Volume 10 (2009), Issue 2, Article 49, 5 pp.
A DISCRETE VERSION OF AN OPEN PROBLEM AND SEVERAL ANSWERS
YU MIAO AND FENG QI
C OLLEGE OF M ATHEMATICS AND I NFORMATION S CIENCE
H ENAN N ORMAL U NIVERSITY
X INXIANG C ITY, H ENAN P ROVINCE
453007, C HINA
yumiao728@yahoo.com.cn
R ESEARCH I NSTITUTE OF M ATHEMATICAL I NEQUALITY T HEORY
H ENAN P OLYTECHNIC U NIVERSITY
J IAOZUO C ITY, H ENAN P ROVINCE
454010, C HINA
qifeng618@gmail.com
Received 10 July, 2008; accepted 29 May, 2009
Communicated by S.S. Dragomir
A BSTRACT. 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.
Key words and phrases: Integral inequality, Discrete version, Open problem.
2000 Mathematics Subject Classification. 26D15; 28A25.
1. I NTRODUCTION
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
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
hold for α and β?
199-08
0
2
Y. M IAO AND F. Q I
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.
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
m
X
(1.3)
xi ≤
i=1
m
X
yi
i=1
for 1 ≤ m ≤ n. Under what conditions does the inequality
n
X
(1.4)
xαi yiβ
≤
i=1
n
X
yiα+β
i=1
hold for α and β?
In the next sections, we shall establish several answers to Open Problem 2.
2. L EMMAS
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
yj (xi − xi+1 ).
i=1 j=1
Proof. Identity (2.1) follows from standard straightforward arguments.
Lemma 2.2 ([2, p. 17]). Let a and b be positive numbers with a + b = 1. Then
(2.2)
ax + by ≥ xa y b
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
yiα
i=1
holds for α ≥ 1 and 1 ≤ m ≤ n.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 49, 5 pp.
http://jipam.vu.edu.au/
A D ISCRETE V ERSION OF AN O PEN P ROBLEM AND S EVERAL A NSWERS
3
Proof. Let xn+1 be a positive number such that xn+1 ≤ xn . From Lemma 2.1 and 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
x2i
i=1
which implies that
m
X
(2.4)
yi2
≥2
i=1
m
X
xi yi −
i=1
m
X
x2i
≥
m
X
i=1
x2i .
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
(yiα − xαi )(yi − xi ) ≥ 0
which leads to
(2.5)
m
X
yiα+1
i=1
≥
m
X
yiα xi
+
m
X
i=1
yi xαi
−
i=1
m
X
xα+1
i
i=1
for 1 ≤ m ≤ n. Further, by virtue of Lemma 2.1, it follows that
m
X
yiα xi
= xm+1
i=1
(2.6)
m
X
yiα
+
i=1
≥ xm+1
m
X
=
yjα (xi − xi+1 )
i=1 j=1
xαi +
i=1
m
X
m X
i
X
m X
i
X
xαj (xi − xi+1 )
i=1 j=1
xα+1
.
i
i=1
A similar argument also yields
(2.7)
m
X
yi xαi
i=1
≥
m
X
xα+1
.
i
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
(2.8)
[α] α α − [α] α
[α] α−[α]
yi +
xi ≥ yi xi
.
α
α
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 49, 5 pp.
http://jipam.vu.edu.au/
4
Y. M IAO AND F. Q I
Summing on both sides of (2.8) and utilizing Lemma 2.1, the conclusion obtained above for
α ∈ N yields
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
=
m
X
α−[α]
xm+1
[α]
yi
i=1
≥
m
X
xαi −
α
i=1
+
m X
i
X
[α]
yj
α−[α]
xi
−
α−[α]
xi+1
i=1 j=1
m
− [α] X α
xi
α
i=1
m
α − [α] X α
−
xi
α
i=1
m
[α] X α
=
x .
α i=1 i
Since
[α]
α
6= 0, the required result is proved.
3. S EVERAL A NSWERS TO O PEN P ROBLEM 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
n
X
X
α β
(3.1)
xi yi ≤
yiα+β
i=1
i=1
holds for α ≥ 1 and β > 0.
Proof. By Hölder’s inequality and Lemma 2.3,
β
α "
" n
# α+β
# α+β
n
n
X
X
X
α+β
α+β
β
β
xαi yi ≤
(xαi ) α
yi β
i=1
i=1
i=1
Pn
≤
≤
α+β
i=1 xi
Pn α+β
i=1 yi
n
X
yiα+β .
i=1
α
! α+β
n
X
yiα+β
i=1
This completes the proof of Theorem 3.1.
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
(3.2)
m
X
xi,l ≤
i=1
m
X
yi,l ,
1 ≤ m ≤ n,
1 ≤ l ≤ k.
i=1
Then
(3.3)
n Y
k
X
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.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 49, 5 pp.
http://jipam.vu.edu.au/
A D ISCRETE V ERSION OF AN O PEN P ROBLEM AND S EVERAL A NSWERS
5
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
k−1
Y
βk
xαj,kk yj,k
i=1 j=1
≤
k−1
Y
βl
l
xαn+1,l
yn+1,l
n
X
n
X
k−1
Y
!
βl
l
xαi+1,l
yi+1,l
l=1
αk +βk
yi,k
i=1
n X
i
X
αk +βk
yj,k
i=1 j=1
=
βl
xαi,ll yi,l
−
l=1
l=1
+
βk
xαi,kk yi,k
i=1
l=1
+
n
X
αk +βk
yj,k
i=1
k−1
Y
βl
xαi,ll yi,l
−
l=1
k−1
Y
k−1
Y
!
βl
l
xαi+1,l
yi+1,l
l=1
βl
xαi,ll yi,l
≤ ··· ≤
n Y
k
X
αl +βl
yi,l
.
i=1 l=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 inequality (1.3) is
satisfied. Then
n
n
X
X
α 1 β1
yiα xβi
(3.4)
yi xi ≤
i=1
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
n
n X
i
X
X
X
β
α β
α
yi +
yjα xβi − xβi+1
yi xi = xn+1
i=1
i=1
≥
xβn+1
n
X
i=1 j=1
1
yiα1 xα−α
i
i=1
=
n
X
1 +β
yiα1 xα−α
=
i
i=1
+
n X
i
X
1
yjα1 xα−α
xβi − xβi+1
j
i=1 j=1
n
X
yiα1 xβi 1
i=1
which completes the proof of Theorem 3.3.
R EFERENCES
[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
sity Press, Cambridge, 1952.
AND
G. PÓLYA, Inequalities, 2nd edition, Cambridge Univer-
[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].
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 49, 5 pp.
http://jipam.vu.edu.au/