Volume 8 (2007), Issue 2, Article 58, 4 pp.
NOTE ON AN OPEN PROBLEM
LAZHAR BOUGOFFA
A L - IMAM M UHAMMAD I BN S AUD I SLAMIC U NIVERSITY
FACULTY OF S CIENCE , D EPARTMENT OF M ATHEMATICS
P.O.B OX 84880, R IYADH 11681, S AUDI A RABIA
bougoffa@hotmail.com
Received 17 December, 2006; accepted 1 April, 2007
Communicated by P.S. Bullen
A BSTRACT. The aim of this short note is to establish an integral inequality and its reverse which
give an affirmative answer to an open problem posed by QUÔC ANH NGÔ, DU DUC THANG,
TRANT TAT DAT, and DANG ANH TUAN, in the paper [Notes on an integral inequality, J.
Ineq. Pure and Appl. Math., 7(4)(2006), Art. 120.]
Key words and phrases: Integral inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
Very recently, in the paper [1] the authors studied some integral inequalities and proposed the
following open problem:
Problem 1.1. Letf be a continuous function on [0, 1] satisfying
Z 1
Z 1
(1.1)
f (t)dt ≥
tdt, ∀x ∈ [0, 1].
x
x
Under what conditions does the inequality
Z 1
Z
α+β
(1.2)
f
(x)dx ≥
0
1
xα f β (x)dx
0
hold for α and β?
This type of integral inequality is a complement, variant and continuation of Qi’s inequality [2]. Before giving an affirmative answer to Problem 1.1 and its reverse, we establish the
following essential lemma:
Lemma 1.1. Let f (x) be nonnegative function, continuous on [a, b] and differentiable on (a, b).
Rb
Rb
If x f (t)dt ≤ x (t − a)dt, ∀x ∈ [a, b], then
(1.3)
319-06
f (x) ≤ x − a.
2
If
L AZHAR B OUGOFFA
Rb
x
f (t)dt ≥
Rb
x
(t − a)dt, ∀x ∈ [a, b], then
f (x) ≥ x − a.
(1.4)
Proof. In order to prove (1.3), set
Z b
G(x) =
[f (t) − (t − a)]dt ≤ 0,
∀x ∈ [a, b],
x
we have
0
G (x) = x − a − f (x), ∀x ∈ [a, b].
0
We shall give an indirect proof, we suppose f (x) ≥ x − a, then G (x) ≤ 0, G(x) decreases,
and G(x) ≥ 0, because of G(b) = 0. This contradiction establishes (1.3).
The proof of (1.4) is the same as the proof of (1.3).
Now, our results can be stated as follows:
2. M AIN R ESULTS
Theorem 2.1. Let f (x) be a nonnegative function, continuous on [a, b] and differentiable on
(a, b), and let α and β be positive numbers.
Rb
Rb
If x f (t)dt ≤ x (t − a)dt, ∀x ∈ [a, b], then
Z b
Z b
α+β
(2.1)
f
(x)dx ≤
(x − a)α f β (x)dx.
a
If
Rb
x
f (t)dt ≥
Rb
(2.2)
x
a
(t − a)dt, ∀x ∈ [a, b], then
Z b
Z b
α+β
f
(x)dx ≥
(x − a)α f β (x)dx.
a
Proof. Set
a
x
Z
α+β
f
(t) − (t − a)α f β (t)dt ,
F (x) =
∀x ∈ [a, b].
a
We can see that
0
F (x) = f α+β (x) − (x − a)α f β (x),
so that
0
F (x) = [f α (x) − (x − a)α ] f β (x).
If
Z
b
Z
b
f (t)dt ≤
x
(t − a)dt,
∀x ∈ [a, b],
x
and from (1.3) of Lemma 1.1, we have, f (x) ≤ (x − a), so that f α (x) ≤ (x − a)α . Thus
0
F (x) ≤ 0, and F (x) is decreasing on [a, b]. Since F (a) = 0, we have F (x) ≤ 0, ∀x ∈ [a, b],
which gives the inequality (2.1).
When
Z
Z
b
b
f (t)dt ≥
x
(t − a)dt,
∀x ∈ [a, b],
x
we have from (1.4) f α (x) ≥ (x − a)α , and the rest of the proof is the same as that of (2.1). Remark 2.2. If f (x) = 0 or f (x) = x − a, the equality in (2.1) holds.
Now we establish new integral inequalities similar to (2.1) and (2.2) involving n functions:
fi (x), i = 1, . . . , n.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 58, 4 pp.
http://jipam.vu.edu.au/
N OTE ON AN O PEN P ROBLEM
3
Theorem 2.3. Let fi (x), i = 1, 2, . . . , n be nonnegative functions, continuous on [a, b] and
differentiable on (a, b), and let αi , βi , i = 1, . . . , n be positive numbers.
If
Z b
Z b
fi (t)dt ≤
(t − a)dt, ∀x ∈ [a, b],
x
x
then
Z bY
n
(2.3)
fiαi +βi (x)dx
Z
b
≤
Pn
(x − a)
a i=1
i=1
αi
a
n
Y
fiβi (x)dx.
i=1
If
b
Z
Z
b
fi (t)dt ≥
(t − a)dt,
x
∀x ∈ [a, b],
x
then
Z bY
n
(2.4)
fiαi +βi (x)dx
Z
b
≥
a i=1
Pn
(x − a)
i=1
αi
a
n
Y
fiβi (x)dx.
i=1
Proof. As in proof of (2.1) and (2.2), we let
#
Z x "Y
n
n
Pn
Y
F (x) =
fiαi +βi (t) − (t − a) i=1 αi
fiβi (t) dt,
a
thus
i=1
"
0
F (x) =
i=1
n
Y
Pn
fiαi (x) − (x − a)
i=1
#
αi
i=1
n
Y
fiβi (x),
i=1
when
Z
b
Z
fi (t)dt ≤
x
b
(t − a)dt
Z
Z
fi (t)dt ≥
resp.
x
b
x
b
(t − a)dt ,
∀x ∈ [a, b],
x
using Lemma 1.1, we have
fi (x) ≤ x − a
(resp. fi (x) ≥ x − a) ,
i = 1, 2, . . . , n,
and
fiαi (x) ≤ (x − a)αi
thus
n
Y
(resp. fiαi (x) ≥ (x − a)αi ) ,
Pn
fiαi (x) ≤ (x − a)
i=1
αi
i=1
resp.
n
Y
i = 1, 2, . . . , n,
Pn
fiαi (x) ≥ (x − a)
i=1
!
αi
.
i=1
The rest of the proof is the same as in Theorem 2.1, and we omit (2.3) and (2.4).
In order to illustrate a possible practical use of this result, we shall give in the following a
simple example in which we can apply Theorem 2.1.
Example 2.1. For α = β = 1 :
i) Let f (t) = cos t + t on [0, π2 ], we see that all the conditions of Theorem 2.1 are fulfilled.
Indeed,
Z π
2
1 π2
2
(cos t + t) dt = 1 − sin x +
−x
2 4
x
Z π
h πi
2
1 π2
2
≥
− x , ∀x ∈ 0,
,
tdt =
2 4
2
x
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 58, 4 pp.
http://jipam.vu.edu.au/
4
L AZHAR B OUGOFFA
and straightforward computation yields
Z π
2
5π π 3
(cos x + x)2 dx =
+
−2
4
24
0
Z π
2
x (cos x + x) dx
>
0
=
That is,
π
2
Z
π π3
+
− 1.
2 24
Z
2
f (x)dx >
π
2
xf (x)dx.
0
0
ii) Let f (t) = t − π2 + cos t on [ π2 , π], all the conditions of Theorem 2.1 be satisfied. We
see that
Z π
Z π
hπ i
π
π
t − + cos t dt ≤
t−
dt, ∀x ∈
,π ,
2
2
2
x
x
which is equivalent to
Z π
hπ i
,π ,
cos tdt = − sin x ≤ 0, x ∈
2
x
and direct computation yields
Z π
2
π π3
π
−2
x − + cos x dx = +
π
2
4 24
2
Z π
π
<
f (x)dx
x−
π
2
2
π3
=
− 1.
24
That is
Z
π
Z
2
π
f (x)dx <
π
2
π
2
x−
π
f (x)dx.
2
R EFERENCES
[1] QUÔC ANH NGÔ, DU DUC THANG, TRAN TAT DAT, AND DANG ANH TUAN, Notes on
an integral inequality, J. Ineq. Pure and Appl. Math., 7(4) (2006), Art. 120. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=737].
[2] FENG QI, Several integral inequalities, J. Ineq. Pure and Appl. Math., 1(2) (2000), Art. 19. [ONLINE: http://jipam.vu.edu.au/article.php?sid=113].
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 58, 4 pp.
http://jipam.vu.edu.au/