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volume 7, issue 4, article 154, 2006.

Received 27 April, 2006;

accepted 09 October, 2006.

Communicated by:B. Yang

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Journal of Inequalities in Pure and Applied Mathematics

ON SEVERAL NEW INEQUALITIES CLOSE TO HILBERT-PACHPATTE’S INEQUALITY

BING HE AND YONGJIN LI

Department of Mathematics Sun Yat-sen University

Guangzhou, 510275 P. R. China EMail:hzs314@163.com EMail:stslyj@mail.sysu.edu.cn

c

2000Victoria University ISSN (electronic): 1443-5756 124-06

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On Several New Inequalities Close to Hilbert-Pachpatte’s

Inequality Bing He and Yongjin Li

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J. Ineq. Pure and Appl. Math. 7(4) Art. 154, 2006

http://jipam.vu.edu.au

Abstract

In this paper we establish several new inequalities similar to Hilbert-Pachpatte’s inequality. Moreover, some new generalizations of Hilbert-Pachpatte’s inequality are presented.

2000 Mathematics Subject Classification:26D15.

Key words: Hilbert’s inequality, Hölder’s inequality, Jensen’s inequality, Power mean inequality.

The work was partially supported by the Foundation of Sun Yat-sen University Ad- vanced Research Centre.

1. Introduction

The well known Hardy-Hilbert’s inequality is (see [1]):

Theorem 1.1. If p > 1, p⁰ = p/(p−1) andP∞

m=1a^p_m ≤ A, P∞

n=1b^p_n⁰ ≤ B, then

(1.1)

∞

X

n=1

∞

X

m=1

a_mb_n

m+n < π sin

π p

(A)¹^p(B)^p¹⁰,

unless the sequence{a_m}or{b_n}is null.

The integral analogue can be stated as follows:

Theorem 1.2. Ifp > 1, p⁰ =p/(p−1)andR∞

0 f^p(x)dx ≤ F,R∞

0 g^p⁰(y)dy≤ G, then

(1.2)

Z ∞

0

Z ∞

0

f(x)g(y)

x+y dxdy < π sin

π p

F¹^pG^p¹⁰

unlessf ≡0org ≡0.

The following two theorems were studied by Pachpatte (see [2])

Theorem 1.3. Let p ≥ 1, q ≥ 1 and f(σ) ≥ 0, g(τ) ≥ 0 for σ ∈ (0, x), τ ∈ (0, y), wherex, y are positive real numbers and defineF(s) =Rs

0 f(σ)dσ,

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G(t) = Rt

0 g(τ)dτ, fors∈(0, x), t∈(0, y). Then Z x

0

Z y

0

F^p(s)G^q(t) s+t dsdt

≤D(p, q, x, y) Z x

0

(x−s)(F^p−1(s)f(s))²ds ¹₂

× Z y

0

(y−t)(G^q−1(t)g(t))²dt ¹₂

,

unlessf ≡0org ≡0, where

D(p, q, x, y) = 1 2pq√

xy.

Theorem 1.4. Let f, g, F, Gbe as in the above theorem, letp(σ)andq(τ)be two positive functions defined for σ ∈ (0, x), τ ∈ (0, y) and define P(s) = Rs

0 p(σ)dσ, Q(t) = Rt

0 q(τ)dτ for s ∈ (0, x), t ∈ (0, y), where x, y are posi- tive real numbers. Let φ andψ be real-valued, nonnegative, convex, and sub- multiplicative functions defined onR+ = [0,∞). Then

(1.3) Z x

0

Z y

0

φ(F(s))ψ(G(t)) s+t dsdt

≤L(x, y) (Z x

0

(x−s)

p(s)φ

f(s) p(s)

2

ds )¹₂

× (Z y

0

(y−t)

q(t)ψ g(t)

q(t) 2

dt )¹₂

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where

L(x, y) = 1 2

Z x

0

φ(P(s)) P(s)

2

ds

!¹₂ Z y

0

ψ(Q(t)) Q(t)

2

dt

!¹₂ .

The inequalities in these theorems were studied extensively and numerous variants, generalizations, and extensions have appeared in the literature (see [3]

– [9]).

The main purpose of the present article is to establish some new inequalities similar to the Hilbert-Pachpatte inequalities.

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2. Main Results

Lemma 2.1. Supposep > 1,¹_p + ¹_q = 1, r > 1,¹_r + _w¹ = 1, λ > 0, define the weight functionsωe₁(w, p, x),eω₂(w, p, x),ωe₃(w, p, x)as

(2.1) ωe₁(w, p, x) :=

Z ∞

0

1

x^λ+y^λ · x^(p−1)(1−^λ^r⁾

y¹⁻^λ^w dy, x∈(0,∞),

(2.2) ωe₂(w, p, x) :=

Z ∞

0

1

max{x^λ, y^λ} · x^(p−1)(1−^λ^r⁾

y¹⁻^λ^w dy, x∈(0,∞),

(2.3) ωe₃(w, p, x) :=

Z ∞

0

lnx/y

x^λ−y^λ · x^(p−1)(1−^λ^r⁾

y¹⁻^w^λ dy, x∈(0,∞).

Then

eω₁(w, p, x) = π

λsin ^π_rx^p(1−^λ^r⁾⁻¹, ωe₂(w, p, x) = rw

λ x^p(1−^λ^r⁾⁻¹,

ωe₃(w, p, x) =

"

π λsin ^π_r

#2

x^p(1−^λ^r⁾⁻¹.

Proof. For fixedx, letu=y^λ/x^λ, then (2.1) turns into ωe₁(w, p, x) = 1

λx^p(1−^λ^r⁾⁻¹ Z ∞

0

1

1 +uu⁻¹⁺^w¹du

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= 1

λx^p(1−^λ^r⁾⁻¹B 1

w,1 r

= π

λsin π

p

x^p(1−^λ^r⁾⁻¹

and similarly, we can prove the others.

Theorem 2.2. Let m ≥ 1, n ≥ 1, p_i > 1, _p¹

i + _q¹

i = 1 for i = 0,1,2,3,4, p₀ =p, p₃ =k, p₄ =r;q₀ =q, q₃ =l, q₄ =wandf(σ)≥0, g(τ)≥0. Define F(s) = Rs

0 f(σ)dσandG(t) =Rt

0 g(τ)dτ such that 0<

Z ∞

0

s^p(1−^λ^r⁾⁻¹F_f^p(s)ds <∞, 0<

Z ∞

0

t^q(1−^w^λ⁾⁻¹G^q_g(t)dt <∞ forσ, τ, s, t∈(0,∞). Then

(2.4) Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p¹ +kt^l/p²)(s^λ+t^λ)dsdt

≤E₁(m, n, k, r, λ) Z ∞

0

s^p(1−^λ^r⁾⁻¹F_f^p(s)ds ¹_p

× Z ∞

0

t^q(1−^w^λ⁾⁻¹G^q_g(t)dt ¹_q

unlessf ≡0org ≡0, whereE₁(m, n, k, r, λ) = _λkl^πmn_sin(π/r),

Ff(s) = Z s

0

(F^m−1(σ)f(σ))^q¹dσ _q¹

1

and

Gg(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q²dτ _q¹

2

.

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Proof. From the hypotheses, we can easily observe that (2.5) F^m(s) =m

Z s

0

F^m−1(σ)f(σ)dσ, s∈(0,∞),

(2.6) Gⁿ(t) =n Z t

0

Gⁿ⁻¹(τ)g(τ)dτ, t∈(0,∞).

From (2.5) and (2.6), applying Hölder’s inequality, equality (2.1) and Young’s inequality:

ab≤ a^p p +b^q

q, wherea≥0, b≥0, p >1, ¹_p +¹_q = 1,we obtain

F^m(s)Gⁿ(t)

=mn Z s

0

F^m−1(σ)f(σ)dσ

Z t

0

Gⁿ⁻¹(τ)g(τ)dτ

≤mns^1/p¹t^1/p² Z s

0

(F^m−1(σ)f(σ))^q¹dσ) _q¹

1 Z t

0

(Gⁿ⁻¹(τ)g(τ))^q²dτ) _q¹

2

≤mn

s^k/p¹

k +t^l/p² l

Z s

0

(F^m−1(σ)f(σ))^q¹dσ) _q¹

1

× Z t

0

(Gⁿ⁻¹(τ)g(τ))^q²dτ) _q¹

2

.

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Note that F_f(s) =

Z s

0

(F^m−1(σ)f(σ))^q¹dσ _q¹

1 , G_g(t) = Z t

0

2

.

Hence Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p¹ +kt^l/p²)(s^λ+t^λ)dsdt

≤ mn kl

Z ∞

0

Z ∞

0

Rs

0(F^m−1(σ)f(σ))^q¹dσ)

1 q1

nRt

0(Gⁿ⁻¹(τ)g(τ))^q²dτ)o_q¹

2

s^λ+t^λ dsdt

= mn kl

Z ∞

0

Z ∞

0

1 s^λ+t^λ

"

F_f(s)s⁽¹⁻^λ^r^)/q t⁽¹⁻^w^λ^)/p

# "

G_g(t)t⁽¹⁻^λ^w^)/p s⁽¹⁻^λ^r^)/q

# dsdt

≤ mn kl

(Z ∞

0

Z ∞

0

F_f^p(s)

s^λ+t^λ · s^(p−1)(1−^λ^r⁾ t⁽¹⁻^w^λ⁾ dsdt

)¹_p

× (Z ∞

0

Z ∞

0

G^q_g(t)

s^λ+t^λ · t^(q−1)(1−^λ^w⁾ s⁽¹⁻^λ^r⁾ dsdt

)¹_q

= mn kl

Z ∞

0 ωe₁(w, p, s)F_f^p(s)ds

¹_p Z ∞

0 ωe₁(r, q, t)G^q_g(t)dt ¹_q

= πmn

λklsin(π/r) Z ∞

0

s^p(1−^λ^r⁾⁻¹F_f^p(s)ds

_p¹ Z ∞

0

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=E₁(m, n, k, r, λ) Z ∞

0

¹_pZ ∞

0

.

This completes the proof.

Theorem 2.3. Letm≥1, n≥1, p_i >1,_p¹

i+_q¹

i = 1fori= 0,1,2,3,4, p₀ =p, p3 =k, p4 =r;q0 =q, q3 =l, q4 =wandf(σ)≥0, g(τ)≥0. DefineF(s) = Rs

0 f(σ)dσ and G(t) = Rt

0 g(τ)dτ such that0 < R∞

0 s^p(1−^λ^r⁾⁻¹F_f^p(s)ds < ∞, 0<R∞

0 t^q(1−^λ^w⁾⁻¹G^q_g(t)dt <∞forσ, τ, s, t∈(0,∞). Then Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p¹ +kt^l/p²) max{s^λ, t^λ}dsdt

≤E₂(m, n, k, r, λ) Z ∞

0

s^p(1−^λ^r⁾⁻¹F_f^p(s)ds ¹_p

× Z ∞

0

,

unlessf ≡0org ≡0, whereE2(m, n, k, r, λ) = ^mnrw_λkl and

F_f(s) = Z s

0

(F^m−1(σ)f(σ))^q¹dσ _q¹

1 ,

G_g(t) = Z t

0

(Gⁿ⁻¹(τ)g(τ))^q²dτ _q¹₂

.

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Proof. By (2.5) and (2.6), using Hölder’s inequality, equality (2.2) and Young’s inequality: ab≤ ^a_p^p + ^b_q^q,wherea≥0, b≥0, p >1, ¹_p +¹_q = 1.We obtain

F^m(s)Gⁿ(t)

=mn Z s

0

F^m−1(σ)f(σ)dσ

Z t

0

Gⁿ⁻¹(τ)g(τ)dτ

≤mns^1/p¹t^1/p² Z s

0

(F^m−1(σ)f(σ))^q¹dσ) _q¹

1 Z t

0

(Gⁿ⁻¹(τ)g(τ))^q²dτ) _q¹

2

≤mn

s^k/p¹

k +t^l/p² l

Z s

0

(F^m−1(σ)f(σ))^q¹dσ) _q¹

1

× Z t

0

(Gⁿ⁻¹(τ)g(τ))^q²dτ) _q¹

2

.

Note

F_f(s) = Z s

0

(F^m−1(σ)f(σ))^q¹dσ _q¹

1 ,

G_g(t) = Z t

0

2

.

Hence Z ∞

0

Z ∞

0

F^m(s)Gⁿ(t)

(ls^k/p¹ +kt^l/p²) max{s^λ, t^λ}dsdt

≤ mn kl

Z ∞

0

Z ∞

0

Rs

0(F^m−1(σ)f(σ))^q¹dσ)

1 q1

nRt

0(Gⁿ⁻¹(τ)g(τ))^q²dτ) o_q¹

2

max{s^λ, t^λ} dsdt

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= mn kl

Z ∞

0

Z ∞

0

1 max{s^λ, t^λ}

"

F_f(s)s⁽¹⁻^λ^r^)/q t⁽¹⁻^w^λ^)/p

# "

G_g(t)t⁽¹⁻^λ^w^)/p s⁽¹⁻^λ^r^)/q

# dsdt

≤ mn kl

(Z ∞

0

Z ∞

0

F_f^p(s) max{s^λ, t^λ}

s^(p−1)(1−^λ^r⁾ t⁽¹⁻^w^λ⁾ dsdt

)_p¹

× (Z ∞

0

Z ∞

0

G^q_g(t) max{s^λ, t^λ}

t^(q−1)(1−^λ^w⁾ s⁽¹⁻^λ^r⁾ dsdt

)¹_q

= mn kl

Z ∞

0 ωe₂(w, p, s)F_f^p(s)ds

¹_p Z ∞

0 ωe₂(r, q, t)G^q_g(t)dt ¹_q

= mnrw λkl

Z ∞

0

¹_pZ ∞

0

t^q(1−^λ^w⁾⁻¹G^q_g(t)dt ¹_q

=E₂(m, n, k, r, λ) Z ∞

0

¹_pZ ∞

0

.

Theorem 2.4. Let m ≥ 1, n ≥ 1, pi > 1, _p¹

i + _q¹

i = 1 for i = 0,1,2,3,4, p₀ =p, p₃ =k, p₄ =r;q₀ =q, q₃ =l, q₄ =wandf(σ)≥0, g(τ)≥0. Define F(s) = Rs

0 f(σ)dσandG(t) =Rt

0 g(τ)dτ such that 0<

Z ∞

0

s^p(1−^λ^r⁾⁻¹F_f^p(s)ds <∞, 0<

Z ∞

0

t^q(1−^w^λ⁾⁻¹G^q_g(t)dt <∞

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forσ, τ, s, t∈(0,∞). Then Z ∞

0

Z ∞

0

ln(s/t)

(s^λ−t^λ)(ls^k/p¹ +kt^l/p²)F^m(s)Gⁿ(t)dsdt≤E₃(m, n, k, r, λ)

× Z ∞

0

¹_p Z ∞

0

,

unlessf ≡0org ≡0, whereE₃(m, n, k, r, λ) = _λ2kl(sin(π/r))^π²^mn ²,

Ff(s) = Z s

0

(F^m−1(σ)f(σ))^q¹dσ _q¹

1

,

Gg(t) = Z t

0

2

.

Proof. From (2.5) and (2.6), by Hölder’s inequality, equality (2.3) and Young’s inequality: ab ≤ ^a_p^p + ^b_q^q, where a ≥ 0, b ≥ 0, p > 1,¹_p + ¹_q = 1, and using a similar method of proof to that of Theorem2.2, the result can be clearly seen.

Theorem 2.5. Let m_i ≥ 1, p_i > 1, Pn i=1

1

pi = 1, _p¹

i + _q¹

i = 1,and f_i(σ_i) ≥ 0 for σ_i ∈ (0, x_i), where x_i are positive real numbers and define F_i(t_i) = Rti

0 f_i(σ_i)dσ_i,fort_i ∈(0, x_i), i = 1,2, . . . , n. Then (2.7)

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i^mⁱ(t_i) Pn

i=1 ti

pi

dt₁dt₂· · ·dt_n

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≤

n

Y

i=1

D(m_i, x_i, p_i) Z xi

0

(x_i−t_i)

Fe_i(t_i)qi

dt_i ¹

qi ,

unlessf ≡0org ≡0, whereD(mi, xi, pi) =mix

1 pi

i ,Fei(ti) = F^mⁱ⁻¹(ti)fi(ti).

Proof. From the hypotheses, we know that (2.8) F_i^mⁱ(t_i) = m_i

Z ti

0

F_i^(mⁱ⁻¹⁾(σ_i)f_i(σ_i)dσ_i, t_i ∈(0, x_i).

Then (2.9)

n

Y

i=1

F_i^mⁱ(t_i) =

n

Y

i=1

m_i Z ti

0

F_i^mⁱ⁻¹(σ_i)f_i(σ_i)dσ_i.

Using Hölder’s inequality, we have Z ti

0

F_i^mⁱ⁻¹(σ_i)dσ_i ≤t

1 pi

i

Z ti

0

(F_i^mⁱ⁻¹(σ_i)f_i(σ_i))^qⁱdσ_i _qi¹ (2.10)

,t

1 pi

i

Z ti

0

Fe_i(σ_i)qi

dσ_i _qi¹

.

By (2.9) and (2.10) it follows that

n

Y

i=1

F_i^mⁱ(t_i)≤

n

Y

i=1

m_it

1 pi

i

Z ti

0

Fe_i(σ_i)qi

dσ_i _qi¹

.

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Applying Young’s inequality

n

Y

k=1

|ak| ≤

n

X

k=1

1

p_k|ak|^p^k,

where1< p_k <∞,Pn k=1

1

pk = 1,we observe that Qn

i=1F_i^mⁱ(t_i) Pn

i=1 ti

pi

≤

n

Y

i=1

mi

Z ti

0

Fei(σi)

qi

dσi

_qi¹ .

Integrating overt_ifrom0tox_i whereiruns from1ton, and using the Hölder’s inequality, we get

Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i^mⁱ(ti) Pn

i=1 ti

pi

≤

n

Y

i=1

m_i

"

Z xi

0

Z ti

0

Fe_i(σ_i)qi

dσ_i _qi¹

dt_i

#

≤

n

Y

i=1

m_ix

1 pi

i

Z xi

0

Z ti

0

Fe_i(σ_i)qi

dσ_i

dt_i _qi¹

=

n

Y

i=1

D(m_i, x_i, p_i) Z xi

0

(x_i −t_i)

Fe_i(t_i)qi

dt_i ¹

qi

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Remark 1. In the special case when p₁ = p₂ = · · · = p_n = n, the inequality (2.7) reduces to the following inequality,

(2.11) Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i^mⁱ(t_i) Pn

i=1t_i dt₁dt₂· · ·dt_n

≤ 1 n

n

Y

i=1

D(m_i, x_i) Z xi

0

(x_i−t_i)

Fe_i(t_i)_n−1ⁿ dt_i

ⁿ⁻¹_n

where D(m_i, x_i) = m_ix

1 n

i . Moreover, (i) (2.11) reduces to Theorem1.3 which belongs to Pachpatte for n = 2; (ii) whenm₁ = m₂ = · · · = m_n = 1, (2.11) turns into

(2.12) Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1F_i(t_i) Pn

i=1t_i dt1dt2· · ·dtn

≤ 1 n

n

Y

i=1

D(m_i, x_i) Z xi

0

(x_i−t_i)(f_i(t_i))ⁿ⁻¹ⁿ dt_i ⁿ⁻¹_n

Theorem 2.6. Let f_i, F_i be as in the above theorem, p > 1, _p¹

i + _q¹

i = 1, let p_i(σ_i) be positive function defined for σ_i ∈ (0, x_i), and define P_i(t_i) = Rti

0 p_i(σ_i)dσ_i for t_i ∈ (0, x_i),where x_i are positive real numbers. Let φ_i be real-valued, nonnegative, convex, and sub-multiplicative function defined on R₊= [0,∞),i= 1,2, . . . , n.Then

(2.13) Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1φi(Fi(ti)) Pn

i=1 ti

pi

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≤

n

Y

i=1

L(x_i, p_i) Z xi

0

(x_i−t_i)

p_i(t_i)φ_i

f_i(t_i) pi(ti)

qi

dt_i _qi¹

,

where

L(x_i, p_i) = Z xi

0

φ_i(P_i(t_i)) P_i(t_i)

pi

dt_i _pi¹

.

Proof. Applying Jensen’s inequality and Hölder’s inequality, it is clear to ob- serve that

φ_i(F_i(t_i)) =φ_i P_i(t_i)Rti

0 p_i(σ_i)^f_pⁱ^(σⁱ⁾

i(σi)dσ_i Rti

0 p_i(σ_i)dσ_i

!

≤ φ_i(P_i(t_i)) P_i(t_i)

Z ti

0

pi(σi)φi

f_i(σ_i) p_i(σ_i)

dσi

≤ φ_i(P_i(t_i)) P_i(t_i) t

1 pi

i

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) p_i(σ_i)

qi

dσ_i _qi¹

.

Using Young’s inequality, we obtain that Qn

i=1φ_i(F_i(t_i)) Pn

i=1 ti

pi

≤

n

Y

i=1

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) p_i(σ_i)

qi

dσ_i _qi¹

.

Integrating both sides of the above inequality overtifrom0toxi withirunning

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fromiton, and applying the Hölder inequality, we get Z x1

0

Z x2

0

· · · Z xn

0

Qn

i=1φ_i(F_i(t_i)) Pn

i=1 ti

pi

≤

n

Y

i=1

"

Z xi

0

Z ti

0

pi(σi)φi

f_i(σ_i) p_i(σ_i)

qi

dσi

_qi¹ dti

#

≤

n

Y

i=1

Z xi

0

pi

dt_i

_pi¹ Z xi

0

Z ti

0

p_i(σ_i)φ_i

f_i(σ_i) p_i(σ_i)

qi

dσ_i

dt_i _qi¹

=

n

Y

i=1

Z xi

0

pi

dt_i

_pi¹ Z xi

0

(x_i−t_i)

p_i(t_i)φ_i

f_i(t_i) p_i(t_i)

qi

dt_i _qi¹

=

n

Y

i=1

L(x_i, p_i) Z xi

0

(x_i−t_i)

p_i(t_i)φ_i

fi(ti) p_i(t_i)

qi

dt_i _qi¹

This completes the theorem.

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References

[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cam- bridge Univ. Press, Cambridge, 1952.

[2] B.G. PACHPATTE, On some new inequalities similar to Hilbert’s inequal- ity, J. Math. Anal. Appl., 226(1) (1998), 166–179.

[3] G.S. DAVIES AND G.M. PETERSEN, On an inequality of Hardy’s, (II), Quart. J. Math. Oxford Ser., 15(2) (1964), 35–40.

[4] D.S. MITRINOVI ´C, Analytic inequalities. In cooperation with P. M. Vasi´c.

Die Grundlehren der mathematischen Wisenschaften, Band 1965, Springer- Verlag, New York-Berlin, 1970.

[5] D.S. MITRINOVI ´C AND J.E. PE ˇCARI ´C, On inequalities of Hilbert and Widder, Proc. Edinburgh Math. Soc. (2), 34(3) (1991), 411–414.

[6] Z. LÜ, Some new inequalities similar to Hilbert-Pachpatte type inequalities, J. Inequal. Pure Appl. Math., 4(2) (2003), Art. 33. [ONLINE: http://

jipam.vu.edu.au/article.php?sid=271].

[7] Y.H. KIM, Some new inverse-type Hilbert-Pachpatte integral inequalities, Acta Math. Sin. (Engl. Ser.), 20(1) (2004), 57–62.

[8] B. YANG, On an extension of Hilbert’s integral inequality with some parameters, Aust. J. Math. Anal. Appl., 1(1) (2004), Art. 11. [ONLINE:

http://ajmaa.org].

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[9] E.F. BECKENBACH AND R. BELLMAN, Inequalities. Second revised printing. Ergebnisse der Mathematik und ihrer Grenzgebiete. Neue Folge, Band 30, Springer-Verlag, New York, Inc. 1965.