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 Lă Zhongxue and Xie Hongzheng u Received 25 January 2007; Revised 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 Introduction The following inequality is well known as Hardy-Hilbert inequality: ∞ ∞ am bn π ≤ m + n sin(π/ p) m=1 n=1 1/ p ∞ 1/q ∞ p an q bn n =1 , (1.1) n =1 where π/sin(π/ p) is the best value (see Hardy et al [1]) Integral analogues of (1.1) are the following inequalities: ∞ ∞ f (x)g(y) dx d y ≤ π x+y ∞ ∞ 0 f (x) dx x+y f (x)dx ∞ 1/2 g (y)d y , (1.2) d y ≤ π2 ∞ f (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 < ln mn sin(π/ p) m=2 n=2 1/ p ∞ 1/q ∞ p n p −1 a n n =2 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 Some lemmas First, we give the β function B(m,n): 1 , = p q B ∞ 1 1+u u 1/q du, (2.1) where p > 1, 1/ p + 1/q = Lemma 2.1 Let b > a ≥ − c, and ω(a,b,x) = b a 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 ln(a + c) ; ln(b + c) ω(0,b,x) = lim ω(a,b,x) ≤ π − 4arctan 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 ln(x + c) (y + c)ln(x + c)(y + c) ln (y + c) ∞ =π− 1 1+u u 1/2 du − ln(x+c)/ ln(b+c) = π − 2arctan 1/2 dy = ∞ 1 ln(b+c)/ ln(x+c) 1+u u 1 1+v v 1/2 ln(b+c)/ ln(x+c) ln(a+c)/ ln(x+c) 1/2 du − ln(a+c)/ ln(x+c) ln(a+c)/ ln(x+c) dv + ln (x + c) + 2arctan ln (b + c) 1 1+u u 1 1+u u 1/2 1 1+u u du 1/2 du 1/2 du ln(a + c) ln(x + c) (2.6) L Zhongxue and X Hongzheng Since arctanx is strictly increasing, then ω(a,b,x) = π − 2arctan ln (x + c)/ ln (b + c) + ln(a + c)/ ln(x + c) − ln(a + c)/ ln(b + c) (2.7) ln (a + c)/ ln(b + c) ≤ π − 2arctan = π − 4arctan − ln (a + c)/ ln (b + c) 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 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 < α < 1, ≤ c < 1, g(s) ∈ C [c,1], g(s) > 0, g (s) > 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) − c1−α (2.9) Proof Let τ = s1−α , then F(x) = x c s−α ds = g(s) 1−α x1−α c1−α g 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) ≤ in [c1−α ,1], and G(y) is concave in [c1−α ,1], then G(y) = G − y 1−α y − c1−α c + = G (1 − λ)c1−α + λ − c1−α − c1−α λ= y − c1−α − c1−α 1− y y − c1−α y − c1−α ≥ G c1−α + G(1) = G(1) − c1−α − c1−α − c1−α (2.11) Thus F(x) = G x1−α ≥ x1−α − c1−α F(1) − c1−α (2.12) The lemma is proved Let F1,r (x) = ln(a+c)/ ln(x+c) u−1/r du, 1+u F2,r (x) = ln(x+c)/ ln(b+c) where r > 1, − c ≤ a ≤ x ≤ b If g(s) = + s and α = 1/r in Lemma 2.2, we get the following u−1/r du, 1+u (2.13) Journal of Inequalities and Applications Lemma 2.3 Let − c < a ≤ x ≤ b < +∞, p > 1, 1/ p + 1/q = Then ln (a + c) ln (x + c) F1,q (x) + F2,p (x) ≥ ln(a + c) ≥ ln (b + c) ln (a + c) ln (x + c) F1,p (x) + F2,q (x) ≥ ln(a + c) ≥ ln (b + c) 1/ p Φ(q) + 1/ pq qΦ(q) 1/q Φ(p) + 1/ pq qΦ(q) ln(x + c) ln(b + c) 1/q Φ(p) (2.14) pΦ(p) ln(x + c) ln(b + c) 1/q 1/q 1/ p 1/ p ; Φ(q) (2.15) pΦ(p) 1/ p , −1/r /1 + u)du (u where Φ(r) = Proof For − c < a ≤ x ≤ b < +∞, by Lemma 2.2, we have ln (a + c) ln (x + c) 1/ p ln (a + c) F1,p (x) + F2,q (x) ≥ ln (x + c) 1/q F1,q (x) + F2,p (x) ≥ ln(x + c) ln(b + c) 1/q ln(x + c) Φ(p) + ln(b + c) 1/ p Φ(q) + Φ(p), (2.16) Φ(q) Let α = 1/ p, β = 1/q, p1 = + α/β, q1 = + β/α, then β 2αβ α + = , p1 q1 α + β 1 + = 1, p1 q1 α + β = (2.17) By Young inequality, we get 1/ p ln(a + c) ln(x + c) = = ≥ ln(a + c) ln(x + c) α 1/ p1 p1 α β α/ p1 α/ p1 β/(α+β) 1+ β α Φ(p) β ln (x + c) ln(b + c) Φ(q) + ln (a + c) ln (x + c) ln(a + c) ln(b + c) 1/q ln(x + c) ln (b + c) 1/ p ln (a + c) p p1 ln (x + c) = 1+ = Φ(q) + p1 Φ(q) Φ(q) α/(α+β) 1/ pq qΦ(q) Φ(p) 1/q 1/ p1 1/ p1 + 1/q1 q1 ln (a + c) ln (b + c) pΦ(p) 1/ p 1/q1 ln(x + c) q q1 ln(b + c) ln(x + c) ln(b + c) αβ/(α+β) β/q1 × Φ(q) Φ(p) β/q1 q1 Φ(p) 1/q1 1/q1 β/(α+β) Φ(p) α/(α+β) (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 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) ψ(y)lnφ(x)ψ(y) lnψ(y) ≤ inf ψ (y) 1/q dy ln ψ(a) π − sin(π/ p) ln ψ(b) (2.19) 1/ pq pΦ(p) 1/ p qΦ(q) 1/q , 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 1/q ln φ(x) ψ(y)lnφ(x)ψ(y) ln ψ(y) a = lnψ(b)/ lnφ(x) lnψ(a)/ lnφ(x) 1 1+u u 1/q dy du ψ (y) lnψ(a)/ lnφ(x) ≤ inf ψ (y) π − sin(π/ p) inf ψ (y) ln ψ(a) π − sin(π/ p) ln φ(x) 1/ p ≤ inf ψ (y) lnψ(a) π − sin(π/ p) ln ψ(b) 1/q 1/ pq ≤ 1 1+u u Φ(q) − pΦ(p) du − lnφ(x) lnψ(b) 1/ p lnφ(x)/ lnψ(b) 1 1+u u 1/ p du 1/q qΦ(q) Φ(p) 1/q (2.20) The lemma is proved Remark 2.5 When a = 1, and b = ∞, we get ∞ lnφ(x) ψ(y)lnφ(x)ψ(y) ln ψ(y) 1/q dy ≤ ≤ inf ψ (y) lnψ(1) π − sin(π/ p) lnφ(x) 1/ p Φ(q) π sin(π/ p) inf ψ (y) (2.21) Journal of Inequalities and Applications 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 < a (x + c) f (x)dx < ∞ and < a (y + c)g (y)d y < ∞ Then b f (x)g(y) dx d y a ln(x + c)(y + c) ≤ π − 4arctan b ln (a + c) ln (b + c) a (x + c) f (x)dx 1/2 1/2 (y + c)g (y)d y a (3.1) 1/2 b Proof By Cauchy-Schwarz inequality and (2.3), we have b f (x)g(y) a ln(x + c)ln(y + c) b = f (x) ln (x + c)(y + c) a × dx d y 1/2 g(y) ln(x + c)(y + c) 1/2 ln (y + c) ln (x + c) b f (x) ln(x + c) ln(x + c)(y + c) ln (y + c) a ≤ b b = a × b (x + c) f (x) b a (y + c)g (y) ≤ π − 4arctan a 1/4 1/2 g (y) ln(y + c) a ln(x + c)(y + c) ln (x + c) × 1/4 ln (x + c) ln (y + c) 1/2 x+c y+c y+c x+c 1/2 x+c dx d y y+c 1/2 dx d y 1/2 y+c dx d y x+c 1/2 ln(x + c) (y + c)ln(x + c)(y + c) ln(y + c) b a ln(y + c) (x + c)ln(x + c)(y + c) ln(x + c) ln(a + c) ln(b + c) b a (x + c) f (x)dx b a d y dx 1/2 1/2 dx d y 1/2 (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 Theorem 3.2 Let − c ≤ a < b < +∞, f is an integrable nonnegative function on [a,b], b such that < a (x + c) f (x)dx < ∞, then b b a a f (x) dx ln(x + c)(y + c) d y ≤ π − 4arctan ln(a + c) ln(b + c) b a (x + c) f (x)dx (3.3) Remark 3.3 Specially, when a = 0, c = 1, and b = ∞ in Theorems 3.1 and 3.2, we get ∞ f (x)g(y) dx d y ≤ π ln(x + 1)(y + 1) ∞ ∞ 0 1/2 ∞ (x + 1) f (x)dx f (x) dx ln(x + 1)(y + 1) d y ≤ π2 ∞ 1/2 ∞ 2 (y + 1)g (y)d y ; (x + 1) f (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 < a φ p−1 (x) f p (x)dx < ∞ and < a ψ q−1 (y)g q (y)d y < ∞ Then 1/ p b × b a b ψ(y) a a f (x) dx ln φ(x)ψ(y) π/sin(π/ p) − φ2 1/ p π/sin(π/ p) − φ1 f (x)g(y) dx d y ≤ lnφ(x)ψ(y) a inf ψ (y) b inf φ (x) 1/ p φ p−1 (x) f p (x)dx p b a 1/q ψ q−1 (y)g q (y)d y inf ψ (y) × b a (3.5) 1/q π/sin(π/ p) − φ1 dy ≤ 1/q ; 1/ p π/sin(π/ p) − φ2 1/ p inf φ (x) 1/q p 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)/ ln(b))1/ pq (p(p))1/ p (q(q))1/q ) Proof By Hă lder inequality and (2.19), we have o b f (x)g(y) a lnφ(x)ψ(y) b = a × dx d y f (x) lnφ(x)ψ(y) 1/ p g(y) lnφ(x)ψ(y) 1/q 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 Journal of Inequalities and Applications 1/q b f p (x) ln φ(x) a lnφ(x)ψ(y) lnψ(y) ≤ 1/ p b g q (y) ln ψ(y) a lnφ(x)ψ(y) lnφ(x) × b = a φ(x) p−1 dx d y ψ(y) 1/ p ψ(y)q−1 dxd y φ(x) 1/q 1/q b ω(φ,ψ, q,x) f p (x)dx 1/ p ω(ψ,φ, p, y)g q (y)d y a ≤ inf ψ (y) b a 1/ pq ln φ(a) π − sin(π/ p) ln φ(b) a × 1/ pq π/sin(π/ p) − φ3 ≤ inf ψ (y) b × a 1/q inf φ (x) ln ψ(a) π − sin(π/ p) ln ψ(b) b × 1/ p 1/ p (pΦ(p))1/ p (qΦ(q))1/q φ p−1 (x) f p (x)dx 1/q (pΦ(p))1/ p (qΦ(q))1/q ψ q−1 (y)g q (y)d y 1/ p π/sin(π/ p) − φ4 1/ p inf φ (x) 1/ p b φ p−1 (x) f p (x)dx a 1/q 1/q 1/q ψ q−1 (y)g q (y)d y (3.7) Hence (3.5) is valid p −1 b Let g(y) = (1/ψ(y))( a ( f (x)/ lnφ(x)ψ(y))dx) > (y ∈ (a,b)) By (4.2), we have b 0< ≤ a ψ(y)q−1 g q (y)d y = b a b ψ(y) a b π/sin(π/ p) inf ψ (y) 1/ p f (x) dx ln φ(x)ψ(y) inf φ (x) 1/q a p b f (x)g(y) dx d y a lnφ(x)ψ(y) dy = 1/ p φ p−1 (x) f p (x)dx b a 1/q ψ q−1 (y)g q (y)d y (3.8) Then we find b a b ψ(y) = a f (x) dx lnφ(x)ψ(y) b a ψ(y) q −1 q g (y)d y ≤ p dy p π/sin(π/ p) inf ψ (y) 1/ p inf φ (x) 1/q b a φ p−1 (x) f p (x)dx (3.9) b Since < a φ p−1 (x) f p (x)dx, it follows that < have (3.6) The theorem is proved b q −1 q g (y)d y a ψ(y) < ∞ Still by (3.5), we L Zhongxue and X Hongzheng Remark 3.5 Specially when a = and b = ∞, we get ∞ f (x)g(y) dx d y lnφ(x)ψ(y) ≤ 1/q (inf {ψ (y)})1/ p (inf {φ (x)}) × × ∞ ln φ(1) π − sin(π/ p) lnψ(y) 1/ p inf ψ (y) × lnψ(1) π − sin(π/ p) lnφ(x) 1/ p 1/ p Φ(q) φ p −1 p (x) f (x)dx (3.10) 1/q 1/q Φ(p) ψ q−1 (y)g q (y)d y π/sin(π/ p) ≤ ∞ ∞ ψ(y) ≤ φ ∞ 1/q 1/ p ∞ inf φ (x) p −1 p (x) f (x)dx f (x) dx lnφ(x)ψ(y) 1/q ∞ ψ q −1 1/ p (y)g (y)d y ; p dy p π/sin(π/ p) inf ψ (y) q inf φ (x) 1/q (3.11) ∞ φ p −1 p (x) f (x)dx, where Φ is as in Lemma 2.3 By Theorem 3.4, we have the following corollary Corollary 3.6 Let − c ≤ a < b < +∞, p > 1, 1/ p + 1/q = 1, f , g are integrable nonnegab 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 ln (a + c) dx d y ≤ B , − p q ln(b + c) a ln(x + c)(y + c) × b a 1/ pq qΦ(q) 1/ p (x + c) p−1 f p (x)dx b a 1/q pΦ(p) 1/ p 1/q (y + c)q−1 g q (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 Journal of Inequalities and Applications Theorem 3.7 Let p > 1, 1/ p + 1/q = 1, {am }, {bn } are nonnegative real sequences, such p q that < ∞ φ p−1 (n)an < ∞, < ∞ ψ q−1 (n)bn < ∞ Then n= n= ∞ ∞ am bn lnφ(m)ψ(n) m=2 n=2 ≤ inf ψ (y) ∞ inf φ (x) 1/q 1/ p ln ψ(1) π − sin(π/ p) ln φ(m) × m=2 ∞ n =2 1/q Φ(p) ψ 1/ p inf φ (x) 1/q q −1 q (n)bn 1/ p ∞ π/sin(π/ p) inf ψ (y) 1/ p p Φ(q) φ p−1 (m)am 1/q ln φ(1) π − sin(π/ p) ln ψ(n) × ≤ 1/ p φ p −1 1/q ∞ p (m)am ψ m=2 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 < p ∞ p −1 (n)an < ∞ Then n =2 φ ∞ ψ(n) n =2 ≤ p ∞ am lnφ(m)ψ(n) m=2 inf ψ (y) 1/ p (3.14) p ∞ π/sin(π/ p) inf φ (x) 1/q 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] 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 p q that < ∞ e(p−1)n an < ∞, < ∞ e(q−1)n bn < ∞ Then n= n= ∞ ∞ ∞ n n =2 am m+n m=2 q e(q−1)n bn n =2 p ∞ π/sin(π/ p) e ≤ 1/q ∞ p e(p−1)m am m=2 p ∞ 1/ p ∞ π/sin(π/ p) am bn ≤ m+n e m=2 n=2 ; (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 < e(p−1)t f p (t)dt < ∞, < e(q−1)t g q (t)dt < ∞ Then ∞ ∞ ∞ y f (x) dx x+y 1/ p ∞ f (x)g(y) π/sin(π/ p) dx d y ≤ x+y e e p (p−1)x f (x)dx p π/sin(π/ p) e dy ≤ ∞ 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 p q that < ∞ n2(p−1) an < ∞, < ∞ e(q−1)n bn < ∞ Then n= n= ∞ ∞ π/sin(π/ p) am bn ≤ 2lnm + n 21/q e1/ p m=2 n=2 ∞ ∞ am 2ln m + n m=2 n n =2 p ∞ 1/ p ∞ p m2(p−1) am m=2 1/q q e(q−1)n bn ; n =2 π/sin(π/ p) ≤ 21/q e1/ p p ∞ (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 < x2(p−1) f p (x)dx < ∞, < e(q−1)x g q (x)dx < ∞ Then ∞ f (x)g(y) π/sin(π/ p) dx d y ≤ 2lnx + y 21/q e1/ p ∞ 1 y ∞ f (x) dx 2ln x + y ∞ p dy ≤ 1/ p x2(p−1) f p (x)dx π/sin(π/ p) 21/q e1/ p ∞ p ∞ 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 Journal of Inequalities and Applications 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 3744, 2003 Lă Zhongxue: School of Mathematical Sciences, Xuzhou Normal University, Xuzhou, u Jiangsu 221116, China Email address: lvzx1@tom.com Xie Hongzheng: Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China Email address: xds@mail.edu.cn ... 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)... 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. ..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 < ln mn sin(π/ p)