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Article

Some New Newton’s Type Integral Inequalities for Co-Ordinated Convex Functions in Quantum Calculus

by
Miguel Vivas-Cortez
1,*,†,
Muhammad Aamir Ali
2,†,
Artion Kashuri
3,†,
Ifra Bashir Sial
4,† and
Zhiyue Zhang
2,†
1
Pontificia Universidad Católica del Ecuador, Facultad de Ciencias Exactas y Naturales, Escuela de Ciencias Matemáticas y Físicas, Av. 12 de octubre 1076, Apartado 17-01-2184, Ecuador
2
Jiangsu Key Laboratory for NSLSCS, School of Mathematical Sciences, Nanjing Normal University, Nanjing 210023, China
3
Department of Mathematics, Faculty of Technical Science, University Ismail Qemali, 9400 Vlora, Albania
4
School of Control Science and Engineering, Jiangsu University, Zhenjiang 212013, China
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2020, 12(9), 1476; https://doi.org/10.3390/sym12091476
Submission received: 23 July 2020 / Revised: 3 August 2020 / Accepted: 6 August 2020 / Published: 8 September 2020

Abstract

:
Some recent results have been found treating the famous Simpson’s rule in connection with the convexity property of functions and those called generalized convex. The purpose of this article is to address Newton-type integral inequalities by associating with them certain criteria of quantum calculus and the convexity of the functions of various variables. In this article, by using the concept of recently defined q 1 q 2 -derivatives and integrals, some of Newton’s type inequalities for co-ordinated convex functions are revealed. We also employ the limits of q 1 , q 2 1 in new results, and attain some new inequalities of Newton’s type for co-ordinated convex functions through ordinary integral. Finally, we provide a thorough application of the newly obtained key outcomes, these new consequences can be useful in the integral approximation study for symmetrical functions, or with some kind of symmetry.

1. Introduction

For numerical integration and approximations of definite integrals, Simpson’s rules are well-known techniques. Thomas Simpson (1710–1761) was the founder of these known techniques. These techniques are also called Kepler’s rules because Johannes Kepler used similar techniques for numerical integration about 100 years ago. Simpson’s rule contains the three-point Newton-Cotes quadrature rule, so estimates depended on the three-step quadratic core are sometimes called Newton-type results.
(1)
Simpson’s quadrature formula (Simpson’s 1 / 3 rule)
1 ε α α ε F ( s ) d s 1 6 F ( α ) + 4 F α + ε 2 + F ( ε ) .
(2)
Simpson’s second formula or Newton–Cotes quadrature formula (Simpson’s 3 / 8 rule).
1 ε α α ε F ( s ) d s 1 8 F ( α ) + 3 F 2 α + ε 3 + 3 F α + 2 ε 3 + F ( ε ) .
There are numerous estimations correlated to these quadrature rules in the literature, one of them is the subsequent estimation identified as Simpson’s inequality:
Theorem 1.
Suppose that F : α , ε R is a four times continuously differentiable mapping on α , ε , and let F 4 = sup s α , ε F 4 ( s ) < . Then, one has the inequality
1 3 F ( α ) + F ( ε ) 2 + 2 F α + ε 2 1 ε α α ε F ( s ) d s 1 2880 F 4 ε α 4 .
In recent years, especially over the past two decades, several authors have been engaged in the study of inequalities, including the Simpson’s various function classes (Symmetric or Asymmetric). Particularly, some mathematicians have dedicated the most to the study of Simpson and Newton-type consequences for functions with several kinds of generalized convexity, given that the theory of convexity it is an appropriate way to solve a huge number of problems appearing in different areas and subareas of applied and pure mathematics. For an instance, In Reference [1], Dragomir et al. proved some new inequalities of Simpson’s type and gave some application of numerical integration using the obtained results. In Reference [2], Alomari et al. used the notion of s–convexity and proved some new inequalities of Simpson’s type with the application of numerical integration. Afterward, Sarikaya et al. observed the variants of Simpson’s type inequalities based on convexity in Reference [3]. On the other hand, Özdemir et al. used the concept of co–ordinated convexity and proved Simpson’s type inequalities for double integrals in Reference [4]. In Reference [5,6], the authors utilized the concept of harmonic and p–harmonic convexities and gave some Newton-type inequalities. Moreover, Iftikhar et al. in Reference [7] proved some new inequalities of Newton’s type for the functions whose local fractional derivatives are generalized convex.
On the other hand, quantum calculus or q–calculus is sometimes referred to as calculus without limits. In this, we gain q–analogs of mathematical items that maybe got back as q 1 . The Nalli-Ward-Al-Salam q–addition (NWA) and the Jackson-Hahn-Cigler q-addition (JHC) are two kinds of q-addition in this subject. The first one is commutative and associative, but at the same time, the second one is not. That’s why from time to time several q–analogs exist. These operators form the basis of the method which associations hypergeometric collection and q–hypergeometric collection and which gives numerous formulations of q–calculus in a usual form. The history of quantum calculus may be traced reverse to Euler ( 1707 1783 ) , who first added the q in the tracks of Newton’s infinite series. In recent decades, numerous researchers have revealed a keen hobby in investigating quantum calculus accordingly it emerges as an interdisciplinary subject. This is, of course, the quantum analysis is extremely useful in numerous fields and has vast applications in different areas of natural sciences such as computer science and particle physics and furthermore acts as a vital tool for researchers working with analytic number theory or in theoretical physics. Quantum calculus can be considered as a link between Mathematics and Physics. Several scientists who employ quantum calculus are physicists, as quantum calculus has numerous applications in quantum group theory. For some recent consequences in quantum calculus concerned readers are referred to References [8,9,10,11,12,13,14].
In recent years, because of the importance of convexity in numerous fields of applied and pure mathematics, it has been significantly investigated. The theory of convexity and inequalities are strongly connected to each other, therefore various inequalities can be established inside the literature which are proved for convex, generalized convex and differentiable convex functions of single and double variables, see References [15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32].
Inspired by these ongoing studies, we establish some new quantum analogs of Newton’s inequalities for q–differentiable co-ordinated convex functions. This is the primary motivation of this paper. The ideas and strategies of the paper may open new venues for the further research in this field.
Throughout in this paper the real numbers q , q 1 , q 2 0 , 1 .

2. Preliminaries of q–Calculus and Some Inequalities

In this section we present some required definitions and related inequalities about q–calculus. For more information about q–calculus, one can refer to References [9,10,11].
Definition 1
([13]). For a continuous function F : a , ε R , the q–derivative of F at s a , ε is characterized by the expression
a d q F s = F s F q s + 1 q a 1 q s a , s a .
The function F is said to be q–differentiable on a , ε if a d q F s exists for all s a , ε . If a = 0 in (1), then 0 d q F s = d q F s , where d q F s is the familiar q-derivative of F at s a , ε defined by the expression (see Reference [12]):
d q F s = F s F q s 1 q s , s 0 .
Definition 2
([13]). Let F : a , ε R be a continuous function. Then, the q a -definite integral on a , ε is defined as
a s F t a d q t = 1 q s a n = 0 q n F q n s + 1 q n a
for s a , ε .
We have to give the following notation which will be used many times in the next sections (see Reference [12]):
n q = q n 1 q 1 .
Moreover, we will need the following lemma in our main results:
Lemma 1
([33]). For α R 1 the following equality holds:
a ε s a α a d q s = ε a α + 1 α + 1 q .
On the other hand, Bermudo et al. gave the following new definition and related Hermite– Hadamard type inequalities:
Definition 3
([17]). Let F : a , ε R be a continuous function. Then, the q ε -definite integral on a , ε is defined as
s ε F t ε d q t = 1 q ε s n = 0 q n F q n s + 1 q n ε ,
for all s a , ε .
Theorem 2
([17]). If F : a , ε R is a convex differentiable function on a , ε and 0 < q < 1 . Then, we have the q-Hermite–Hadamard inequalities
F a + q ε 2 q 1 ε a a ε F s ε d q s F a + q F ε 2 q .
In Reference [20], Latif defined q a c -integral and partial q-derivatives for two variables functions as follows:
Definition 4.
Suppose that F : a , ε × c , d R 2 R is continuous function. Then, the definite q a c -integral on a , ε × c , d is defined by
a s c y F t , s c d q 2 s a d q 1 t = 1 q 1 1 q 2 s a y c × n = 0 m = 0 q 1 q 2 F q 1 n s + 1 q 1 n a , q 2 m y + 1 q 2 m c ,
for all s , y a , ε × c , d .
Lemma 2
([18]). If the assumptions of Definition 4 holds, then
y 1 y s 1 s F t , s a d q 1 t c d q 2 s = y 1 y a s F t , s a d q 1 t c d q 2 s y 1 y a s 1 F t , s a d q 1 t c d q 2 s = c y a s F t , s a d q 1 t c d q 2 s c y 1 a s F t , s a d q 1 t c d q 2 s c y a s 1 F t , s a d q 1 t c d q 2 s + c y 1 a s 1 F t , s a d q 1 t c d q 2 s .
Definition 5
([20]). Let F : a , b × c , d R 2 R be a continuous function of two variables. Then the partial q 1 -derivatives, q 2 -derivatives and q 1 q 2 -derivatives at s , y a , b × c , d can be given as follows:
a q 1 F s , y a q 1 s = F q 1 s + 1 q 1 a , y F s , y 1 q 1 s a , s b ; c q 1 F s , y c q 2 y = F s , q 2 y + 1 q 2 c F s , y 1 q 2 y c , y c ; a , c q 1 , q 2 2 F s , y a q 1 s c q 2 y = 1 s a y c 1 q 1 1 q 2 F q 1 s + 1 q 1 a , q 2 y + 1 q 2 c F q 1 s + 1 q 1 a , y F s , q 2 y + 1 q 2 c + F s , y , s a , y c .
For more details related to q-integrals and derivatives for the functions of two variables one can see Reference [20].
On the other hand, Budak et al. gave the following definitions of q a d , q b c and q b d integrals and related inequalities of Hermite–Hadamard type:
Definition 6
([34]). Suppose that F : a , b × c , d R 2 R is continuous function. Then the following q a d , q c b and q b d integrals on a , b × c , d are defined by
a s y d F t , s d d q 2 s a d q 1 t = 1 q 1 1 q 2 s a d y × n = 0 m = 0 q 1 q 2 F q 1 n s + 1 q 1 n a , q 2 m y + 1 q 2 m d ,
s b c y F t , s c d q 2 s b d q 1 t = 1 q 1 1 q 2 b s y c × n = 0 m = 0 q 1 q 2 F q 1 n s + 1 q 1 n b , q 2 m y + 1 q 2 m c
and
s b y d F t , s d d q 2 s b d q 1 t = 1 q 1 1 q 2 b s d y × n = 0 m = 0 q 1 q 2 F q 1 n s + 1 q 1 n b , q 2 m y + 1 q 2 m d ,
respectively, for all s , y a , b × c , d .
Theorem 3
([34]). Let F : a , b × c , d R 2 R be a coordinated convex function on a , b × c , d . Then we have the following inequalities:
F q 1 a + b 2 q 1 , c + q 2 d 2 q 2 1 2 1 b a a b F s , c + q 2 d 2 q 2 a d q 1 s + 1 d c c d F q 1 a + b 2 q 1 , y d d q 2 y 1 b a d c a b c d F s , y d d q 2 y a d q 1 s q 1 2 2 q 1 d c c d F a , y d d q 2 y + 1 2 2 q 1 d c c d F b , y d d q 2 y + 1 2 2 q 2 b a a b F s , c a d q 1 s + q 2 2 2 q 2 b a a b F s , d a d q 1 s q 1 F a , c + q 1 q 2 F a , d + F b , c + q 2 F b , d 2 q 1 2 q 2
for all q 1 , q 2 0 , 1 .
Theorem 4
([34]). Let F : a , b × c , d R 2 R be a coordinated convex function on a , b × c , d . Then we have the following inequalities:
F a + q 1 b 2 q 1 , q 2 c + d 2 q 2 1 2 1 b a a b F x , q 2 c + d 2 q 2 b d q 1 x + 1 d c c d F a + q 1 b 2 q 1 , y c d q 2 y 1 b a d c a b c d F x , y c d q 2 y b d q 1 x 1 2 2 q 1 d c c d F a , y c d q 2 y + q 1 2 2 q 1 d c c d F b , y c d q 2 y + q 2 2 2 q 2 b a a b F x , c b d q 1 x + 1 2 2 q 2 b a a b F x , d b d q 1 x q 2 F a , c + F a , d + q 1 q 2 F b , c + q 1 F b , d 2 q 1 2 q 2
for all q 1 , q 2 0 , 1 .
Theorem 5
([34]). Let F : a , b × c , d R 2 R be a coordinated convex function on a , b × c , d . Then we have the following inequalities:
F a + q 1 b 2 q 1 , c + q 2 d 2 q 2 1 2 1 b a a b F x , c + q 2 d 2 q 2 b d q 1 x + 1 d c c d F a + q 1 b 2 q 1 y , d d q 2 y 1 b a d c a b c d F x , y d d q 2 y b d q 1 x 1 2 2 q 1 d c c d F a , y d d q 2 y + q 1 2 2 q 1 d c c d F b , y d d q 2 y + 1 2 2 q 2 b a a b F x , c d d q 2 y + q 2 2 2 q 2 b a a b F x , d b d q 1 x F a , c + q 2 F a , d + q 1 F b , c + q 1 q 2 F b , d 2 q 1 2 q 2
for all q 1 , q 2 0 , 1 .
Theorem 6.
( q 1 q 2 -Hölder’s inequality for two variables functions, [20]). Let x , y > 0 , 0 < q 1 , q 2 < 1 , p 1 > 1 such that 1 p 1 + 1 r 1 = 1 . Then
0 x 0 y F x , y G x , y d q 1 x d q 2 y 0 x 0 y F x , y p 1 d q 1 x d q 2 y 1 p 1 0 x 0 y G x , y r 1 d q 1 x d q 2 y 1 r 1 .

3. New q–Derivatives for the Functions of Two Variables

In this section, some new partial q-derivatives for functions of two variables are given.
Definition 7
([35]). Let F : a , b × c , d R 2 R be a continuous function of two variables. Then the partial q 1 -derivatives, q 2 -derivatives and q 1 q 2 -derivatives at x , y a , b × c , d can be given as follows:
b q 1 F x , y b q 1 x = F q 1 x + 1 q 1 b , y F x , y 1 q 1 b x , x b ; d q 1 F x , y b q 2 y = F x , q 2 y + 1 q 2 d F x , y 1 q 2 d y , d y ; a d q 1 , q 2 2 F x , y a q 1 x d q 2 y = 1 x a d y 1 q 1 1 q 2 F q 1 x + 1 q 1 a , q 2 y + 1 q 2 d F q 1 x + 1 q 1 a , y F x , q 2 y + 1 q 2 d + F x , y , x a , y d ; c b q 1 , q 2 2 F x , y b q 1 x c q 2 y = 1 b x y c 1 q 1 1 q 2 F q 1 x + 1 q 1 b , q 2 y + 1 q 2 c F q 1 x + 1 q 1 b , y F x , q 2 y + 1 q 2 c + F x , y , x b , y c ; b , d q 1 , q 2 2 F x , y b q 1 x d q 2 y = 1 b x d y 1 q 1 1 q 2 F q 1 x + 1 q 1 b , q 2 y + 1 q 2 d F q 1 x + 1 q 1 b , y F x , q 2 y + 1 q 2 d + F x , y , x b , y d .

4. New Identity

We deal with an identity which is necessary to attain our main estimations in this section.
Let’s start with the following useful Lemma 3:
Lemma 3.
Let F : Δ R 2 R be a twice partially q 1 q 2 -differentiable function on Δ . If the partial q 1 q 2 -derivative b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s is continuous and integrable on a , b × c , d Δ , then the following identity holds for q 1 q 2 -integrals:
b , d I q 1 , q 2 = b a d c 0 1 0 1 Λ q 1 t Λ q 2 s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s ,
where
b , d I q 1 , q 2 = 3 64 F a + 2 b 3 , d + F b , c + 2 d 3 + 3 F a + 2 b 3 , c + 2 d 3 + 3 F a + 2 b 3 , 2 c + d 3 + F 2 a + b 3 , d + 3 F 2 a + b 3 , c + 2 d 3 + F b , 2 c + d 3 + 3 F 2 a + b 3 , 2 c + d 3 + F a + 2 b 3 , c + F 2 a + b 3 , c + F a , c + 2 d 3 + F a , 2 c + d 3 + F a , c + F a , d + F b , c + F b , d 64 1 8 b a a b F x , c + 3 F x , c + 2 d 3 + 3 F x , 2 c + d 3 + F x , d b d q 1 x 1 8 d c c d F a , y + 3 F a + 2 b 3 , y + 3 F 2 a + b 3 , y + F b , y d d q 2 y + 1 b a d c a b c d F x , y b d q 1 x d d q 2 y
and
Λ q 1 t = q 1 t 1 8 , t 0 , 1 3 q 1 t 1 2 , t 1 3 , 2 3 q 1 t 7 8 , t 2 3 , 1 ,
Λ q 2 s = q 2 s 1 8 , s 0 , 1 3 q 2 s 1 2 , s 1 3 , 2 3 q 2 s 7 8 , s 2 3 , 1 .
Proof. 
Using Lemma 2, the definition of Λ q 1 t and Λ q 2 s , it is easy to see that
0 1 0 1 Λ q 1 t Λ q 2 s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 9 64 0 1 3 0 1 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 1 3 0 2 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 2 3 0 1 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 2 3 0 2 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 3 8 0 1 3 0 1 q 2 s 7 8 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 2 3 0 1 q 2 s 7 8 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 1 0 1 3 q 1 t 7 8 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 1 0 2 3 q 1 t 7 8 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s + 0 1 0 1 q 1 t 7 8 q 1 t 7 8 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = I 1 + I 2 + I 3 + I 4 + I 5 + I 6 + I 7 + I 8 + I 9 .
From Definition 7, we have
b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s = 1 1 q 1 1 q 2 b a d c t s F t q 1 a + 1 t q 1 b , s q 2 c + 1 s q 2 d F t q 1 a + 1 t q 1 b , s c + 1 s d F t a + 1 t b , s q 2 c + 1 s q 2 d + F t a + 1 t b , s c + 1 s d .
We need to calculate the integrals in the right side of (12) in order to finish this proof. By using the definition of q 1 q 2 -integrals, we obtain that
0 1 3 0 1 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 1 q 1 1 q 2 b a d c 0 1 3 0 1 3 1 t s F t q 1 a + 1 t q 1 b , s q 2 c + 1 s q 2 d F t q 1 a + 1 t q 1 b , s c + 1 s d F t a + 1 t b , s q 2 c + 1 s q 2 d + F t a + 1 t b , s c + 1 s d d q 1 t d q 2 s = 1 b a d c n = 0 m = 0 F q 1 n + 1 3 a + 1 q 1 n + 1 3 b , q 2 m + 1 3 c + 1 q 2 m + 1 3 d n = 0 m = 0 F q 1 n + 1 3 a + 1 q 1 n + 1 3 b , q 2 m 3 c + 1 q 2 m 3 d n = 0 m = 0 F q 1 n 3 a + 1 q 1 n 3 b , q 2 m + 1 3 c + 1 q 2 m + 1 3 d + n = 0 m = 0 F q 1 n 3 a + 1 q 1 n 3 b , q 2 m 3 c + 1 q 2 m 3 d = 1 b a d c F b , d F a + 2 b 3 , d F b , c + 2 d 3 + F a + 2 b 3 , c + 2 d 3 .
Similarly, we get
0 1 3 0 2 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F a + 2 b 3 , d F b , 2 c + d 3 + F a + 2 b 3 , 2 c + d 3 ,
0 2 3 0 1 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F 2 a + b 3 , d F b , c + 2 d 3 + F 2 a + b 3 , c + 2 d 3 ,
0 2 3 0 2 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F 2 a + b 3 , d F b , 2 c + d 3 + F 2 a + b 3 , 2 c + d 3 ,
0 1 3 0 1 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F a + 2 b 3 , d F b , c + F a + 2 b 3 , c ,
0 1 0 1 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F b , c + 2 d 3 F a , d + F a , c + 2 d 3 ,
0 2 3 0 1 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F 2 a + b 3 , d F b , c + F 2 a + b 3 , c ,
0 1 0 2 3 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F a , d F b , 2 c + d 3 + F a , 2 c + d 3 ,
0 1 0 1 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c F b , d F a , d F b , c + F a , c .
Now from Definition 6, we obtain the following
0 1 3 0 1 s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 1 q 1 1 q 2 b a d c 0 1 3 0 1 1 t F t q 1 a + 1 t q 1 b , s q 2 c + 1 s q 2 d F t q 1 a + 1 t q 1 b , s c + 1 s d F t a + 1 t b , s q 2 c + 1 s q 2 d + F t a + 1 t b , s c + 1 s d d q 1 t d q 2 s
= 1 b a d c n = 0 m = 0 q 2 m F q 1 n + 1 3 a + 1 q 1 n + 1 3 b , q 2 m + 1 c + 1 q 2 m + 1 d n = 0 m = 0 q 2 m F q 1 n + 1 3 a + 1 q 1 n + 1 3 b , q 2 m c + 1 q 2 m d n = 0 m = 0 q 2 m F q 1 n 3 a + 1 q 1 n 3 b , q 2 m + 1 c + 1 q 2 m + 1 d + n = 0 m = 0 q 2 m F q 1 n 3 a + 1 q 1 n 3 b , q 2 m c + 1 q 2 m d = 1 b a d c m = 0 q 2 m n = 0 F q 1 n + 1 3 a + 1 q 1 n + 1 3 b , q 2 m + 1 c + 1 q 2 m + 1 d n = 0 F q 1 n 3 a + 1 q 1 n 3 b , q 2 m + 1 c + 1 q 2 m + 1 d + m = 0 q 2 m n = 0 F q 1 n 3 a + 1 q 1 n 3 b , q 2 m c + 1 q 2 m d n = 0 F q 1 n + 1 3 a + 1 q 1 n + 1 3 b , q 2 m c + 1 q 2 m d = 1 b a d c m = 0 q 2 m F b , q 2 m + 1 c + 1 q 2 m + 1 d m = 0 q 2 m F b , q 2 m c + 1 q 2 m d + m = 0 q 2 m F a + 2 b 3 , q 2 m c + 1 q 2 m d m = 0 q 2 m F a + 2 b 3 , q 2 m + 1 c + 1 q 2 m + 1 d = 1 b a d c 1 q 2 q 2 m = 0 q 2 m F b , q 2 m c + 1 q 2 m d 1 q 2 F b , c 1 q 2 q 2 m 0 q 2 m F a + 2 b 3 , q 2 m c + 1 q 2 m d + 1 q 2 F a + 2 b 3 , c = 1 b a d c 1 q 2 d c c d F b , y d d q 2 y 1 q 2 d c c d F a + 2 b 3 , y d d q 2 y 1 q 2 F b , c + 1 q 2 F a + 2 b 3 , c .
Similarly, we get
0 2 3 0 1 s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c 1 q 2 d c c d F b , y d d q 2 y 1 q 2 d c c d F 2 a + b 3 , y d d q 2 y 1 q 2 F b , c + 1 q 2 F 2 a + b 3 , c ,
0 1 0 1 3 t b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c 1 q 1 b a a b F x , d b d q 1 x 1 q 1 b a a b F x , c + 2 d 3 b d q 1 x F a , d + F a , c + 2 d 3 ,
0 1 0 2 3 t b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c 1 q 1 b a a b F x , d b d q 1 x 1 q 1 b a a b F x , 2 c + 2 3 b d q 1 x F a , d + F a , 2 c + d 3 .
Also, we have
0 1 0 1 s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c 1 q 2 d c c d F b , y d d q 2 y 1 q 2 d c c d F a , y d d q 2 y 1 q 2 F b , c + 1 q 2 F a , c ,
0 1 0 1 t b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c 1 q 1 b a a b F x , d b d q 1 x 1 q 1 b a a b F x , c b d q 1 x 1 q 1 F a , d + 1 q 1 F a , c
and
0 1 0 1 t s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s = 1 b a d c m = 0 n = 0 q 1 n q 2 m F q 1 n + 1 a + 1 q 1 n + 1 b , q 2 m + 1 c + 1 q 2 m + 1 d m = 0 n = 0 q 1 n q 2 m F q 1 n + 1 a + 1 q 1 n + 1 b , q 2 m c + 1 q 2 m d m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m + 1 c + 1 q 2 m + 1 d + m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m c + 1 q 2 m d
= 1 b a d c 1 q 1 q 2 m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m c + 1 q 2 m d m = 0 q 2 m F a , q 2 m c + 1 q 2 m d n = 0 q 1 n F q 1 n a + 1 q 1 n b , c + F a , c 1 q 1 m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m c + 1 q 2 m d m = 0 q 2 m F a , q 2 m c + 1 q 2 m d 1 q 2 m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m c + 1 q 2 m d n = 0 q 1 n F q 1 n a + 1 q 1 n b , c + m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m c + 1 q 2 m d = 1 b a d c 1 q 1 1 q 2 q 1 q 2 m = 0 n = 0 q 1 n q 2 m F q 1 n a + 1 q 1 n b , q 2 m c + 1 q 2 m d 1 q 2 q 1 q 2 m = 0 q 2 m F a , q 2 m c + 1 q 2 m d 1 q 1 q 1 q 2 n = 0 q 1 n F q 1 n a + 1 q 1 n b , c + 1 q 1 q 2 F a , c = 1 b a d c 1 q 1 q 2 b a d c a b c d F x , y b d q 1 x d d q 2 y 1 q 1 q 2 b a a b F x , c b d q 1 x 1 q 1 q 2 d c c d F a , x d d q 2 y + 1 q 1 q 2 F a , c .
From (13)–(29), we attain that
I 1 = 9 64 b a d c F b , d F a + 2 b 3 , d F b , c + 2 d 3 + F a + 2 b 3 , c + 2 d 3 ,
I 2 = 9 64 b a d c F b , d F a + 2 b 3 , d F b , 2 c + d 3 + F a + 2 b 3 , 2 c + d 3 ,
I 3 = 3 8 b a d c 1 d c c d F b , y d d q 2 y 1 d c c d F a + 2 b 3 , y d d q 2 y F b , c + F a + 2 b 3 , c 21 64 b a d c F b , d F a + 2 b 3 , d F b , c + F a + 2 b 3 , c ,
I 4 = 9 64 b a d c F b , d F 2 a + b 3 , d F b , c + 2 d 3 + F 2 a + b 3 , c + 2 d 3 ,
I 5 = 9 64 b a d c F b , d F 2 a + b 3 , d F b , 2 c + d 3 + F 2 a + b 3 , 2 c + d 3 ,
I 6 = 3 8 b a d c 1 d c c d F b , y d d q 2 y 1 d c c d F 2 a + b 3 , y d d q 2 y F b , c + F 2 a + b 3 , c 21 64 b a d c F b , d F 2 a + b 3 , d F b , c + F 2 a + b 3 , c ,
I 7 = 3 8 b a d c 1 b a a b F x , d b d q 1 x 1 b a a b F x , c + 2 d 3 b d q 1 x F a , d + F a , c + 2 d 3 21 64 b a d c F b , d F b , c + 2 d 3 F a , d + F a , c + 2 d 3 ,
I 8 = 3 8 b a d c 1 b a a b F x , d b d q 1 x 1 b a a b F x , 2 c + d 3 b d q 1 x F a , d + F a , 2 c + d 3 21 64 b a d c F b , d F b , 2 c + d 3 F a , d + F a , c + 2 d 3 , I 9 = 1 b a d c 1 b a d c a b c d F x , y b d q 1 x d d q 2 y 1 b a a b F x , c b d q 1 x 1 d c c d F a , y d d q 2 y + F a , c 7 8 b a d c 1 b a a b F x , d b d q 1 x 1 b a a b F x , c b d q 1 x + F a , c F a , d 7 8 b a d c 1 d c c d F b , y d d q 2 y 1 d c c d F a , y d d q 2 y + F a , c F b , c + 49 64 b a d c F b , d F a , d F b , c + F a , c .
Now using the calculated integrals ( I 1 ) ( I 9 ) in (12) and multiplying the resulting one with b a d c , we have the desired equality (11) which accomplishes the proof. □
Remark 1.
Under the given conditions of Lemma 3 with q 1 , q 2 1 , we have the following new identity:
I F = b a d c 0 1 0 1 Λ t Λ s 2 F t a + 1 t b , s c + 1 s d t s d t d s ,
where
I F = 3 64 F a + 2 b 3 , d + F b , c + 2 d 3 + 3 F a + 2 b 3 , c + 2 d 3 + 3 F a + 2 b 3 , 2 c + d 3 + F 2 a + b 3 , d + 3 F 2 a + b 3 , c + 2 d 3 + F b , 2 c + d 3 + 3 F 2 a + b 3 , 2 c + d 3 + F a + 2 b 3 , c + F 2 a + b 3 , c + F a , c + 2 d 3 + F a , 2 c + d 3 + F a , c + F a , d + F b , c + F b , d 64 1 8 b a a b F x , c + 3 F x , c + 2 d 3 + 3 F x , 2 c + d 3 + F x , d d x 1 8 d c c d F a , y + 3 F a + 2 b 3 , y + 3 F 2 a + b 3 , y + F b , y d y + 1 b a d c a b c d F x , y d x d y
and
Λ t = t 1 8 , t 0 , 1 3 t 1 2 , t 1 3 , 2 3 t 7 8 , t 2 3 , 1 ,
Λ s = s 1 8 , s 0 , 1 3 s 1 2 , s 1 3 , 2 3 s 7 8 , s 2 3 , 1 .

5. Some New q 1 q 2 –Newton’s Type Inequalities

For brevity, we give some calculated integrals before giving new estimates.
Ψ 1 ( q ) = 0 1 3 q t 1 8 t d q t = 3 5 q 5 q 2 216 1 + q 1 + q + q 2 , 0 < q < 3 8 160 q 2 + 160 q 69 6912 1 + q 1 + q + q 2 , 3 8 q < 1 ,
Ψ 2 ( q ) = 0 1 3 q t 1 8 1 t d q t = 6 q q 2 15 q 3 216 1 + q 1 + q + q 2 , 0 < q < 3 8 480 q 3 + 248 q 2 + 248 q 3 6912 1 + q 1 + q + q 2 , 3 8 q < 1 ,
Ψ 3 ( q ) = 1 3 2 3 q t 1 2 t d q t = 9 5 q 5 q 2 54 1 + q 1 + q + q 2 , 0 < q < 3 4 6 q 2 + 6 q 3 108 1 + q 1 + q + q 2 , 3 4 q < 1 ,
Ψ 4 ( q ) = 1 3 2 3 q t 1 2 1 t d q t = 5 q + 5 q 2 9 q 3 54 1 + q 1 + q + q 2 , 0 < q < 3 4 6 q 3 + 3 108 1 + q 1 + q + q 2 , 3 4 q < 1 ,
Ψ 5 ( q ) = 2 3 1 q t 7 8 t d q t = 105 47 q 47 q 2 216 1 + q 1 + q + q 2 , 0 < q < 7 8 224 q 2 + 224 q + 525 6912 1 + q 1 + q + q 2 , 7 8 q < 1 ,
Ψ 6 ( q ) = 2 3 1 q t 7 8 1 t d q t = 42 + 53 q + 53 q 2 57 q 3 216 1 + q 1 + q + q 2 , 0 < q < 7 8 96 q 3 + 184 q 2 + 184 q 21 6912 1 + q 1 + q + q 2 , 7 8 q < 1 ,
Ψ 7 ( q ) = 0 1 3 q t 1 8 d q t = 3 5 q 72 1 + q , 0 < q < 3 8 20 q 3 288 1 + q , 3 8 q < 1 ,
Ψ 8 ( q ) = 1 3 2 3 q t 1 2 d q t = 3 3 q 18 1 + q , 0 < q < 3 4 q 18 1 + q , 3 4 q < 1
and
Ψ 9 ( q ) = 2 3 1 q t 7 8 d q t = 21 19 q 72 1 + q , 0 < q < 7 8 21 4 q 288 1 + q , 7 8 q < 1 .
Now we give some new quantum estimates by using the identity in Lemma 3.
Theorem 7.
Let F : Δ R 2 R be a twice partially q 1 q 2 -differentiable function on Δ such that partial q 1 q 2 -derivative b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s is continuous and integrable on a , b × c , d Δ . If b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s is convex on a , b × c , d , then we have the following inequality:
b , d I q 1 , q 2 b a d c × Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 Ψ 1 q 2 + Ψ 3 q 2 + Ψ 5 q 2 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s + Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 Ψ 2 q 2 + Ψ 4 q 2 + Ψ 6 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s + Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 Ψ 1 q 2 + Ψ 3 q 2 + Ψ 5 q 2 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s + Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 Ψ 2 q 2 + Ψ 4 q 2 + Ψ 6 q 2 b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s ,
where q 1 , q 2 0 , 1 .
Proof. 
Taking properties of modulus in Lemma 3, we find that
b , d I q 1 , q 2 b a d c 0 1 0 1 Λ q 1 t Λ q 2 s b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s
Now using the convexity of b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s , (41) becomes
b , d I q 1 , q 2 b a d c 0 1 Λ q 2 s 0 1 Λ q 1 t t b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s + 1 t b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s d q 1 t d q 2 s .
Now we compute the integrals that appear in right side of inequality (42).
0 1 Λ q 1 t t b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s + 1 t b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s d q 1 t = 0 1 3 t q 1 t 1 8 b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s d q 1 t + 1 3 2 3 t q 1 t 1 2 b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s d q 1 t + 2 3 1 t q 1 t 7 8 b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s d q 1 t + 0 1 3 1 t q 1 t 1 8 b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s d q 1 t + 1 3 2 3 1 t q 1 t 1 2 b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s d q 1 t + 2 3 1 1 t q 1 t 7 8 b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s d q 1 t
From (31)–(36), we obtain
0 1 Λ q 1 t t b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s + 1 t b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s d q 1 t = b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 + b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 .
Thus, we have
b , d I q 1 , q 2 b a d c 0 1 Λ q 2 s b , d q 1 , q 2 2 F a , s c + 1 s d b q 1 t d q 2 s Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 + b , d q 1 , q 2 2 F b , s c + 1 s d b q 1 t d q 2 s Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 d q 2 s
b a d c 0 1 Λ q 2 s s b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s + 1 s b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s × Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 + s b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s + 1 s b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s × Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 d q 2 s = b a d c Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s 0 1 3 s q 2 s 1 8 d q 2 s + 1 3 2 3 s q 2 s 1 2 d q 2 s + 2 3 1 s q 2 s 7 8 d q 2 s + b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s 0 1 3 1 s q 2 s 1 8 d q 2 s + 1 3 2 3 1 s q 2 s 1 2 d q 2 s + 2 3 1 1 s q 2 s 7 8 d q 2 s + b a d c Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s 0 1 3 s q 2 s 1 8 d q 2 s + 1 3 2 3 s q 2 s 1 2 d q 2 s + 2 3 1 s q 2 s 7 8 d q 2 s + b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s 0 1 3 1 s q 2 s 1 8 d q 2 s + 1 3 2 3 1 s q 2 s 1 2 d q 2 s + 2 3 1 1 s q 2 s 7 8 d q 2 s .
From (31)–(36), we have
b , d I q 1 , q 2 b a d c × Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 Ψ 1 q 2 + Ψ 3 q 2 + Ψ 5 q 2 b , d q 1 , q 2 2 F a , c b q 1 2 t d q 2 s + Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 Ψ 2 q 2 + Ψ 4 q 2 + Ψ 6 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s + Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 Ψ 1 q 2 + Ψ 3 q 2 + Ψ 5 q 2 b , d q 1 , q 2 2 F b , c b q 1 2 t d q 2 s + Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 Ψ 2 q 2 + Ψ 4 q 2 + Ψ 6 q 2 b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s ,
which completes the proof. □
Remark 2.
Under the given conditions of Theorem 7 with q 1 , q 2 1 , we attain the following new inequality:
I F 625 b a d c 576 2 F a , c t s + 2 F a , d t s + 2 F b , c t s + 2 F b , d t s 576 .
Corollary 1.
Under the given conditions of Theorem 7 if b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s K , we get
b , d I q 1 , q 2 K b a d c × Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 Ψ 1 q 2 + Ψ 3 q 2 + Ψ 5 q 2 + Ψ 1 q 1 + Ψ 3 q 1 + Ψ 5 q 1 Ψ 2 q 2 + Ψ 4 q 2 + Ψ 6 q 2 + Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 Ψ 1 q 2 + Ψ 3 q 2 + Ψ 5 q 2 + Ψ 2 q 1 + Ψ 4 q 1 + Ψ 6 q 1 Ψ 2 q 2 + Ψ 4 q 2 + Ψ 6 q 2 .
Theorem 8.
Let F : Δ R 2 R be a twice partially q 1 q 2 -differentiable function on Δ such that the partial q 1 q 2 -derivative b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s is continuous and integrable on a , b × c , d Δ . If b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s p 1 is convex on a , b × c , d for some p 1 > 1 and 1 r 1 + 1 p 1 = 1 , then we have the following inequality:
b , d I q 1 , q 2 b a d c 0 1 0 1 Λ q 1 t Λ q 2 s r 1 d q 1 t d q 2 s 1 r 1 × 1 2 q 1 2 q 2 b , d q 1 , q 2 2 F ( a , c ) b q 1 t d q 2 s p 1 + q 2 2 q 1 2 q 2 b , d q 1 , q 2 2 F ( a , d ) b q 1 t d q 2 s p 1 + q 1 2 q 1 2 q 2 b , d q 1 , q 2 2 F ( b , c ) b q 1 t d q 2 s p 1 + q 1 q 2 2 q 1 2 q 2 b , d q 1 , q 2 2 F ( b , d ) b q 1 t d q 2 s p 1 1 p 1 ,
where q 1 , q 2 ( 0 , 1 ) .
Proof. 
Applying well–known Hölder’s inequality for q 1 q 2 -integrals to the integrals in right side of (41), it is found that
b , d I q 1 , q 2 b a d c 0 1 0 1 Λ q 1 t Λ q 2 s r 1 d q 1 t d q 2 s 1 r 1 × 0 1 0 1 b , d q 1 , q 2 2 F t a + 1 t b , s c + 1 s d b q 1 t d q 2 s p 1 d q 1 t d q 2 s 1 p 1 .
By applying convexity of b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s p 1 , (44) becomes
b , d I q 1 , q 2 b a d c 0 1 0 1 Λ q 1 t Λ q 2 s r 1 d q 1 t d q 2 s 1 r 1 × 0 1 0 1 t s b , d q 1 , q 2 2 F ( a , c ) b q 1 t d q 2 s p 1 + t 1 s b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s p 1 + 1 t s b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s p 1 + 1 t 1 s b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s p 1 d q 1 t d q 2 s 1 p 1 .
Now, if we apply the concept of Lemma 1 for a = 0 to the above quantum integrals, we attain
0 1 0 1 t s d q 1 t d q 2 s = 0 1 t d q 1 t 0 1 s d q 2 s = 1 2 q 1 2 q 2 ,
0 1 0 1 t 1 s d q 1 t d q 2 s = q 2 2 q 1 2 q 2 ,
0 1 0 1 1 t s d q 1 t d q 2 s = q 1 2 q 1 2 q 2 ,
0 1 0 1 1 t 1 s d q 1 t d q 2 s = q 1 q 2 2 q 1 2 q 2 .
By substituting the calculated integrals (45)–(49) in (45), we obtain the desired inequality (43) which finishes the proof. □
Corollary 2.
Under the given conditions of Theorem 8 if b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s K , we get
b , d I q 1 , q 2 K b a d c 2 q 1 2 q 2 p 1 0 1 0 1 Λ q 1 t Λ q 2 s r 1 d q 1 t d q 2 s 1 r 1 .
Theorem 9.
Let F : Δ R 2 R be a twice partially q 1 q 2 -differentiable function on Δ such that the partial q 1 q 2 -derivative b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s is continuous and integrable on a , b × c , d Δ . If b , d q 1 , q 2 2 F t , s b q 1 t d q 2 s p is convex on a , b × c , d for some p 1 . Then, we have following inequality
b , d I q 1 , q 2 ( F ) b a d c Ψ 7 1 1 p q 1 Ψ 7 1 1 p q 2 × Ψ 1 q 1 Ψ 1 q 2 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s p + Ψ 2 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s p + Ψ 2 q 1 Ψ 1 q 2 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s p + Ψ 2 q 2 b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s p 1 p + Ψ 7 1 1 p q 1 Ψ 8 1 1 p q 2 × Ψ 1 q 1 Ψ 3 q 2 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s p + Ψ 4 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s p + Ψ 2 q 1 Ψ 3 q 2 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s p + Ψ 4 q 2 b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s p 1 p + Ψ 7 1 1 p q 1 Ψ 9 1 1 p q 2 × Ψ 1 q 1 Ψ 5 q 2 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s p + Ψ 6 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s p + Ψ 2 q 1 Ψ 5 q 2 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s p + Ψ 6 q 2 b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s p 1 p + Ψ 8 1 1 p q 1 Ψ 7 1 1 p q 2 × Ψ 3 q 1 Ψ 1 q 2 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s p + Ψ 2 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s p + Ψ 4 q 1 Ψ 1 q 2 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s p + Ψ 2 q 2 b , d q 1 , q 2 2 F b , d b q 1 t d q 2 s p 1 p + Ψ 8 1 1 p q 1 Ψ 8 1 1 p q 2 × Ψ 3 q 1 Ψ 3 q 2 b , d q 1 , q 2 2 F a , c b q 1 t d q 2 s p + Ψ 4 q 2 b , d q 1 , q 2 2 F a , d b q 1 t d q 2 s p + Ψ 4 q 1 Ψ 3 q 2 b , d q 1 , q 2 2 F b , c b q 1 t d q 2 s p + Ψ 4 q 2 b , d