Journal of Inequalities and Special Functions
ISSN: 2217-4303, URL: http://ilirias.com/jiasf
Volume 7 Issue 3(2016), Pages 114-120.
OPEN PROBLEM RELATED TO INTEGRAL INEQUALITIES
ARTION KASHURI, ROZANA LIKO, AKLI FUNDO, MIFTAR RAMOSAÇO
Abstract. In this paper, we have solved an open problem, and as consequence some interesting integral inequalities are obtained. At the end, an
open problem is proposed.
1. Introduction
In the paper (see [2]), Qi proposed the following open problem, which has attracted much attention from some mathematicians (cf. [[3],[4],[5],[6],[7],[8],[9],[10],
[11],[12],[13],[14],[15],[16],[17]]).
Open Problem 1. Under what conditions does the inequality
!t−1
Z b
Z b
t
[f (x)] dx ≥
f (x)dx
a
(1.1)
a
hold for t > 1?
More recently, Liu et al. (see [1]) obtained the following theorems.
Theorem 1.1. Let f (x) ≥ 0 be a continuous function on [a, b] and satisfies
[(y − a)α f α (x) − (x − a)α f α (y)] · [f β−γ (x) − f β−γ (y)] ≥ 0, ∀x, y ∈ [a, b]. (1.2)
Then the inequality
Rb
a
Rb
a
f α+β (x)dx
f α+γ (x)dx
Rb
≥ Rab
a
(x − a)α · f β (x)dx
(x − a)α · f γ (x)dx
(1.3)
holds for every positive real number α > 0 and β ≥ γ > 0. If (1.1) reverses, then
(1.3) reverses.
Theorem 1.2. Let f (x), g(x) > 0 be continuous functions on [a, b] and satisfies
[g α (y)f α (x) − g α (x)f α (y)] · [f β−γ (x) − f β−γ (y)] ≥ 0, ∀x, y ∈ [a, b].
(1.4)
Then the inequality
Rb
a
Rb
a
f α+β (x)dx
f α+γ (x)dx
Rb
≥ Rab
a
g α (x) · f β (x)dx
g α (x) · f γ (x)dx
2010 Mathematics Subject Classification. 26D10, 26D15.
Key words and phrases. Integral inequality, Qi’s inequality.
c
2016
Ilirias Publications, Prishtinë, Kosovë.
Submitted August 22, 2016. Published September 22, 2016.
114
(1.5)
OPEN PROBLEM RELATED TO INTEGRAL INEQUALITIES
115
holds for every positive real number α > 0 and β ≥ γ > 0. If (1.4) reverses, then
(1.5) reverses.
Liu et al. (see [1]) presented the following interesting open problem.
Open Problem 2. Under what conditions does the inequality
R
δ
b
R b α+β
α
β
(x
−
a)
·
f
(x)dx
f
(x)dx
a
a
≥ R
Rb
λ
α+γ
b
f
(x)dx
α · f γ (x)dx
(x
−
a)
a
a
(1.6)
hold for α, β, γ, δ and λ?
2. Main Results
Theorem 2.1. Let f (x) ≥ 0 be a continuous function on [a, b] and satisfies
[(y − a)α f α (x) − (x − a)α f α (y)] · [f β−γ (x) − f β−γ (y)] ≥ 0, ∀x, y ∈ [a, b]
(2.1)
and f β (x) ≤ f γ (x), ∀x ∈ [a, b]. Then inequality (2.2) for every positive real number
α > 0 and β ≥ γ > 0
R
δ
b
R b α+β
(x − a)α · f β (x)dx
f
(x)dx
a
a
(2.2)
≥ R
Rb
λ
b
f α+γ (x)dx
α · f γ (x)dx
(x
−
a)
a
a
holds under each of the following conditions:
(1) λ = δ = 0 and β = γ, ∀x ∈ [a, b];
(2) λ = δ ∈ [1, +∞), ∀x ∈ [a, b];
Z b
1
for 1 ≤ δ < λ and ∀x ∈ [a, b];
(3) If
f β (t)dt ≥
(b
−
a)α
Zxb
1
(4) If
f β (t)dt ≤
for 1 ≤ λ < δ and ∀x ∈ [a, b].
(b − a)α
x
If (2.1) reverses and f β (x) ≥ f γ (x), ∀x ∈ [a, b], then (2.2) reverses.
Proof. We only prove the case ”≥”.
(1) If λ = δ = 0 and β = γ, for all x ∈ [a, b] then inequality (2.2) turns into an
equality.
(2) If λ = δ = 1 inequality (2.2) coincides with theorem 1.1.
R b α+β
f
(x)dx
.
Now let λ = δ > 1 and denote by d = Rab
f α+γ (x)dx
a
Since f β (x) ≤ f γ (x), ∀x ∈ [a, b] then d ∈ [0, 1]. By theorem 1.1, we have
the following inequalities
!δ
!δ
Rb
R b α+β
R b α+β
α
β
(x
−
a)
·
f
(x)dx
f
(x)dx
f
(x)dx
a
a
≤ Rb
≤ Rab
(2.3)
Rb
α
γ
α+γ
α+γ
(x − a) · f (x)dx
f
(x)dx
f
(x)dx
a
a
a
since d ∈ [0, 1], for all δ > 1. So inequality (2.2) follows.
116
ARTION KASHURI, ROZANA LIKO, AKLI FUNDO, MIFTAR RAMOSAÇO
(3) For 1 ≤ δ < λ there exists a real positive number r such that λ = δ + r.
Using case (2) for λ = δ ∈ [1, +∞) we have
R
δ
!δ
b
Rb
(x − a)α · f β (x)dx
(x − a)α · f β (x)dx
a
1
a
r
· R
R
λ = R b
b
b
α · f γ (x)dx
(x − a)α · f γ (x)dx
α · f γ (x)dx
(x
−
a)
(x
−
a)
a
a
a
Rb
≤
a
Rb
a
f α+β (x)dx
f α+γ (x)dx
!
1
· R
b
Rb
r ≤ Rab
f α+β (x)dx
f α+γ (x)dx
(x − a)α · f γ (x)dx
a
R
r
b
The last inequality follows by the fact that a (x − a)α · f γ (x)dx ≥ 1
Z b
1
, ∀x ∈ [a, b].
for r > 0, since we have assumed that
f β (t)dt ≥
(b − a)α
x
Also we have used the following relation
!
Z
Z
Z
a
b
b
(x − a)α · f γ (x)dx = α
a
b
(x − a)α−1 ·
a
f γ (t)dt dx
x
where f β (t) ≤ f γ (t), ∀t ∈ [x, b]. So inequality (2.2) holds.
(4) For 1 ≤ λ < δ there exists a real positive number r1 such that δ = λ + r1 .
Using case (2) for λ = δ ∈ [1, +∞) we have
R
δ
!λ Z
!r1
b
Rb
α
β
b
α
β
(x
−
a)
·
f
(x)dx
(x
−
a)
·
f
(x)dx
a
a
·
(x − a)α · f β (x)dx
R
λ = R b
α · f γ (x)dx
b
a
(x
−
a)
α
γ
(x − a) · f (x)dx
a
a
!
!r1
R b α+β
R b α+β
Z b
f
(x)dx
f
(x)dx
α
β
a
·
≤ Rb
(x − a) · f (x)dx
≤ Rab
α+γ
α+γ
a
f
(x)dx
f
(x)dx
a
a
R
r1
b
In the last inequality we have used the fact that a (x − a)α f β (x)dx
≤1
Z b
1
, ∀x ∈ [a, b].
for r1 > 0, since we have assumed that
f β (t)dt ≤
(b
−
a)α
x
Also we have used the following relation
!
Z b
Z b
Z b
α
β
α−1
β
(x − a) · f (x)dx = α
(x − a)
·
f (t)dt dx
a
a
β
x
γ
where f (t) ≤ f (t), ∀t ∈ [x, b]. So inequality (2.2) follows.
Theorem 2.2. Let f (x), g(x) > 0 be continuous functions on [a, b] and satisfies
[g α (y)f α (x) − g α (x)f α (y)] · [f β−γ (x) − f β−γ (y)] ≥ 0, ∀x, y ∈ [a, b]
β
γ
(2.4)
and f (x) ≤ f (x), ∀x ∈ [a, b]. Then the inequality (2.5) for every positive real
number α > 0 and β ≥ γ > 0
R
δ
b α
R b α+β
g (x) · f β (x)dx
f
(x)dx
a
a
≥ R
(2.5)
Rb
λ
b α
f α+γ (x)dx
γ (x)dx
g
(x)
·
f
a
a
OPEN PROBLEM RELATED TO INTEGRAL INEQUALITIES
117
holds under each of the following conditions:
(1) λ = δ = 0 and β = γ, ∀x ∈ [a, b];
(2) λ = δ ∈ [1, +∞), ∀x ∈ [a, b];
1
(3) If g α (a) ≥
for 1 ≤ δ < λ;
(b − a)f γ (a)
1
(4) If g α (b) ≤
for 1 ≤ λ < δ.
(b − a)f β (b)
If (2.4) reverses and f β (x) ≥ f γ (x), ∀x ∈ [a, b], then (2.5) reverses.
Proof. We only prove the case ”≥”.
(1) If λ = δ = 0 and β = γ, for all x ∈ [a, b] then inequality (2.5) turns into an
equality.
(2) If λ = δ = 1 inequality (2.5) coincides with theorem 1.2.
R b α+β
f
(x)dx
.
Now let λ = δ > 1 and denote by d = Rab
α+γ
f
(x)dx
a
Since f β (x) ≤ f γ (x), ∀x ∈ [a, b] then d ∈ [0, 1]. By theorem 1.2, we have
the following inequalities
!δ
!δ
Rb α
R b α+β
R b α+β
g (x) · f β (x)dx
f
(x)dx
f
(x)dx
a
a
≤ Rb
≤ Rab
(2.6)
Rb
g α (x) · f γ (x)dx
f α+γ (x)dx
f α+γ (x)dx
a
a
a
since d ∈ [0, 1], for all δ > 1. So inequality (2.5) follows.
(3) For 1 ≤ δ < λ there exists a real positive number r such that λ = δ + r.
Using case (2) for λ = δ ∈ [1, +∞) we have
R
δ
!δ
b α
Rb α
g (x) · f β (x)dx
g (x) · f β (x)dx
a
1
a
r
· R
R
λ = R b α
b
b α
α (x) · f γ (x)dx
g (x) · f γ (x)dx
γ (x)dx
g
g
(x)
·
f
a
a
a
Rb
≤
a
Rb
a
f α+β (x)dx
!
· R
f α+γ (x)dx
Rb
f α+β (x)dx
r ≤ Rab
b α
f α+γ (x)dx
g (x) · f γ (x)dx
a
a
1
r
g α (x) · f γ (x)dx ≥ 1 for
1
r > 0, since we have assumed that g α (a) ≥
. So inequality
(b − a)f γ (a)
(2.5) holds.
(4) For 1 ≤ λ < δ there exists a real positive number r1 such that δ = λ + r1 .
Using case (2) for λ = δ ∈ [1, +∞) we have
R
δ
!λ
!r1
b α
Rb α
β
Z b
g
(x)
·
f
(x)dx
g (x) · f β (x)dx
a
α
β
a
·
g (x) · f (x)dx
R
λ = R b α
γ (x)dx
b α
a
g
(x)
·
f
γ
g (x) · f (x)dx
a
a
R
b
a
The last inequality follows by the fact that
Rb
≤
a
Rb
a
f α+β (x)dx
f α+γ (x)dx
!
Z
·
!r1
b
α
β
g (x) · f (x)dx
a
Rb
≤ Rab
a
f α+β (x)dx
f α+γ (x)dx
118
ARTION KASHURI, ROZANA LIKO, AKLI FUNDO, MIFTAR RAMOSAÇO
r1
g α (x) · f β (x)dx
≤1
1
for r1 > 0, since we have assumed that g α (b) ≤
. So inequality
(b − a)f β (b)
(2.5) follows.
In the last inequality we have used the fact that
R
b
a
3. Applications
Corollary 3.1. Let f (x) ≥ 0 be a continuous function on [a, b] and satisfies
[(y − a)α f α (x) − (x − a)α f α (y)] · [f α−β (x) − f α−β (y)] ≥ 0, ∀x, y ∈ [a, b]
(3.1)
and f α (x) ≤ f β (x), ∀x ∈ [a, b]. Then inequality (3.2) for every positive real number
α≥β>0
R
δ
b
R b 2α
α
((x
−
a)
·
f
(x))
dx
f (x)dx
a
a
≥ R
(3.2)
Rb
λ
α+β
b
f
(x)dx
α · f β (x)dx
(x
−
a)
a
a
holds under each of the following conditions:
(1) λ = δ = 0 and α = β, ∀x ∈ [a, b];
(2) λ = δ ∈ [1, +∞), ∀x ∈ [a, b];
Z b
1
(3) If
f α (t)dt ≥
for 1 ≤ δ < λ and ∀x ∈ [a, b];
(b
−
a)α
Zxb
1
(4) If
f α (t)dt ≤
for 1 ≤ λ < δ and ∀x ∈ [a, b].
(b − a)α
x
If (3.1) reverses and f α (x) ≥ f β (x), ∀x ∈ [a, b], then (3.2) reverses.
Proof. Let α = β and use theorem 2.1.
Corollary 3.2. Let f (x) ≥ 0 be a continuous function on [a, b] and satisfies
[(y − a)α f α (x) − (x − a)α f α (y)] · [f β−α (x) − f β−α (y)] ≥ 0, ∀x, y ∈ [a, b]
(3.3)
and f β (x) ≤ f α (x), ∀x ∈ [a, b]. Then inequality (3.4) for every positive real number
β≥α>0
R
δ
b
R b α+β
(x − a)α · f β (x)dx
f
(x)dx
a
a
≥ R
(3.4)
Rb
λ
b
f 2α (x)dx
α dx
((x
−
a)
·
f
(x))
a
a
holds under each of the following conditions:
(1) λ = δ = 0 and α = β, ∀x ∈ [a, b];
(2) λ = δ ∈ [1, +∞), ∀x ∈ [a, b];
Z b
1
(3) If
f β (t)dt ≥
for 1 ≤ δ < λ and ∀x ∈ [a, b];
(b
−
a)α
Zxb
1
(4) If
f β (t)dt ≤
for 1 ≤ λ < δ and ∀x ∈ [a, b].
(b − a)α
x
If (3.3) reverses and f β (x) ≥ f α (x), ∀x ∈ [a, b], then (3.4) reverses.
Proof. Let α = γ and use theorem 2.1.
OPEN PROBLEM RELATED TO INTEGRAL INEQUALITIES
119
Corollary 3.3. Let f (x), g(x) > 0 be continuous functions on [a, b] and satisfies
[g α (y)f α (x) − g α (x)f α (y)] · [f α−β (x) − f α−β (y)] ≥ 0, ∀x, y ∈ [a, b]
α
(3.5)
β
and f (x) ≤ f (x), ∀x ∈ [a, b]. Then the inequality (3.6) for every positive real
number α ≥ β > 0
R
δ
b
R b 2α
α
(g(x)
·
f
(x))
dx
f (x)dx
a
a
≥ R
(3.6)
Rb
λ
α+β
b α
f
(x)dx
β
g (x) · f (x)dx
a
a
holds under each of the following conditions:
(1) λ = δ = 0 and α = β, ∀x ∈ [a, b];
(2) λ = δ ∈ [1, +∞), ∀x ∈ [a, b];
1
for 1 ≤ δ < λ;
(3) If g α (a) ≥
(b − a)f β (a)
1
for 1 ≤ λ < δ.
(4) If g α (b) ≤
(b − a)f α (b)
If (3.5) reverses and f α (x) ≥ f β (x), ∀x ∈ [a, b], then (3.6) reverses.
Proof. Let α = β and use theorem 2.2.
Corollary 3.4. Let f (x), g(x) > 0 be continuous functions on [a, b] and satisfies
[g α (y)f α (x) − g α (x)f α (y)] · [f β−α (x) − f β−α (y)] ≥ 0, ∀x, y ∈ [a, b]
β
(3.7)
α
and f (x) ≤ f (x), ∀x ∈ [a, b]. Then the inequality (3.8) for every positive real
number β ≥ α > 0
R
δ
b α
R b α+β
β
g
(x)
·
f
(x)dx
f
(x)dx
a
a
≥ R
(3.8)
Rb
λ
2α
b
f (x)dx
α dx
(g(x)
·
f
(x))
a
a
holds under each of the following conditions:
(1) λ = δ = 0 and α = β, ∀x ∈ [a, b];
(2) λ = δ ∈ [1, +∞), ∀x ∈ [a, b];
1
for 1 ≤ δ < λ;
(3) If g α (a) ≥
(b − a)f α (a)
1
(4) If g α (b) ≤
for 1 ≤ λ < δ.
(b − a)f β (b)
If (3.7) reverses and f β (x) ≥ f α (x), ∀x ∈ [a, b], then (3.8) reverses.
Proof. Let α = γ and use theorem 2.2.
Lastly, we propose the following open problem.
Open Problem 3. Can the conditions in the Theorem 2.1 (or in the Theorem
2.2) be improved? It is possible to find strongly conditions such that they include
our conditions?
Acknowledgements. The authors would like to express sincere gratitude to Professor Valmir Krasniqi for pointing out the reference [2].
120
ARTION KASHURI, ROZANA LIKO, AKLI FUNDO, MIFTAR RAMOSAÇO
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Artion Kashuri
Department of Mathematics, Faculty of Technical Science, University ”Ismail Qemali”,
Vlora, Albania
E-mail address: [email protected]
Rozana Liko
Department of Mathematics, Faculty of Technical Science, University ”Ismail Qemali”,
Vlora, Albania
E-mail address: [email protected]
Akli Fundo
Department of Mathematics, Polytechnic University of Tirana,
Tirana, Albania
E-mail address: [email protected]
Miftar Ramosaço
Department of Mathematics, Faculty of Technical Science, University ”Ismail Qemali”,
Vlora, Albania
E-mail address: [email protected]