INVERSES OF NEW HILBERT-PACHPATTE-TYPE INEQUALITIES CHANG-JIAN ZHAO AND WING-SUM CHEUNG Received 7 February 2006; Revised 3 June 2006; Accepted 5 June 2006 We establish a new inequality of Hilbert type for a finite double number of nonnegative sequences of real numbers and some interrelated results, which are inverse and general forms of Pachpatte’s and Handley’s results. An integral version and some interrelated results are also obtained. These results provide some new estimates on such types of in- equalities. Copyright © 2006 C J. Zhao and W S. Cheung. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis- tribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction The various generalizations and shar penings of Hilbert’s double series inequality and its integral version are obtained by Pachpatte, Handley et al., Gao et al. (see [1, 3, 5, 7, 10, 12, 16–18]). An elegant survey on this kind of inequalities was provided by Yang and Rassias (see [13]). Moreover, Pachpatte [9] established a new Hilbert-type inequality and its integral version as follows. Theorem 1.1. Let {a m }, {b n } be two nonnegative sequences of real numbers defined for m = 1,2, ,k and n = 1,2, ,r with a 0 = b 0 = 0,andlet{p m }, {q n } be two positive sequences of real numbers defined for m = 1,2, ,k, n = 1,2, ,r,wherek, r are natural numbers. Define P m = m s =1 p s and Q n = n t =1 q t .Letφ and ψ be two real-valued nonnegative, convex, and submultiplicative functions defined on R + = [0,∞). Then k m=1 r n=1 φ a m ψ b n m + n ≤ M(k, r) k m=1 (k − m +1) p m φ ∇ a m p m 2 1/2 × r n=1 (r − n+1) q n ψ ∇ b n q n 2 1/2 , (1.1) Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 97860, Pages 1–11 DOI 10.1155/JIA/2006/97860 2 Inverses of new Hilbert-Pachpatte-type inequalities where M(k, r) = 1 2 k m=1 φ P m P m 2 1/2 r n=1 φ Q n Q n 2 1/2 , (1.2) and ∇a m = a m − a m−1 , ∇b n = b n − b n−1 . Theorem 1.2. Let f ∈ C 1 [[0,x), R + ], g ∈ C 1 [[0, y),R + ] with f (0) = g(0) = 0 and let p(σ), q(τ) be two positive functions defined for σ ∈ [0,x) and τ ∈ [0, y).LetP(s) = s 0 p(σ)dσ and Q(t) = t 0 q(τ)dτ for s ∈ [0,x) and t ∈ [0, y),wherex, y are positive real numbers. Let φ,andψ be as in Theorem 1.1. Then x 0 y 0 φ f (s) ψ g(t) s + t dsdt ≤ L(x, y) x 0 (x − s) p(s)φ f (s) p(s) 2 ds 1/2 × y 0 (y − t) q(y)φ g (t) q(t) 2 dt 1/2 , (1.3) where L(x, y) = 1 2 x 0 φ P(s) P(s) 2 ds 1/2 y 0 φ Q(t) Q(t) 2 dt 1/2 , (1.4) and denotes the derivative of a function. In [4], Handley et al. gave general versions of inequalities (1.1)and(1.3)asfollows. Theorem 1.3. Let {a i,m i } (i = 1,2, ,n) be n sequences of nonnegative real numbers defined for m i = 1,2, ,k i with a 1,0 = a 2,0 = ··· =a n,0 = 0,andlet{p i,m i } be n sequences of posi- tive real numbers defined for m i = 1,2, ,k i ,wherek i (i = 1,2, ,n) are natural numbers. Set P i,m i = m i s i =1 p i,s i .Letφ i (i = 1,2, ,n) be n real-valued nonnegative convex and sub- multiplicative functions de fined on R + .Letα i ∈ (0,1),andsetα i = 1 − α i (i = 1,2, ,n), α = n i =1 α i ,andα = n i =1 α i = n− α. Then k 1 m 1 =1 ··· k n m n =1 n i =1 φ i a i,m i n i =1 α i m i α ≤ M k 1 , ,k n n i=1 k i m i =1 k i − m i +1 p i,m i φ i ∇ a i,m i p i,m i 1/α i α i , (1.5) where M k 1 , ,k n = 1 (α ) α n i=1 k i m i =1 φ i P i,m i P i,m i 1/α i α i , (1.6) and ∇a i,m i = a i,m i − a i,m i −1 (i = 1,2, ,n). C J. Zhao and W S. Cheung 3 Theorem 1.4. Let f i ∈ C 1 [[0,k i ],R + ], i = 1,2, ,n,with f i (0) = 0(i = 1,2, ,n),let p i (σ i ) be n positive functions defined for σ i ∈ [0,x i ](i = 1, ,n).SetP i (s i ) = s i 0 p i (σ i )dσ i for s i ∈ [0,x i ],wherex i are positive real numbers. Let φ i , α i , α i , α,andα be as in Theorem 1.3. Then x 1 0 ··· x n 0 n i =1 φ i f i s i n i =1 α i s i α ds 1 ···ds n ≤ L x 1 , ,x n n i=1 x i 0 x i − s i p i s i φ i f i s i p i s i 1/α i ds i α i , (1.7) where L x 1 , ,x n = 1 (α ) α n i=1 x i 0 φ i P i s i P i s i 1/α i ds i α i . (1.8) The main purpose of the present paper is to establish their reverse versions, which are more extensive results for this type of inequalities. Our main results are given in the following theorems. Theorem 1.5. Let {a s i ,t i ,m s i ,m t i } (i = 1,2, ,n) be n sequences of nonne gative numbers de- fined for m s i = 1,2, ,k s i , m t i = 1,2, ,k t i ,witha s i ,t i ,0,m t i = a s i ,t i ,m s i ,0 = 0(i = 1,2, ,n), where k s i and k t i (i = 1, 2, , n) are natural numbers. Let {p s i ,t i ,m s i ,m t i } be n sequences of positive real numbers defined for m s i , m t i .Set P s i ,t i ,m s i ,m t i = m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i (i = 1,2, ,n). (1.9) Let φ i (i = 1,2, ,n) be n real-valued nonnegat ive concave and supermultiplicative func- tions defined on R + .Letα i ∈ (1,∞).Setα i = 1 − α i (i = 1,2, ,n), α = n i =1 α i ,andα = n i =1 α i = n− α.Defineoperators ∇ 1 a s i ,t i ,m s i ,m t i = a s i ,t i ,m s i ,m t i − a s i ,t i ,m s i −1,m t i , ∇ 2 a s i ,t i ,m s i ,m t i = a s i ,t i ,m s i ,m t i − a s i ,t i ,m s i ,m t i −1 , ∇ 2 ∇ 1 a s i ,t i ,m s i ,m t i =∇ 2 ∇ 1 a s i ,t i ,m s i ,m t i . (1.10) Then k s 1 m s 1 =1 k t 1 m t =1 ··· k s n m s n =1 k t n m t n =1 n i =1 φ i a s i ,t i ,m s i ,m t i (1/α ) n i =1 α i m s i m t i α ≥ C k s 1 k t 1 , ,k s n k t n × n i=1 k s i m s i =1 k t i m t i =1 k s i −m s i +1 k t i −m t i +1 p s i ,t i ,m s i ,m t i φ i ∇ 2 ∇ 1 a s i ,t i ,m s i ,m t i p s i ,t i ,m s i ,m t i 1/α i α i , (1.11) 4 Inverses of new Hilbert-Pachpatte-type inequalities where C k s 1 k t 1 , ,k s n k t n = n i=1 k s i m s i =1 k t i m t i =1 φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i 1/α i α i . (1.12) Theorem 1.6. Let f i (s i ,t i )(i = 1,2, ,n) be real-valued continuous functions defined on [0,x i )×[0, y i ),wherex i ∈ (0,∞), y i ∈ (0,∞),andwith f i (0,t i )= f i (s i ,0)= 0(i =1,2, ,n). Let p i (σ i ) and q i (τ i ) be positive continuous functions defined for σ i ∈ (0,s i ), τ i ∈ (0,t i ).Set P i s i ,t i = t i 0 s i 0 p i σ i q i τ i dσ i dτ i . (1.13) For the function f i (s i ,t i ), denote the partial derivatives (∂/∂s i ) f i (s i ,t i ), (∂/∂t i ) f i (s i ,t i ),and (∂ 2 /∂s i ∂t i ) f i (s i ,t i ) by D 1 f i (s i ,t i ), D 2 f i (s i ,t i ),andD 2 D 1 f i (s i ,t i ) = D 1 D 2 f i (s i ,t i ),respectively. Let φ i , α i , α i , α,andα be as in Theorem 1.5. Then x 1 0 y 1 0 ··· x n 0 y n 0 n i=1 φ i f i s i ,t i (1/α ) n i =1 α i s i t i α ds 1 dt 1 ···ds n dt n = G x 1 y 1 , ,x n y n × n i=1 x i 0 y i 0 x i − s i y i − t i p i s i q i t i · φ i D 2 D 1 f i s i ,t i p i s i q i t i 1/α i ds i dt i α i , (1.14) where G x 1 y 1 , ,x n y n = n i=1 x i 0 y i 0 φ i P i s i ,t i P i s i ,t i 1/α i ds i dt i α i . (1.15) These results provide new estimates on these types of inequalities. 2. Proofs and remarks Proof of Theorem 1.5. Fro m the operators defined in Theorem 1.5,weobtain a s i ,t i ,m s i ,m t i = m s i m ξ i =1 a s i ,t i ,m ξ i ,m t i − a s i ,t i ,m ξ i −1,m t i = m s i m ξ i =1 m t i m η i =1 a s i ,t i ,m ξ i ,m η i − a s i ,t i ,m ξ i −1,m η i − a s i ,t i ,m ξ i ,m η i −1 + a s i ,t i ,m ξ i −1,m η i −1 = m s i m ξ i =1 m t i m η i =1 ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i . (2.1) C J. Zhao and W S. Cheung 5 From (2.1) and the hypotheses of Theorem 1.5, and in view of Jensen’s inequality and inverse H ¨ older’s inequality [6], we obtain φ i a s i ,t i ,m s i ,m t i = φ i P s i ,t i ,m s i ,m t i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i ≥ φ i P s i ,t i ,m s i ,m t i · φ i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i ≥ φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i φ i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i ≥ φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i m s i m t i α i × m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i φ i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i 1/α i α i . (2.2) On the other hand, noticing α i < 0(i = 1,2, ,n), λ i > 0(i = 1,2, ,n), and applying the well-known means inequality, we have n i=1 λ α i i ≥ 1 α n i=1 α i λ i α . (2.3) Hence n i =1 φ i a s i ,t i ,m s i ,m t i (1/α ) n i =1 α i m s i m t i α ≥ n i=1 φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i φ i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i 1/α i α i . (2.4) Taking the sum of both sides of (2.4)overm s i , m t i (i = 1,2, ,n)from1tok s i , k s i ,respec- tively, and in view of inverse H ¨ older’s inequality, we get k s 1 m s 1 =1 k t 1 m t =1 ··· k s n m s n =1 k t n m t n =1 n i =1 φ i a s i ,t i ,m s i ,m t i (1/α ) n i =1 α i m s i m t i α ≥ n i=1 k s i m s i =1 k t i m t i =1 φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i φ i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i 1/α i α i 6 Inverses of new Hilbert-Pachpatte-type inequalities ≥ n i=1 k s i m s i =1 k t i m t i =1 φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i 1/α i α i × k s i m s i =1 k t i m t i =1 m s i m ξ i =1 m t i m η i =1 p s i ,t i ,m ξ i ,m η i φ i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i 1/α i α i = C k s 1 k t 1 , ,k s n k t n × n i=1 k s i m ξ i =1 k t i m η i =1 k s i −m ξ i +1 k t i −m η i +1 p s i ,t i ,m ξ i ,m η i φ i ∇ 2 ∇ 1 a s i ,t i ,m ξ i ,m η i p s i ,t i ,m ξ i ,m η i 1/α i α i = C k s 1 k t 1 , ,k s n k t n × n i=1 k s i m s i =1 k t i m t i =1 k s i −m s i +1 k t i − m t i +1 p s i ,t i ,m s i ,m t i φ i ∇ 2 ∇ 1 a s i ,t i ,m s i ,m t i p s i ,t i ,m s i ,m t i 1/α i α i . (2.5) This completes the proof. Remark 2.1. Let a s i ,t i ,m s i ,m t i , m s i , m t i , k s i , k t i , a s i ,t i ,0,m t i , a s i ,t i ,m s i ,0 , P s i ,t i ,m s i ,m t i , ∇ 1 ,and∇ 2 be as in Theorem 1.5.Letφ i , α i , α i , α,andα (i = 1,2, ,n)beasinTheorem 1.3. Similar to the proof of Theorem 1.5,weobtain k s 1 m s 1 =1 k t 1 m t =1 ··· k s n m s n =1 k t n m t n =1 n i =1 φ i a s i ,t i ,m s i ,m t i n i =1 α i m s i m t i α ≤ C k s 1 k t 1 , ,k s n k t n × n i=1 k s i m s i =1 k t i m t i =1 k s i −m s i +1 k t i − m t i +1 p s i ,t i ,m s i ,m t i φ i ∇ 2 ∇ 1 a s i ,t i ,m s i ,m t i p s i ,t i ,m s i ,m t i 1/α i α i , (2.6) where C k s 1 k t 1 , ,k s n k t n = 1 (α ) α n i=1 k s i m s i =1 k t i m t i =1 φ i P s i ,t i ,m s i ,m t i P s i ,t i ,m s i ,m t i 1/α i α i . (2.7) This is just a general form of inequality (1.5) and the inverse form of inequality (1.11). C J. Zhao and W S. Cheung 7 Moreover , let {a i,m i }, m i , k i , a i,0 , p i,m i , ∇,andP i,m i (i = 1,2, ,n)beasinTheorem 1.3. Let φ i , α i , α i , α,andα (i = 1, 2, , n)beasinTheorem 1.5. Similar to the proof of Theorem 1.5,weobtain k 1 m 1 =1 ··· k n m n =1 n i =1 φ i a i,m i (1/α ) n i =1 α i m i α ≥ M k 1 , ,k n n i=1 k i m i =1 k i − m i +1 p i,m i φ i ∇ a i,m i p i,m i 1/α i α i , (2.8) where M k 1 , ,k n = n i=1 k i m i =1 φ i P i,m i P i,m i 1/α i α i . (2.9) This is just an inverse form of inequality (1.5) which was given by Handley et al. [4]. Let n = 2, α 1 = α 2 = 2, then α 1 = α 2 =−1, putting them in inequality (2.8), we have k 1 m 1 =1 k 2 m 2 =1 φ 1 a 1,m 1 φ 2 a 2,m 2 m 1 + m 2 −2 ≥ M k 1 ,k 2 k 1 m 1 =1 k 1 − m 1 +1 p 1,m i φ 1 ∇ a 1,m 1 p 1,m 1 1/2 2 × k 2 m 2 =1 k 2 − m 2 +1 p 2,m 2 φ 2 ∇ a 2,m 2 p 2,m 2 1/2 2 , (2.10) where M k 1 ,k 2 = 4 k 1 m 1 =1 φ 1 P 1,m 1 P 1,m 1 −1 −1 k 2 m 2 =1 φ 2 P 2,m 2 P 2,m 2 −1 −1 . (2.11) This is just an inverse form of inequality (1.1) which was given by Pachpatte [9]. Proof of Theorem 1.6. From the hypotheses of Theorem 1.6,weobtain f i s i ,t i = s i 0 t i 0 D 2 D 1 f i σ i ,τ i dσ i dτ i . (2.12) 8 Inverses of new Hilbert-Pachpatte-type inequalities From (2.12) and using Jensen’s integral inequality and H ¨ older’s integ ral inequality [6], we obtain φ i f i s i ,t i = φ i P i s i ,t i t i 0 s i 0 p i σ i q i τ i D 2 D 1 f i σ i ,τ i p i σ i q i τ i dσ i dτ i t i 0 s i 0 p i σ i q i τ i dσ i dτ i ≥ φ i P i s i ,t i · φ i t i 0 s i 0 p i σ i q i τ i D 2 D 1 f i σ i ,τ i /p i σ i q i τ i dσ i dτ i t i 0 s i 0 p i σ i q i τ i dσ i dτ i ≥ φ i (P i s i ,t i P i s i ,t i t i 0 s i 0 p i σ i q i τ i · φ i D 2 D 1 f i σ i ,τ i p i σ i q i τ i dσ i dτ i ≥ φ i P i s i ,t i P i s i ,t i s i t i α i t i 0 s i 0 p i σ i q i τ i · φ i D 2 D 1 f i σ i ,τ i p i σ i q i τ i 1/α i dσ i dτ i α i . (2.13) From the well-known inequality for means, we have n i=1 φ i f i s i ,t i (1/α ) n i =1 α i s i t i α ≥ n i=1 φ i P i s i ,t i P i s i ,t i t i 0 s i 0 p i σ i q i τ i · φ i D 2 D 1 f i σ i ,τ i p i σ i q i τ i 1/α i dσ i dτ i α i . (2.14) Integrating both sides of (2.14)overs i , t i from 1 to x i , y i (i = 1,2, ,n)andinviewof H ¨ older’s integral inequality and Fubini’s theorem, we observe that x 1 0 y 1 0 ··· x n 0 y n 0 n i=1 φ i f i s i ,t i (1/α ) n i =1 α i s i t i α ds 1 dt 1 ···ds n dt n ≥ n i=1 x i 0 y i 0 φ i P i s i ,t i P i s i ,t i × t i 0 s i 0 p i σ i q i τ i · φ i D 2 D 1 f i σ i ,τ i p i σ i q i τ i 1/α i dσ i dτ i α i ds i dt i ≥ n i=1 x i 0 y i 0 φ i P i s i ,t i P i s i ,t i 1/α i ds i dt i α i × x i 0 y i 0 t i 0 s i 0 p i σ i q i τ i · φ i D 2 D 1 f i σ i ,τ i p i σ i q i τ i 1/α i dσ i dτ i ds i dt i α i C J. Zhao and W S. Cheung 9 = G x 1 y 1 , ,x n y n × n i=1 x i 0 y i 0 x i − s i y i − t i p i s i q i t i · φ i D 2 D 1 f i s i ,t i p i s i q i t i 1/α i ds i dt i α i . (2.15) The proof is complete. Remark 2.2. Let f i (s i ,t i ), x i , y i , f i (0,t i ), f i (s i ,0), D 1 f i (s i ,t i ), D 2 f i (s i ,t i ), D 2 D 1 f i (s i ,t i ), p i (σ i ), q i (τ i ), and P i (s i ,t i )beasinTheorem 1.6.Letφ i , α i , α i , α,andα be as in Theorem 1.3. Similar to the proof of Theorem 1.6,wehave x 1 0 y 1 0 ··· x n 0 y n 0 n i=1 φ i f i s i ,t i n i =1 α i s i t i α ds 1 dt 1 ···ds n dt n ≤ G x 1 y 1 , ,x n y n × n i=1 x i 0 y i 0 x i − s i y i − t i p i s i q i t i · φ i D 2 D 1 f i s i ,t i p i s i q i t i 1/α i ds i dt i α i , (2.16) where G x 1 y 1 , ,x n y n = 1 (α ) α n i=1 x i 0 y i 0 φ i P i s i ,t i P i s i ,t i 1/α i ds i dt i α i . (2.17) This is just a general form of inequality (1.7)andaninverse form of the inequality (1.14). On the other hand, let f i , f i (0), f i (s i ), p i (σ i ), σ i , k i , s i , x i , P i (s i )beasinTheorem 1.4, and let φ i , α i , α i , α, α be as in Theorem 1.5.Then x 1 0 ··· x n 0 n i =1 φ i f i s i (1/α ) n i =1 α i s i α ds 1 ···ds n ≥ L x 1 , ,x n n i=1 x i 0 x i − s i p i s i φ i f i s i p i s i 1/α i ds i α i , (2.18) where L x 1 , ,x n = n i=1 x i 0 φ i P i s i P i s i 1/α i ds i α i . (2.19) This is just an inverse form of inequality (1.7) which was given by Handley et al. [4]. 10 Inverses of new Hilbert-Pachpatte-type inequalities Let n = 2, α 1 = α 2 = 2, then α 1 = α 2 =−1, taking them in inequality (2.18), we have x 1 0 x 2 0 φ 1 f 1 s 1 φ 2 f 2 s 2 s 1 + s 2 −2 ds 1 ds 2 ≥ L x 1 ,x 2 x 1 0 x 1 − s 1 p 1 s 1 φ 1 f 2 s 1 p 1 s 1 1/2 ds 1 2 × x 2 0 x 2 − s 2 p 2 s 2 φ 2 f 2 s 2 p 2 s 2 1/2 ds 2 2 , (2.20) where L x 1 ,x 2 = 4 x 1 0 φ 1 P 1 s 1 P 1 s 1 −1 ds 1 −1 x 2 0 φ 2 P 2 s 2 P 2 s 2 −1 ds 2 −1 . (2.21) This is just an inverse form of inequality (1.3) which was given by Pachpatte [9]. For interrelated research on similar inequalities, one is referred to [2, 8, 11, 14, 15], and the references cited therein. Acknowledgments Research of the first author is supported by Zhejiang Provincial Natural Science Foun- dation of China (Y605065), National Natural Sciences Foundation of China (10271071), Foundation of the Education Department of Zhe jiang Province of China (20050392), and the Academic Mainstay of Middle-age and Youth Foundation of Shandong Province of China (200203). Research of the second author is partially supported by the Research Grants Council of the Hong Kong SAR, China (Project no. HKU7040/03P, and HKU 7017/05P). References [1] W S. Cheung, ˇ Z. Hanj ˇ s, and J. Pe ˇ cari ´ c, Some Hardy-type inequalities, Journal of Mathematical Analysis and Applications 250 (2000), no. 2, 621–634. [2] W S. Cheung and Q H. 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Weight characterizations for the discrete Hardy inequality with kernel, Journal of Inequalities and Applications 2006 (2006), Article ID 18030, 14 pages [9] B G Pachpatte, A note on Hilbert type inequality, Tamkang Journal of Mathematics 29 (1998), no 4, 293–298 , On some new inequalities similar to Hilbert’s inequality, Journal of Mathematical Anal[10] ysis and Applications 226 (1998), no 1, 166–179 ˇ . INVERSES OF NEW HILBERT-PACHPATTE-TYPE INEQUALITIES CHANG-JIAN ZHAO AND WING-SUM CHEUNG Received 7 February 2006; Revised 3 June 2006; Accepted 5 June 2006 We establish a new inequality of. Education Department of Zhe jiang Province of China (20050392), and the Academic Mainstay of Middle-age and Youth Foundation of Shandong Province of China (200203). Research of the second author. ,x n y n = n i=1 x i 0 y i 0 φ i P i s i ,t i P i s i ,t i 1/α i ds i dt i α i . (1.15) These results provide new estimates on these types of inequalities. 2. Proofs and remarks Proof of Theorem 1.5. Fro m the operators defined in Theorem 1.5,weobtain a s i ,t i ,m s i ,m t i = m s i m ξ i =1 a s i ,t i ,m ξ i ,m t i −