**REFINEMENTS OF INEQUALITIES BETWEEN THE SUM OF SQUARES AND**
**THE EXPONENTIAL OF SUM OF A NONNEGATIVE SEQUENCE**

YU MIAO, LI-MIN LIU, AND FENG QI

COLLEGE OFMATHEMATICS ANDINFORMATIONSCIENCE

HENANNORMALUNIVERSITY

HENANPROVINCE, 453007, P.R. CHINA

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

RESEARCHINSTITUTE OFMATHEMATICALINEQUALITYTHEORY

HENANPOLYTECHNICUNIVERSITY

JIAOZUOCITY, HENANPROVINCE

454010, P.R. CHINA

qifeng618@gmail.com

*Received 12 November, 2007; accepted 07 March, 2008*
*Communicated by A. Sofo*

ABSTRACT. Using probability theory methods, the following sharp inequality is established:

e^{k}
k^{k}

n

X

i=1

xi

!k

≤exp

n

X

i=1

xi

! ,

wherek∈N,n∈Nandx_{i} ≥0for1≤i≤n. Upon takingk= 2in 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.**

*Key words and phrases: Inequality, Exponential of sum, Nonnegative sequence, Normal random variable.*

*2000 Mathematics Subject Classification. 26D15; 60E15.*

**1. I****NTRODUCTION**

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

* Theorem A. For*(x

_{1}, x

_{2}, . . . , x

_{n})∈[0,∞)

^{n}

*and*n ≥2, the inequality

(1.1) e^{2}

4

n

X

i=1

x^{2}_{i} ≤exp

n

X

i=1

x_{i}

!

*is valid. Equality in (1.1) holds if*x_{i} = 2*for some given*1≤i≤n*and*x_{j} = 0*for all*1≤j ≤n
*with*j 6=i. Thus, the constant ^{e}_{4}^{2} *in (1.1) is the best possible.*

337-07

* Theorem B. Let*{x

_{i}}

^{∞}

_{i=1}

*be a nonnegative sequence such that*P∞

i=1x_{i} <∞. Then

(1.2) e^{2}

4

∞

X

i=1

x^{2}_{i} ≤exp

∞

X

i=1

x_{i}

! .

*Equality in (1.2) holds if*x_{i} = 2 *for some given*i ∈ N *and*x_{j} = 0 *for all* j ∈ N*with*j 6= i.

*Thus, the constant* ^{e}_{4}^{2} *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,∞)

^{n}

*and*n≥1, the inequality

(1.3) e^{k}

k^{k}

n

X

i=1

x_{i}

!k

≤exp

n

X

i=1

x_{i}

!

*holds for all*k ∈N*. Equality in (1.3) holds if*Pn

i=1xi =k.

* Theorem 1.2. Let* {x

_{i}}

^{∞}

_{i=1}

*be a nonnegative sequence such that*P∞

i=1x_{i} < ∞. Then the in-
*equality*

(1.4) e^{k}

k^{k}

∞

X

i=1

x_{i}

!k

≤exp

∞

X

i=1

x_{i}

!

*is valid for all*k∈N*. Equality in (1.4) holds if*P∞

i=1x_{i} =k.

Our original ideas stem from probability theory, so we will prove the above theorems 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 Section 3.

**2. P****ROOFS OF****T****HEOREM****1.1** **AND****T****HEOREM****1.2**

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

* Lemma 2.1. Let*S

*be a normal random variable with mean*µ

*and variance*σ

^{2}

*. Then*

(2.1) E exp{tS}

= exp

µt+t^{2}σ^{2}
2

, t ∈R

*and*

(2.2) E(S−µ)^{2k} =σ^{2k}(2k−1)!!, k∈N.

*Proof. The proof is straightforward.*

* Lemma 2.2. Let*S

*be a normal random variable with mean*0

*and variance*σ

^{2}

*. Then*

(2.3) e^{k}

(2k)^{k}(2k−1)!!E(S^{2k})≤E e^{S}

, 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 Lemma 2.1, we know that
E(S^{2k}) =σ^{2k}(2k−1)!!, and E e^{S}

=e^{σ}^{2}^{/2}.

Therefore, for the function
g(σ^{2}) = e^{k}

(2k)^{k}(2k−1)!!σ^{2k}(2k−1)!!−e^{σ}^{2}^{/2} = e^{k}

(2k)^{k}(2k−1)!!E(S^{2k})−E e^{S}
,
it is easy to check thatg(σ^{2})≤0and at the pointσ^{2} = 2k, the equality holds.

Now we are in a position to prove Theorem 1.1 and Theorem 1.2.

*Proof of Theorem 1.1. Let* {ξ_{i}}1≤i≤n be a sequence of independent normal random variables
with mean zero and varianceσ_{i}^{2} = 2x_{i} for alli = 1, . . . , n. Furthermore, letS_{n} = Pn

i=1ξ_{i},
and it is well known thatS_{n} is a normal random variable with mean zero and variance σ^{2} =
2Pn

i=1x_{i}. Therefore, we have

Ee^{S}^{n} =e^{σ}^{2}^{/2} = exp

n

X

i=1

x_{i}

!

and

E(S_{n}^{2}) =σ^{2} = 2

n

X

i=1

x_{i}.
From Lemma 2.2, we have

e^{k}

(2k)^{k}(2k−1)!!E(S_{n}^{2k})≤E e^{S}^{n}
,

that is,

(2.4) e^{k}

k^{k}

n

X

i=1

xi

!k

≤exp

n

X

i=1

xi

! , where the equality holds in (2.4) ifPn

i=1x_{i} =k.

*Proof of Theorem 1.2. This can be concluded by letting*n→ ∞in Theorem 1.1.

**3. F****URTHER****D****ISCUSSION**

In this section, we will give the general results of Theorem 1.1 and Theorem 1.2 by a simpler proof.

* Theorem 3.1. For*(x1, x2, . . . , xn)∈[0,∞)

^{n}

*and*n≥1, the inequality

(3.1) e^{k}

k^{k}

n

X

i=1

x_{i}

!k

≤exp

n

X

i=1

x_{i}

!

*holds for all* k ∈ (0,∞). Equality in (3.1) holds if Pn

i=1x_{i} = k. For (x_{1}, x_{2}, . . . , x_{n}) ∈
(−∞,0]^{n}*and*n≥1, the inequality

(3.2) e^{k}

|k|^{k} −

n

X

i=1

x_{i}

!k

≥exp

n

X

i=1

x_{i}

!

*holds for all*k ∈(−∞,0). Equality in (3.2) holds ifPn

i=1x_{i} =k.

*Proof. For all*C > 0, s > 0and k > 0, letf(s) = logC+klogs−s. It is easy to see that
the function f(s) takes its maximum at the points = k. If we lets = k, then we can obtain
C = _{k}^{e}^{k}k. 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* {x

_{i}}

^{∞}

_{i=1}

*be a nonnegative sequence such that*P∞

i=1x_{i} < ∞. Then the in-
*equality*

(3.3) e^{k}

k^{k}

∞

X

i=1

x_{i}

!k

≤exp

∞

X

i=1

x_{i}

!

*is valid for all*k ∈(0,∞). Equality in (3.3) holds ifP∞

i=1x_{i} =k. In addition, let{x_{i}}^{∞}_{i=1} *be a*
*non-positive sequence such that*P∞

i=1x_{i} >−∞. Then the inequality

(3.4) e^{k}

|k|^{k} −

∞

X

i=1

x_{i}

!k

≥exp

∞

X

i=1

x_{i}

!

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

i=1x_{i} =k.

**4. R****EMARKS**

After proving Theorem 1.1 and Theorem 1.2, we would like to state several remarks and an open problem posed in [1].

**Remark 1. If we take**k = 2in inequality (1.3), then
(4.1) e^{2}

4

n

X

i=1

x^{2}_{i} ≤ e^{2}
4

n

X

i=1

x^{2}_{i} + 2 X

1≤i<j≤n

x_{i}x_{j}

!

= e^{2}
4

n

X

i=1

x_{i}

!2

≤exp

n

X

i=1

x_{i}

!

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

**Remark 2. If we let**k = 2in inequality (1.4), then
(4.2) e^{2}

4

∞

X

i=1

x^{2}_{i} ≤ e^{2}
4

∞

X

i=1

x^{2}_{i} + 2 X

j>i≥1

x_{i}x_{j}

!

= e^{2}
4

∞

X

i=1

x_{i}

!2

≤exp

∞

X

i=1

x_{i}

! ,

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

**Remark 3. If we let**Pn

i=1x_{i} = y ≥ 0in (1.3) orP∞

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

(4.3) e^{k}

k^{k}y^{k}≤e^{y}
which is equivalent to

(4.4)

y k

k

≤e^{y−k} and y

k ≤e^{y}^{k}^{−1}.
Taking ^{y}_{k} =sin the above inequality yields

(4.5) s≤e^{s−1}.

It is clear that inequality (4.5) is valid for alls ∈ Rand the equality in it holds if and only if
s = 1. This implies that inequalities (1.3) and (1.4) hold fork ∈ (0,∞)andx_{i} ∈ Rsuch that
Pn

i=1xi ≥ 0 forn ∈ N orP∞

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

i=1xi =korP∞

i=1xi =k respectively.

**Remark 4. In [1], Open Problem 1, was posed: For** (x_{1}, x_{2}, . . . , x_{n}) ∈ [0,∞)^{n} and n ≥ 2,
determine the best possible constantsα_{n}, λ_{n} ∈Rand0< β_{n}, µ_{n}<∞such that

(4.6) β_{n}

n

X

i=1

x^{α}_{i}^{n} ≤exp

n

X

i=1

x_{i}

!

≤µ_{n}

n

X

i=1

x^{λ}_{i}^{n}.

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

(4.7) β_{n} = e^{α}^{n}

α_{n}^{α}^{n}
ifα_{n}≥1. This means that the inequality

(4.8) exp 2

n

X

i=1

xi

!

≤n^{1−λ}^{n}

n

X

i=1

xi

!λn n

X

i=1

x^{λ}_{i}^{n}

holds forλ_{n}∈R. Ifx_{i} = 1for1≤i≤n, then the above inequality becomes

(4.9) e^{2n}≤n^{1−λ}^{n}n^{λ}^{n}n =n^{2}

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

x^{λ}_{i}^{n} ≤µn
n

X

i=1

xi

!λn

, which is equivalent to

(4.11) e^{x} ≤µnx^{λ}^{n}

for x ≥ 0 and two constants λ_{n} ∈ R and 0 < µ_{n} < ∞. This must lead to a contradiction.

Therefore, Open Problem 1 in [1] should and can be modified as follows.

* Open Problem 1. For*(x

_{1}, x

_{2}, . . . , x

_{n})∈R

^{n}

*and*n ∈N

*, determine the best possible constants*α

_{n}, λ

_{n}∈R

*and*0< β

_{n}, µ

_{n}<∞

*such that*

(4.12) β_{n}

n

X

i=1

|x_{i}|^{α}^{n} ≤exp

n

X

i=1

x_{i}

!

≤µ_{n}

n

X

i=1

|x_{i}|^{λ}^{n}.

**R****EFERENCES**

[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. [ONLINE:*http://jipam.vu.edu.au/

article.php?sid=895].