Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 154, 2006
ON SEVERAL NEW INEQUALITIES CLOSE TO HILBERT-PACHPATTE’S
INEQUALITY
BING HE AND YONGJIN LI
D EPARTMENT OF M ATHEMATICS
S UN YAT- SEN U NIVERSITY
G UANGZHOU , 510275 P. R. C HINA
hzs314@163.com
stslyj@mail.sysu.edu.cn
Received 27 April, 2006; accepted 09 October, 2006
Communicated by B. Yang
A BSTRACT. 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.
Key words and phrases: Hilbert’s inequality, Hölder’s inequality, Jensen’s inequality, Power mean inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
The well known Hardy-Hilbert’s inequality is (see [1]):
P
P∞ p0
p
Theorem 1.1. If p > 1, p0 = p/(p − 1) and ∞
m=1 am ≤ A,
n=1 bn ≤ B, then
∞ X
∞
X
1
1
am bn
π
(A) p (B) p0 ,
(1.1)
<
m+n
sin πp
n=1 m=1
unless the sequence {am } or {bn } is null.
The integral analogue can be stated as follows:
R∞
R∞ 0
Theorem 1.2. If p > 1, p0 = p/(p − 1) and 0 f p (x)dx ≤ F, 0 g p (y)dy ≤ G, then
Z ∞Z ∞
1
1
f (x)g(y)
π
F p G p0
(1.2)
dxdy <
x+y
0
0
sin πp
unless f ≡ 0 or g ≡ 0.
The following two theorems were studied by Pachpatte (see [2])
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
The work was partially supported by the Foundation of Sun Yat-sen University Advanced Research Centre.
124-06
2
B ING H E AND YONGJIN L I
Theorem 1.3. Let p ≥ 1, q ≥ 1 and f (σ) ≥ 0, Rg(τ ) ≥ 0 for σ ∈ R(0, x), τ ∈ (0, y), where
s
t
x, y are positive real numbers and define F (s) = 0 f (σ)dσ, G(t) = 0 g(τ )dτ , for s ∈ (0, x),
t ∈ (0, y). Then
Z x
21
Z xZ y p
F (s)Gq (t)
p−1
2
dsdt ≤ D(p, q, x, y)
(x − s)(F (s)f (s)) ds
s+t
0
0
0
Z y
12
q−1
2
×
(y − t)(G (t)g(t)) dt ,
0
unless f ≡ 0 or g ≡ 0, where
1 √
D(p, q, x, y) = pq xy.
2
Theorem 1.4. Let f, g, F, G be as in the above theorem, let p(σ)
R s and q(τ ) be twoR tpositive
functions defined for σ ∈ (0, x), τ ∈ (0, y) and define P (s) = 0 p(σ)dσ, Q(t) = 0 q(τ )dτ
for s ∈ (0, x), t ∈ (0, y), where x, y are positive real numbers. Let φ and ψ be real-valued,
nonnegative, convex, and sub-multiplicative functions defined on R+ = [0, ∞). Then
(Z
2 ) 12
Z xZ y
x
f (s)
φ(F (s))ψ(G(t))
(1.3)
dsdt ≤ L(x, y)
(x − s) p(s)φ
ds
s+t
p(s)
0
0
0
(Z
2 ) 12
y
g(t)
×
(y − t) q(t)ψ
dt
q(t)
0
where
x
Z
1
L(x, y) =
2
0
φ(P (s))
P (s)
! 12
2
Z
y
ds
0
ψ(Q(t))
Q(t)
2
! 12
dt
.
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.
2. M AIN R ESULTS
Lemma 2.1. Suppose p > 1, p1 + 1q = 1, r > 1, 1r + w1 = 1, λ > 0, define the weight functions
ω
e1 (w, p, x), ω
e2 (w, p, x), ω
e3 (w, p, x) as
Z ∞
λ
1
x(p−1)(1− r )
dy,
x ∈ (0, ∞),
(2.1)
ω
e1 (w, p, x) :=
·
λ
xλ + y λ
y 1− w
0
∞
Z
(2.2)
ω
e2 (w, p, x) :=
0
Z
(2.3)
ω
e3 (w, p, x) :=
λ
1
x(p−1)(1− r )
dy,
·
λ
max {xλ , y λ }
y 1− w
∞
0
ω
e1 (w, p, x) =
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
λ
ln x/y x(p−1)(1− r )
dy,
·
λ
xλ − y λ
y 1− w
Then
x ∈ (0, ∞),
π
λ sin
x ∈ (0, ∞).
λ
π
r
xp(1− r )−1 ,
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O N I NEQUALITIES C LOSE TO H ILBERT-PACHPATTE ’ S I NEQUALITY
3
rw p(1− λ )−1
r
x
,
λ
#2
"
λ
π
xp(1− r )−1 .
ω
e3 (w, p, x) =
π
λ sin r
ω
e2 (w, p, x) =
Proof. For fixed x, let u = y λ /xλ , then (2.1) turns into
Z
1
1 p(1− λ )−1 ∞ 1
r
u−1+ w du
ω
e1 (w, p, x) = x
λ
1+u
0
1 1
1 p(1− λ )−1
r
= x
B
,
λ
w r
λ
π
xp(1− r )−1
=
λ sin πp
and similarly, we can prove the others.
Theorem 2.2. Let m ≥ 1, n ≥ 1, pi > 1, p1i + q1i = 1 for i = 0, 1, 2, 3, 4, p0 = p, p3 = k,
Rs
p4 = r; q0 = q, q3 = l, q4 = w and f (σ) ≥ 0, g(τ ) ≥ 0. Define F (s) = 0 f (σ)dσ and
Rt
G(t) = 0 g(τ )dτ such that
Z ∞
Z ∞
λ
p(1− λ
)−1 p
r
0<
s
Ff (s)ds < ∞,
0<
tq(1− w )−1 Gqg (t)dt < ∞
0
0
for σ, τ, s, t ∈ (0, ∞). Then
Z ∞Z ∞
F m (s)Gn (t)
(2.4)
dsdt
(lsk/p1 + ktl/p2 )(sλ + tλ )
0
0
p1 Z
Z ∞
p
p(1− λ
)−1
r
Ff (s)ds
≤ E1 (m, n, k, r, λ)
s
unless f ≡ 0 or g ≡ 0, where E1 (m, n, k, r, λ) =
Ff (s) =
(F
t
1q
Gqg (t)dt
πmn
,
λkl sin(π/r)
q1
s
m−1
λ
q(1− w
)−1
0
0
Z
∞
q1
1
(σ)f (σ)) dσ
and
0
t
Z
(Gn−1 (τ )g(τ ))q2 dτ
Gg (t) =
q1
2
.
0
Proof. From the hypotheses, we can easily observe that
Z s
m
(2.5)
F (s) = m
F m−1 (σ)f (σ)dσ,
s ∈ (0, ∞),
0
(2.6)
n
Z
G (t) = n
t
Gn−1 (τ )g(τ )dτ,
t ∈ (0, ∞).
0
From (2.5) and (2.6), applying Hölder’s inequality, equality (2.1) and Young’s inequality:
ab ≤
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
ap bq
+ ,
p
q
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4
B ING H E AND YONGJIN L I
where a ≥ 0, b ≥ 0, p > 1,
1
p
+
1
q
= 1, we obtain
F m (s)Gn (t)
Z t
Z s
m−1
n−1
= mn
F
(σ)f (σ)dσ
G (τ )g(τ )dτ
0
≤ mns
0
1/p1 1/p2
Z
s
t
(F
m−1
q1 Z t
q1
2
1
n−1
q2
(σ)f (σ)) dσ)
(G (τ )g(τ )) dτ )
q1
0
≤ mn
sk/p1
k
0
Z s
q1 Z t
q1
2
1
tl/p2
+
(F m−1 (σ)f (σ))q1 dσ)
(Gn−1 (τ )g(τ ))q2 dτ )
.
l
0
0
Note that
Z
Ff (s) =
q1
s
(F
m−1
q1
(σ)f (σ)) dσ
1
Z
,
Gg (t) =
(G
n−1
q2
(τ )g(τ )) dτ
q1
2
.
0
0
Hence
Z ∞Z
t
∞
F m (s)Gn (t)
dsdt
(lsk/p1 + ktl/p2 )(sλ + tλ )
0
0
n
o1
1
Z ∞ Z ∞ R s (F m−1 (σ)f (σ))q1 dσ) q1 R t (Gn−1 (τ )g(τ ))q2 dτ ) q2
0
0
mn
≤
dsdt
λ
λ
kl 0
s +t
0
#"
#
"
Z Z
λ
λ
t(1− w )/p
1
s(1− r )/q
mn ∞ ∞
Gg (t)
dsdt
Ff (s)
=
λ
λ
kl 0
sλ + tλ
t(1− w )/p
s(1− r )/q
0
(Z Z
) p1 (Z Z
) 1q
λ
∞
∞ F p (s)
∞
∞ Gq (t)
(p−1)(1− λ
)
(q−1)(1− w
)
r
mn
s
t
f
g
≤
·
dsdt
·
dsdt
λ
λ
λ + tλ
λ + tλ
(1− w
)
kl
s
s
t
s(1− r )
0
0
0
0
Z ∞
p1 Z ∞
1q
mn
p
q
=
ω
e1 (w, p, s)Ff (s)ds
ω
e1 (r, q, t)Gg (t)dt
kl
0
0
p1 Z ∞
1q
Z ∞
λ
πmn
p
)−1
q(1−
)−1
q
p(1− λ
r
w
t
=
s
Ff (s)ds
Gg (t)dt
λkl sin(π/r)
0
0
1q
p1 Z ∞
Z ∞
λ
)−1 q
p(1− λ
)−1 p
q(1− w
r
t
Gg (t)dt .
= E1 (m, n, k, r, λ)
s
Ff (s)ds
0
0
This completes the proof.
Theorem 2.3. Let m ≥ 1, n ≥ 1, pi > 1, p1i + q1i = 1 for i = 0, 1, 2, 3, 4, p0 = p, p3 = k,
Rs
p4 = r; q0 = q, q3 = l, q4 = w and f (σ) ≥ 0, g(τ ) ≥ 0. Define F (s) = 0 f (σ)dσ and
R∞
Rt
R∞
λ
λ
G(t) = 0 g(τ )dτ such that 0 < 0 sp(1− r )−1 Ffp (s)ds < ∞, 0 < 0 tq(1− w )−1 Gqg (t)dt < ∞
for σ, τ, s, t ∈ (0, ∞). Then
Z ∞Z ∞
F m (s)Gn (t)
dsdt
(lsk/p1 + ktl/p2 ) max{sλ , tλ }
0
0
1q
Z ∞
p1 Z ∞
λ
)−1 q
p(1− λ
)−1 p
q(1− w
r
Gg (t)dt ,
Ff (s)ds
t
≤ E2 (m, n, k, r, λ)
s
0
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
0
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O N I NEQUALITIES C LOSE TO H ILBERT-PACHPATTE ’ S I NEQUALITY
unless f ≡ 0 or g ≡ 0, where E2 (m, n, k, r, λ) =
mnrw
λkl
and
q1
s
Z
Ff (s) =
(F
m−1
(G
n−1
5
q1
1
(σ)f (σ)) dσ
,
0
t
Z
Gg (t) =
q2
q1
2
(τ )g(τ )) dτ
.
0
Proof. By (2.5) and (2.6), using Hölder’s inequality, equality (2.2) and Young’s inequality:
p
q
ab ≤ ap + bq , where a ≥ 0, b ≥ 0, p > 1, p1 + 1q = 1. We obtain
F m (s)Gn (t)
Z s
Z t
m−1
n−1
= mn
F
(σ)f (σ)dσ
G (τ )g(τ )dτ
0
≤ mns
0
1/p1 1/p2
Z
s
t
(F
m−1
q1 Z t
q1
2
1
n−1
q2
(σ)f (σ)) dσ)
(G (τ )g(τ )) dτ )
q1
0
≤ mn
sk/p1
k
0
Z s
q1 Z t
q1
2
1
tl/p2
+
(F m−1 (σ)f (σ))q1 dσ)
(Gn−1 (τ )g(τ ))q2 dτ )
.
l
0
0
Note
Z
q1
s
Ff (s) =
(F
m−1
(G
n−1
q1
1
(σ)f (σ)) dσ
,
0
Z
Gg (t) =
t
q2
(τ )g(τ )) dτ
q1
2
.
0
Hence
Z ∞Z
∞
F m (s)Gn (t)
dsdt
(lsk/p1 + ktl/p2 ) max{sλ , tλ }
0
0
n
o1
1
Z ∞ Z ∞ R s (F m−1 (σ)f (σ))q1 dσ) q1 R t (Gn−1 (τ )g(τ ))q2 dτ ) q2
0
0
mn
≤
dsdt
λ
kl 0
max{s , tλ }
0
#
#"
"
Z Z
λ
λ
t(1− w )/p
mn ∞ ∞
1
s(1− r )/q
dsdt
Gg (t)
=
Ff (s)
λ
λ
kl 0
max{sλ , tλ }
s(1− r )/q
t(1− w )/p
0
) 1q
) p1 (Z Z
(Z Z
λ
λ
∞
∞
∞
∞
Ffp (s) s(p−1)(1− r )
Gqg (t)
mn
t(q−1)(1− w )
dsdt
dsdt
≤
max{sλ , tλ } s(1− λr )
kl
max{sλ , tλ } t(1− wλ )
0
0
0
0
Z ∞
p1 Z ∞
1q
mn
p
q
=
ω
e2 (w, p, s)Ff (s)ds
ω
e2 (r, q, t)Gg (t)dt
kl
0
0
1q
p1 Z ∞
Z ∞
λ
mnrw
p
)−1
q
)−1
q(1−
p(1− λ
w
r
Gg (t)dt
Ff (s)ds
t
=
s
λkl
0
0
Z ∞
p1 Z ∞
1q
λ
λ
p
tq(1− w )−1 Gqg (t)dt .
= E2 (m, n, k, r, λ)
sp(1− r )−1 Ff (s)ds
0
This completes the proof.
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
0
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6
B ING H E AND YONGJIN L I
Theorem 2.4. Let m ≥ 1, n ≥ 1, pi > 1, p1i + q1i = 1 for i = 0, 1, 2, 3, 4, p0 = p, p3 = k,
Rs
p4 = r; q0 = q, q3 = l, q4 = w and f (σ) ≥ 0, g(τ ) ≥ 0. Define F (s) = 0 f (σ)dσ and
Rt
G(t) = 0 g(τ )dτ such that
Z ∞
λ
0<
sp(1− r )−1 Ffp (s)ds < ∞,
Z0 ∞
λ
0<
tq(1− w )−1 Gqg (t)dt < ∞
0
for σ, τ, s, t ∈ (0, ∞). Then
Z
∞
0
Z
0
∞
ln(s/t)
(sλ
−
tλ )(lsk/p1
F m (s)Gn (t)dsdt ≤ E3 (m, n, k, r, λ)
+ ktl/p2 )
Z ∞
p1 Z ∞
1q
λ
p
p(1− λ
)−1
)−1
q
q(1−
r
w
×
s
Ff (s)ds
Gg (t)dt ,
t
0
0
unless f ≡ 0 or g ≡ 0, where E3 (m, n, k, r, λ) =
Z
π 2 mn
,
λ2 kl(sin(π/r))2
q1
s
Ff (s) =
(F
m−1
(G
n−1
q1
1
(σ)f (σ)) dσ
,
0
Z
q1
t
Gg (t) =
2
q2
(τ )g(τ )) dτ
.
0
Proof. From (2.5) and (2.6), by Hölder’s inequality, equality (2.3) and Young’s inequality: ab ≤
q
ap
+ bq , where a ≥ 0, b ≥ 0, p > 1, p1 + 1q = 1, and using a similar method of proof to that of
p
Theorem 2.2, the result can be clearly seen.
P
Theorem 2.5. Let mi ≥ 1, pi > 1, ni=1 p1i = 1, p1i + q1i = 1, and fi (σi ) ≥ 0 for σi ∈
Rt
(0, xi ), where xi are positive real numbers and define Fi (ti ) = 0 i fi (σi )dσi , for ti ∈ (0, xi ),
i = 1, 2, . . . , n. Then
Z x1 Z x2
Z xn Q n
mi
i=1 Fi (ti )
Pn ti dt1 dt2 · · · dtn
(2.7)
···
0
0
0
i=1 pi
≤
n
Y
Z
D(mi , xi , pi )
xi
qi q1i
e
(xi − ti ) Fi (ti ) dti
,
0
i=1
1
p
unless f ≡ 0 or g ≡ 0, where D(mi , xi , pi ) = mi xi i , Fei (ti ) = F mi −1 (ti )fi (ti ).
Proof. From the hypotheses, we know that
Z ti
(m −1)
mi
(2.8)
Fi (ti ) = mi
Fi i (σi )fi (σi )dσi ,
ti ∈ (0, xi ).
0
Then
(2.9)
n
Y
Fimi (ti )
i=1
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
=
n
Y
i=1
Z
mi
ti
Fimi −1 (σi )fi (σi )dσi .
0
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O N I NEQUALITIES C LOSE TO H ILBERT-PACHPATTE ’ S I NEQUALITY
7
Using Hölder’s inequality, we have
ti
Z
Fimi −1 (σi )dσi
(2.10)
1
pi
Z
1
pi
Z
ti
≤ ti
0
(Fimi −1 (σi )fi (σi ))qi dσi
q1
i
0
ti
, ti
q1
qi
i
e
Fi (σi ) dσi
.
0
By (2.9) and (2.10) it follows that
n
Y
Fimi (ti )
≤
i=1
n
Y
ti
Z
1
pi
mi ti
q1
qi
i
e
Fi (σi ) dσi
.
0
i=1
Applying Young’s inequality
n
Y
n
X
1
|ak |pk ,
p
k
k=1
|ak | ≤
k=1
where 1 < pk < ∞,
Pn
1
k=1 pk
= 1, we observe that
Qn
Z
n
mi
Y
i=1 Fi (ti )
Pn ti
≤
mi
i=1 pi
ti
q1
qi
i
.
Fei (σi ) dσi
0
i=1
Integrating over ti from 0 to xi where i runs from 1 to n, and using the Hölder’s inequality, we
get
Z x1 Z x2
Z xn Q n
mi
i=1 Fi (ti )
Pn ti dt1 dt2 · · · dtn
···
0
0
0
i=1 pi
≤
n
Y
"Z
≤
i=1
=
n
Y
i=1
ti
Z
mi
0
i=1
n
Y
xi
1
pi
q1
qi
i
Fei (σi ) dσi
dti
#
0
Z
xi
ti
Z
mi xi
0
q1
qi
i
Fei (σi ) dσi dti
0
Z
D(mi , xi , pi )
xi
qi q1i
e
(xi − ti ) Fi (ti ) dti
0
This completes the proof.
Remark 2.6. In the special case when p1 = p2 = · · · = pn = n, the inequality (2.7) reduces to
the following inequality,
Z
x1
Z
x2
Z
···
(2.11)
0
0
0
xn
Qn
F mi (ti )
i=1
Pn i
dt1 dt2 · · · dtn
i=1 ti
n−1
Z xi
n
n
n−1
n
1Y
e
≤
dti
D(mi , xi )
(xi − ti ) Fi (ti )
n i=1
0
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
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8
B ING H E AND YONGJIN L I
1
where D(mi , xi ) = mi xin . Moreover, (i) (2.11) reduces to Theorem 1.3 which belongs to
Pachpatte for n = 2; (ii) when m1 = m2 = · · · = mn = 1, (2.11) turns into
x1
Z
Z
x2
Z
xn
···
(2.12)
0
0
0
Qn
i=1 Fi (ti )
P
dt1 dt2 · · · dtn
n
i=1 ti
n−1
Z xi
n
n
n
1Y
≤
D(mi , xi )
(xi − ti )(fi (ti )) n−1 dti
n i=1
0
Theorem 2.7. Let fi , Fi be as in the above theorem, p > 1, p1i + q1i = 1, let pi (σi ) be positive
Rt
function defined for σi ∈ (0, xi ), and define Pi (ti ) = 0 i pi (σi )dσi for ti ∈ (0, xi ),where xi
are positive real numbers. Let φi be real-valued, nonnegative, convex, and sub-multiplicative
function defined on R+ = [0, ∞), i = 1, 2, . . . , n. Then
x1
Z
Z
x2
Z
xn
Qn
0
0
0
≤
φ (F (t ))
Pni tii i dt1 dt2 · · · dtn
i=1
···
(2.13)
i=1 pi
n
Y
Z
L(xi , pi )
0
i=1
xi
qi q1
i
fi (ti )
(xi − ti ) pi (ti )φi
dti
,
pi (ti )
where
Z
xi
L(xi , pi ) =
0
φi (Pi (ti ))
Pi (ti )
p1
pi
dti
i
.
Proof. Applying Jensen’s inequality and Hölder’s inequality, it is clear to observe that
!
Rt
i)
Pi (ti ) 0 i pi (σi ) pfii (σ
dσi
(σi )
φi (Fi (ti )) = φi
R ti
pi (σi )dσi
0
Z ti
φi (Pi (ti ))
fi (σi )
≤
pi (σi )φi
dσi
Pi (ti )
pi (σi )
0
Z ti qi
q1
i
φi (Pi (ti )) p1i
fi (σi )
≤
ti
pi (σi )φi
dσi
.
Pi (ti )
pi (σi )
0
Using Young’s inequality, we obtain that
Qn
n
Y φi (Pi (ti ))
i=1 φi (Fi (ti ))
Pn ti
≤
Pi (ti )
i=1 pi
i=1
J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
Z
0
ti
qi
q1
i
fi (σi )
pi (σi )φi
dσi
.
pi (σi )
http://jipam.vu.edu.au/
O N I NEQUALITIES C LOSE TO H ILBERT-PACHPATTE ’ S I NEQUALITY
9
Integrating both sides of the above inequality over ti from 0 to xi with i running from i to n,
and applying the Hölder inequality, we get
Z x1 Z x2
Z xn Q n
i=1 φi (Fi (ti ))
Pn ti
···
dt1 dt2 · · · dtn
0
0
0
i=1 pi
#
qi
q1
ti i
φi (Pi (ti ))
fi (σi )
≤
pi (σi )φi
dσi
dti
P
(t
)
p
(σ
)
i
i
i
i
0
0
i=1
qi
q1
p
p1 Z xi Z ti n Z xi Y
i
i
fi (σi )
φi (Pi (ti )) i
dσi dti
≤
dti
pi (σi )φi
pi (σi )
Pi (ti )
0
0
0
i=1
1
p
p Z xi
qi q1
n Z xi Y
i
i
φi (Pi (ti )) i
fi (ti )
=
(xi − ti ) pi (ti )φi
dti
dti
Pi (ti )
pi (ti )
0
0
i=1
Z xi
qi q1
n
Y
i
fi (ti )
dti
=
L(xi , pi )
(xi − ti ) pi (ti )φi
pi (ti )
0
i=1
n
Y
"Z
xi
Z
This completes the theorem.
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271].
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J. Inequal. Pure and Appl. Math., 7(4) Art. 154, 2006
http://jipam.vu.edu.au/