REFINEMENTS OF INEQUALITIES BETWEEN THE
SUM OF SQUARES AND THE EXPONENTIAL OF
SUM OF A NONNEGATIVE SEQUENCE
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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
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EMail: qifeng618@gmail.com
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Received:
12 November, 2007
Accepted:
07 March, 2008
Communicated by:
A. Sofo
2000 AMS Sub. Class.:
26D15, 60E15.
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Key words:
Inequality, Exponential of sum, Nonnegative sequence, Normal random
variable.
Abstract:
Using probability theory methods, the following sharp inequality is established:
!k
!
n
n
X
ek X
xi
≤ exp
xi ,
k k i=1
i=1
where k ∈ N, n ∈ N and xi ≥ 0 for 1 ≤ i ≤ n. Upon taking k = 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.
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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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
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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1.
Introduction
In [1], the following two inequalities were found.
Theorem A. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n ≥ 2, the inequality
!
n
n
X
e2 X 2
(1.1)
x ≤ exp
xi
4 i=1 i
i=1
is valid. Equality in (1.1) holds if xi = 2 for some given 1 ≤ i ≤ n and xj = 0 for
2
all 1 ≤ j ≤ n with j 6= i. Thus, the constant e4 in (1.1) is the best possible.
P∞
Theorem B. Let {xi }∞
i=1 be a nonnegative sequence such that
i=1 xi < ∞. Then
!
∞
∞
X
e2 X 2
(1.2)
x ≤ exp
xi .
4 i=1 i
i=1
Equality in (1.2) holds if xi = 2 for some given i ∈ N and xj = 0 for all j ∈ N with
2
j 6= i. Thus, the constant e4 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
!k
!
n
n
X
ek X
xi
≤ exp
xi
(1.3)
k k i=1
i=1
P
holds for all k ∈ N. Equality in (1.3) holds if ni=1 xi = k.
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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Theorem 1.2. Let {xi }∞
i=1 be a nonnegative sequence such that
the inequality
!k
!
∞
∞
X
ek X
(1.4)
xi
≤ exp
xi
k k i=1
i=1
is valid for all k ∈ N. Equality in (1.4) holds if
P∞
i=1
P∞
i=1
xi < ∞. Then
xi = 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.
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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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. Let S be a normal random variable with mean µ and variance σ 2 . Then
(2.1)
t2 σ 2
E exp{tS} = exp µt +
2
,
t∈R
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
and
(2.2)
Refinements of Inequalities
Yu Miao, Li-Min Liu
E(S − µ)2k = σ 2k (2k − 1)!!,
k ∈ N.
Proof. The proof is straightforward.
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Lemma 2.2. Let S be a normal random variable with mean 0 and variance σ 2 . Then
Contents
k
(2.3)
e
(2k)k (2k
− 1)!!
E(S 2k ) ≤ E eS ,
k ∈ N.
Proof. Putting
x
f (x) = k log x − ,
x ∈ (0, ∞),
2
it is easy to check that f (x) takes the maximum value at x = 2k. By Lemma 2.1, we
know that
2
E(S 2k ) = σ 2k (2k − 1)!!, and E eS = eσ /2 .
Therefore, for the function
ek
ek
2
2k
σ 2 /2
2k
S
g(σ ) =
σ
(2k
−
1)!!
−
e
=
E(S
)
−
E
e
,
(2k)k (2k − 1)!!
(2k)k (2k − 1)!!
it is easy to check that g(σ 2 ) ≤ 0 and at the point σ 2 = 2k, the equality holds.
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Now we are in a position to prove Theorem 1.1 and Theorem 1.2.
Proof of Theorem 1.1. Let {ξi }1≤i≤n be a sequence of independent normal random
2
variables with
Pn mean zero and variance σi = 2xi for all i = 1, . . . , n. Furthermore,
that Sn is a normal random variable with
let Sn = i=1 ξi , and it is well
Pknown
n
2
mean zero and variance σ = 2 i=1 xi . Therefore, we have
!
n
X
2
xi
EeSn = eσ /2 = exp
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
i=1
vol. 9, iss. 2, art. 53, 2008
and
E(Sn2 )
2
=σ =2
n
X
xi .
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i=1
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From Lemma 2.2, we have
ek
E(Sn2k ) ≤ E eSn ,
k
(2k) (2k − 1)!!
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that is,
(2.4)
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ek
kk
n
X
!k
≤ exp
xi
i=1
where the equality holds in (2.4) if
n
X
!
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xi ,
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i=1
Pn
i=1 xi = k.
Proof of Theorem 1.2. This can be concluded by letting n → ∞ in Theorem 1.1.
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3.
Further Discussion
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
!k
!
n
n
X
ek X
xi
≤ exp
xi
(3.1)
k k i=1
i=1
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
P
holds for all k ∈ (0, ∞). Equality in (3.1) holds if ni=1 xi = k. For (x1 , x2 , . . . , xn ) ∈
(−∞, 0]n and n ≥ 1, the inequality
!k
!
n
n
X
X
ek
(3.2)
−
xi
≥ exp
xi
|k|k
i=1
i=1
holds for all k ∈ (−∞, 0). Equality in (3.2) holds if
Pn
i=1
xi = k.
Proof. For all C > 0, s > 0 and k > 0, let f (s) = log C + k log s − s. It is easy to
see that the function f (s) takes its maximum at the point s = k. If we let s = k, then
k
we can obtain C = kek . The remainder of the proof is easy and thus omitted.
By similar arguments to those above, we can further obtain the following result.
P∞
Theorem 3.2. Let {xi }∞
i=1 be a nonnegative sequence such that
i=1 xi < ∞. Then
the inequality
!k
!
∞
∞
X
ek X
(3.3)
xi
≤ exp
xi
k k i=1
i=1
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P
is valid for all k ∈ (0, ∞). Equality in (3.3) P
holds if ∞
i=1 xi = k. In addition, let
∞
{xi }∞
be
a
non-positive
sequence
such
that
x
>
−∞. Then the inequality
i=1
i=1 i
!k
!
∞
∞
k
X
X
e
(3.4)
−
xi
≥ exp
xi
|k|k
i=1
i=1
is valid for all k ∈ (−∞, 0). Equality in (3.4) holds if
P∞
i=1
xi = k.
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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4.
Remarks
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 = 2 in inequality (1.3), then
(4.1)
!
n
n
X
e2 X 2 e2 X 2
x ≤
x +2
xi xj
4 i=1 i
4 i=1 i
1≤i<j≤n
!2
!
n
n
X
e2 X
=
xi
≤ exp
xi
4 i=1
i=1
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
Title Page
which means that inequality (1.3) refines inequality (1.1).
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Remark 2. If we let k = 2 in inequality (1.4), then
(4.2)
!
∞
∞
X
e2 X 2 e2 X 2
x ≤
x +2
xi xj
4 i=1 i
4 i=1 i
j>i≥1
!2
!
∞
∞
X
e2 X
=
xi
≤ exp
xi ,
4 i=1
i=1
which means that inequality (1.4) refines inequality (1.2).
P
P
Remark 3. If we let ni=1 xi = y ≥ 0 in (1.3) or ∞
i=1 xi = y ≥ 0 in (1.4), then
inequalities (1.3) and (1.4) can be rewritten as
(4.3)
ek k
y ≤ ey
k
k
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which is equivalent to
k
y
≤ ey−k
k
(4.4)
Taking
y
k
and
y
y
≤ e k −1 .
k
= s in the above inequality yields
s ≤ es−1 .
(4.5)
It is clear that inequality (4.5) is valid for all s ∈ R and the equality in it holds if
and only if s = 1. This
that inequalities (1.3)
Pimplies
P∞ and (1.4) hold for k ∈ (0, ∞)
n
and xi ∈ R such that i=1 xi ≥ 0 for n ∈ N or i=1
Pxi ≥ 0 respectively
P and that
the equalities in (1.3) and (1.4) hold if and only if ni=1 xi = k or ∞
i=1 xi = k
respectively.
Remark 4. In [1], Open Problem 1, was posed: For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and
n ≥ 2, determine the best possible constants αn , λn ∈ R and 0 < βn , µn < ∞ such
that
!
n
n
n
X
X
X
αn
(4.6)
βn
xi ≤ exp
x i ≤ µn
xλi n .
i=1
i=1
i=1
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
eαn
βn = αn
αn
(4.7)
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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if αn ≥ 1. This means that the inequality
!
!λn n
n
n
X
X
X
(4.8)
exp 2
xi ≤ n1−λn
xi
xλi n
i=1
Refinements of Inequalities
Yu Miao, Li-Min Liu
i=1
i=1
holds for λn ∈ R. If xi = 1 for 1 ≤ i ≤ n, then the above inequality becomes
e2n ≤ n1−λn nλn n = n2
(4.9)
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
!λn
!
n
n
n
X
X
X
(4.10)
exp
x i ≤ µn
xλi n ≤ µn
xi
,
i=1
i=1
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
i=1
vol. 9, iss. 2, art. 53, 2008
which is equivalent to
ex ≤ µn xλn
(4.11)
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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 ) ∈ Rn and n ∈ N, determine the best possible
constants αn , λn ∈ R and 0 < βn , µn < ∞ such that
!
n
n
n
X
X
X
(4.12)
βn
|xi |αn ≤ exp
x i ≤ µn
|xi |λn .
i=1
i=1
i=1
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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. [ONLINE: http://jipam.vu.edu.au/article.php?sid=895].
Refinements of Inequalities
Yu Miao, Li-Min Liu
and Feng Qi
vol. 9, iss. 2, art. 53, 2008
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