AN IMPROVEMENT OF SOME INEQUALITIES SIMILAR TO HILBERT ’ S INEQUALITY

We give an improvement of some inequalities similar to Hilbert’s inequality involving series of nonnegative terms. The integral analogies of the main results are also given. 2000 Mathematics Subject Classification. 26D15.


Introduction.
It is well known that the following Hilbert's double series inequality (see [3, page 226]) plays an important role in many branches of mathematics.unless all the sequence {a m } or {b n } is null.
Recently, Pachpatte [9] gave new inequalities similar to Hilbert's inequalities given in the above theorems, involving a series of nonnegative terms as follows.
unless {a m } or {b n } is null, where An integral analogue of Theorem 1.3 is given in the following theorem.
Theorem 1.4.Let p ≥ 1, q ≥ 1 and f (σ ) ≥ 0, g(τ) ≥ 0 for σ ∈ (0,x), τ ∈ (0,y), where x, y are positive real numbers and define unless f ≡ 0 or g ≡ 0, where In this paper, we give an improvement of the inequalities given in Theorems 1.3 and 1.4 similar to Hilbert's double series inequality and its integral analogue, involving a series of nonnegative terms.In addition, we obtain some new Hilbert type inequalities.These inequalities improve the results obtained by Pachpatte [9].

Main results.
Our main results are given in the following theorems.
unless {a m } or {b n } is null, where Proof.By using the following inequality (see [1,7]), where β ≥ 1 is a constant and z m ≥ 0, (m = 1, 2,...), it is easy to observe that , n= 1, 2,...,r . (2.4) From (2.4) and using the Schwarz inequality and the elementary inequality (for a i , i = 1, 2,...,n, nonnegative real numbers) we observe that Dividing both sides of (2.6) by (m α + n α ) 1/α , and then taking the sum over n from 1 to r first and then the sum over m from 1 to k and using the Schwarz inequality and then interchanging the order of the summations (see [7,8]) we observe that This completes the proof.
Remark 2.2.In Theorem 2.1, setting α ≡ 1, we have Theorem 1.3.If we take p = q = 1 in Theorem 2.1, then the inequality of the result of Theorem 2.1 reduces to the following inequality: where (2.10) Proof.From the hypotheses of φ and ψ and by using Jensen's inequality and the Schwarz inequality (see [5]), it is easy to observe that (2.12) From (2.11) and (2.12) and using the elementary inequality (2.14) Dividing both sides of the above inequality by (m α +n α ) 1/α , and then taking the sum over n from 1 to r first and then the sum over m from 1 to k and using the Schwarz inequality and then interchanging the order of the summations we observe that (2.15) The proof is complete.
Proof.From the hypotheses and by using Jensen's inequality and the Schwarz inequality, it is easy to observe that (2.20) The rest of the proof can be completed by following the same steps as in the proofs of Theorems 2.1 and 2.3 with suitable changes and hence we omit the details.
Proof.From the hypotheses and by using Jensen's inequality and the Schwarz inequality, it is easy to observe that The rest of the proof can be completed by following the same steps as in the proofs of Theorems 2.1 and 2.3 with suitable changes and hence we omit the details.

Integral analogues.
In this section, we present the integral analogues of the inequalities given in Theorems 2.1, 2.3, 2.5, and 2.6, which in fact are motivated by the integral analogue of Hilbert's inequality given in Theorem 1.2.
An integral analogue of Theorem 2.1 is given in the following theorem.
(3.3) From (3.3) and using the Schwarz inequality and the elementary inequality (for a i , i = 1, 2,...,n, nonnegative real numbers) we observe that Dividing both sides of the above inequality by (s α + t α ) 1/α , and then integrating over t from 0 to y first and then integrating the resulting inequality over s from 0 to x and using the Schwarz inequality we observe that where . This completes the proof.

Theorem 1 . 3 .
Let p ≥ 1, q ≥ 1, and let {a m } and {b n } be two nonnegative sequences of real numbers defined for m = 1, 2,...,k and n = 1, 2,...,r , where k, r are the natural numbers and define A m = m s=1 a s , B n =

Theorem 2 . 6 .
Let {a m }, {b n }, {p m }, {q n }, P m , and Q n be as in Theorem 2.3, and define A m = 1/P m m s=1 p s a s and B n = 1/Q n n t=1 q t b t , for m = 1, 2,...,k and n = 1, 2,...,r , where k, r are the natural numbers.Let φ and ψ be as defined in Theorem 2 2).Let {a m }, {b n }, A m , and B n be as defined in Theorem 2.1.Let {p m } and {q n } be two nonnegative sequences for m = 1, 2,...,k and n = 1, 2,...,r , and define P m = n t=1 q t .Let φ and ψ be two real-valued, nonnegative, convex, and submultiplicative functions defined on .13) (for a i , i = 1, 2,...,n, nonnegative real numbers) we observe that . ThenProof.From the hypotheses and by using Jensen's inequality and the Schwarz inequality, it is easy to observe that .(3.13)