ON OPEN PROBLEMS OF F. QI
BENHARRAT BELAÏDI, ABDALLAH EL FARISSI AND
ZINELAÂBIDINE LATREUCH
Department of Mathematics
Laboratory of Pure and Applied Mathematics
University of Mostaganem
B. P. 227 Mostaganem, Algeria
EMail: belaidi@univ-mosta.dz elfarissi.abdallah@yahoo.fr z.latreuch@gmail.com
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
Title Page
Contents
Received:
07 May, 2008
Accepted:
28 September, 2009
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
26D15.
Key words:
Inequality, Sum of power, Exponential of sum, Nonnegative sequence, Integral
Inequality.
Abstract:
In this paper, we give a complete answer to Problem 1 and a partial answer to
Problem 2 posed by F. Qi in [2] and we propose an open problem.
Acknowledgements:
The authors would like to thank the referees for their helpful remarks and suggestions to improve the paper.
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Contents
1
Introduction
3
2
Lemmas
9
3
Proofs of the Theorems
4
Open Problem
12
15
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
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1.
Introduction
Before, we state our results, for our own convenience, we introduce the following
notations:
(1.1)
4
[0, ∞)n = [0, ∞) × [0, ∞) × ... × [0, ∞)
|
{z
}
n times
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
and
(1.2)
4
(0, ∞)n = (0, ∞) × (0, ∞) × ... × (0, ∞)
|
{z
}
vol. 10, iss. 3, art. 90, 2009
n times
for n ∈ N, where N denotes the set of all positive integers.
In [2], F. Qi proved the following:
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Contents
n
Theorem A. For (x1 , x2 , . . . , xn ) ∈ [0, ∞) and n > 2, inequality
!
n
n
X
e2 X 2
(1.3)
x 6 exp
xi
4 i=1 i
i=1
is valid. Equality in (1.3) holds if xi = 2 for some given 1 6 i 6 n and xj = 0 for
2
all 1 6 j 6 n with j 6= i. Thus, the constant e4 in (1.3) is the best possible.
P∞
Theorem B. Let {xi }∞
be
a
nonnegative
sequence
such
that
i=1
i=1 xi < ∞. Then
!
∞
∞
X
e2 X 2
(1.4)
x 6 exp
xi .
4 i=1 i
i=1
Equality in (1.4) 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 (1.4) is the best possible.
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In the same paper, F. Qi posed the following two open problems:
Problem 1.1. For (x1 , x2 , ..., xn ) ∈ [0, ∞)n and n > 2, determine the best possible
constants αn , λn ∈ R and βn > 0, µn < ∞ such that
!
n
n
n
X
X
X
αn
xi 6 exp
(1.5)
βn
x i ≤ µn
xλi n .
i=1
i=1
i=1
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
Problem 1.2. What is the integral analogue of the double inequality (1.5)?
Recently, Huan-Nan Shi gave a partial answer in [3] to Problem 1.1. The main
purpose of this paper is to give a complete answer to this problem. Also, we give a
partial answer to Problem 1.2. The method used in this paper will be quite different
from that in the proofs of Theorem 1.1 of [2] and Theorem 1 of [3]. For some related
results, we refer the reader to [1]. We will prove the following results.
Theorem 1.1. Let p > 1 be a real number. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and
n > 2, the inequality
!
n
n
X
ep X p
(1.6)
x 6 exp
xi
pp i=1 i
i=1
is valid. Equality in (1.6) holds if xi = p for some given 1 6 i 6 n and xj = 0 for
p
all 1 6 j 6 n with j 6= i. Thus, the constant pep in (1.6) is the best possible.
n
Theorem 1.2. Let 0 < p 6 1 be a real number. For (x1 , x2 , . . . , xn ) ∈ [0, ∞) and
n > 2, the inequality
!
n
n
p X
X
e
p
(1.7)
np−1 p
x 6 exp
xi
p i=1 i
i=1
vol. 10, iss. 3, art. 90, 2009
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is valid. Equality in (1.7) holds if xi =
p
np−1 pep in (1.7) is the best possible.
p
n
for all 1 6 i 6 n. Thus, the constant
Theorem 1.3. Let {xi }∞
i=1 be a nonnegative sequence such that
p > 1 be a real number. Then
!
∞
∞
X
ep X p
x 6 exp
xi .
(1.8)
pp i=1 i
i=1
P∞
i=1
xi < ∞ and
Equality in (1.8) holds if xi = p for some given i ∈ N and xj = 0 for all j ∈ N with
p
j 6= i. Thus, the constant pep in (1.8) is the best possible.
Remark 1. In general, we cannot find 0 < µn < ∞ and λn ∈ R such that
!
n
n
X
X
exp
x i 6 µn
xλi n .
i=1
i=1
Proof. We suppose that there exists 0 < µn < ∞ and λn ∈ R such that
!
n
n
X
X
exp
x i 6 µn
xλi n .
i=1
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
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Then for (x1 , 1, ..., 1), we obtain as x1 → +∞,
1−n
16e
This is a contradiction.
µn n − 1 +
xλ1 n
Close
−x1
e
→ 0.
Theorem 1.4. Let p > 0 be a real number, (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n > 2
such that 0 < xi 6 p for all 1 6 i 6 n. Then the inequality
!
n
n
X
pp np X −p
(1.9)
exp
xi
xi 6 e
n
i=1
i=1
is valid. Equality in (1.9) holds if xi = p for all 1 6 i 6 n. Thus, the constant
is the best possible.
pp np
e
n
Remark 2. Let p > 0 be a real number, (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n > 2 such
that 0 < xi 6 p for all 1 6 i 6 n. Then
(i) if 0 < p ≤ 1, we have
(1.10)
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
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n
n
p X
X
p
p−1 e
n
x 6 exp
xi
pp i=1 i
i=1
!
n
X −p
pp
6 enp
xi ;
n
i=1
(ii) if p ≥ 1, we have
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n
n
X
e X p
x 6 exp
xi
pp i=1 i
i=1
p
(1.11)
!
p
6
p np
e
n
n
X
x−p
i .
i=1
Remark 3. Taking p = 2 in Theorems 1.1 and 1.3 easily leads to Theorems A and B
respectively.
Remark 4. Inequality (1.6) can be rewritten as either
(1.12)
n
n
ep X p Y xi
x 6
e
pp i=1 i
i=1
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or
ep
kxkpp 6 exp kxk1 ,
p
p
(1.13)
where x = (x1 , x2 , ..., xn ) and k·kp denotes the p-norm.
Remark 5. Inequality (1.8) can be rewritten as
∞
∞
ep X p Y xi
x 6
e
pp i=1 i
i=1
(1.14)
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
∞
which is equivalent to inequality (1.12) for x = (x1 , x2 , ...) ∈ [0, ∞) .
Remark 6. Taking xi = 1i for i ∈ N in (1.6) and rearranging gives
!
n
n
X
X
1
1
(1.15)
p − p ln p + ln
6
.
p
i
i
i=1
i=1
(1.16)
1
is
for i ∈ N and s > 1 in (1.8) and rearranging gives
!
∞
∞
X
X
1
1
p − p ln p + ln
= p − p ln p + ln ς (ps) 6
= ς (s) ,
ps
i
is
i=1
i=1
Taking xi =
where ς denotes the well-known Riemann Zêta function.
In the following, we give a partial answer to Problem 1.2.
Theorem 1.5. Let 0 < p 6 1 be a real number, and let f be a continuous function
on [a, b] . Then the inequality
Z b
Z b
ep
p−1
p
(1.17)
(b − a)
|f (x)| dx ≤ exp
|f (x)| dx
pp
a
a
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is valid. Equality in (1.17) holds if f (x) =
(1.17) is the best possible.
p
.
b−a
Thus, the constant
ep
pp
(b − a)p−1 in
Theorem 1.6. Let x > 0. Then
(1.18)
Γ(x) 6
2x+1 xx−1
ex
is valid, where Γ denotes the well-known Gamma function.
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
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2.
Lemmas
Lemma 2.1. For x ∈ [0, ∞) and p > 0, the inequality
ep p
x 6 ex
p
p
(2.1)
is valid. Equality in (2.1) holds if x = p. Thus, the constant
possible.
ep
pp
in (2.1) is the best
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
Proof. Letting f (x) = p ln x − x on the set (0, ∞) , it is easy to obtain that the
function f has a maximal point at x = p and the maximal value equals f (p) =
p ln p − p. Then, we obtain (2.1). It is clear that the inequality (2.1) also holds at
x = 0.
vol. 10, iss. 3, art. 90, 2009
Lemma 2.2. Let p > 0 be a real number. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n > 2,
we have:
Contents
(i) If p > 1, then the inequality
n
X
(2.2)
xpi
6
i=1
n
X
!p
xi
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(ii) If 0 < p 6 1, then inequality
np−1
n
X
i=1
is valid.
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i=1
is valid.
(2.3)
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xpi 6
n
X
i=1
!p
xi
Proof. (i) For the proof, we use mathematical induction. First, we prove (2.2) for
n = 2. We have for any (x1 , x2 ) 6= (0, 0)
(2.4)
x1
≤1
x1 + x2
x2
≤ 1.
x1 + x2
and
Then, by p > 1 we get
p
x1
x1
(2.5)
6
x1 + x2
x1 + x2
and
x2
x1 + x2
p
6
x2
.
x1 + x2
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
By addition from (2.5), we obtain
p p
x2
x1
x2
x1
+
6
+
.
x1 + x2
x1 + x2
x1 + x2 x1 + x2
So,
(2.6)
xp1
+
xp2
p
6 (x1 + x2 ) .
It is clear that inequality (2.6) holds also at the point (0, 0) .
Now we suppose that
!p
n
n
X
X
(2.7)
xpi 6
xi
i=1
i=1
(2.8)
i=1
xpi
6
n+1
X
i=1
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and we prove that
n+1
X
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!p
xi
.
We have by (2.6)
(2.9)
n+1
X
!p
xi
=
i=1
n
X
!p
xi + xn+1
>
i=1
n
X
!p
xi
+ xpn+1
i=1
and by (2.7) and (2.9), we obtain
(2.10)
n+1
X
xpi
i=1
=
n
X
xpi
+
i=1
xpn+1
6
n
X
i=1
!p
xi
+
xpn+1
6
n+1
X
!p
xi
.
i=1
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
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vol. 10, iss. 3, art. 90, 2009
Then for all n > 2, (2.2) holds.
(ii) For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n , 0 < p 6 1 and n > 2, we have
!p
!p
n
n
X
X
xi
(2.11)
xi
=
n
.
n
i=1
i=1
By using the concavity of the function x 7→ xp (x > 0, 0 < p 6 1) , we obtain from
(2.11)
!p
!p
n
n
n
n
X
X
X
X
xi
np xpi
p−1
(2.12)
xi
=
n
>
=n
xpi .
n
n
i=1
i=1
i=1
i=1
Hence (2.3) holds.
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3.
Proofs of the Theorems
We are now in a position to prove our theorems.
n
Proof
Pn of Theorem 1.1. For (x1 , x2 , . . . , xn ) ∈ [0, ∞) and p > 1, we put x =
i=1 xi . Then by (2.1), we have
!p
!
n
n
X
ep X
(3.1)
xi
6 exp
xi
pp i=1
i=1
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
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vol. 10, iss. 3, art. 90, 2009
and by (2.2) we obtain (1.6).
Proof of Theorem 1.2. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and 0 < p 6 1, we put x =
P
n
i=1 xi . Then by (2.1), we have
!p
!
n
n
X
ep X
(3.2)
xi
6 exp
xi
pp i=1
i=1
and by (2.3) we obtain (1.7).
Proof of Theorem 1.3. This can be concluded by letting n → +∞ in Theorem 1.1.
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Proof of Theorem 1.4. By the condition of Theorem 1.4, we haveP
0 < xi 6 p for all
n
−p
−p
1 6 i 6 n. Then, x−p
>
p
for
all
1
6
i
6
n.
It
follows
that
> np−p .
i
i=1 xi
Then we obtain
!
n
n
X
X
1
−p
(3.3)
xi − ln
xi
6 np − ln np−p = np + ln + p ln p.
n
i=1
i=1
Close
It follows that
n
X
exp
!
xi
i=1
n
X −p
pp
6 enp
xi .
n
i=1
The proof of Theorem 1.4 is completed.
Proof of Theorem 1.5. Let 0 < p 6 1. By Hölder’s inequality, we have
Z b
p
Z b
p
(3.4)
|f (x)| dx 6
|f (x)| dx (b − a)1−p .
a
a
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B. Belaïdi, A. El Farissi and
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vol. 10, iss. 3, art. 90, 2009
It follows that
(3.5)
p−1
Z
(b − a)
b
Z
p
|f (x)| dx 6
a
p
b
|f (x)| dx
.
By (3.5) and (3.6), we get (1.17).
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Proof of Theorem 1.6. Let x > 0 and t > 0. Then by Lemma 2.1, we have
et >
ex x
t .
xx
So,
(3.8)
Contents
a
On the other hand, by Lemma 2.1, we have
Z b
p
Z b
ep
(3.6)
|f (x)| dx ≤ exp
|f (x)| dx .
pp
a
a
(3.7)
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e−t >
ex x −2t
t e .
xx
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It is clear that
(3.9)
ex
1> x
x
Z
∞
x −2t
t e
0
ex
dt = x+1 x−1 Γ(x).
2 x
The proof of Theorem 1.6 is completed.
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
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vol. 10, iss. 3, art. 90, 2009
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4.
Open Problem
Problem 4.1. For p ≥ 1 a real number, determine the best possible constant α ∈ R
such that
Z b
Z b
ep
p
α
|f (x)| dx ≤ exp
|f (x)| dx .
pp a
a
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
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vol. 10, iss. 3, art. 90, 2009
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References
[1] Y. MIAO, L.-M. LIU AND F. QI, Refinements of inequalities between the sum of
squares and the exponential of sum of a nonnegative sequence, J. Inequal. Pure
and Appl. Math., 9(2) (2008), Art. 53. [ONLINE: http://jipam.vu.edu.
au/article.php?sid=985].
[2] 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].
[3] H.N. SHI, Solution of an open problem proposed by Feng Qi, RGMIA Research
Report Collection, 10(4) (2007), Art. 9. [ONLINE: http://www.staff.
vu.edu.au/RGMIA/v10n4.asp].
On Open Problems of F. Qi
B. Belaïdi, A. El Farissi and
Z. Latreuch
vol. 10, iss. 3, art. 90, 2009
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