Generalizations of Hu-type inequalities and their applications

Available online at www.isr-publications.com/jnsa J. Nonlinear Sci. Appl., 10 (2017), 1971–1985 Research Article

Journal Homepage: www.tjnsa.com - www.isr-publications.com/jnsa

Generalizations of Hu-type inequalities and their applications Jingfeng Tiana,∗, Zhen-Hang Yangb a College

of Science and Technology, North China Electric Power University, Baoding, Hebei Province, 071051, P. R. China.

b Customer

Service Center, State Grid Zhejiang Electric Power Research Institute, Hangzhou, Zhejiang Province, 310009, P. R.

China.

Communicated by C. Zaharia

Abstract In this paper, we present some new generalizations of Hu-type inequalities, and then we obtain some new generalizations c ¨ and refinements of Holder’s inequality. 2017 All rights reserved. ¨ Keywords: Holder’s inequality, Hu-type inequality, generalization, refinement. 2010 MSC: 26D15, 26D10.

1. Introduction 1 p

¨ The classical Holder’s inequality states that if ak > 0, bk > 0 (k = 1, 2, · · · , n), p > 0, q > 0 and 1 + q = 1, then q1 X p1 X n n n X q p bk . (1.1) ak b k 6 ak k=1

k=1

k=1

The inequality (1.1) is reversed for p < 1 (p 6= 0); (For p < 1, we assume that ak , bk > 0). ¨ Holder’s inequality plays a very important role in both theory and applications. This classical inequality has been widely studied by many authors, and it has motivated a large number of research papers involving different proofs, various generalizations, variations and applications (see e.g., [1, 6–14, 16–18] and the references therein). Among various refinements of (1.1), Hu in [4] established the following interesting theorems. 1 p

Theorem 1.1. Let p > q > 0, (r, s = 1, 2, · · · , n). Then n X

Ak Bk 6

+

X n

k=1

1 q

Bq k

1 − 1 X n q

k=1

−

X n k=1

∗ Corresponding

= 1, let Ak , Bk > 0 (k = 1, 2, · · · , n), and let 1 − er + es > 0 p

Bq k

X n

k=1

Bq k ek

X n k=1

Ap k

Ap k

2

k=1

−

X n k=1

Bq k

X n

Ap k ek

k=1

author Email addresses: [email protected] (Jingfeng Tian), [email protected] (Zhen-Hang Yang)

doi:10.22436/jnsa.010.04.55 Received 2016-10-27

2

(1.2)

1 2p

.

J. F. Tian, Z.-H. Yang, J. Nonlinear Sci. Appl., 10 (2017), 1971–1985

1972

The integral form is as follows: Theorem 1.2. Let f(x), g(x), e(x) be integrable functions defined on [a, b] and f(x), g(x) > 0, 1 − e(x) + e(y) > 0 for all x, y ∈ [a, b], and let p > q > 0, p1 + q1 = 1. Then Zb

Zb f(x)g(x)dx 6 a

1 − 1 Z b 2 Zb q p gq (x)dx fp (x)dx gq (x)dx

a

Zb −

p

Zb

q

f (x)e(x)dx a

a

a

Zb

g (x)dx −

Zb

p

f (x)dx

a

a

(1.3)

2 1 2p g (x)e(x)dx . q

a

Later, Tian in [7] gave the reversed versions of Hu’s inequalities (1.2) and (1.3). Theorem 1.3. Let p < 0, q > 0, (r, s = 1, 2, · · · , n). Then n X

Ak Bk >

1 p

+

X n

k=1

1 q

= 1, let Ak > 0, Bk > 0 (k = 1, 2, · · · , n), and let 1 − er + es > 0

Bq k

1 − 1 X n q

p

k=1

−

Bq k

X n

k=1

X n

Bq k ek

X n

k=1

Ap k

Ap k

2

k=1

−

X n

k=1

Bq k

X n

k=1

Ap k ek

2

(1.4)

1 2p

.

k=1

The integral form is as follows: Theorem 1.4. Let f(x), g(x), e(x) be integrable functions defined on [a, b] and f(x) > 0, g(x) > 0, 1 − e(x) + e(y) > 0 for all x, y ∈ [a, b], and let p < 0, q > 0, p1 + q1 = 1. Then Zb

Zb f(x)g(x)dx >

a

2 1 − 1 Z b Zb q p fp (x)dx gq (x)dx gq (x)dx

a

Zb −

a

Zb

p

f (x)e(x)dx a

a

Zb

q

g (x)dx −

Zb

p

f (x)dx

a

a

(1.5)

2 1 2p g (x)e(x)dx . q

a

In 2007, Wu [16] presented the generalizations of Hu’s results, as follows: Theorem 1.5. Let Ar > 0, Br > 0 (r = 1, 2, · · · , n), let 1 − er + es > 0 (r, s = 1, 2, · · · , n), and let p > q > 0, µ = min{ p1 + q1 , 1}. Then X q1 − p1 X X 2 n n n n X 1−µ q q p Ar Br 6 n Br Br Ar r=1

r=1

−

X n

r=1

Bq r er

X n r=1

r=1

Ap − r

r=1

X n

Bq r

X n

Ap r er

2 2p1

(1.6) .

r=1

r=1

Theorem 1.6. Let f(x), g(x), e(x) be integrable functions defined on [a, b] and f(x) > 0, g(x) > 0, 1 − e(x) + e(y) > 0 for all x, y ∈ [a, b], and let p > q > 0, p1 + q1 6 1. Then Zb f(x)g(x)dx 6 (b − a) a

Zb − a

1− p1 − q1

Zb a

q1 − p1 Z b 2 Zb g (x)dx gq (x)dx fp (x)dx q

Zb

gq (x)e(x)dx

Zb fp (x)dx −

a

a

Zb

gq (x)dx a

a

fp (x)e(x)dx

(1.7)

2 2p1 .

a

In 2012, Tian [8] proved the following reversed versions of inequalities (1.6) and (1.7). Theorem 1.7. Let Ar > 0, Br > 0 (r = 1, 2, · · · , n), let 1 − er + es > 0 (r, s = 1, 2, · · · , n), and let q
0, µ = max{ p1 + q1 , 1}, λ = max{ q1 , −1}. Then n X

1−µ

Ar Br 6 n

X n

r=1

Ap r

p1 X n

r=1

Bq r

q1

r=1

Pn Pn 2 λ2 Bq Ar Br er r er r=1 r=1 Pn 1− − Pn . q r=1 Ar Br r=1 Br

Theorem 1.8. Let f(x), g(x), e(x) be integrable functions defined on [a, b] and f(x) > 0, g(x) > 0, 1 − e(x) + e(y) > 0 for all x, y ∈ [a, b], and let q < 0, p1 + q1 > 1, λ = max{−1, q1 }. Then Zb f(x)g(x)dx > (b − a) a

1− p1 − q1

Zb a

Rb

q1 p1 Z b q g (x)dx f (x)dx p

Rb

gq (x)e(x)dx Rb q a g (x)dx

f(x)g(x)e(x)dx − Rb a f(x)g(x)dx

a

× 1−

a

a

2 λ2 .

Later, in 2013, Tian and Hu [13] presented another reversed versions of inequalities (1.6) and (1.7). Theorem 1.9. Let Ar > 0, Br > 0 (r = 1, 2, · · · , n), let 1 − er + es > 0 (r, s = 1, 2, · · · , n), and let q < 0, p > 0, ρ = max{ p1 + q1 , 1}. Then X X p1 − q1 X 2 n n n n X 1−ρ p p q Ar Br > n Ar Ar Br r=1

r=1

r=1

−

X n

Ap r er

X n

r=1

Bq − r

r=1

X n

r=1

Ap r

X n

r=1

Bq r er

2 2q1

(1.8) .

r=1

Theorem 1.10. Let f(x), g(x), e(x) be integrable functions defined on [a, b] and f(x), g(x) > 0, 1 − e(x) + e(y) > 0 for all x, y ∈ [a, b], and let q < 0, p1 + q1 > 1. Then Zb f(x)g(x)dx > (b − a) a

1− p1 − q1

p1 − q1 Z b 2 Zb p q f (x)dx f (x)dx g (x)dx p

a

Zb −

Zb

Zb

p

q

f (x)e(x)dx a

a

Zb

p

g (x)dx − a

Zb a

In 2011, Tian in [7] gave the following generalizations of inequalities (1.2) and (1.3). P P λ Theorem 1.11. Let Anj > 0, n Anjj < ∞ (j = 1, 2, · · · , k), λ1 > λ2 > · · · > λk > 0, k j=1 P 1 − en + em > 0, n |en | < ∞. If k is even number, then k !2 λ 2 − λ2 " X X X k 2 Y XY 2j 2j−1 λ2j−1 λ2j−1 λ2j An(2j−1) An(2j−1) An(2j) Anj 6 n j=1

n

j=1

−

X

n λ

2j−1 An(2j−1) en

X

n

λ

2j An(2j)

(1.9)

2 2q1 . g (x)e(x)dx q

f (x)dx a

a

1 λj

= 1, and let

n

X !2 # λ12j X λ λ 2j−1 2j − An(2j−1) An(2j) en .

n

n

(1.10)

n

If k is odd number, then k XY

Anj 6

X

n j=1

k Aλ nk

n

×

2 k−1 2 λk

Y

λ2j−1 An(2j−1)

λ2j−1 An(2j−1)

2 2 λ2j−1 − λ2j

n

j=1

" X

X

X

n

n

λ2j An(2j)

!2 −

X

!2 # λ12j X λ2j−1 X λ2j . − An(2j−1) An(2j) en n

n

n

λ2j−1 An(2j−1) en

X n

λ2j An(2j)

(1.11)


1974

Theorem 1.12. Let Ak1 > 0, Akj > 0, (j = 1, 2, · · · , m, k = 1, 2, · · · , n), 0 (r, s = 1, 2, · · · , n). If λ1 > 0, λj < 0, (j = 2, 3, · · · , m), then n Y m X

Akj >

X n

k=1 j=1

Aλk11

λ1

1

P − m j=2

1 λj

m X n Y j=2

k=1

−

X n

Aλk11 ek

X n

k=1

Aλk11

λ

Akjj −

k=1

1 j=1 λj

X n

k=1

Pm

λ

Akjj

= 1, and let 1 − er + es >

2

k=1

X n

Aλk11

X n

k=1

λ

Akjj ek

(1.12)

2 2λ1

j

.

k=1

¨ The classic Holder inequality is an important cornerstone in different branches of modern mathematics such as classical real and complex analysis, probability and statistics, numerical analysis, qualitative theory of differential equations. It is also a bridge to help solve problems into depth. The Hu’s inequality ¨ (1.2), which was put forward by Hu in [4], improves the Holder inequality exquisitely. The mathematical ¨ reviews [5] calls it “an extraordinary, outstanding and new inequality”. The classic Holder inequality is playing a basic role in mathematics and can be applied in a wide range of areas, while the function of the Hu’s inequality is the same. The purpose of this work is to give some new generalizations of the above Hu-type inequalities (1.2), (1.3), (1.4), (1.5), (1.6), (1.7) and (1.8), (1.9), (1.10), (1.11), (1.12). Moreover, the obtained results will be ¨ applied to improve Holder’s inequality and Popoviciu-type inequality which is due to Wu and Debnath. 2. Main results We begin this section with some lemmas, which will be used in the sequel. Lemma 2.1 ([2]). If x > −1, α > 1 or α < 0, then (1 + x)α > 1 + αx. The inequality is reversed for 0 < α < 1. Lemma 2.2 ([3]). If xi > 0, λi > 0, i = 1, 2, · · · , n, 0 < p 6 1, then X 1−p X p n n n X p λi xi 6 λi λi xi . i=1

i=1

(2.1)

i=1

The inequality is reversed for p > 1 or p < 0. ¨ Lemma 2.3 (Generalized Holder’s inequality [15]). (a) Let Aij > 0 (i = 1, 2, · · · , n, j = 1, 2, · · · , m), and let λj > 0 with n Y m X

Aij 6

i=1 j=1

m X n Y j=1

Pm

1 j=1 λj

> 1. Then

λ1 Aij

j

.

(b) Let Aij > 0 (i = 1, 2, · · · , n, j = 1, 2, · · · , m), and let λ1 > 0, λj < 0 (j = 2, 3, · · · , m) with Then λ1 n n Y m m X Y X j Aij > Aij . i=1 j=1

j=1

i=1

(c) Let Aij > 0 (i = 1, 2, · · · , n, j = 1, 2, · · · , m), and let λj < 0 (j = 1, 2, · · · , m). Then n Y m X i=1 j=1

Aij >

m X n Y j=1

(2.2)

i=1

i=1

λ1 Aij

j

.

Pm

1 j=1 λj

6 1. (2.3)


1975

Next, we generalize the inequalities (1.2), (1.3), (1.6), (1.7), (1.10) and (1.11) as follows. Theorem 2.4. Let Arj > 0, (r = 1, 2, · · · , n, j = 1, 2, · · · , k), λ1 > λ2 > · · · > λk > 0, let ρ = min{ and let 1 − er + es > 0, (s = 1, 2, · · · , n). If k is even, then n Y k X

Arj 6 n1−ρ

r=1 j=1

k X n 2 Y

j=1

" n X

1 − λ1 2j 2j−1

λ

r=1

X n

−

λ2j−1 Ar(2j−1)

λ2j−1 Ar(2j−1)

X n

r=1

λ

2j−1 er Ar(2j−1)

X n

λ

2j Ar(2j) −

r=1

r=1

X n

λ2j Ar(2j)

λ

2j−1 Ar(2j−1)

r=1

1 j=1 λj , 1},

!2

r=1

X n

Pk

λ

2j Ar(2j) er

(2.4)

!2 # 2λ12j .

r=1

If k is odd, then n Y k X

1−ρ

Arj 6 n

X n

r=1 j=1

Aλrkk

k−1 X λ1 Y n 2 k

j=1

r=1

" n X

×

λ2j−1 Ar(2j−1)

−

λ2j Ar(2j)

r=1

λ

2j−1 Ar(2j−1)

X n

r=1

1 − λ1 2j 2j−1

λ

r=1

X n

r=1

X n

λ2j−1 Ar(2j−1)

λ

2j Ar(2j) er

!2

X n

−

λ2j−1 er Ar(2j−1)

λ2j Ar(2j)

(2.5)

r=1

r=1

!2 # 2λ12j

X n

.

r=1

Proof. Preforming some simple computations, we have n Y k X r=1

Arj

X n Y k

j=1

s=1

=

Asi (1 − er + es )

i=1

n X n Y k X r=1 s=1

+

j=1

X n Y k

Asi −

i=1

n X n Y k X r=1 s=1

=

Arj

Y k

Arj

Y k

j=1

n X n Y k X r=1 s=1

Arj

Y k

j=1

Asi er

i=1

(2.6)

Asi es

i=1

2 .

Arj

r=1 j=1

Case 1. Let λ1 > λ2 > · · · > λk > 0. P Subcase (1): When k is even, and k j=1 n Y k X r=1

Arj

X n Y k

j=1

=

s=1

>

=

Pk 1 Asi (1 − er + es ) j=1 λj

Arj

Y k

j=1

X n X n Y k r=1 s=1


i=1

Arj

Y k

j=1

i=1

X n X n Y k

Y k

r=1 s=1

> 1. From Lemma 2.2, we have

i=1

n X n Y k X r=1 s=1

1 λj

j=1

Arj

i=1

Asi

1−Pkj=1 λ1 X n X n Y k j r=1 s=1

Asi

1−Pkj=1 λ1 X n X n Y k j r=1 s=1

Arj

j=1

j=1

Y k

Pkj=1 λ1 j Asi (1 − er + es ) (2.7)

i=1

Arj

Y k i=1

Pkj=1 λ1 Asi

j


=

n X n Y k X r=1 s=1

=

j=1

X n Y k r=1

r=1

Arj

=

s=1

n Y k X r=1

6

r=1

=

n X

2 .

Arj

Pk

1 j=1 λj

Arj

i=1 1

Asi (1 − er + es ) λi

s=1 i=1

Arj

" Y k X n

j=1

i=1

Aλsii (1 − er + es )

λ1 # i

(2.8)

s=1

k " λ 1 − λ1 λ1 n n 2 Y X X 2j 2j 2j−1 λ2j−1 λ2j−1 λ2j−1 λ2j Ar(2j−1) As(2j−1) (1 − er + es ) Ar(2j−1) As(2j) (1 − er + es )

j=1

r=1

> 1, from inequality (2.2), we deduce


X n Y k

j=1

n Y k X

Asi

i=1

X n Y k

j=1

j=1

Moreover, in view of n Y k X

Arj

Y k

1976

s=1

s=1

λ1 # n X 2j λ(2j−1) λ2j × Ar(2j) As(2j−1) (1 − er + es ) . s=1

Hence, according to λ11 − λ12 + λ12 + λ12 + λ13 − λ14 + inequality (2.2) on the right side of (2.8), we have n Y k X r=1

Arj

X n Y k

j=1

s=1

k 2

6

" n Y X j=1

×

=

Y

λ2j−1 Ar(2j−1)

j=1

λ

×

n X

1 λk−1

−

1 λk

+

1 λk

+

1 λk

> 1, applying

1 − λ1 2j 2j−1

λ

λ2j−1 As(2j−1) (1 − er

+ es )

n X

λ

2j As(2j) (1 − er + es )

λ1 X n 2j

s=1 λ2j−1 Ar(2j−1)

λ

2j Ar(2j)

r=1 2 − λ2 2j 2j−1

λ

X n X n

r=1

X n X n

+···+

s=1

2j−1 Ar(2j−1)

n X

1 λ4

i=1

r=1

k 2

+


r=1

X n

1 λ4

n X

λ

2j−1 As(2j−1) (1 − er + es )

λ1 # 2j

s=1

λ2j−1 λ2j Ar(2j−1) As(2j) (1 − er

+ es )

r=1 s=1 λ2j−1 λ2j Ar(2j) As(2j−1) (1 − er

λ1 2j

+ es )

(2.9)

r=1 s=1

=

k X n 2 Y

j=1 n X

−

×

λ2j−1 Ar(2j−1)

2 − λ2 2j 2j−1

λ

X n

r=1

r=1

λ2j−1 Ar(2j−1) er

r=1 X n r=1

n X

λ2j As(2j)

+

s=1 λ2j Ar(2j)

λ

2j−1 Ar(2j−1)

n X s=1

λ2j−1 As(2j−1)

−

n X r=1 n X r=1

λ2j−1 Ar(2j−1)

λ2j Ar(2j) er

n X

λ

2j As(2j)

s=1 n X s=1 n X s=1

λ2j As(2j) es

λ2j−1 As(2j−1)

+

n X r=1

λ2j Ar(2j)

n X s=1

λ2j−1 As(2j−1) es

λ1 2j

J. F. Tian, Z.-H. Yang, J. Nonlinear Sci. Appl., 10 (2017), 1971–1985 k X n 2 Y

=

j=1

2 − λ2 2j 2j−1

λ

" n X

r=1

X n

−

λ2j−1 Ar(2j−1)

λ2j−1 Ar(2j−1)

X n

r=1

λ2j−1 er Ar(2j−1)

X n

λ2j Ar(2j)

−

X n

λ2j−1 Ar(2j−1)

r=1

Arj

X n Y k

j=1

=

s=1

n Y k X r=1

6

r=1

=

n X

λ2j Ar(2j) er

!2 # λ12j .

r=1

1 λj

= β (0 < β < 1), which implies that

i=1

Arj

1

Asi (1 − er + es ) βλi

s=1 i=1

Arj

" Y k X n

j=1

i=1

i Aβλ si (1 − er + es )

βλ1 # i

(2.10)

s=1

k " βλ 1 − βλ1 n 2 Y X 2j 2j−1 βλ2j−1 βλ2j−1 Ar(2j−1) As(2j−1) (1 − er + es )

j=1

r=1

!2

Asi (1 − er + es )

X n Y k

j=1

n Y k X

X n

r=1

Combining inequalities (2.7) and (2.9) yields inequality (2.4). P Pk 1 Subcase (2): When k is even, and k j=1 λj < 1. Let j=1 Pk 1 j=1 βλj = 1. From inequality (2.2) we find n Y k X

λ2j Ar(2j)

r=1

r=1

r=1

1977

s=1

βλ1 βλ1 # n n X X 2j 2j βλ2j−1 βλ2j−1 βλ2j βλ2j . × Ar(2j−1) As(2j) (1 − er + es ) Ar(2j) As(2j−1) (1 − er + es ) s=1

Consequently, in view of 1 βλk

s=1

1 βλ1

1 1 1 + βλ + βλ + − βλ 2 2 2

1 βλ3

1 1 1 − βλ + βλ + βλ +···+ 4 4 4

1 βλk−1

1 1 − βλ + βλ + k k

= 1, and applying inequality (2.2) on the right side of (2.10), we have n Y k X r=1

Arj

X n Y k

j=1

s=1

k 2

6

" n Y X j=1

×

i=1

βλ2j−1 Ar(2j−1)

r=1

X n

βλ2j−1 Ar(2j−1)

×

βλ 1

βλ2j−1 As(2j−1) (1 − er

2j−1

+ es )

n X

βλ2j As(2j) (1 − er

+ es )

βλ1 X n 2j

s=1

k " X n n X 2 Y

j=1

n X

− βλ1

2j

s=1

r=1

=

Asi (1 − er + es )

βλ2j Ar(2j)

r=1

βλ2j−1 βλ2j−1 Ar(2j−1) As(2j−1) (1 − er

βλ 1

2j−1

+ es )

n X

βλ1 # + es )

2j

s=1 − βλ1

2j

r=1 s=1

X n X n

βλ2j−1 βλ2j Ar(2j−1) As(2j) (1 − er

+ es )

βλ1 X n X n 2j

r=1 s=1

βλ2j βλ2j−1 Ar(2j) As(2j−1) (1 − er

r=1 s=1

Additionally, applying Lemma 2.2 on the right side of (2.11), we find k " X n X n 2 Y

j=1

βλ2j−1 As(2j−1) (1 − er

r=1 s=1

βλ2j−1 βλ2j−1 Ar(2j−1) As(2j−1) (1 − er

βλ 1 + es )

2j−1

− βλ1

2j

βλ1 # 2j + es ) .

(2.11)

J. F. Tian, Z.-H. Yang, J. Nonlinear Sci. Appl., 10 (2017), 1971–1985 X n X n

×

βλ2j−1 βλ2j As(2j) (1 − er Ar(2j−1)

+ es )

βλ1 X n X n 2j

r=1 s=1

1978

βλ2j βλ2j−1 Ar(2j) As(2j−1) (1 − er

βλ1 # 2j

+ es )

r=1 s=1

k 2

" n n (1−β)( βλ 1 − βλ1 ) Y X X 2j 2j−1 6 (1 − er + es ) j=1

r=1 s=1

X n X n

×

λ2j−1 λ2j−1 (1 − er As(2j−1) Ar(2j−1)

1 − λ1 2j 2j−1

λ + es )

r=1 s=1 1−β n X λ1 βλ n X n n XX 2j 2j λ2j−1 λ2j × (1 − er + es ) Ar(2j−1) As(2j) (1 − er + es )

r=1 s=1

r=1 s=1

1−β n λ1 # X βλ n n X n XX 2j 2j λ2j−1 λ2j × Ar(2j) As(2j−1) (1 − er + es ) (1 − er + es )

r=1 s=1

r=1 s=1 k 2

X λ 1 − λ1 1−β Y " X n X n X n n 2j 2j−1 λ2j−1 λ2j−1 = (1 − er + es ) Ar(2j−1) As(2j−1) (1 − er + es ) j=1

r=1 s=1

X n X n

×

λ2j−1 λ2j Ar(2j−1) As(2j) (1 − er

r=1 s=1

+ es )

λ1 X n X n 2j

= n2−2β

Y

j=1

2−2β

=n

Y j=1

n X

−

n X

λ

2j−1 Ar(2j−1)

+ es )

X n X n

×

k 2

2−2β

Y

j=1

−

λ2j−1 Ar(2j−1)

2 − λ2 2j 2j−1

λ

n X

λ2j As(2j)

+

s=1

r=1

=n

n X

X n r=1

X n

n X

λ2j−1 As(2j−1)

−

s=1 n X

n X r=1

s=1

n X

n X

λ2j Ar(2j) er

2j−1 Ar(2j−1) er

λ2j As(2j) es

2j

+ es )

r=1

λ

" n X

2j Ar(2j) −

X n r=1

λ

2j As(2j)

λ2j−1 As(2j−1)

+

n X

λ2j Ar(2j)

r=1 λ2j−1 Ar(2j−1)

r=1

X n

n X

s=1

2 2 λ2j−1 − λ2j

r=1 λ

λ1

s=1

λ2j−1 Ar(2j−1)

r=1 λ2j−1 Ar(2j−1)

λ2j−1 Ar(2j−1)

r=1 n X

r=1

λ2j Ar(2j)

λ2j−1 λ2j Ar(2j) As(2j−1) (1 − er

r=1 s=1

λ2j−1 Ar(2j−1) er

r=1 X n

(2.12)

2 − λ2 2j 2j−1

λ2j−1 λ2j Ar(2j−1) As(2j) (1 − er

2j

λ

r=1 s=1 k 2

+ es )

r=1

X n X n

×

λ1 #

r=1 s=1

r=1 s=1 k 2

λ2j−1 λ2j Ar(2j) As(2j−1) (1 − er

n X

λ2j Ar(2j)

λ

2j

!2

r=1

2j−1 Ar(2j−1)

λ1

s=1

X n

X n

λ2j−1 As(2j−1) es

λ

2j Ar(2j) er

!2 # λ12j .

r=1

Combining inequalities (2.6), (2.11), and (2.12) leads to inequality (2.4) immediately. P 1 Subcase (3): When k is odd, and k j=1 λj > 1. By the same method as in the above Subcase (1), we have the inequality (2.5). P 1 Subcase (4): When k is odd, and k j=1 λj < 1. By the same method as in the above Subcase (2), we have the inequality (2.5). Case 2. When λ1 > λ2 > · · · > λk > 0, at least one of the equalities holds. By the same way as in Case 1, we can obtain the desired results. The proof of Theorem 2.4 is completed.


1979

From Theorem 2.4 and Lemma 2.1, we obtain the generalizations and refinements of the generalized ¨ Holder’s inequality (2.2) as follows. P Pn λj 1 Corollary 2.5. Let Arj , λj , er be as in Theorem 2.4, let ρ = min{ k j=1 λj , 1}, and let r=1 Arj 6= 0. If k is even, then n Y k X

" Arj 6 n1−ρ

r=1 j=1

k X n Y j=1

λ Arjj

k λ1 # Y 2 j

j=1

r=1

1 1− 2λ2j

Pn

λ2j−1 r=1 Ar(2j−1) er Pn λ2j−1 r=1 Ar(2j−1)

λ2j 2 r=1 Ar(2j) er . Pn λ2j r=1 Ar(2j)

Pn −

(2.13)

If k is odd, then n Y k X

" 1−ρ

Arj 6 n

r=1 j=1

k X n Y j=1

λ Arjj

k−1 λ1 # Y 2 j

j=1

r=1

1 1− 2λ2j

Pn


λ2j 2 r=1 Ar(2j) er . Pn λ2j r=1 Ar(2j)

Pn −

(2.14)

Proof. We only need to prove the inequality (2.13). The proof of inequality (2.14) is similar. From inequality (2.4), we obtain n Y k X

" 1−ρ

Arj 6 n

r=1 j=1

k X n Y j=1

λ Arjj

k λ1 # Y 2 j

Pn 1−

j=1

r=1

λ2j 2 2λ1 2j r=1 Ar(2j) er . Pn λ2j A r=1 r(2j)

Pn


−

(2.15)

Furthermore, preforming some simple computations, we have Pn λ2j−1 λ2j Pn r=1 Ar(2j−1) er r=1 Ar(2j) er P < 1. − Pn λ2j−1 λ2j n A A r=1 r(2j) r=1 r(2j−1)

(2.16)

Consequently, from Lemma 2.1 and the inequalities (2.15) and (2.16), we have the desired inequality (2.13). The proof of Corollary 2.5 is complete. P 1 Theorem 2.6. Let λ1 > λ2 > · · · > λk > 0, k j=1 λj 6 1, let Fj (x), e(x) be nonnegative integrable functions defined on [a, b], and let 1 − e(x) + e(y) > 0. If k is even, then Zb Y k

Fj (x)dx 6 (b − a)

P 1− k j=1

1 λj

a j=1

k Zb 2 Y

a

j=1

Zb

Zb

λ

2j−1 F2j−1 (x)e(x)dx

− a

λ

" Z b a

Zb

F2j2j (x)dx −

a

1 − λ1 2j 2j−1

λ

λ2j−1 F2j−1 (x)dx

a

λ

2j−1 F2j−1 (x)dx

Zb a


Zb a

2

λ F2j2j (x)dx

(2.17)

2 # 2λ12j λ2j . F2j (x)e(x)dx

If k is odd, then Zb Y k

Fj (x)dx 6 (b − a)

a j=1

P 1− k j=1

1 λj

Zb a

×

k−1 Zb 2 Y

j=1

Zb − a

a

λ1

Fλkk (x)dx

1 − λ1 2j 2j−1

λ


λ2j−1 F2j−1 (x)e(x)dx

k

" Z b a

Zb a

λ F2j2j (x)dx −

Zb a



Zb a

Zb a

2

λ F2j2j (x)dx

(2.18) 2 # 2λ12j

λ F2j2j (x)e(x)dx

.


1980

Proof. For any positive integer n, we choose an equidistant partition of [a, b] as b−a b−a b−a < ··· < a+ r < ··· < a+ (n − 1) < b, n n n b−a b−a xr = a + r, ∆xr = , r = 1, 2, · · · , n. n n Case (a). When k is even, by the inequality (2.4) we have a < a+

n Y k X

P 1− k j=1

Fj (xr ) 6 n

1 λj

k X n 2 Y

j=1

r=1 j=1

−

X n

λ2j−1 (xr ) F2j−1

1 − λ1 2j 2j−1

λ

" n X

r=1

λ

2j−1 (xr )er F2j−1

X n

r=1

X n

λ

F2j2j (xr ) −

r=1

r=1

λ2j−1 (xr ) F2j−1

X n

λ F2j2j (xr )

r=1

λ

2j−1 (xr ) F2j−1

r=1

X n

λ

F2j2j (xr )er

!2

!2 # 2λ12j .

r=1

Hence, we have k 1 1 X n P 2 1 Y b − a λ2j−1 − λ2j b−a 1− k λ2j−1 j=1 λj 6 (b − a) F2j−1 (xr ) Fj (xr ) n n j=1 r=1 j=1 r=1 " n X !2 n X λ b − a b − a λ 2j−1 F2j2j (xr ) × F2j−1 (xr ) n n r=1 r=1 X n n X b−a b−a λ2j−1 λ2j F2j (xr ) − F2j−1 (xr )e(xr ) n n r=1 r=1 X X !2 # 2λ12j n n b−a b−a λ2j−1 λ2j − F2j−1 (xr ) . F2j (xr )e(xr ) n n

n Y k X

r=1

(2.19)

r=1

In view of the hypotheses that Fj (x), e(x) are positive Riemann integrable functions on [a, b], we λ λ conclude that Fj j (x), Fj j (x)e(x) are also integrable on [a, b]. Passing the limit as n → ∞ in both sides of inequality (2.19), we obtain inequality (2.17). Case (b). When k is odd, by the same method as in the above Case (a), we have the inequality (2.18). The proof of Theorem 2.6 is completed. ¨ From Theorem 2.6 we obtain the following generalizations and refinements of the generalized Holder’s inequality. Rb λ Corollary 2.7. Let Fj (x), λj , e(x) be as in Theorem 2.6, and let a Fj j (x)dx 6= 0. If k is even, then " k Z λ1 # Zb Y k P b Y 1 j 1− k λj j=1 λj Fj (x)dx 6 (b − a) Fj (x)dx a j=1

×

j=1 k

2 Y

j=1

1 1− 2λ2j

a

Rb

λ2j−1 a F2j−1 (x)e(x)dx Rb λ2j−1 a F2j−1 (x)dx

−

If k is odd, then Zb Y k a j=1

Fj (x)dx 6 (b − a)

P 1− k j=1

1 λj

"

k Zb Y j=1

a

λ2j 2 a F2j (x)e(x)dx . Rb λ2j F (x)dx a 2j

Rb

λ1 # j λ Fj j (x)dx

J. F. Tian, Z.-H. Yang, J. Nonlinear Sci. Appl., 10 (2017), 1971–1985 ×

Rb

k−1

2 Y

j=1

1981

λ2j−1 a F2j−1 (x)e(x)dx Rb λ2j−1 a F2j−1 (x)dx

1 1− 2λ2j

λ2j 2 a F2j (x)e(x)dx . Rb λ2j a F2j (x)dx

Rb −

Now, we present the following generalizations of inequalities (1.4), (1.5), (1.8), (1.9) and (1.12). P Theorem 2.8. Let Arj > 0, (r = 1, 2, · · · , n, j = 1, 2, · · · , m), λ1 > 0, λi < 0 (i = 2, 3, · · · , m), and m j=1 Pm 1 0, let τ = max{ j=1 λj , 1}, and let 1 − er + es > 0 (s = 1, 2, · · · , n). Then n Y m X

1−τ

Arj > n

X n

r=1 j=1

Aλr11

λ1

1

P − m j=2

1 λj

j=2

r=1

−

X n

m X n Y

Aλr11 er

X n

r=1

λ

Arjj

−

r=1

r=1

Arj

X n Y k

j=1

=

s=1

+

=

Arj

r=1

Y k

Asi −

i=1

n X n Y k X

1 λj

λ

Arjj er

(2.20)

2 2λ1

j

.

r=1

= t (t > 1), which implies

Arj

n X n Y k X r=1 s=1

Y k

j=1

X n Y k

r=1

X n

2

Pm

1 j=1 tλj

= 1.

Asi (1 − er + es )

j=1

r=1 s=1

Aλr11

λ Arjj

>

i=1

n X n Y k X r=1 s=1

X n

r=1

X n

Pm P 1 Proof. We first consider the case (I) m j=1 j=1 λj > 1. Let Preforming some simple computations, we have n Y k X

Aλr11

1 λj

Arj

j=1

Y k

Asi er

i=1

(2.21)

Asi es

i=1

2 Arj

.

r=1 j=1

From inequality (2.3), we have m n Y X s=1

X n Y m

Asi i=1 r=1 n m X Y

=

>

j=1

Asi

s=1

i=1

n Y m X s=1

Arj (1 − er + es )

X n Y m

1

Arj (1 − er + es ) tλj

r=1 j=1

Asi

" Y m X n

i=1

j=1

tλ Arj j (1 − er

tλ1 # + es )

j

(2.22)

r=1

1 m tλ1 −Pm tλ1 n n n j=2 tλj Y X X X j 1 tλj tλ1 tλ1 tλ1 = As1 Ar1 (1 − er + es ) As1 Arj (1 − er + es ) s=1

r=1

Y tλ1 m n X j tλj tλ1 × Asj Ar1 (1 − er + es ) . j=2

j=2

r=1

r=1

P 1 1 1 1 1 1 1 Consequently, in view of tλ1 1 − m j=2 tλj + tλ2 + tλ3 + · · · + tλm + tλ2 + tλ3 + · · · + tλm = 1, applying inequality (2.3) on the right side of (2.22), we find

J. F. Tian, Z.-H. Yang, J. Nonlinear Sci. Appl., 10 (2017), 1971–1985 n Y m X s=1

Asi

X n Y m

i=1

>

r=1

X n X n

Arj (1 − er + es )

j=1

tλ tλ As1 1 Ar1 1 (1 − er

Pm 1 1 tλ1 − j=2 tλj

+ es )

Y m X n X n j=2

s=1 r=1

×

Y m X n X n j=2

1982

tλ tλ Asj j Ar1 1 (1 − er

+ es )

1 tλj

tλ

tλ

As1 1 Arj j (1 − er + es )

1 tλj

(2.23)

s=1 r=1

.

s=1 r=1

Moreover, using Lemma 2.2 together with t > 1, we find X n X n

tλ1

tλ1 1 Atλ s1 Ar1 (1 − er

1

+ es )

P − m j=2

1 tλj

Y m X n X n j=2

s=1 r=1

Y m X n X n

×

j=2

tλ

1 Asj j Atλ r1 (1 − er + es )

tλ

j 1 Atλ s1 Arj (1 − er + es )

tλ1 j

s=1 r=1

tλ1 j

s=1 r=1

1 1 X X (1−t)( tλ1 −Pm λ1 −Pm n X n X n n j=2 tλj ) j=2 λj 1 1 λ1 λ1 > (1 − er + es ) As1 Ar1 (1 − er + es )

s=1 r=1

s=1 r=1

Y 1−t X λ1 m X n X n n X n tλj j λ1 λj × (1 − er + es ) As1 Arj (1 − er + es ) j=2

s=1 r=1

s=1 r=1

Y 1−t X λ1 m X n X n n X n tλj j λj λ1 × (1 − er + es ) Asj Ar1 (1 − er + es ) j=2

s=1 r=1

s=1 r=1

1 λ1 −Pm 1−t X X n X n n X n j=2 λj 1 λ1 λ1 (1 − er + es ) = As1 Ar1 (1 − er + es )

s=1 r=1

s=1 r=1

Y m X n X n

×

j=2

= n2−2t

λ Aλs11 Arjj (1 − er

s=1 r=1

X n

Aλr11

λ2

1

P − m j=2

2 λj

×

s=1 r=1 X n 2−2t

=n

×

λ Asjj Aλr11 (1 − er

Aλr11

λ2

1

P − m j=2

X n

= n2−2t

Aλs11

s=1 λ Asjj

s=1

s=1 r=1


+ es )

s=1 r=1

j

+ es )

n X

2 λj

Aλr11

λ Arjj

−

r=1

Aλr11

r=1

X n

n X

λ2

1

−

n X

X n r=1

Aλr11

r=1

Aλs11

s=1 λ Asjj

s=1 Pm 2 − j=2 λ

X n

n X

j

λ Arjj er

n X

λ Arjj er

r=1

n X

Aλr11 er

+

r=1

" n X

j=2

r=1

−

X n r=1

Aλs11 es

s=1 n X

λ Asjj es

s=1

m Y

+

n X

Aλr11

Aλr11 er

n X

λ Arjj

r=1

n X

Aλr11

r=1

X n

λ Arjj

r=1

X n r=1

λ Arjj

λ1 j

!2

!2 # λ1j .

λ1 + es )

j

(2.24)

λ1

r=1

−


r=1 m X n Y j=2

×

j=2 m X n X n Y j=2

r=1

X n X n

+ es )

λ1 Y m X n X n j


1983

Combining inequalities (2.21), (2.23) andP (2.24) leads to inequality (2.20) immediately. 1 Nextly, we consider the case (II) 0 < m j=1 λj 6 1. On the one hand, by the inequality (2.1), we have n Y m X s=1

X n Y m

Pm 1 Arj (1 − er + es ) j=1 λj

Asi i=1 r=1 j=1 Y n n m m XX Y

=

Asi

s=1 r=1

i=1

j=1

X n X n Y m

6

s=1 r=1

×

Asi

×

=

=

Y m

Asi

Asi

Y m j=1

Y m

Asi

s=1 r=1 i=1 n n m XX Y

Arj −

j=1

i=1

Arj es

Arj

n X n Y m X

s=1 r=1 Pm 1

Y m

Asi

Y m

i=1

(2.25)

Arj er

j=1

j=1 λj

1 n 1−Pm n Y m j=1 λj XX

s=1 r=1

j=1

Asi

Asi

i=1

Y m

1 Pm j=1 λ

Arj

j

j=1

Arj

j=1

2 .

Arj

r=1

j

j=1

X n X n Y m

s=1 r=1 i=1 X n Y m

1 1−Pm j=1 λ

Arj

Y m

i=1

n X n Y m X

1 Pm j=1 λj Arj (1 − er + es )

j=1

Asi

s=1 r=1 i=1 X n X n Y m

s=1 r=1

=

Y m

i=1

s=1 r=1

+

Asi

j

Arj

j=1

X n X n Y m

=

1 1−Pm j=1 λ

Y m

i=1

X n X n Y m s=1 r=1


j=1

On the other hand, by the inequality (2.3), we obtain n Y m X s=1

X n Y m

Asi i=1 r=1 n m X Y

=

>

j=1

X n Y m

Asi

s=1

i=1

n Y m X s=1

Pm 1 Arj (1 − er + es ) j=1 λj 1

Arj (1 − er + es ) λj

r=1 j=1

Asi

" Y m X n

i=1

j=1

λ Arjj (1 − er

λ1 # + es )

j

(2.26)

r=1

1 m λ1 −Pm λ1 n n n j=2 λj X X Y X j 1 λj λ1 λ1 λ1 As1 Arj (1 − er + es ) = As1 Ar1 (1 − er + es ) s=1

r=1

Y λ1 m n X j λj λ1 × Asj Ar1 (1 − er + es ) . j=2

r=1

j=2

r=1

J. F. Tian, Z.-H. Yang, J. Nonlinear Sci. Appl., 10 (2017), 1971–1985 P 1 1 1 Consequently, according to λ11 − m j=2 λj + λ2 + λ3 + · · · + inequality (2.3) on the right side of (2.26), we observe that n Y m X s=1

Asi

X n Y m

i=1

r=1

X n X n

>

Aλs11 Aλr11 (1 − er

P − m j=2

λ1

1

+ es )

1 λj

Y m X n X n j=2

s=1 r=1

×

j=2

X n

=

λ

Asjj Aλr11 (1 − er + es )

s=1 r=1 λ2 −Pm j=2 1

Aλr11

2 λj

×


s=1 r=1 X λ2 −Pm n j=2 1 Aλr11 r=1

=

×

X n s=1

X n

=

λ

Asjj

Aλr11

n X

2 λj

−

Aλr11 −

r=1 P − m j=2

λ2

1

1 λm

6 1, by using the

λ Arjj er

Aλs11

s=1 λ

Asjj

m Y

λ1 + es )

j

s=1 r=1

j


+ es )

(2.27)

m X n Y

n X


j

+ es )

n X

n X

λ Arjj

−

r=1

Aλr11 er +

n X

Aλs11

s=1 n X

λ

Asjj es

r=1

s=1

r=1

X n

!2

Aλr11

r=1

−

X n

r=1

Aλr11 er

X n

λ Arjj er

+

λ Arjj

Aλr11

λ1

n X

Aλs11 es

s=1

n X

λ Arjj

r=1

j

!2 # λ1j .

r=1

r=1

r=1

λ

Arjj

n X r=1

n X

" n X

s=1

2 λj

X n

r=1

+ ··· +

λ1

j=2

Aλr11

1 λ3

s=1 r=1

j=2

r=1

X n

+

λ1

m X n X n Y j=2

r=1

X n X n

1 λ2

+


j=1

Y m X n X n

1 λm

1984

Combining inequalities (2.25) and (2.27) leads to inequality (2.20) immediately. The proof of Theorem 2.8 is completed. From Theorem 2.8 and Lemma 2.1, we obtain the following generalizations and refinements of gener¨ alized Holder’s inequality (2.3). Corollary 2.9. Let Arj , λj , er be as in Theorem 2.8, let τ = max{ n Y m X

" 1−τ

Arj > n

r=1 j=1

m X n Y j=1

λ Arjj

λ1 # Y m j j=2

r=1

1 1− 2λj

Pm

1 j=1 λj , 1},

and let

λj r=1 Arj

6= 0. Then

λj 2 r=1 Arj er . Pn λj A r=1 rj

Pn

Pn

λ1 r=1 Ar1 er Pn λ1 r=1 Ar1

Pn

−

Theorem 2.10. Let FP j (x), e(x) be integrable functions defined on [a, b] and Fj (x) > 0, 1 − e(x) + e(y) > 0 for all 1 x, y ∈ [a, b], and let m j=1 λj > 1. If λ1 > 0, λj < 0 (j = 2, 3, · · · , m), then Zb Y m

Fj (x)dx > (b − a)

a j=1

P 1− m j=1

1 λj

Zb a

Zb − a

Fλ1 1 (x)e(x)dx

λ1

Fλ1 1 (x)dx

Zb a

λ

1

P − m j=2

Fj j (x)dx −

Zb a

1 λj

m Z b Y j=2

Fλ1 1 (x)dx

a

Zb a

Fλ1 1 (x)dx

Zb a

2 λ Fj j (x)dx

2 2λ1 j λj Fj (x)e(x)dx .

(2.28)


1985

Proof. Making similar method as in the proof of Theorem 2.6 by using inequality (2.20), we have the desired inequality (2.28). Rb λ Corollary 2.11. Let Fj (x), λj , e(x) be as in Theorem 2.10, and let a Fj j (x)dx 6= 0. Then, we have the following generalization and refinement of generalized Hölder’s inequality (1.9). Zb Y m

Fj (x)dx > (b − a)

a j=1

×

P 1− m j=1

1 λj

"

m Zb Y j=1

m Y j=2

1 1− 2λj

a

λ1 #

λ Fj j (x)dx

(2.29)

λj 2 a Fj (x)e(x)dx . Rb λj F (x)dx a j

Rb

Rb

λ1 a F1 (x)e(x)dx Rb λ1 a F1 (x)dx

j

−

Proof. Making similar arguments as in the proof of Corollary 2.5, we have the desired inequality (2.29). Acknowledgment The authors would like to express their sincere thanks to the anonymous referees for their great efforts to improve this paper. This work was supported by the Application Basic Research Plan Key Basic Research Project of Hebei Province of China (No. 16964213D), the Fundamental Research Funds for the Central Universities (No. 2015ZD29, 13ZD19) and the Higher School Science Research Funds of Hebei Province of China (No. Z2015137). References ˇ [1] H. Agahi, Y. Ouyang, R. Mesiar, E. Pap, M. Strboja, Hölder and Minkowski type inequalities for pseudo-integral, Appl. Math. Comput., 217 (2011), 8630–8639. 1 [2] E. F. Beckenbach, R. Bellman, Inequalities, Second revised printing, Ergebnisse der Mathematik und ihrer Grenzgebiete, Neue Folge, Band 30 Springer-Verlag, New York, Inc., (1965). 2.1 ´ [3] G. H. Hardy, J. E. Littlewood, G. Polya, Inequalities, Reprint of the 1952 edition, Cambridge Mathematical Library, Cambridge University Press, Cambridge, (1988). 2.2 [4] K. Hu, On an inequality and its applications, Sci. Sinica, 24 (1981), 1047–1055. 1, 1 [5] J.-C. Kuang, Applied inequalities, Shandong Science and Technology Press, Jinan, China, (2004). 1 [6] J. Matkowski, A converse of the Hölder inequality theorem, Math. Inequal. Appl., 12 (2009), 21–32. 1 [7] J.-F. Tian, Extension of Hu Ke’s inequality and its applications, J. Inequal. Appl., 2011 (2011), 14 pages. 1, 1 [8] J.-F. Tian, Reversed version of a generalized sharp Hölder’s inequality and its applications, Inform. Sci., 201 (2012), 61–69. 1 [9] J.-F. Tian, Property of a Hölder-type inequality and its application, Math. Inequal. Appl., 16 (2013), 831–841. [10] J.-F. Tian, New property of a generalized Hölder’s inequality and its applications, Inform. Sci., 288 (2014), 45–54. [11] J.-F. Tian, Properties of generalized Hölder’s inequalities, J. Math. Inequal., 9 (2015), 473–480. [12] J.-F. Tian, M.-H. Ha, Properties of generalized sharp Hölder’s inequalities, J. Math. Inequal., 11 (2017), 511–525. [13] J.-F. Tian, X.-M. Hu, A new reversed version of a generalized sharp Hölder’s inequality and its applications, Abstr. Appl. Anal., 2013 (2013), 9 pages. 1 [14] J.-F. Tian, W. Pedrycz, New refinements of generalized Hölder’s inequality and their applications, Math. Inequal. Appl., 19 (2016), 805–822. 1 [15] P. M. Vasić, J. E. Peˇcarić, On the Jensen inequality for monotone functions, An. Univ. Timis¸oara Ser. S¸tiint¸. Mat., 17 (1979), 95–104. 2.3 [16] S.-H. Wu, Generalization of a sharp Hölder’s inequality and its application, J. Math. Anal. Appl., 332 (2007), 741–750. 1, 1 [17] C.-J. Zhao, W.-S. Cheung, On reverse Minkowski-type inequalities, Mediterr. J. Math., 12 (2015), 1085–1094. [18] C.-J. Zhao, W. S. Cheung, Hölder’s reverse inequality and its applications, Publ. Inst. Math. (Beograd) (N.S.), 99 (2016), 211–216. 1