on some generalizations of the hilbert–hardy type discrete inequalities

Journal of the Egyptian Mathematical Society (2013) xxx, xxx–xxx

Egyptian Mathematical Society

Journal of the Egyptian Mathematical Society

www.etms-eg.orgwww.elsevier.com/locate/joems

On some generalizations of the Hilbert–Hardy type

discrete inequalities

S.A.A. El-Marouf *

Permanent address: Department of Mathematics, Faculty of Science, Minoufiya University, Shebin El-Koom, EgyptCurrent address: Department of Mathematics, Faculty of Science, Taibah University, Kingdom of Saudi Arabia

Received 8 November 2012; revised 16 December 2012; accepted 10 October 2013

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KEYWORDS

Hilbert’s inequality;

Holder’s inequality;

Hardy’s inequality;

Beta and gamma functions;

Weight functions

Current address: Departmen

ibah University, Kingdom

3486398.

mail address: sobhy_2000_99

er review under responsibilit

Production an

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10-256X ª 2013 Production

tp://dx.doi.org/10.1016/j.joem

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Abstract New Hilbert-type discrete inequalities are presented by using new techniques in proof. By

specializing the weight coefficient functions in the hypothesis and the parameters, we obtain many

special cases which include, in particular, the discrete inequality derived by Hilbert and Hardy.

Many improvements and generalizations of known results are given in this paper.

2010 MATHEMATICAL SUBJECT CLASSIFICATION: Primary 26D15; Secondary 47A07

ª 2013 Production and hosting by Elsevier B.V. on behalf of Egyptian Mathematical Society.

1. Introduction

The Hilbert’s double series inequality is given as follows: (see

[1,2]).Let p > 1; q > 1; 1

pþ 1

q¼ 1 and am, bn > 0. If

0 <P1

m¼1apm <1, and 0 <

P1n¼1b

qn <1, then we have

X1m¼1

X1n¼1

ambnmþ n

6p

sin pp

� � X1m¼1

apm

!1p X1

n¼1bqn

!1q

; ð1:1Þ

where p/sin(p/p) is the best possible constant.

hematics, Faculty of Science,

udi Arabia. Tel./fax: +20

.com

tian Mathematical Society.

g by Elsevier

S.A.A. El-Marouf, On some geical Society (2013), http://dx.d

ing by Elsevier B.V. on behalf of E

0.007

Many inequalities, in general and different versions of theHilbert inequality, in particular play a major role in mathemat-

ical analysis and applications. In recent years, considerableattention has been given to various extensions and improve-ments of theHilbert inequality (1.1) (see Refs. [3–10]). The main

purpose of this paper is to obtain some extensions of (1.1).

2. The main results

First, we introduce some lemmas.

Lemma 2.1. For p > 1; a P 0; b > 1cp and a > 0 such that

1pþ 1

q ¼ 1, define the weight coefficient function w1(m) asfollows:

w1ðmÞ ¼X1n¼1

1

aþmanbð Þcm

n

� �1q

: ð2:1Þ

Then we get

w1ðmÞ 61

ba

1bp�cm

bp�b�abp B

1

bp; c� 1

bp

� �; ð2:2Þ

where B(a,b) is the Beta function for a> 0 and b> 0.

neralizations of the Hilbert–Hardy type discrete inequalities,oi.org/10.1016/j.joems.2013.10.007

gyptian Mathematical Society.

mailto:[email protected]

http://dx.doi.org/10.1016/j.joems.2013.10.007


http://www.sciencedirect.com/science/journal/1110256X



2 S.A.A. El-Marouf

Proof. From (2.1), we have

w1ðmÞ ¼X1n¼1

1

ac 1þ manb

a

� �c m

n

� �1q

61

ac

Z 1

0

1

1þ mayb

a

� �cm

y

� �1q

dy:

Using the change of variable u ¼ mayb

a, we have dy ¼ 1

ba1bu

1b�1

ma=b duand 0 6 u<1.

Substituting u and dy in the right hand side of the aboveinequality, we get

w1ðmÞ 61

ac

Z 1

0

1

ð1þ uÞcm1þa

b

a1bu

1b

!1q1

ba

1bu

1b�1

mab

du

¼ 1

ba

1b�

1bq�cm

1qþ a

bq�ab

Z 1

0

u1b�

1bq�1

ð1þ uÞc du�

It follows from [9] and 1pþ 1

q¼ 1, that

w1ðmÞ 61

ba

1bp�cm

bp�b�abp B

1

bp; c� 1

bp

� �:

Hence the lemma is proved. h

By a similar manner we can prove the following lemma.

Lemma 2.2. For p > 1; a > 1cq ; b P 0 and a> 0 such that

1pþ 1

q ¼ 1, define the weight coefficient

w2(n) as:

w2ðnÞ ¼X1m¼1

1

ðaþmanbÞcn

m

� �1p

: ð2:3Þ

Then we get

w2ðnÞ 61

aa

1aq�cn

aq�a�baq B

1

aq; c� 1

aq

� �: ð2:4Þ

Theorem 2.3. If a > 1cq ; b >

1cp and (p > 1), such that

1pþ 1

q¼ 1; famg and {bn} P 0, satisfy that

0 <X1m¼1

mbp�b�a

bp apm <1 and 0 <X1n¼1

naq�a�b

aq bqn <1:

Then for (a,c > 0).

X1m¼1

X1n¼1

ambnaþmanbð Þc6 a

a q�cbq�1ð Þþb p�cap�1ð Þabpq

1

bBð 1

bp;c� 1

bpÞ

� �1p 1

aB

1

aq;c� 1

aq

� �� 1q

�X1m¼1

mbp�b�a

bp apm

!1p X1

n¼1n

aq�a�baq bqn

!1q

: ð2:5Þ

Proof. Using Holder’s inequality, we have

X1m¼1

X1n¼1

ambnaþmanbð Þc

¼X1m¼1

X1n¼1

am

aþmanbð Þcp

m

n

� � 1pq bn

aþmanbð Þcq

n

m

� � 1pq

6

X1m¼1

X1n¼1

apmaþmanbð Þc

m

n

� �1q

!1p

�X1m¼1

X1n¼1

bqnaþmanbð Þc

n

m

� �1p

!1q

¼X1m¼1

apm

X1n¼1

1

aþmanbð Þcm

n

� �1q

! !1p X1

n¼1bqn

X1m¼1

1

aþmanbð Þcn

m

� �1p

! !1q

:

ð2:6Þ

Please cite this article in press as: S.A.A. El-Marouf, On some gJournal of the Egyptian Mathematical Society (2013), http://dx.d

By (2.1), (2.3) and (2.6), we get

X1m¼1

X1n¼1

ambnðaþmanbÞc 6

X1m¼1

apmw1ðmÞ !1

p X1n¼1

bqnw2ðnÞ !1

q

� ð2:7Þ

Substituting by (2.2) and (2.4) in (2.7), we obtain

X1m¼1

X1n¼1

ambnðaþmanbÞc 6

X1m¼1

apm1

ba

1bp�cm

bp�b�abp B

1

bp; c� 1

bp

� � !1p

�X1n¼1

bqn1

aa

1aq�cn

aq�a�baq B

1

aq; c� 1

aq

� � !1q

:

Since 1pþ 1

q¼ 1, we have

X1m¼1

X1n¼1

ambnðaþmanbÞc 6 a

aðq�cbq�1Þþb p�cap�1ð Þabpq

1

bB

1

bp; c� 1

bp

� �� 1p

1

aB

1

aq; c� 1

aq

� �� 1q

�X1m¼1

mbp�b�a

bp apm

!1p X1

n¼1n

aq�a�baq bqn

!1q

:

This completes the proof. h

Remark 2.1.

1. Let p= q= 2 in (2.5), then we have

X1m¼1

X1n¼1

ambnðaþmanbÞc6

1

ab

� �1=2

aa�4abcþb

4ab B1

2b;c� 1

2b

� �B

1

2a;c� 1

2a

� �� 12

�X1m¼1

mb�a2b a2m

!12 X1

n¼1n

a�b2a b2n

!12

:

ð2:8Þ

2. Let c = 1 in (2.5). Then we get

X1m¼1

X1n¼1

ambnaþmanb

6pa

aðq�bq�1Þþbðp�ap�1Þabpq

b sin pbp

� �1p

a sin paq

� �1q

X1m¼1

mbp�b�a

bp apm

!1p X1

n¼1n

aq�a�baq bqn

!1q

;

ð2:9Þ

which is a new Hilbert-type inequality.

3. Let a = b = 1 in (2.9), then we obtain

X1m¼1

X1n¼1

ambnaþmn

6pa

�2pq

sin pp

X1m¼1

m1�2papm

!1p X1

n¼1n1�

2qbqn

!1q

: ð2:10Þ

4. Let a= 1 in (2.9), then we have

X1m¼1

X1n¼1

ambn1þmanb

6p

b sin pbp

� �1p

a sin paq

� �1q

X1m¼1

mbp�b�a

bp apm

!1p X1

n¼1n

aq�a�baq bqn

!1q

:

ð2:11Þ

5. Let a = b = 1 in (2.11), then we find

X1m¼1

X1n¼1

ambn1þmn

6p

sin pp

X1m¼1

m1�2papm

!1p X1

n¼1n1�

2qbqn

!1q

:

6. Let a= 4 in (2.10), then we get

X1m¼1

X1n¼1

ambn4þmn

6p4

�2pq

sin pp

X1m¼1

m1�2papm

!1p X1

n¼1n1�

2qbqn

!1q

:

eneralizations of the Hilbert–Hardy type discrete inequalities,oi.org/10.1016/j.joems.2013.10.007


On some generalizations of the Hilbert–Hardy type discrete inequalities 3

7. Let a = b = 1, a = 1 in (2.8), then we have

X1m¼1

X1n¼1

ambnð1þmnÞc 6

ffiffiffipp

C c� 12

� �CðcÞ

X1m¼1

a2m

!12 X1

n¼1b2n

!12

: ð2:12Þ

8. Let c = 2 in (2.12), then we find

X1m¼1

X1n¼1

ambn

ð1þmnÞ26

p2

X1m¼1

a2m

!12 X1

n¼1b2n

!12

:

Lemma 2.4. For a P 0; b > 1cp ; c > 0 and p> 1 such that

1pþ 1

q¼ 1, define the weight coefficient w1(m) as follows:

w1ðmÞ ¼X1n¼1

1

ðma þ nbÞcm

n

� �1q

: ð2:13Þ

Then we get

w1ðmÞ 61

bm

bp�bþa�acbp B

1

bp; c� 1

bp

� �: ð2:14Þ

Proof. From (2.13), we have

w1ðmÞ ¼X1n¼1

1

macð1þ nbÞcm

n

� �1q

61

mac

Z 1

0

1

1þ yb

ma

� �cm

y

� �1q

dy:

Let u ¼ yb

ma, then we have dy ¼ 1bm

abu

1b�1du and 0 6 u <1.

Hence

w1ðmÞ 61

mac

Z 1

0

1

ð1þ uÞcm1�a

b

u1b

!1q1

bm

abu

1b�1du

¼ 1

bm

1q� a

bqþab�ac

Z 1

0

u1b�

1bq�1

1þ uð Þc du�

Using definition of Beta function and 1pþ 1

q¼ 1, we have

w1ðmÞ 61

bm

bp�bþa�abcpbp B

1

bp; c� 1

bp

� �:

Hence the lemma is proved. h

Lemma 2.5. For b P 0; a > 1cq ; c > 0 and p> 1 such that

1pþ 1

q¼ 1; p > 1; q > 1; 1

pþ 1

q¼ 1, define the weight coefficient

w2 (n) as:

w2ðnÞ ¼X1m¼1

1

ðma þ nbÞcn

m

� �1p

: ð2:15Þ

Then we get

w2ðnÞ 61

an

aq�aþb�abcqaq B

1

aq; c� 1

aq

� �; ð2:16Þ

where B(a,b) is the Beta function, a > 0 and b > 0.

Proof. The proof is similar to the proof of Lemma 2.4, so it isomitted. h

Theorem 2.6. If ðp > 1Þ; a P 1cq and b P 1

cp such that 1pþ 1

q¼ 1,

and f(x) P 0, g(y) P 0, satisfy that 0 <P1

m¼1

mbp�bþa�abcp

bp apm <1 and <P1

n¼1naq�aþb�abcq

aq bqn <1�

Please cite this article in press as: S.A.A. El-Marouf, On some geJournal of the Egyptian Mathematical Society (2013), http://dx.d

Then for (c > 0)

X1m¼1

X1n¼1

ambnðma þ nbÞc 6

1

b

� �1p 1

a

� �1q

B1

bp; c� 1

bp

� �� 1p

B1

aq; c� 1

aq

� �� 1q

�X1m¼1

mbðp�acp�1Þþa

bp apm

!1p X1

n¼1n

aðq�bcq�1Þþbaq bqn

!1q

: ð2:17Þ

Proof. Put the left hand side of the inequality (2.17) in theform:

X1m¼1

X1n¼1

ambnðma þ nbÞc ¼

X1m¼1

X1n¼1

am

ðma þ nbÞcp

m

n

� � 1pq bn

ðma þ nbÞcq

n

m

� � 1pq

:

Applying Holder’s inequality to get the right hand side of theabove inequality as follows:

X1m¼1

X1n¼1

ambnðmaþnbÞc6

X1m¼1

X1n¼1

apmðmaþnbÞc

m

n

� �1q

!1p

X1m¼1

X1n¼1

bqnðmaþnbÞc

n

m

� �1p

!1q

¼X1m¼1

apm

X1n¼1

1

ðmaþnbÞcm

n

� �1q

! !1p

X1n¼1

bqnX1m¼1

1

ðmaþnbÞcn

m

� �1p

! !1q

: ð2:18Þ

By (2.13), (2.15) and (2.18), we get

X1m¼1

X1n¼1

ambnðma þ nbÞc 6

X1m¼1

apmw1ðmÞ !1

p X1n¼1

bqnw2ðnÞ !1

q

: ð2:19Þ

Substituting (2.14) and (2.16) of Lemmas 2.4 and 2.5 in (2.19),we obtain

X1m¼1

X1n¼1

ambnðma þ nbÞc 6

X1m¼1

apm1

bm

bp�bþa�abcpbp B

1

bp; c� 1

bp

� � !1p

�X1n¼1

bqn1

an

aq�aþb�abcqaq B

1

aq; c� 1

aq

� � !1q

:

Then

X1m¼1

X1n¼1

ambnðma þ nbÞc 6

1

b

� �1p 1

a

� �1q

B1

bp; c� 1

bp

� �� 1p

B1

aq; c� 1

aq

� �� 1q

�X1m¼1

mbðp�acp�1Þþa

bp apm

!1p X1

n¼1n

aðq�bcq�1Þþbaq bqn

!1q

:


Now, we discuss some special values for the parametersinequality (2.17) in order to obtain some known inequalitiesas special cases from our result.

Remark 2.2.

1. Let p= q= 2 in (2.17), then we get

X1m¼1

X1n¼1

ambnðma þ nbÞc 6

1

ab

� �12

B1

2b; c� 1

2b

� �� 12

B1

2a; c� 1

2a

� �� 12

�X1m¼1

mbþa�2abc

2b a2m

!12 X1

n¼1n

aþb�2abc2a b2n

!12

:



4 S.A.A. El-Marouf

2. Let a = b = 1 in (2.17), then we obtain

X1m¼1

X1n¼1

ambnðmþ nÞc 6 B

1

p; c� 1

p

� �� 1p

B1

q; c� 1

q

� �� 1q

X1m¼1

m1�capm

!1p X1

n¼1n1�cbqn

!1q

: ð2:20Þ

3. Let p= q= 2 in (2.20), then we get

X1m¼1

X1n¼1

ambnðmþ nÞc 6 B

1

2; c� 1

2

� �� X1m¼1

m1�ca2m

!12 X1

n¼1n1�cb2n

!12

¼ffiffiffipp

C c� 12

� �CðcÞ

X1m¼1

m1�ca2m

!12 X1

n¼1n1�cb2n

!12

:

ð2:21Þ

4. Let c = 1 in (2.21), then we have

X1m¼1

X1n¼1

ambnmþ n

6 pX1m¼1

a2m

!12 X1

n¼1b2n

!12

;

which is Hilbert’s double series inequality.5. Let c = 2 in (2.21), then we have

X1m¼1

X1n¼1

ambn

ðmþ nÞ26

p2

X1m¼1

m�1a2m

!12 X1

n¼1n�1b2n

!12

:

6. Let a = b = 2 in (2.17), then we have

X1m¼1

X1n¼1

ambnðm2 þ n2Þc 6

1

2B

1

2p; c� 1

2p

� �� 1p

B1

2q; c� 1

2q

� �� 1q

�X1m¼1

m1�2capm

!1p X1

n¼1n1�2cbqn

!1q

:

ð2:22Þ

7. Let p= q= 2 in (2.22), then we have

X1m¼1

X1n¼1

ambnðm2þn2Þc6

1

2B

1

4;c�1

4

� �� X1m¼1

m1�2ca2m

!12 X1

n¼1n1�2cb2n

!12

:

ð2:23Þ

8. Let c = 1 in (2.23), then we get

X1m¼1

X1n¼1

ambnm2 þ n2ð Þ 6

pffiffiffi2p

X1m¼1

m�1a2m

!12 X1

n¼1n�1b2n

!12

:

9. Let a = b = l in (2.17), then we have

X1m¼1

X1n¼1

ambnðmlþnlÞc6

1

l

� �B

1

lp;c� 1

lp

� �� 1p

� B1

lq;c� 1

lq

� �� 1q X1

m¼1m1�lcapm

!1p X1

n¼1n1�lcbqn

!1q

:

ð2:24Þ10. Let p= q= 2 in (2.24), then we find

X1m¼1

X1n¼1

ambnðmlþnlÞc6

1

l

� �B

1

2l;c� 1

2l

� �� X1m¼1

m1�lca2m

!12 X1

n¼1n1�lcb2n

!12

:

ð2:25Þ



X1m¼1

X1n¼1

ambnml þ nl

6p

l sin p2l

X1m¼1

m1�la2m

!12 X1

n¼1n1�lb2n

!12

:

12. Let c = 1 in (2.17), then we get

X1m¼1

X1n¼1

ambnma þ nb

6p

b sin pbp

� �1p

a sin paq

� �1q

X1m¼1

mbp�bþa�abp

bp apm

!1p

�X1n¼1

naq�aþb�abq

aq bqn

!1q

:

13. Let a = 1 in (2.17), then we have

X1m¼1

X1n¼1

ambnðmþ nbÞc 6

1

b

� �1p

Bð 1bp; c� 1

bpÞ

� �1p

B1

q; c� 1

q

� �� 1q

�X1m¼1

mbp�b�bcpþ1

bp apm

!1p X1

n¼1nqþb�bcq�1

q bqn

!1q

:

ð2:26Þ

14. Let p = q= 2 in (2.26), then we getX1m¼1

X1n¼1

ambnðmþ nbÞc 6

1

b

� �12

B1

2b; c� 1

2b

� �� 12

B1

2; c� 1

2

� �� 12

�X1m¼1

mb�2bcþ1

2b a2m

!12 X1

n¼1n

b�2bcþ12 b2n

!12

: ð2:27Þ


X1m¼1

X1n¼1

ambnmþnb

6p

bsin p2b

� �12

X1m¼1

m1�b2b a2m

!12 X1

n¼1n1�b2 b2n

!12

: ð2:28Þ

16. Let b = 2 in (2.28), then we have

X1m¼1

X1n¼1

ambnmþ n2

6p

ð2Þ14

X1m¼1

m�14 a2m

!12 X1

n¼1n�12 b2n

!12

:

By introducing some parameters, a new form of Hardy–Hilbert’s inequality is given as follows:

Theorem 2.7. If a, b, c> 0, 0 < a < pq, 0 < b < qr,

0< c < pr and (p > 1), such that 1pþ 1

qþ 1r ¼ 1. Also, if 0 <P1

0 mabþa

c�akapm<1;0<P1

0 nbcþ

ba�bkbqn<1 and 0<

P10 t

caþ

cb�ck

crt <1. Then, for any k>max 1bþ 1

c� 1qr ;

1cþ 1

a� 1pr ;

1aþ 1

b� 1pq

n o,

X1m¼1

X1n¼1

X1t¼1

ambnct

amaþbnbþctc� �k6 a

1bþ

1c�k

bb1bcc

1c

!1p

b1cþ1

a�k

aa1acc

1c

!1q

c1aþ1

b�k

aa1abb

1c

!1r

� B1

b� 1

qr;k� 1

bþ 1

qr

� �B

1

qr� 1

b�1

cþk;

1

c

� �� 1p

� B1

c� 1

pr;k�1

cþ 1

pr

� �B

1

pr�1

c�1

aþk;

1

a

� �� 1q

� B1

a� 1

pq;k�1

aþ 1

pq

� �B

1

pq�1

a� 1

bþk;

1

b

� �� 1r

�X1m¼1

mabþ

ac�akapm

!1p X1

n¼1n

bcþ

ba�bkbqn

!1q X1

t¼1t

caþ

cb�ckcrt

!1r

: ð2:29Þ




Proof. Apply Holder’s inequality to estimate the right hand

side of the above inequality as follows:

X1m¼1

X1n¼1

X1t¼1

ambnct

ðama þ bnb þ ctcÞk

¼X1m¼1

X1n¼1

X1t¼1

am

ðama þ bnb þ ctcÞkp

ama

bnb

� � 1pqr

� bn

ama þ bnb þ ctc� �k

q

bnb

ctc

� � 1pqr ct

ðama þ bnb þ ctcÞkr

ctc

ama

� � 1pqr

�

6

X1m¼1

X1n¼1

X1t¼1

apm

ðama þ bnb þ ctcÞkama

bnb

� � 1qr

!1p

�X1m¼1

X1n¼1

X1t¼1

bqn

ðama þ bnb þ ctcÞkbnb

ctc

� � 1pr

!1q

�X1m¼1

X1n¼1

X1t¼1

crt

ðama þ bnb þ ctcÞkctc

ama

� � 1pq

!1r

¼ S1=pT1=qR1=r: ð2:30Þ

Such that here

S ¼X1m¼1ðamaÞ

1qrapmX1t¼1

X1n¼1

ðbnbÞ�1qr

ðama þ bnb þ ctcÞk:

SinceP1

n¼1fðnÞ 6R10

fðyÞdy, then

X1n¼1

ðbnbÞ�1qr

ðama þ bnb þ ctcÞk6

Z 1

0

ðbybÞ�1qr

ðama þ byb þ ctcÞkdy

¼ 1

ðama þ ctcÞkþ1qr

Z 1

0

byb

amaþctc

� �� 1qr

1þ byb

amaþctc

� �k dy:

By putting u ¼ byb

amaþctc, then we find dy ¼ 1b

amaþctcb

� �u

� �1b�1

amaþctcb

� �du and 0 6 u <1. Then

Z 1

0

ðbybÞ�1qr

ðamaþbybþ ctcÞkdy¼ 1

bb1bðamaþ ctcÞkþ

1qr�1

b

Z 1

0

u1b�

1qr�1

ð1þuÞkdu:

From definition of Beta function, we get

Z 1

0

ðbybÞ�1qr

ðamaþbybþctcÞkdy¼ 1

bb1bðamaþ ctcÞkþ

1qr�1

b

B1

b� 1

qr;k� 1

bþ 1

qr

� �:

Therefore, we have

S¼ 1

bb1b

B1

b� 1

qr;k� 1

bþ 1

qr

� �X1m¼1

amað Þ1qrapm

X1t¼1

1

ðamaþ ctcÞkþ1qr�1

b

:

Now,

X1t¼1

1

ðama þ ctcÞkþ1qr�1

b

6

Z 1

0

1

ðama þ czcÞkþ1qr�1

b

dz

¼Z 1

0

ðczcÞ1b�

1qr�k

1þ ama

czc

� �kþ 1qr�1

b

dz:


Using the change in variables u ¼ ama

czc; dz ¼

� 1c

ama

cu�1

� �1c�1 ama

cu�2du.

Hence

Z 1

0

1

ðama þ czcÞkþ1qr�1

b

dz ¼ amað Þ1b�

1qrþ1

c�k

cc1c

Z 1

0

u1qr�1

b�1cþk�1

ð1þ uÞkþ1qr�1

b

du:

Using definition of Beta function, we have

Z 1

0

1

ðaxa þ czcÞkþ1qr�1

b

dz ¼ ðamaÞ

1b�

1qrþ1

c�k

cc1c

B1

qr� 1

b� 1

cþ k;

1

c

� �:

Then

S ¼ a1bþ

1c�k

bb1bcc

1c

B1

b� 1

qr; k� 1

bþ 1

qr

� �B

1

qr� 1

b� 1

cþ k;

1

c

� �X1m¼1

mabþ

ac�akapm:

Similarly, we can write T and R as follows:

T ¼ b1cþ1

a�k

aa1acc

1c

B1

c� 1

pr; k� 1

cþ 1

pr

� �B

1

pr� 1

c� 1

aþ k;

1

a

� �X1n¼1

nbcþ

ba�bkbqn:

R ¼ c1aþ1

b�k

aa1abb

1c

B1

a� 1

pq; k� 1

aþ 1

pq

� �B

1

pq� 1

a� 1

bþ k;

1

b

� �X1t¼1

tcaþ

cb�ckcrt :

X1m¼1

X1n¼1

X1t¼1

ambnct

ðamaþbnbþctcÞk6

a1bþ1c�k

bb1bcc

1c

� �1p

b1cþ

1a�k

aa1acc

1c

� �1q

c1aþ1b�k

aa1abb

1c

� �1r

� B 1b� 1

qr; k� 1

bþ 1qr

� �B 1

qr� 1

b� 1c þ k; 1c

� �� 1p

� B 1c � 1

pr; k� 1

c þ 1pr

� �B 1

pr� 1

c � 1aþ k; 1a

� �� 1q

� B 1a� 1

pq; k� 1

aþ 1pq

� �B 1

pq� 1

a� 1bþ k; 1b

� �� 1r

�X1m¼1

mabþ

ac�akapm

!1p X1

n¼1n

bcþ

ba�bkbqn

!1q X1

t¼1t

caþ

cb�ckcrt

!1r

:

Or

X1m¼1

X1n¼1

X1t¼1

ambnct

amaþbnbþctcð Þk 61

CðkÞa1bþ1c�k

bb1bcc

1c

� �1p

b1cþ1a�k

aa1acc

1c

� �1q

c1aþ

1b�k

aa1abb

1b

� �1r

� C 1c

� �C 1

b� 1qr

� �C 1

qr� 1

b� 1c þ k

� �� 1p

� C 1a

� �C 1

c � 1pr

� �C 1

pr� 1

c � 1aþ k

� �� 1q

� C 1b

� �C 1

a� 1pq

� �C 1

pq� 1

a� 1bþ k

� �� 1r

�X1m¼1

mabþ

ac�akapm

!1p X1

n¼1n

bcþ

ba�bkbqn

!1q X1

t¼1t

caþ

cb�ckcrt

!1r

:

ð2:31Þ


Remark 2.3.

1. Setting p= q= r = 3 in (2.29), we get



6 S.A.A. El-Marouf

X1m¼1

X1n¼1

X1t¼1

ambnct

ðamaþbnbþctcÞk6

1C kð Þ

a1bþ1c�k

bb1bcc

1c

� �13

b1cþ1a�k

aa1acc

1c

� �13

c1aþ1b�k

aa1abb

1b

� �13

� C 1c

� �C 1

b� 19

� �C 1

9� 1

b� 1cþk

� �� 13

� C 1a

� �C 1

c� 19

� �C 1

9� 1

c� 1aþk

� �� 13

� C 1b

� �C 1

a� 19

� �C 1

9� 1

a� 1bþk

� �� 13

�X1m¼1

mabþ

ac�aka3m

!13 X1

n¼1n

bcþ

ba�bkb3n

!13 X1

t¼1t

caþ

cb�ckc3t

!13

: ð2:32Þ

2. Let a = b = c = 1 in (2.29), then we get

X1m¼1

X1n¼1

X1t¼1

ambnct

amþbnþ ctð Þk

6 a3�kp �1b

3�kq �1c

3�kr �1 B 1� 1

qr;k�1þ 1

qr

� �B

1

qrþk�2;1

� �� 1p

;

� B 1� 1

pr;k�1þ 1

pr

� �B

1

prþk�2;1

� �� 1q

� B 1� 1

pq;k�1þ 1

pq

� �B

1

pqþk�2;1

� �� 1r

�X1m¼1

m2�kapm

!1p X1

n¼1n2�kbqn

!1q X1

t¼1t2�kcrt

!1r

: ð2:33Þ

3. For a= b= c = k = 1 in (2.33), one has the following

inequality:

X1m¼1

X1n¼1

X1t¼1

ambnct

mþnþ t6

pðqrÞ1pðprÞ

1qðpqÞ

1r

ð1�qrÞsin pqr

� �1p ð1�prÞsin p

pr

� �1q ð1�pqÞsin p

pq

� �1r

�X1m¼1

mapm

!1p X1

n¼1nbqn

!1q X1

t¼1tcrt

!1r

:

4. Let k = 1 in (2.29), then we get

X1m¼1

X1n¼1

X1t¼1

ambnctamaþbnbþctc 6

a1bþ1c�1

bb1bcc

1c

� �1p

b1cþ1a�1

aa1acc

1c

� �1q

c1aþ1b�1

aa1abb

1b

� �1r

� B 1b� 1

qr; 1� 1

bþ 1qr

� �B 1

qr� 1

b� 1c þ 1; 1c

� �� 1p

� B 1c � 1

pr; 1� 1

c þ 1pr

� �B 1

pr� 1

c � 1aþ 1; 1a

� �� 1q

� B 1a� 1

pq; 1� 1

aþ 1pq

� �B 1

pq� 1

a� 1bþ 1; 1b

� �� 1r

�X1m¼1

mabþ

ac�aapm

!1p X1

n¼1y

bcþ

ba�bbqn

!1q X1

t¼1t

caþ

cb�ccrt

!1r

:

5. For a = b = c= 1 in (2.29), we obtain

X1m¼1

X1n¼1

X1t¼1

ambnct

ðmaþnbþtcÞk6

1bc

� �1p 1

ac

� �1q 1

ab

� �1r

� B 1b� 1

qr; k� 1

bþ 1qr

� �B 1

qr� 1

b� 1c þ k; 1c

� �� 1p

� B 1c � 1

pr; k� 1

c þ 1pr

� �B 1

pr� 1

c � 1aþ k; 1a

� �� 1q

� B 1a� 1

pq; k� 1

aþ 1pq

� �B 1

pq� 1

a� 1bþ k; 1bþ k; 1b

� �� 1r

�X1m¼1

mabþ

ac�akapm

!1p X1

n¼1n

bcþ

ba�bk

!1q X1

t¼1t

caþ

cb�ckcrt

!1r

:

ð2:34Þ


6. Substituting k = 1 in (2.34), then we obtain,

X1m¼1

X1n¼1

X1t¼1

ambnctmaþnbþ tc

61

bc

� �1p 1

ac

� �1q 1

ab

� �1r

B1

b� 1

qr;1� 1

bþ 1

qr

� �B

1

qr� 1

b�1

cþ1;

1

c

� �� 1p

� B1

c� 1

pr;1�1

cþ 1

pr

� �B

1

pr�1

c�1

aþ1;

1

a

� �� 1q

� B1

a� 1

pq;1�1

aþ 1

pq

� �B

1

pq�1

a� 1

bþ1;

1

b

� �� 1r

�X1m¼1

mabþ

ac�aapm

!1p X1

n¼1n

bcþ

ba�b

!1q X1

t¼1t

caþ

cb�ccrt

!1r

:

7. Let a = b = 1 in (2.29), then we get

X1m¼1

X1n¼1

X1t¼1

ambnctamþbnþctcð Þk 6

a1�kþ1c

bcc1c

� �1p

b1�kþ1c

acc1c

� �1q

c2�k

ab

� �1r

� B 1� 1qr; k� 1þ 1

qr

� �B 1

qr� 1

c � 1þ k; 1c

� �� 1p

� B 1c � 1

pr; k� 1

c þ 1pr

� �B 1

pr� 1

c � 1þ k; 1� �� 1

q

� B 1� 1pq; k� 1þ 1

pq

� �B 1

pq� 2þ k; 1

� �� 1r

�X1m¼1

m1cþ1�kapm

!1p X1

n¼1n1cþ1�kbqn

!1q X1

t¼1tð2�kÞccrt

!1r

:

ð2:35Þ

8. For k = 2,c = 2 in (2.35), we have

X1m¼1

X1n¼1

X1t¼1

ambnct

amþbnþct2ð Þ26 2

1�rr a

1�2p2p b

1�2q2q c

1�r2r B 1� 1

qr;1þ 1

qr

� �B 1

qrþ 1

2; 12

� �� 1p

� B 12� 1

pr; 32þ 1

pr

� �B 1

prþ 1

2;1

� �� 1q

B 1� 1pq;1þ 1

pq

� �B 1

pq;1

� �� 1r

�X1m¼1

m�12apm

!1p X1

n¼1n�

12bqn

!1q X1

t¼1crt

!1r

:

Acknowledgment

This work is supported by Deanship of Scientific Research,Taibah University, Kingdom of Saudi Arabia No. (3090-1434).

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on some generalizations of the hilbert–hardy type discrete inequalities

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