New Inequalities Similar to Hardy-Hilbert Inequality ...

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 90641, 12 pages doi:10.1155/2007/90641

Research Article New Inequalities Similar to Hardy-Hilbert Inequality and their Applications Lü Zhongxue and Xie Hongzheng Received 25 January 2007; Revised 7 July 2007; Accepted 22 November 2007 Recommended by Lars-Erik Persson

Two classes of new inequalities similar to Hardy-Hilbert inequality are showed by introducing some parameters a,b,c and two real functions φ(x) and ψ(x). Some applications are obtained. Copyright © 2007 L. Zhongxue and X. Hongzheng. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction The following inequality is well known as Hardy-Hilbert inequality:

∞ ∞ am bn

∞ π p ≤ an m + n sin(π/ p) m=1 n=1 n =1

1/ p

∞

1/q

q bn n =1

,

(1.1)

where π/sin(π/ p) is the best value (see Hardy et al. [1]). Integral analogues of (1.1) are the following inequalities: ∞ 0

f (x)g(y) dx d y ≤ π x+y ∞ ∞ 0

0

f (x) dx x+y

∞

0

f 2 (x)dx

2

d y ≤ π2

∞

∞ 0

0

1/2

g 2 (y)d y

, (1.2)

f 2 (x)dx,

where π is the best value (cf., [1, Chapter 9]). In recent years, Gao [2], Yang [3–5], Yang and Debnath [6], Kuang [7], and Kuang and Debnath [8] gave some distinct improvements and generalizations of (1.1)-(1.2).

2

Journal of Inequalities and Applications

Yang and Rassias [9] gave a new inequality with a best constant factor similar to (1.1) as

∞ ∞ am bn

∞ π p n p −1 a n < ln mn sin(π/ p) m=2 n=2 n =2

1/ p

1/q

∞

q nq−1 bn n =2

,

(1.3)

where π/sin(π/ p) is the best possible. In this paper, we have two major objectives. One is motivated by [10], to give a generalization of (1.3) by introducing two real functions φ(x) and ψ(x). The other is to build a class of new inequalities similar to Hardy-Hilbert inequality (1.2) by introducing some parameters a, b, and c. 2. Some lemmas First, we give the β function B(m,n):

B

1 1 , = p q

∞ 0

1/q

1 1 1+u u

du,

(2.1)

where p > 1, 1/ p + 1/q = 1. Lemma 2.1. Let b > a ≥ 1 − c, and ω(a,b,x) =

b a

1 ln(x + c) (y + c) ln (x + c)(y + c) ln(y + c)

1/2

d y,

(2.2)

provided the generalized integral exists. Then

ω(a,b,x) ≤ π − 4arctan

4

ln(a + c) ; ln(b + c)

ω(0,b,x) = lim ω(a,b,x) ≤ π − 4arctan

4

a→0

(2.3)

lnc ; ln(b + c)

(2.4)

ln(a + c) . ln(x + c)

(2.5)

ω(a, ∞,x) = lim ω(a,b,x) ≤ π − 2arctan b→∞

Proof. Putting u = ln (y + c)/ ln (x + c), we have ω(a,b,x) = =

b a

1 ln(x + c) (y + c)ln(x + c)(y + c) ln (y + c)

∞ 0

1/2

1 1 1+u u

=π−

du −

ln(x+c)/ ln(b+c) 0

= π − 2arctan

1/2

dy =

∞

ln(b+c)/ ln(x+c) ln(a+c)/ ln(x+c)

1/2

1 1 ln(b+c)/ ln(x+c) 1+u u

ln(a+c)/ ln(x+c) 0

ln(a+c)/ ln(x+c)

1/2

1 1 1+v v

du −

dv +

ln (x + c) + 2arctan ln (b + c)

0

1/2

1 1 1+u u

1/2

1 1 1+u u

du

1/2

1 1 1+u u

du

du

ln(a + c) . ln(x + c) (2.6)

L. Zhongxue and X. Hongzheng 3 Since arctanx is strictly increasing, then

ln (x + c)/ ln (b + c) + ln(a + c)/ ln(x + c) ω(a,b,x) = π − 2arctan 1 − ln(a + c)/ ln(b + c)

2 4 ln (a + c)/ ln(b + c) ≤ π − 2arctan = π − 4arctan 1 − ln (a + c)/ ln (b + c)

4

(2.7) ln(a + c) . ln(b + c)

Relation (2.3) is valid. By (2.3) as a→0, we have

ω(0,b,x) = lim ω(a,b,x) ≤ π − 4arctan

4

a→0

lnc . ln(b + c)

(2.8)

Relation (2.4) is valid. Similarly, (2.5) is also valid. The lemma is proved.

Lemma 2.2. Let 0 < α < 1, 0 ≤ c < 1, g(s) ∈ C 1 [c,1], g(s) > 0, g (s) > 0 for all s ∈ [c,1], x and F(x) = c (s−α /g(s))ds for all x ∈ [c,1]. Then F(x) ≥

x1−α − c1−α F(1). 1 − c1−α

(2.9)

Proof. Let τ = s1−α , then F(x) =

x c

s−α 1 ds = g(s) 1−α

x1−α c1−α

g

1

τ 1/(1−α)

dτ.

(2.10)

y

Let G(y) = (1/1 − α) c1−α (1/g(τ 11−α ))dτ. Since G (y) > 0, G (x) ≤ 0 in [c1−α ,1], and G(y) is concave in [c1−α ,1], then

G(y) = G

1 − y 1−α y − c1−α c + = G (1 − λ)c1−α + λ 1 − c1−α 1 − c1−α

λ=

y − c1−α 1 − c1−α

1−α y − c1−α 1− y y − c1−α ≥ G c + G(1) = G(1). 1 − c1−α 1 − c1−α 1 − c1−α

(2.11)

Thus

F(x) = G x1−α ≥

x1−α − c1−α F(1). 1 − c1−α

(2.12)

The lemma is proved. Let F1,r (x) =

ln(a+c)/ ln(x+c) 0

u−1/r du, 1+u

F2,r (x) =

ln(x+c)/ ln(b+c) 0

where r > 1, 1 − c ≤ a ≤ x ≤ b. If g(s) = 1 + s and α = 1/r in Lemma 2.2, we get the following.

u−1/r du, 1+u

(2.13)

4


Lemma 2.3. Let 1 − c < a ≤ x ≤ b < +∞, p > 1, 1/ p + 1/q = 1. Then

F1,q (x) + F2,p (x) ≥

ln (a + c) ln (x + c)

ln(a + c) ≥ ln (b + c)

F1,p (x) + F2,q (x) ≥

ln (a + c) ln (x + c)

ln(a + c) ≥ ln (b + c) where Φ(r) =

1

0 (u

−1/r

1/ p

Φ(q) +

1/ pq 1/q

qΦ(q)

Φ(p) +

1/ pq

1/q

qΦ(q)

1/q

ln(x + c) ln(b + c)

pΦ(p)

1/ p

1/ p

ln(x + c) ln(b + c)

1/q

Φ(p)

pΦ(p)

(2.14) ;

Φ(q)

1/ p

(2.15) ,

/1 + u)du.

Proof. For 1 − c < a ≤ x ≤ b < +∞, by Lemma 2.2, we have

F1,q (x) + F2,p (x) ≥

ln (a + c) ln (x + c)

ln (a + c) F1,p (x) + F2,q (x) ≥ ln (x + c)

1/ p 1/q

Φ(q) +

ln(x + c) ln(b + c)

ln(x + c) Φ(p) + ln(b + c)

1/q

Φ(p),

1/ p

(2.16) Φ(q).

Let α = 1/ p, β = 1/q, p1 = 1 + α/β, q1 = 1 + β/α, then β 2αβ α + = , p1 q1 α + β

1 1 + = 1, p1 q1

α + β = 1.

(2.17)

By Young inequality, we get

ln(a + c) ln(x + c)

=

1/ p

Φ(q) +

ln(a + c) ln(x + c)

α

ln(x + c) ln (b + c)

Φ(q) +

1 1/ p ln (a + c) = p 1 p1 1 ln (x + c) 1/ p p1 1

≥

ln (a + c) ln (x + c)

Φ(p)

ln (x + c) ln(b + c)

α/ p1

α/ p1

1/q

Φ(q)

Φ(q)

β

Φ(p)

1/ p1

1/ p1

p1

1 1/q1 ln(x + c) + q q1 1 ln(b + c)

1/q q1 1

ln(x + c) ln(b + c)

β/q1

β/q1

Φ(p)

1/q1

Φ(p)

1/q1

q1

β/(α+β) α/(α+β) β α/(α+β) ln (a + c) αβ/(α+β) α β/(α+β) 1+ × Φ(q) Φ(p) = 1+

β

=

ln(a + c) ln(b + c)

α

1/ pq

qΦ(q)

ln (b + c)

1/q

pΦ(p)

1/ p

. (2.18)

Then (2.14) is valid. In the same way, (2.15) can be obtained. This completes the proof.

L. Zhongxue and X. Hongzheng 5 Lemma 2.4. Let p > 1, 1/ p + 1/q = 1, φ(x) and ψ(x) are continuously differentiable functions on (a,b), φ(a) ≥ 1, φ (x) > 0, ψ(a) ≥ 1, ψ (x) > 0, inf x φ (x) = 0, and inf x ψ (x) = 0, provided that the generalized integral exists. Then b a

ln φ(x) 1 ψ(y)lnφ(x)ψ(y) lnψ(y)

1/q

dy

ln ψ(a) 1 π ≤ − ln ψ(b) inf ψ (y) sin(π/ p)

1/ pq

pΦ(p)

1/ p

qΦ(q)

1/q

(2.19)

,

where Φ is as in Lemma 2.3. Proof. Putting u = ln ψ(y)/ lnφ(x), by Lemma 2.2 and the proof of Lemma 2.3, we have b

ln φ(x) 1 ψ(y)lnφ(x)ψ(y) ln ψ(y)

a

=

lnψ(b)/ lnφ(x) lnψ(a)/ lnφ(x)

1/q

dy

1/q

1 1 1+u u

1 du ψ (y)

lnψ(a)/ lnφ(x)

≤

1 π − inf ψ (y) sin(π/ p)

≤

ln ψ(a) 1 π − ln φ(x) inf ψ (y) sin(π/ p)

≤

lnψ(a) 1 π − ln ψ(b) inf ψ (y) sin(π/ p)

0

1/q

1 1 1+u u

1/ p

1/ pq

Φ(q) −

du −

pΦ(p)

lnφ(x)/ lnψ(b)

lnφ(x) lnψ(b)

1/ p

0

1/q

qΦ(q)

1/ p

1 1 1+u u

du

Φ(p)

1/q

. (2.20)

The lemma is proved. Remark 2.5. When a = 1, and b = ∞, we get ∞ 1

lnφ(x) 1 ψ(y)lnφ(x)ψ(y) ln ψ(y)

1/q

lnψ(1) 1 π dy ≤ − lnφ(x) inf ψ (y) sin(π/ p)

≤

1/ p

Φ(q)

1 π . sin(π/ p) inf ψ (y)

(2.21)

6


3. Main results Now, we introduce main results. Theorem 3.1. Let −c ≤ a < b < +∞, f , g are integrable nonnegative functions on [a,b] b b such that 0 < a (x + c) f 2 (x)dx < ∞ and 0 < a (y + c)g 2 (y)d y < ∞. Then b

f (x)g(y) dx d y a ln(x + c)(y + c)

≤ π − 4arctan

4

ln (a + c) ln (b + c)

b a

(x + c) f (x)dx

a

(3.1)

1/2

b

2

2

(y + c)g (y)d y

.

Proof. By Cauchy-Schwarz inequality and (2.3), we have b

f (x)g(y)

a ln(x + c)ln(y + c)

=

b

a

×

≤

dx d y

f (x) ln (x + c)(y + c)

1/2

g(y) ln(x + c)(y + c)

1/2

b

ln (x + c) ln (y + c)

ln (y + c) ln (x + c)

f 2 (x) ln(x + c) ln(x + c)(y + c) ln (y + c) a

×

b

b a

2

(x + c) f (x)

×

b

a

b

a

≤ π − 4arctan

4

x+c y+c

y+c x+c

1/2

1/2

dx d y

x+c dx d y y+c

1/2

1/2

y+c dx d y x+c

1/2

1 ln(x + c) (y + c)ln(x + c)(y + c) ln(y + c)

(y + c)g 2 (y)

1/4

1/2

g 2 (y) ln(y + c) a ln(x + c)(y + c) ln (x + c)

=

1/4

b a

ln(y + c) 1 (x + c)ln(x + c)(y + c) ln(x + c)

ln(a + c) ln(b + c)

b a

2

(x + c) f (x)dx

b a

1/2

1/2

d y dx 1/2

1/2

dx d y 1/2 2

(y + c)g (y)d y

. (3.2)

Then relation (3.1) is valid. Theorem 3.1 is proved. In a similar way to the proof of Theorem 3.1, we can prove the following theorem.

L. Zhongxue and X. Hongzheng 7 Theorem 3.2. Let 1 − c ≤ a < b < +∞, f is an integrable nonnegative function on [a,b], b such that 0 < a (x + c) f 2 (x)dx < ∞, then b b a

a

f (x) dx ln(x + c)(y + c)

2

d y ≤ π − 4arctan

4

ln(a + c) ln(b + c)

2

b

a

(x + c) f 2 (x)dx. (3.3)

Remark 3.3. Specially, when a = 0, c = 1, and b = ∞ in Theorems 3.1 and 3.2, we get ∞ 0

f (x)g(y) dx d y ≤ π ln(x + 1)(y + 1) ∞ ∞ 0

0

1/2

∞

0

2

(x + 1) f (x)dx

f (x) dx ln(x + 1)(y + 1)

2

d y ≤ π2

0

∞ 0

1/2

∞

2

(y + 1)g (y)d y

;

(x + 1) f 2 (x)dx. (3.4)

Theorem 3.4. Let p > 1, 1/ p + 1/q = 1, f , g are integrable nonnegative functions on [a,b], b b such that 0 < a φ p−1 (x) f p (x)dx < ∞ and 0 < a ψ q−1 (y)g q (y)d y < ∞. Then 1/ p

b

π/sin(π/ p) − φ1 π/sin(π/ p) − φ2 f (x)g(y) dx d y ≤

1/ p 1/q lnφ(x)ψ(y) a inf ψ (y) inf φ (x) ×

b a

1 ψ(y)

b a

b a

f (x) dx ln φ(x)ψ(y)

φ p−1 (x) f p (x)dx

p

dy ≤

b a

π/sin(π/ p) − φ1

×

1/ p

b a

inf ψ (y)

1/q

(3.5)

1/q

ψ q−1 (y)g q (y)d y

;

1/ p

1/q p


1/ p

inf φ (x)

1/q

φ p−1 (x) f p (x)dx, (3.6)

where the φ(x) and ψ(y) are as in Lemma 2.4 (φ1 = (lnψ(a)/ lnψ(b))1/ pq (pΦ(p))1/ p × (qΦ(q))1/q , φ2 = (lnφ(a)/ lnφ(b))1/ pq (pΦ(p))1/ p (qΦ(q))1/q ). Proof. By Hölder inequality and (2.19), we have b

f (x)g(y)

a lnφ(x)ψ(y)

=

b a

×

dx d y

f (x)

1/ p

lnφ(x)ψ(y) g(y)

1/q

lnφ(x)ψ(y)

lnφ(x) ln ψ(y)

ln ψ(y) lnφ(x)

1/ pq

1/ pq

φ(x)1/q ψ(y)1/ p

ψ(y)1/ p dxd y φ(x)1/q

8

Journal of Inequalities and Applications ≤

×

f p (x) ln φ(x) a lnφ(x)ψ(y) lnψ(y)

b

b a

φ(x) p−1 dx d y ψ(y)

1/ p

1/ p

ω(φ,ψ, q,x) f p (x)dx 1

inf ψ (y) b

×

a

b

×

a

1/ p

ln φ(a) π − sin(π/ p) ln φ(b)

×

inf ψ (y) b

a

1/q

b

ω(ψ,φ, p, y)g q (y)d y

1/ pq

(pΦ(p)) 1/ pq

φ p−1 (x) f p (x)dx

1/ p

1/q

1/ p

φ

(qΦ(q))

p −1

p

(x) f (x)dx

π/sin(π/ p) − φ4 inf φ (x)

1/ p

1/q

(pΦ(p))1/ p (qΦ(q))1/q ψ q−1 (y)g q (y)d y

1/ p

1/ p

1/q

1/q


inf φ (x)

1/ p

ψ(y)q−1 dxd y φ(x)

a

ln ψ(a) π − sin(π/ p) ln ψ(b)

≤

1/q

g q (y) ln ψ(y) a lnφ(x)ψ(y) lnφ(x)

= ≤

b

1/q

1/q

b

a

1/q

ψ q−1 (y)g q (y)d y (3.7)

Hence (3.5) is valid. b p −1 Let g(y) = (1/ψ(y))( a ( f (x)/ lnφ(x)ψ(y))dx) > 0 (y ∈ (a,b)). By (4.2), we have b

0
1, 1/ p + 1/q = 1, f , g are integrable nonnega b b tive functions on [a,b], such that 0< a (x + c) p−1 f p (x)dx < ∞ and 0< a (y + c)q−1 g q (y)d y < ∞. Then b

f (x)g(y) 1 1 ln (a + c) dx d y ≤ B , − p q ln(b + c) a ln(x + c)(y + c) ×

b a

(x + c)

p −1

1/ pq

qΦ(q)

1/ p p

f (x)dx

1/q

1/q

b a

pΦ(p)

1/ p

(y + c)

q −1 q

g (y)d y

, (3.12)

where Φ is as in Lemma 2.3. In what follows, we give the associated discrete inequalities. The proofs should be omitted.

10


Theorem 3.7. Let p > 1, 1/ p + 1/q = 1, {am }, {bn } are nonnegative real sequences, such p p −1 q−1 (n)bq < ∞. Then that 0 < ∞ (n)an < ∞, 0 < ∞ n n =2 φ n =2 ψ ∞ ∞

am bn lnφ(m)ψ(n) m=2 n=2 ≤

1

inf ψ (y)

×

∞

m=2

×

∞

n =2

≤

1/ p

inf φ (x)

1/q

ln ψ(1) π − sin(π/ p) ln φ(m)

ln φ(1) π − sin(π/ p) ln ψ(n)

1/ p

1/q

inf φ (x)

1/ p p

Φ(q) φ p−1 (m)am

Φ(p) ψ

π/sin(π/ p)

inf ψ (y)

1/ p

1/q

∞

1/q q −1

q (n)bn

1/ p

φ

p −1

p (m)am

m=2

∞

1/q

ψ

q −1

n=2

q (n)bn

. (3.13)

where φ(x) and ψ(y) are as in Lemma 2.4, and Φ is as in Lemma 2.3. Theorem 3.8. Let p > 1, 1/ p + 1/q = 1, {am } is nonnegative real sequence, such that 0 < p p −1 (n)an < ∞. Then n =2 φ

∞

∞

∞ 1 am ψ(n) lnφ(m)ψ(n) n =2 m=2

≤

p

p

π/sin(π/ p)

inf ψ (y)

1/ p

inf φ (x)

1/q

∞

(3.14) φ

p −1

p (m)am ,

m=2

where φ(x) and ψ(y) are as in Lemma 2.4. Remark 3.9. When φ(x) = x and ψ(y) = y, then inequalities (3.10), (3.11), (3.13), and (3.14) change to (2.4), (2.10), (3.3), and (3.4) in [10], respectively, hence inequalities (3.10), (3.11), (3.13), and (3.14) are generalizations of related results in [10]. 4. Some corollaries By Theorems 3.4, 3.7, and 3.8, some inequalities can also be obtained. For example, we take φ(x) and ψ(y) as φ(x) = ex ,

ψ(y) = e y ,

then by Theorems 3.4, 3.7, and 3.8, we get the following corollaries.

(4.1)

L. Zhongxue and X. Hongzheng 11 Corollary 4.1. Let p > 1, 1/ p + 1/q = 1, {am }, {bn } are nonnegative real sequences, such ∞ (q−1)n q (p−1)n a p < ∞, 0 < that 0 < ∞ bn < ∞. Then n n =2 e n =2 e

∞ ∞ am bn

∞ π/sin(π/ p) p ≤ e(p−1)m am m + n e m=2 n=2 m=2 ∞ 1

n n =2

∞

am m +n m=2

p

π/sin(π/ p) e

≤

1/ p

1/q

∞

q e(q−1)n bn n =2

p ∞

; (4.2)

p

e(p−1)m am .

m=2

Corollary 4.2. Let ∞p > 1, 1/ p + 1/q = 1, f , g∞ are integrable nonnegative functions on [a,b], such that 0 < 0 e(p−1)t f p (t)dt < ∞, 0 < 0 e(q−1)t g q (t)dt < ∞. Then ∞ 1

f (x)g(y) π/sin(π/ p) dx d y ≤ x+y e ∞ 1

1 y

∞

1

f (x) dx x+y

∞

1

p

e

dy ≤

(p−1)x

1/ p

f (x)dx

π/sin(π/ p) e

1

p

∞

1

1/q

∞

p

e

(q−1)y q

g (y)d y

;

e(p−1)x f p (x)dx. (4.3)

We take φ(x) and ψ(y) as ψ(y) = e y .

φ(x) = x2 ,

(4.4)

Then we have the following corollary. Corollary 4.3. Let p > 1, 1/ p + 1/q = 1, {am }, {bn } are nonnegative real sequences, such ∞ (q−1)n q 2(p−1) a p < ∞, 0 < that 0 < ∞ bn < ∞. Then n n =2 n n =2 e

∞ ∞

∞ π/sin(π/ p) am bn p ≤ m2(p−1) am 1/q e1/ p 2lnm + n 2 m=2 n=2 m=2 ∞ 1

∞

am 2ln m+n m=2

n n =2

p

π/sin(π/ p) ≤ 21/q e1/ p

1/ p

p

1/q

∞

q e(q−1)n bn n =2 ∞

; (4.5)

p

m2(p−1) am .

m=2

Corollary 4.4. Let ∞p > 1, 1/ p + 1/q = 1, f , g ∞are integrable nonnegative functions on [a,b], such that 0 < 1 x2(p−1) f p (x)dx < ∞, 0 < 0 e(q−1)x g q (x)dx < ∞. Then ∞ 1

f (x)g(y) π/sin(π/ p) dx d y ≤ 2lnx + y 21/q e1/ p ∞ 1

1 y

∞

1

f (x) dx 2ln x + y

p

∞

1

x2(p−1) f p (x)dx

π/sin(π/ p) dy ≤ 21/q e1/ p

1/ p

p

∞

1 ∞

1

1/q

e(q−1)y g q (y)d y

;

x2(p−1) f p (x)dx. (4.6)

Remark 4.5. Inequalities (4.2)–(4.6) are also new results.

12


Acknowledgment The authors thank the referees for their help and patience in improving the paper. This work is supported by the Natural Science Foundation of China, Project no. 10771181 and the Natural Science Foundation of Jiangsu Higher Education Bureau, Project no. 07KJD110206. References ´ [1] G. H. Hardy, J. E. Littlewood, and G. Polya, Inequalities, Cambridge University Press, London, UK, 2nd edition, 1952. [2] M. Gao, “On Hilbert’s inequality and its applications,” Journal of Mathematical Analysis and Applications, vol. 212, no. 1, pp. 316–323, 1997. [3] B. Yang, “Some generalizations of the Hardy-Hilbert integral inequalities,” Acta Mathematica Sinica, vol. 41, no. 4, pp. 839–844, 1998 (Chinese). [4] B. Yang, “On Hilbert’s integral inequality,” Journal of Mathematical Analysis and Applications, vol. 220, no. 2, pp. 778–785, 1998. [5] B. Yang, “A generalized Hilbert’s integral inequality with the best const,” Chinese Annals of Mathematics, vol. 21A, no. 4, pp. 401–408, 2000. [6] B. Yang and L. Debnath, “On a new generalization of Hardy-Hilbert’s inequality and its applications,” Journal of Mathematical Analysis and Applications, vol. 245, no. 1, pp. 248–265, 2000. [7] J. Kuang, “Note on new extensions of Hilbert’s integral inequality,” Journal of Mathematical Analysis and Applications, vol. 235, no. 2, pp. 608–614, 1999. [8] J. Kuang and L. Debnath, “On new generalizations of Hilbert’s inequality and their applications,” Journal of Mathematical Analysis and Applications, vol. 245, no. 1, pp. 248–265, 2000. [9] B. Yang and T. M. Rassias, “On the way of weight coefficient and research for the Hilbert-type inequalities,” Mathematical Inequalities & Applications, vol. 6, no. 4, pp. 625–658, 2003. [10] B. Yang, “On a new inequality similar to Hardy-Hilbert’s inequality,” Mathematical Inequalities & Applications, vol. 6, no. 1, pp. 37–44, 2003. Lü Zhongxue: School of Mathematical Sciences, Xuzhou Normal University, Xuzhou, Jiangsu 221116, China Email address: [email protected] Xie Hongzheng: Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China Email address: [email protected]