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Article

Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality

1
Croatian Academy of Sciences and Arts, Trg Nikole Šubića Zrinskog 11, 10000 Zagreb, Croatia
2
Faculty of Civil Engineering, University of Rijeka, Radmile Matejčić 3, 51000 Rijeka, Croatia
3
Faculty of Textile Technology, University of Zagreb, Prilaz Baruna Filipovića 28a, 10000 Zagreb, Croatia
*
Author to whom correspondence should be addressed.
Mathematics 2022, 10(9), 1505; https://doi.org/10.3390/math10091505
Received: 30 March 2022 / Revised: 26 April 2022 / Accepted: 29 April 2022 / Published: 1 May 2022

Abstract

:
In this paper, we obtain some new weighted Hermite–Hadamard-type inequalities for ( n + 2 ) convex functions by utilizing generalizations of Steffensen’s inequality via Taylor’s formula.

1. Introduction

The Hermite–Hadamard inequality is one of the most important mathematical inequalities. It was discovered independently first by Hermite [1] and later by Hadamard [2]. The classical Hermite–Hadamard inequality provides an estimate from below and above the mean value of convex function f: [ a , b ] R . More precisely, we have the following.
f a + b 2 1 b a a b f ( x ) d x f ( a ) + f ( b ) 2 .
To illustrate the importance of the Hermite–Hadamard inequality, let us mention that the Hermite–Hadamard inequality can be considered as the necessary and sufficient condition for convexity of a function. Furthermore, the Hermite–Hadamard inequality has an important role in numerical analysis, mathematical analysis and functional analysis. Various generalizations, extensions and applications of the Hermite-Hadamard inequality have appeared in the literature (see [3,4,5,6,7,8]).
In this paper, we consider the weighted Hermite–Hadamard inequality for convex functions given in following theorem (see [8,9,10]).
Theorem 1.
Let p: [ a , b ] R be a non-negative function. If f: [ a , b ] R is a convex function, then we have the following:
f ( m ) 1 P ( b ) a b p ( x ) f ( x ) d x b m b a f ( a ) + m a b a f ( b )
or
P ( b ) f ( m ) a b p ( x ) f ( x ) d x P ( b ) b m b a f ( a ) + m a b a f ( b ) ,
where the following is the case.
P ( t ) = a t p ( x ) d x a n d m = 1 P ( b ) a b p ( x ) x d x .
In 1918, Steffensen proved the following inequality (see [11]).
Theorem 2
([11]). Suppose that f is non-increasing and g is integrable on [ a , b ] with 0 g 1 and λ = a b g ( t ) d t . Then, we have the following.
b λ b f ( t ) d t a b f ( t ) g ( t ) d t a a + λ f ( t ) d t .
The inequalities are reversed for f non-decreasing.
Many papers have been devoted to generalizations and refinements of Steffensen’s inequality and its connection to other well-known inequalities such as Gauss–Steffensen’s, Hölder’s, Jenssen-=Steffensen’s and other inequalities. A complete overview of the results related to Steffensen’s inequality can be found in monographs [12,13].
By using the Mitrinović [14] result in which the inequalities in (2) follow from identities:
a a + λ f ( t ) d t a b f ( t ) g ( t ) d t = a a + λ [ f ( t ) f ( a + λ ) ] [ 1 g ( t ) ] d t + a + λ b [ f ( a + λ ) f ( t ) ] g ( t ) d t
and
a b f ( t ) g ( t ) d t b λ b f ( t ) d t = a b λ [ f ( t ) f ( b λ ) ] g ( t ) d t + b λ b [ f ( b λ ) f ( t ) ] [ 1 g ( t ) ] d t
and using Taylor’s formulae in points a and b
f ( x ) = i = 0 n 1 f ( i ) ( a ) i ! ( x a ) i + 1 ( n 1 ) ! a x f ( n ) ( t ) ( x t ) n 1 d t
f ( x ) = i = 0 n 1 f ( i ) ( b ) i ! ( x b ) i 1 ( n 1 ) ! x b f ( n ) ( t ) ( x t ) n 1 d t
in paper [15], the authors proved the following identities related to generalizations of Steffensen’s inequality.
Theorem 3
([15]). Let f: [ a , b ] R be such that f ( n 1 ) is absolutely continuous for some n 2 and let g: [ a , b ] R be an integrable function such that 0 g 1 . Let λ = a b g ( t ) d t and let the function G 1 be defined by the following.
G 1 ( x ) = a x ( 1 g ( t ) ) d t , x [ a , a + λ ] , x b g ( t ) d t , x [ a + λ , b ] .
Then, we have the following:
a a + λ f ( t ) d t a b f ( t ) g ( t ) d t + i = 0 n 2 f ( i + 1 ) ( a ) i ! a b G 1 ( x ) ( x a ) i d x = 1 ( n 2 ) ! a b t b G 1 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t
and the following is obtained.
a a + λ f ( t ) d t a b f ( t ) g ( t ) d t + i = 0 n 2 f ( i + 1 ) ( b ) i ! a b G 1 ( x ) ( x b ) i d x = 1 ( n 2 ) ! a b a t G 1 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t .
Theorem 4
([15]). Let f: [ a , b ] R be such that f ( n 1 ) is absolutely continuous for some n 2 and let g: [ a , b ] R be an integrable function such that 0 g 1 . Let λ = a b g ( t ) d t and let the function G 2 be defined by the following.
G 2 ( x ) = a x g ( t ) d t , x [ a , b λ ] , x b ( 1 g ( t ) ) d t , x [ b λ , b ] .
Then, we have the following:
a b f ( t ) g ( t ) d t b λ b f ( t ) d t + i = 0 n 2 f ( i + 1 ) ( a ) i ! a b G 2 ( x ) ( x a ) i d x = 1 ( n 2 ) ! a b t b G 2 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t
and the following is obtained.
a b f ( t ) g ( t ) d t b λ b f ( t ) d t + i = 0 n 2 f ( i + 1 ) ( b ) i ! a b G 2 ( x ) ( x b ) i d x = 1 ( n 2 ) ! a b a t G 2 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t .
Since, in this paper, we will deal with n convex functions, let us recall the definition of the n convex function. For more details on convex functions, we refer the interested reader to [6,8].
Let f be a real-valued function defined on the segment [ a , b ] . The divided difference of order n of the function f at distinct points x 0 , , x n [ a , b ] is defined recursively (see [8]) by the following.
f [ x i ] = f ( x i ) , ( i = 0 , , n )
f [ x 0 , , x n ] = f [ x 1 , , x n ] f [ x 0 , , x n 1 ] x n x 0 .
The value f [ x 0 , , x n ] is independent of the order of the points x 0 , , x n .
The definition may be extended to include the case in which some (or all) of the points coincide. Assuming that f ( j 1 ) ( x ) exists, we define the following.
f [ x , , x j t i m e s ] = f ( j 1 ) ( x ) ( j 1 ) ! .
Definition 1
([8]). A function f : [ a , b ] R is said to be n-convex on [ a , b ] , n 0 , if for all choices of ( n + 1 ) distinct points in [ a , b ] , the n t h order divided difference of f satisfies the following.
f [ x 0 , , x n ] 0 .
Note that 1 convex functions are non-decreasing functions and 2 convex functions are convex functions. An n convex function need not to be n times differentiable; however, if f ( n ) exists, then f is n convex if and only if f ( n ) 0 . The following property also holds: if f is an ( n + 2 ) convex function, then there exists the n th derivative f ( n ) , which is a convex function.
The aim of this paper is to use identities related to generalizations of Steffensen’s inequality, obtained by using Taylor’s formula, to prove new weighted Hermite–Hadamard-type inequalities for ( n + 2 ) convex functions.

2. Main Results

In this section, applying identities given in Theorems 3 and 4 and the properties of n convex functions, we derive new weighted Hermite–Hadamard-type inequalities.
Theorem 5.
Let f: [ a , b ] R be ( n + 2 ) convex on [ a , b ] and f ( n 1 ) absolutely continuous for n 2 . Let g: [ a , b ] R be an integrable function such that 0 g 1 and λ = a b g ( t ) d t . Let function G 1 be defined by the following.
G 1 ( x ) = a x ( 1 g ( t ) ) d t , x [ a , a + λ ] , x b g ( t ) d t , x [ a + λ , b ] .
Then, we have the following:
P 1 ( b ) · f ( n ) m 1 ( n 2 ) ! a b f ( t ) g ( t ) d t a a + λ f ( t ) d t i = 0 n 2 f ( i + 1 ) ( a ) i ! a b G 1 ( x ) ( x a ) i d x P 1 ( b ) · b m 1 b a f ( n ) ( a ) + m 1 a b a f ( n ) ( b ) ,
where the following is the case:
P 1 ( b ) = 1 ( n 1 ) · n a b g ( x ) ( x a ) n d x λ n + 1 n + 1
and the following is obtained.
m 1 = a + 1 ( n 1 ) · n · ( n + 1 ) · P 1 ( b ) a b g ( x ) ( x a ) n + 1 d x λ n + 2 n + 2 .
Proof. 
Since f ( n 1 ) is absolutely continuous, function f satisfies the conditions of Theorem 3. Therefore, identity (3) holds.
From condition 0 g 1 , function G 1 defined by (7) is non-negative. Hence, for every n 2 , we have the following.
t b G 1 ( x ) ( x t ) n 2 d x 0 , t a , b .
Define
p ( t ) = t b G 1 ( x ) ( x t ) n 2 d x .
Since the function f is ( n + 2 ) convex, function f ( n ) is convex. Furthermore, function p is non-negative, so we can apply Theorem 1 and obtain the following inequality:
P 1 ( b ) · f ( n ) m 1 a b t b G 1 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t P 1 ( b ) · b m 1 b a f ( n ) ( a ) + m 1 a b a f ( n ) ( b ) ,
where P 1 ( b ) and m 1 are given by
P 1 ( b ) = a b t b G 1 ( x ) ( x t ) n 2 d x d t
and
m 1 = 1 P 1 ( b ) a b t b G 1 ( x ) ( x t ) n 2 d x t d t .
By calculating P 1 ( b ) and m 1 , we obtain the following:
P 1 ( b ) = a b t b G 1 ( x ) ( x t ) n 2 d x d t = a a + λ a x ( 1 g ( s ) ) d s ( x a ) n 1 n 1 d x + a + λ b x b g ( s ) d s ( x a ) n 1 n 1 d x = a a + λ ( x a ) n n 1 d x + λ · a + λ b ( x a ) n 1 n 1 d x a b a x g ( s ) d s ( x a ) n 1 n 1 d x = λ n + 1 ( n 1 ) · n · ( n + 1 ) + a b g ( x ) ( x a ) n ( n 1 ) · n d x
and
m 1 = 1 P 1 ( b ) a b t b G 1 ( x ) ( x t ) n 2 d x t d t = 1 P 1 ( b ) a b G 1 ( x ) a x ( x t ) n 2 · t d t d x = 1 P 1 ( b ) a b G 1 ( x ) t · ( x t ) n 1 n 1 | a x + a x ( x t ) n 1 n 1 d t d x = 1 P 1 ( b ) a b G 1 ( x ) a · ( x a ) n 1 n 1 + ( x a ) n ( n 1 ) · n d x = a + 1 P 1 ( b ) a b G 1 ( x ) ( x a ) n ( n 1 ) · n d x = a + 1 P 1 ( b ) λ n + 2 ( n 1 ) · n · ( n + 1 ) · ( n + 2 ) + a b g ( x ) ( x a ) n + 1 ( n 1 ) · n · ( n + 1 ) d x .
Using identity (3) for the middle part of the inequality (11), inequality (11) becomes inequality (8). Hence, the proof is completed. □
Theorem 6.
Let f: [ a , b ] R be ( n + 2 ) convex on [ a , b ] and f ( n 1 ) absolutely continuous for n 2 . Let g: [ a , b ] R be an integrable function such that 0 g 1 and λ = a b g ( t ) d t . Let function G 1 be defined by (7). If the following is the case:
a t G 1 ( x ) ( x t ) n 2 d x 0 , t [ a , b ] ,
then we have the following:
P 2 ( b ) · f ( n ) m 2 ( n 2 ) ! a b f ( t ) g ( t ) d t a a + λ f ( t ) d t i = 0 n 2 f ( i + 1 ) ( b ) i ! a b G 1 ( x ) ( x b ) i d x P 2 ( b ) · b m 2 b a f ( n ) ( a ) + m 2 a b a f ( n ) ( b ) ,
where
P 2 ( b ) = 1 ( n 1 ) · n ( a b ) n + 1 ( a + λ b ) n + 1 n + 1 + a b g ( x ) ( x b ) n d x
and
m 2 = b + 1 ( n 1 ) · n · ( n + 1 ) · P 2 ( b ) × ( a b ) n + 2 ( a + λ b ) n + 2 n + 2 + a b g ( x ) ( x b ) n + 1 d x .
Proof. 
If we assume the following:
a t G 1 ( x ) ( x t ) n 2 d x 0 , t [ a , b ]
then we have the following.
a t G 1 ( x ) ( x t ) n 2 d x 0 , t [ a , b ] .
Now similarly to the proof of Theorem 5 using the following non-negative function:
p ( t ) = a t G 1 ( x ) ( x t ) n 2 d x
and identity (4), we obtain inequality (12). Similarly, we calculate the expressions for P 2 ( b ) and m 2 and obtain the following:
P 2 ( b ) = a b a t G 1 ( x ) ( x t ) n 2 d x d t = a a + λ a x ( 1 g ( s ) ) d s ( x b ) n 1 n 1 d x + a + λ b x b g ( s ) d s ( x b ) n 1 n 1 d x = a a + λ ( x a ) ( x b ) n 1 n 1 d x + λ · a + λ b ( x b ) n 1 n 1 d x a b a x g ( s ) d s ( x b ) n 1 n 1 d x = ( a b ) n + 1 ( n 1 ) · n · ( n + 1 ) ( a + λ b ) n + 1 ( n 1 ) · n · ( n + 1 ) + a b g ( x ) ( x b ) n ( n 1 ) · n d x
and
m 2 = 1 P 2 ( b ) a b a t G 1 ( x ) ( x t ) n 2 d x t d t = 1 P 2 ( b ) a b G 1 ( x ) x b ( x t ) n 2 · t d t d x = 1 P 2 ( b ) a b G 1 ( x ) t · ( x t ) n 1 n 1 | x b + x b ( x t ) n 1 n 1 d t d x = 1 P 2 ( b ) a b G 1 ( x ) b · ( x b ) n 1 n 1 ( x b ) n ( n 1 ) · n d x = b + 1 P 2 ( b ) a b G 1 ( x ) ( x b ) n ( n 1 ) · n d x = b + 1 ( n 1 ) · n · ( n + 1 ) · P 1 ( b ) × ( a b ) n + 2 n + 2 ( a + λ b ) n + 2 n + 2 + a b g ( x ) ( x b ) n + 1 d x .
Hence, the proof is completed. □
Theorem 7.
Let f: [ a , b ] R be ( n + 2 ) convex on [ a , b ] and f ( n 1 ) absolutely continuous for n 2 . Let g: [ a , b ] R be an integrable function such that 0 g 1 and λ = a b g ( t ) d t . Let function G 2 be defined by the following.
G 2 ( x ) = a x g ( t ) d t , x [ a , b λ ] , x b ( 1 g ( t ) ) d t , x [ b λ , b ] .
Then, the following is obtained:
P 3 ( b ) · f ( n ) m 3 ( n 2 ) ! b λ b f ( t ) d t i = 0 n 2 f ( i + 1 ) ( a ) i ! a b G 2 ( x ) ( x a ) i d x a b f ( t ) g ( t ) d t P 3 ( b ) · b m 3 b a f ( n ) ( a ) + m 3 a b a f ( n ) ( b ) ,
where
P 3 ( b ) = 1 ( n 1 ) · n ( b a ) n + 1 ( b λ a ) n + 1 n + 1 a b g ( x ) ( x a ) n d x
and
m 3 = a + 1 ( n 1 ) · n · ( n + 1 ) · P 3 ( b ) × ( b a ) n + 2 ( b λ a ) n + 2 n + 2 a b g ( x ) ( x a ) n + 1 d x .
Proof. 
We follow the similar arguments as in the proof of Theorem 5. As function f ( n 1 ) is absolutely continuous, the identity (5) holds. The inequality (14) follows directly from Theorem 1, substituting the non-negative function p by a non-negative function of the following:
p ( t ) = t b G 2 ( x ) ( x t ) n 2 d x
and a convex function f by a convex function f ( n ) , and then using identity (5) for integral a b t b G 2 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t . Furthermore, we calculate P 3 ( b ) and m 3 as follows.
P 3 ( b ) = a b t b G 2 ( x ) ( x t ) n 2 d x d t = a b λ a x g ( s ) d s ( x a ) n 1 n 1 d x + b λ b x b ( 1 g ( s ) ) d s ( x a ) n 1 n 1 d x = b λ b ( b x ) ( x a ) n 1 n 1 d x λ · b λ b ( x a ) n 1 n 1 d x + a b a x g ( s ) d s ( x a ) n 1 n 1 d x = ( b a ) n + 1 ( b λ a ) n + 1 ( n 1 ) · n · ( n + 1 ) a b g ( x ) ( x a ) n ( n 1 ) · n d x ,
m 3 = 1 P 3 ( b ) a b t b G 2 ( x ) ( x t ) n 2 d x t d t = 1 P 3 ( b ) a b G 2 ( x ) a x ( x t ) n 2 · t d t d x = 1 P 3 ( b ) a b G 2 ( x ) t · ( x t ) n 1 n 1 | a x + a x ( x t ) n 1 n 1 d t d x = 1 P 3 ( b ) a b G 2 ( x ) a · ( x a ) n 1 n 1 + ( x a ) n ( n 1 ) · n d x = a + 1 P 3 ( b ) a b G 2 ( x ) ( x a ) n ( n 1 ) · n d x = a + 1 P 3 ( b ) ( b a ) n + 2 ( b λ a ) n + 2 ( n 1 ) · n · ( n + 1 ) · ( n + 2 ) a b g ( x ) ( x a ) n + 1 ( n 1 ) · n · ( n + 1 ) d x .
Hence, the proof is completed. □
Theorem 8.
Let f: [ a , b ] R be ( n + 2 ) convex on [ a , b ] and f ( n 1 ) absolutely continuous for n 2 . Let g: [ a , b ] R be an integrable function such that 0 g 1 and λ = a b g ( t ) d t . Let function G 2 be defined by (13). If the following is the case:
a t G 2 ( x ) ( x t ) n 2 d x 0 , t [ a , b ]
then we obtain the following:
P 4 ( b ) · f ( n ) m 4 ( n 2 ) ! b λ b f ( t ) d t i = 0 n 2 f ( i + 1 ) ( b ) i ! a b G 2 ( x ) ( x b ) i d x a b f ( t ) g ( t ) d t P 4 ( b ) · b m 4 b a f ( n ) ( a ) + m 4 a b a f ( n ) ( b ) ,
where
P 4 ( b ) = 1 ( n 1 ) · n ( λ ) n + 1 n + 1 + a b g ( x ) ( x b ) n d x
and
m 4 = b 1 ( n 1 ) · n · ( n + 1 ) · P 4 ( b ) ( λ ) n + 2 n + 2 + a b g ( x ) ( x b ) n + 1 d x .
Proof. 
Under the assumption that a t G 2 ( x ) ( x t ) n 2 d x 0 , it is obvious that the following is the case:
p ( t ) = a t G 2 ( x ) ( x t ) n 2 d x
where it is a non-negative function. Again, replacing p ( t ) in Theorem 1 by (16) and f by f ( n ) and then using the identity (6) for
a b a t G 2 ( x ) ( x t ) n 2 d x f ( n ) ( t ) d t ,
we obtain the required inequalities (15). Finally, a simple calculation yields the following:
P 4 ( b ) = a b a t G 2 ( x ) ( x t ) n 2 d x d t = a b λ a x g ( s ) d s ( x b ) n 1 n 1 d x + b λ b x b ( 1 g ( s ) ) d s ( x b ) n 1 n 1 d x = b λ b ( x b ) n n 1 d x λ · b λ b ( x b ) n 1 n 1 d x + a b a x g ( s ) d s ( x b ) n 1 n 1 d x = ( λ ) n + 1 ( n 1 ) · n · ( n + 1 ) a b g ( x ) ( x b ) n ( n 1 ) · n d x
and
m 4 = 1 P 4 ( b ) a b a t G 2 ( x ) ( x t ) n 2 d x t d t = 1 P 4 ( b ) a b G 2 ( x ) x b ( x t ) n 2 · t d t d x = 1 P 4 ( b ) a b G 2 ( x ) t · ( x t ) n 1 n 1 | x b + x b ( x t ) n 1 n 1 d t d x = 1 P 4 ( b ) a b G 2 ( x ) b · ( x b ) n 1 n 1 ( x b ) n ( n 1 ) · n d x = b + 1 P 4 ( b ) a b G 2 ( x ) ( x b ) n ( n 1 ) · n d x = b 1 P 4 ( b ) ( λ ) n + 2 ( n 1 ) · n · ( n + 1 ) · ( n + 2 ) + a b g ( x ) ( x b ) n + 1 ( n 1 ) · n · ( n + 1 ) d x .
Remark 1.
If function f is ( n + 2 ) concave, the inequalities in Theorems 5–8 are reversed. This follows from the fact that for ( n + 2 ) concave function, we have f ( n + 2 ) 0 . Hence, f ( n ) is convex and we can apply inequality (1) to function f ( n ) .
Remark 2.
The expressions P i ( b ) and m i for i = 1 , , 4 can also be achieved by the method introduced in [16]. By this method, we calculate P 1 ( b ) and m 1 . Other expressions can be recaptured in a similar manner.
The value of P 1 ( b ) can be obtained from (3) by taking f ( t ) = ( t a ) n n ! . Then, f ( n ) ( t ) = 1 . Thus, we have the following.
P 1 ( b ) = ( n 2 ) ! a a + λ ( x a ) n n ! d t a b ( x a ) n n ! g ( t ) d t = λ n + 1 ( n 1 ) · n · ( n + 1 ) + a b ( x a ) n ( n 1 ) · n g ( t ) d t .
Hence, we obtained expression (9).
From Theorem 1, we previously obtained the following.
m 1 = 1 P 1 ( b ) a b t b G 1 ( x ) ( x t ) n 2 d x t d t .
To calculate m 1 , we take function f ( t ) = ( t a ) n + 1 ( n + 1 ) ! . Then, f ( n ) ( t ) = t a . Hence, from the identity (3), we obtain expression (10).

3. Conclusions

In this paper, we obtained new weighted Hermite–Hadamard-type inequalities for higher order convex functions. We used previously obtained identities related to the generalizations of Steffensen’s inequality. Results obtained in this paper can be considered as a starting point for some future work.

Author Contributions

Conceptualization, J.P., A.P.P. and K.S.K.; Writing — original draft, J.P., A.P.P. and K.S.K. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

Croatian Science Foundation (HRZZ 7926) “Separation of parameter influence in engineering modeling and parameter identification”.

Conflicts of Interest

The authors declare no conflict of interest.

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Pečarić, J.; Perušić Pribanić, A.; Smoljak Kalamir, K. Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality. Mathematics 2022, 10, 1505. https://doi.org/10.3390/math10091505

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Pečarić J, Perušić Pribanić A, Smoljak Kalamir K. Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality. Mathematics. 2022; 10(9):1505. https://doi.org/10.3390/math10091505

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Pečarić, Josip, Anamarija Perušić Pribanić, and Ksenija Smoljak Kalamir. 2022. "Weighted Hermite–Hadamard-Type Inequalities by Identities Related to Generalizations of Steffensen’s Inequality" Mathematics 10, no. 9: 1505. https://doi.org/10.3390/math10091505

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