A DISCRETE VERSION OF AN OPEN PROBLEM
AND SEVERAL ANSWERS
Open Problem
YU MIAO
FENG QI
College of Mathematics and Information Science
Henan Normal University
Xinxiang City, Henan Province
453007, China
EMail: yumiao728@yahoo.com.cn
Research Instit. of Mathematical Inequality Theory
Henan Polytechnic University
Jiaozuo City, Henan Province
454010, China
EMail: qifeng618@gmail.com
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
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Received:
10 July, 2008
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Accepted:
29 May, 2009
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Communicated by:
S.S. Dragomir
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2000 AMS Sub. Class.:
26D15; 28A25
Key words:
Integral inequality, Discrete version, Open problem.
Abstract:
In this article, a discrete version of an open problem in [Q. A. Ngô, D. D.
Thang, T. T. Dat, and D. A. Tuan, Notes on an integral inequality, J. Inequal.
Pure Appl. Math. 7 (2006), no. 4, Art. 120; Available online at http:
//jipam.vu.edu.au/article.php?sid=737] is posed and several answers are provided.
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Contents
1
Introduction
3
2
Lemmas
5
3
Several Answers to Open Problem 2
8
Open Problem
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
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1.
Introduction
In [4], some integral inequalities were obtained and the following open problem was
posed.
Open Problem 1. Let f be a continuous function on [0, 1] satisfying the following
condition
Z 1
Z 1
(1.1)
f (t) dt ≥
t dt
x
x
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
for x ∈ [0, 1]. Under what conditions does the inequality
Z 1
Z 1
α+β
(1.2)
f
(t) dt ≥
tβ f α (t) dt
0
Open Problem
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0
hold for α and β?
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In [1], some affirmative answers to Open Problem 1 and the reversed inequality
of (1.2) were given.
In [3], an abstract version of Open Problem 1 was posed, respective answers to
these two open problems were presented, and the results in [1] were extended.
Now we would like to further pose the following discrete version of the open
problems in [1, 3] as follows.
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Open Problem 2. For n ∈ N, let {x1 , x2 , . . . , xn } and {y1 , y2 , . . . , yn } be two positive sequences satisfying x1 ≥ x2 ≥ · · · ≥ xn , y1 ≥ y2 ≥ · · · ≥ yn and
(1.3)
m
X
i=1
xi ≤
m
X
i=1
yi
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for 1 ≤ m ≤ n. Under what conditions does the inequality
(1.4)
n
X
i=1
xαi yiβ
≤
n
X
yiα+β
i=1
hold for α and β?
In the next sections, we shall establish several answers to Open Problem 2.
Open Problem
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
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2.
Lemmas
In order to establish several answers to Open Problem 2, the following lemmas are
necessary.
Lemma 2.1. For n ∈ N, let {x1 , x2 , . . . , xn , xn+1 } and {y1 , y2 , . . . , yn } be two real
sequences. Then
(2.1)
n
X
xi yi = xn+1
i=1
n
X
yi +
i=1
n X
i
X
Open Problem
yj (xi − xi+1 ).
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
i=1 j=1
Proof. Identity (2.1) follows from standard straightforward arguments.
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Lemma 2.2 ([2, p. 17]). Let a and b be positive numbers with a + b = 1. Then
(2.2)
a b
ax + by ≥ x y
is valid for positive numbers x and y.
Lemma 2.3. For n ∈ N, let {x1 , x2 , . . . , xn } and {y1 , y2 , . . . , yn } be two positive
sequences satisfying x1 ≥ x2 ≥ · · · ≥ xn , y1 ≥ y2 ≥ · · · ≥ yn and inequality (1.3).
Then
(2.3)
m
X
i=1
xαi
≤
m
X
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yiα
i=1
holds for α ≥ 1 and 1 ≤ m ≤ n.
Proof. Let xn+1 be a positive number such that xn+1 ≤ xn . From Lemma 2.1 and
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using inequality (1.3), it is easy to see that, for α = 2 and 1 ≤ m ≤ n,
m
X
xi yi = xm+1
m
X
i=1
m X
i
X
yi +
i=1
≥ xm+1
m
X
m X
i
X
xi +
i=1
=
m
X
yj (xi − xi+1 )
i=1 j=1
xj (xi − xi+1 )
i=1 j=1
Open Problem
Yu Miao and Feng Qi
x2i
vol. 10, iss. 2, art. 49, 2009
i=1
which implies that
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m
X
(2.4)
yi2 ≥ 2
i=1
m
X
xi yi −
i=1
m
X
x2i ≥
m
X
i=1
i=1
Suppose that inequality (2.3) holds for some integer α > 2. Since {x1 , x2 , . . . , xn }
and {y1 , y2 , . . . , yn } are two positive sequences, then
m
X
yiα+1
≥
m
X
i=1
yiα xi
i=1
+
m
X
i=1
yi xαi
−
m
X
i=1
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xα+1
i
i=1
for 1 ≤ m ≤ n. Further, by virtue of Lemma 2.1, it follows that
m
X
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which leads to
(2.6)
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(yiα − xαi )(yi − xi ) ≥ 0
(2.5)
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x2i .
yiα xi = xm+1
m
X
i=1
yiα +
m X
i
X
i=1 j=1
yjα (xi − xi+1 )
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≥ xm+1
m
X
xαi
i=1
+
m X
i
X
xαj (xi
− xi+1 ) =
i=1 j=1
m
X
xα+1
.
i
i=1
A similar argument also yields
m
X
(2.7)
yi xαi
≥
m
X
i=1
xα+1
.
i
Open Problem
i=1
Substituting (2.6) and (2.7) into (2.5) gives inequality (2.3) for α + 1.
By induction, this means that inequality (2.3) holds for all α ∈ N.
Let [α] denote the integral part of a real number α ≥ 1. By inequality (2.2) in
Lemma 2.2, we have
[α] α α − [α] α
[α] α−[α]
yi +
x i ≥ yi xi
.
α
α
Summing on both sides of (2.8) and utilizing Lemma 2.1, the conclusion obtained
above for α ∈ N yields
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
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(2.8)
m
m
m
[α] X α X [α] α−[α] α − [α] X α
y ≥
yi xi
−
xi
α i=1 i
α
i=1
i=1
α−[α]
= xm+1
m
X
i=1
≥
m
X
i=1
Since
[α]
α
xαi −
[α]
yi +
α
m X
i
X
i=1 j=1
m
− [α] X α
xi
α
i=1
[α]
yj
=
6= 0, the required result is proved.
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α−[α]
xi
[α]
α
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m
X
i=1
α−[α]
− xi+1
xαi .
m
−
α − [α] X α
xi
α
i=1
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3.
Several Answers to Open Problem 2
Now we establish several answers to Open Problem 2.
Theorem 3.1. For n ∈ N, let {x1 , x2 , . . . , xn } and {y1 , y2 , . . . , yn } be two positive
sequences such that x1 ≥ x2 ≥ · · · ≥ xn , y1 ≥ y2 ≥ · · · ≥ yn and inequality (1.3)
is satisfied. Then
n
X
(3.1)
xαi yiβ ≤
i=1
n
X
Open Problem
yiα+β
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
i=1
holds for α ≥ 1 and β > 0.
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Proof. By Hölder’s inequality and Lemma 2.3,
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n
X
"
xαi yiβ ≤
i=1
n
X
(xαi )
α+β
α
α "
# α+β
i=1
yiβ
α+β
β
β
# α+β
i=1
Pn
≤
n
X
α+β
i=1 xi
Pn α+β
i=1 yi
α
! α+β
n
X
yiα+β ≤
i=1
n
X
yiα+β .
i=1
This completes the proof of Theorem 3.1.
(3.2)
i=1
xi,l ≤
m
X
i=1
yi,l ,
1 ≤ m ≤ n,
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Theorem 3.2. Let {x1,l , x2,l , . . . , xn,l } and {y1,l , y2,l , . . . , yn,l } for n ∈ N, k > 0 and
1 ≤ l ≤ k be positive sequences such that x1,l ≥ x2,l ≥ · · · ≥ xn,l , y1,l ≥ y2,l ≥
· · · ≥ yn,l and
m
X
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1 ≤ l ≤ k.
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Then
n Y
k
X
(3.3)
βl
xαi,ll yi,l
≤
i=1 l=1
n Y
k
X
αl +βl
yi,l
i=1 l=1
for αl ≥ 1 and βl > 0, 1 ≤ l ≤ k.
Proof. As in the proof of Lemma 2.3, let xn+1,l be positive numbers such that xn+1,l ≤
xn,l for 1 ≤ l ≤ k. By Lemma 2.1 and Theorem 3.1, it is shown that
n Y
k
X
βl
xαi,ll yi,l
=
i=1 l=1
k−1
Y
βl
l
xαn+1,l
yn+1,l
n X
i
X
βk
xαj,kk yj,k
i=1 j=1
≤
k−1
Y
βl
l
xαn+1,l
yn+1,l
n X
i
X
αk +βk
yj,k
i=1 j=1
=
n
X
i=1
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k−1
Y
βl
xαi,ll yi,l
−
k−1
Y
!
βl
l
xαi+1,l
yi+1,l
l=1
αk +βk
yi,k
i=1
l=1
+
βk
xαi,kk yi,k
l=1
n
X
αk +βk
yj,k
k−1
Y
βl
xαi,ll yi,l
−
l=1
k−1
Y
l=1
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
i=1
l=1
+
n
X
Open Problem
βl
xαi,ll yi,l
≤ ··· ≤
k−1
Y
!
βl
l
xαi+1,l
yi+1,l
l=1
n Y
k
X
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αl +βl
yi,l
.
i=1 l=1
The proof of Theorem 3.2 is completed.
Theorem 3.3. For n ∈ N, let {x1 , x2 , . . . , xn } and {y1 , y2 , . . . , yn } be two positive
sequences with the properties that x1 ≥ x2 ≥ · · · ≥ xn , y1 ≥ y2 ≥ · · · ≥ yn and
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inequality (1.3) is satisfied. Then
n
X
(3.4)
yiα1 xβi 1
≤
n
X
i=1
yiα xβi
i=1
if α ≥ α1 ≥ 1, β > 0 and β + α = β1 + α1 .
Proof. Let xn+1 be a positive number such that xn+1 ≤ xn . By Lemma 2.1 and
Theorem 3.1, we have
n
X
yiα xβi
=
xβn+1
n
X
yiα
i=1
i=1
≥ xβn+1
n
X
+
n X
i
X
i=1
=
n
X
i=1
1 +β
yiα1 xα−α
=
i
yjα xβi − xβi+1
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n X
i
X
i=1 j=1
n
X
yiα1 xβi 1
i=1
which completes the proof of Theorem 3.3.
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
i=1 j=1
1
yiα1 xα−α
+
i
Open Problem
1
yjα1 xα−α
xβi − xβi+1
j
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References
[1] L. BOUGOFFA, Note on an open problem, J. Inequal. Pure Appl. Math., 8(2)
(2007), Art. 58. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=871].
[2] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, 2nd edition,
Cambridge University Press, Cambridge, 1952.
[3] Y. MIAO AND F. QI, Another answer to an open problem, submitted.
[4] Q.A. NGÔ, D.D. THANG, T.T. DAT AND D.A. TUAN, Notes on an integral inequality, J. Inequal. Pure Appl. Math., 7(4) (2006), Art. 120. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=737].
Open Problem
Yu Miao and Feng Qi
vol. 10, iss. 2, art. 49, 2009
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