, . . . , xn)∈ (−∞,0]nandn ≥1, the inequality

(3.2) ek

|k|k

n

X

i=1

xi

!k

≥exp

n

X

i=1

xi

!

holds for allk ∈(−∞,0). Equality in (3.2) holds ifPn

i=1xi =k.

Proof. For allC >0,s > 0andk >0, letf(s) = logC+klogs−s. It is easy to see that the functionf(s)takes its maximum at the points =k. If we lets=k, then we can obtainC = kekk. The remainder of the proof is easy and thus omitted.

By similar arguments to those above, we can further obtain the following result.

Theorem 3.2. Let{xi}i=1be a nonnegative sequence such thatP

i=1xi <∞. Then the inequality

(3.3) ek

kk

X

i=1

xi

!k

≤exp

X

i=1

xi

!

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is valid for allk ∈ (0,∞). Equality in (3.3) holds if P

i=1xi = k. In addition, let {xi}i=1 be a non-positive sequence such thatP

i=1xi >−∞. Then the inequality

(3.4) ek

|k|k

X

i=1

xi

!k

≥exp

X

i=1

xi

!

is valid for allk ∈(−∞,0). Equality in (3.4) holds ifP

i=1xi =k.

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4. Remarks

After proving Theorem1.1and Theorem1.2, we would like to state several remarks and an open problem posed in [1].

Remark 1. If we takek = 2in inequality (1.3), then e2

4

n

X

i=1

x2i ≤ e2 4

n

X

i=1

x2i + 2 X

1≤i<j≤n

xixj

! (4.1)

= e2 4

n

X

i=1

xi

!2

≤exp

n

X

i=1

xi

!

which means that inequality (1.3) refines inequality (1.1).

Remark 2. If we letk= 2in inequality (1.4), then e2

4

X

i=1

x2i ≤ e2 4

X

i=1

x2i + 2 X

j>i≥1

xixj

! (4.2)

= e2 4

X

i=1

xi

!2

≤exp

X

i=1

xi

! ,

which means that inequality (1.4) refines inequality (1.2).

Remark 3. If we let Pn

i=1xi = y ≥ 0in (1.3) or P

i=1xi = y ≥ 0 in (1.4), then inequalities (1.3) and (1.4) can be rewritten as

(4.3) ek

kkyk≤ey

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which is equivalent to (4.4)

y k

k

≤ey−k and y

k ≤eyk−1. Taking ky =sin the above inequality yields

(4.5) s≤es−1.

It is clear that inequality (4.5) is valid for all s ∈ R and the equality in it holds if and only ifs = 1. This implies that inequalities (1.3) and (1.4) hold fork ∈(0,∞) andxi ∈ Rsuch thatPn

i=1xi ≥ 0forn ∈ NorP

i=1xi ≥ 0respectively and that the equalities in (1.3) and (1.4) hold if and only if Pn

i=1xi = k or P

i=1xi = k respectively.

Remark 4. In [1], Open Problem 1, was posed: For(x1, x2, . . . , xn) ∈ [0,∞)n and n ≥2, determine the best possible constantsαn, λn∈ Rand0< βn, µn <∞such that

(4.6) βn

n

X

i=1

xαin ≤exp

n

X

i=1

xi

!

≤µn

n

X

i=1

xλin.

Recently, in a private communication with the third author, Huan-Nan Shi proved using majorization that the best constant in the left hand side of (4.6) is

(4.7) βn = eαn

αnαn ifαn ≥1. This means that the inequality

(4.8) exp 2

n

X

i=1

xi

!

≤n1−λn

n

X

i=1

xi

!λn n

X

i=1

xλin

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holds forλn∈R. Ifxi = 1for1≤i≤n, then the above inequality becomes

(4.9) e2n≤n1−λnnλnn =n2

which is not valid. This prompts us to check the validity of the right-hand inequality in (4.6): If the right-hand inequality in (4.6) is valid, then it is clear that

(4.10) exp

n

X

i=1

xi

!

≤µn

n

X

i=1

xλin ≤µn

n

X

i=1

xi

!λn

,

which is equivalent to

(4.11) ex ≤µnxλn

for x ≥ 0 and two constants λn ∈ R and 0 < µn < ∞. This must lead to a contradiction. Therefore, Open Problem 1 in [1] should and can be modified as follows.

Open Problem 1. For(x1, x2, . . . , xn)∈Rnandn∈N, determine the best possible constantsαn, λn∈Rand0< βn, µn <∞such that

(4.12) βn

n

X

i=1

|xi|αn ≤exp

n

X

i=1

xi

!

≤µn

n

X

i=1

|xi|λn.

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References

[1] F. QI, Inequalities between the sum of squares and the exponential of sum of a nonnegative sequence, J. Inequal. Pure Appl. Math., 8(3) (2007), Art. 78. [ON- LINE:http://jipam.vu.edu.au/article.php?sid=895].

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