Volume 10 (2009), Issue 3, Article 90, 7 pp.
ON OPEN PROBLEMS OF F. QI
BENHARRAT BELAÏDI, ABDALLAH EL FARISSI, AND ZINELAÂBIDINE LATREUCH
D EPARTMENT OF M ATHEMATICS
L ABORATORY OF P URE AND A PPLIED M ATHEMATICS
U NIVERSITY OF M OSTAGANEM
B. P. 227 M OSTAGANEM , A LGERIA
belaidi@univ-mosta.dz
elfarissi.abdallah@yahoo.fr
z.latreuch@gmail.com
Received 07 May, 2008; accepted 28 September, 2009
Communicated by S.S. Dragomir
A BSTRACT. 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.
Key words and phrases: Inequality, Sum of power, Exponential of sum, Nonnegative sequence, Integral Inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
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
and
(1.2)
4
(0, ∞)n = (0, ∞) × (0, ∞) × ... × (0, ∞)
|
{z
}
n times
for n ∈ N, where N denotes the set of all positive integers.
In [2], F. Qi proved the following:
Theorem A. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n 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 all
2
1 6 j 6 n with j 6= i. Thus, the constant e4 in (1.3) is the best possible.
The authors would like to thank the referees for their helpful remarks and suggestions to improve the paper.
146-08
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B ENHARRAT B ELAÏDI , A BDALLAH E L FARISSI , AND Z INELAÂBIDINE L ATREUCH
Theorem B. Let {xi }∞
i=1 be a nonnegative sequence such that
!
∞
∞
X
e2 X 2
(1.4)
x 6 exp
xi .
4 i=1 i
i=1
P∞
i=1
xi < ∞. Then
Equality in (1.4) holds if xi = 2 for some given i ∈ N and xj = 0 for all j ∈ N with j 6= i.
2
Thus, the constant e4 in (1.4) is the best possible.
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
(1.5)
βn
xi 6 exp
x i ≤ µn
xλi n .
i=1
i=1
i=1
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
x 6 exp
xi
(1.6)
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 all 1 6 j 6 n
p
with j 6= i. Thus, the constant pep in (1.6) is the best possible.
Theorem 1.2. Let 0 < p 6 1 be a real number. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n > 2, the
inequality
!
n
n
p X
X
e
(1.7)
np−1 p
xp 6 exp
xi
p i=1 i
i=1
is valid. Equality in (1.7) holds if xi =
is the best possible.
p
n
p
for all 1 6 i 6 n. Thus, the constant np−1 pep in (1.7)
Theorem 1.3. Let {xi }∞
i=1 be a nonnegative sequence such that
real number. Then
!
∞
∞
X
ep X p
(1.8)
x 6 exp
xi .
pp i=1 i
i=1
P∞
i=1
xi < ∞ and p > 1 be a
Equality in (1.8) holds if xi = p for some given i ∈ N and xj = 0 for all j ∈ N with j 6= i.
p
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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 90, 7 pp.
i=1
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O N O PEN P ROBLEMS OF F. Q I
3
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
i=1
Then for (x1 , 1, ..., 1), we obtain as x1 → +∞,
1 6 e1−n µn n − 1 + xλ1 n e−x1 → 0.
This is a contradiction.
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
possible.
pp np
e
n
is the best
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)
n
n
p X
X
p
p−1 e
n
x 6 exp
xi
pp i=1 i
i=1
!
n
pp np X −p
6 e
xi ;
n
i=1
(ii) if p ≥ 1, we have
(1.11)
n
n
X
ep X p
x
6
exp
xi
pp i=1 i
i=1
!
n
6
pp np X −p
e
xi .
n
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
or
(1.13)
ep
kxkpp 6 exp kxk1 ,
pp
where x = (x1 , x2 , ..., xn ) and k·kp denotes the p-norm.
Remark 5. Inequality (1.8) can be rewritten as
(1.14)
∞
∞
ep X p Y xi
x
6
e
pp i=1 i
i=1
which is equivalent to inequality (1.12) for x = (x1 , x2 , ...) ∈ [0, ∞)∞ .
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 90, 7 pp.
http://jipam.vu.edu.au/
4
B ENHARRAT B ELAÏDI , A BDALLAH E L FARISSI , AND Z INELAÂBIDINE L ATREUCH
Remark 6. Taking xi =
(1.15)
Taking xi =
(1.16)
1
i
for i ∈ N in (1.6) and rearranging gives
!
n
n
X
X
1
1
6
.
p − p ln p + ln
p
i
i
i=1
i=1
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
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
is valid. Equality in (1.17) holds if f (x) =
the best possible.
p
.
b−a
Thus, the constant
ep
pp
(b − a)p−1 in (1.17) is
Theorem 1.6. Let x > 0. Then
2x+1 xx−1
(1.18)
Γ(x) 6
ex
is valid, where Γ denotes the well-known Gamma function.
2. L EMMAS
Lemma 2.1. For x ∈ [0, ∞) and p > 0, the inequality
ep p
x 6 ex
pp
(2.1)
is valid. Equality in (2.1) holds if x = p. Thus, the constant
ep
pp
in (2.1) is the best possible.
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.
Lemma 2.2. Let p > 0 be a real number. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and n > 2, we have:
(i) If p > 1, then the inequality
!p
n
n
X
X
p
(2.2)
xi 6
xi
i=1
i=1
is valid.
(ii) If 0 < p 6 1, then inequality
np−1
(2.3)
n
X
i=1
xpi
6
n
X
!p
xi
i=1
is valid.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 90, 7 pp.
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O N O PEN P ROBLEMS OF F. Q I
5
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)
x2
x1
(2.4)
≤1
and
≤ 1.
x1 + x2
x1 + x2
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
By addition from (2.5), we obtain
p p
x2
x1
x1
x2
+
6
+
.
x1 + x2
x1 + x2
x1 + x2 x1 + x2
So,
xp1 + xp2 6 (x1 + x2 )p .
(2.6)
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
and we prove that
n+1
X
(2.8)
xpi
i=1
6
n+1
X
!p
xi
.
i=1
We have by (2.6)
n+1
X
(2.9)
!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
i=1
xpi =
n
X
xpi + xpn+1 6
i=1
n
X
!p
xi
+ xpn+1 6
i=1
n+1
X
!p
xi
.
i=1
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.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 90, 7 pp.
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6
B ENHARRAT B ELAÏDI , A BDALLAH E L FARISSI , AND Z INELAÂBIDINE L ATREUCH
3. P ROOFS OF THE T HEOREMS
We are now in a position to prove our theorems.
Proof of Theorem 1.1. For (x1 , x2 , . . . , xn ) ∈ [0, ∞)n and p > 1, we put x =
by (2.1), we have
!p
!
n
n
X
ep X
(3.1)
xi
6 exp
xi
pp i=1
i=1
Pn
i=1
xi . Then
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 =
Then by (2.1), we have
!p
!
n
n
X
ep X
(3.2)
xi
6 exp
xi
pp i=1
i=1
Pn
i=1
xi .
and by (2.3) we obtain (1.7).
Proof of Theorem 1.3. This can be concluded by letting n → +∞ in Theorem 1.1.
Proof of Theorem 1.4. By the condition of Theorem
we have 0 < xi 6 p for all 1 6 i 6 n.
P1.4,
n
−p
−p
−p
Then, x−p
>
p
for
all
1
6
i
6
n.
It
follows
that
x
i
i=1 i > np . Then we obtain
!
n
n
X
X
1
(3.3)
xi − ln
x−p
6 np − ln np−p = np + ln + p ln p.
i
n
i=1
i=1
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
It follows that
(3.5)
p−1
Z
(b − a)
b
p
Z
|f (x)| dx 6
a
p
b
|f (x)| dx
.
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
By (3.5) and (3.6), we get (1.17).
Proof of Theorem 1.6. Let x > 0 and t > 0. Then by Lemma 2.1, we have
ex
(3.7)
et > x tx .
x
So,
ex
(3.8)
e−t > x tx e−2t .
x
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 90, 7 pp.
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O N O PEN P ROBLEMS OF F. Q I
7
It is clear that
Z
ex ∞ x −2t
ex
(3.9)
1> x
t e dt = x+1 x−1 Γ(x).
x 0
2 x
The proof of Theorem 1.6 is completed.
4. O PEN P ROBLEM
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
R EFERENCES
[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].
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 90, 7 pp.
http://jipam.vu.edu.au/