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Article

# Generalizations of the Jensen–Mercer Inequality via Fink’s Identity

Faculty of Electrical Engineering, Mechanical Engineering and Naval Architecture, University of Split, Ruđera Boškovića 32, 21000 Split, Croatia
Mathematics 2021, 9(19), 2406; https://doi.org/10.3390/math9192406
Received: 25 August 2021 / Revised: 20 September 2021 / Accepted: 23 September 2021 / Published: 27 September 2021

## Abstract

:
We generalize an integral Jensen–Mercer inequality to the class of n-convex functions using Fink’s identity and Green’s functions. We study the monotonicity of some linear functionals constructed from the obtained inequalities using the definition of n-convex functions at a point.

## 1. Introduction

Jensen’s inequality,
$f 1 W n ∑ i = 1 n w i x i ≤ 1 W n ∑ i = 1 n w i f x i ,$
for a convex function $f : I → R$, real numbers $x 1 , … , x n ∈ I$ and positive real numbers $w 1 , … , w n$, where $W n = ∑ i = 1 n w i$, is one of the most important inequalities in many areas of mathematics and other areas of science. Many other inequalities can be derived from it and there are numerous of its variants, generalizations and refinements (see, for example [1,2]). One of these variants is the so-called Jensen–Mercer inequality,
$f α + β − 1 W n ∑ i = 1 n w i x i ≤ f α + f β − 1 W n ∑ i = 1 n w i f x i ,$
for a convex function $f : α , β → R$, real numbers $x 1 , … , x n ∈ α , β$ and positive real numbers $w 1 , … , w n$, which was introduced in  by A. McD. Mercer. It has been a main topic of our research (see, e.g., [4,5,6,7]) in different settings and for different classes of functions.
In the paper , we proved the following integral version.
Theorem 1.
Let $g : a , b → R$ be a continuous and monotonic function and $α , β$ be an interval such that $g a , b ⊂ α , β$. Let the function $λ : a , b → R$ be either continuous or of bounded variation, and satisfying
$λ a ≤ λ t ≤ λ b f o r a l l t ∈ α , β , λ b − λ a > 0 .$
Then, for every continuous convex function $φ : α , β → R$ the inequality
$φ α + β − ∫ a b g x d λ x ∫ a b d λ x ≤ φ α + φ β − ∫ a b φ g x d λ x ∫ a b d λ x$
holds. If φ is concave, then the reversed inequality (4) holds.
Remark 1.
Theorem 1 is also valid when the condition (3) is replaced with the more strict condition that λ is a nondecreasing function such that $λ a ≠ λ b$.
The main goal of this paper is to present generalizations of inequality (4) to the class of n-convex functions. We achieve this by means of Fink’s identity and Green’s functions, which we introduce below.
A real-valued function $φ$ defined on an interval $α , β$ is called n-convex if its nth order divided differences are nonnegative for all choices of $n + 1$ distinct points in $α , β$. Thus, 0-convex functions are nonnegative functions, 1-convex functions are increasing functions and 2-convex functions are convex functions. An n-convex function need not be n-times differentiable, however, if $φ n$ exists then $φ$ is n-convex if and only if $φ n ≥ 0$. For more information about n-convex functions, see  and also .
Fink’s identity ,
$φ s = n β − α ∫ α β φ t d t − ∑ k = 1 n − 1 n − k k ! · φ k − 1 α s − α k − φ k − 1 β s − β k β − α + 1 n − 1 ! β − α ∫ α β s − t n − 1 K t , s φ n t d t ,$
with
holds for every function $φ : α , β → R$ such that $φ n − 1$ is absolutely continuous on $α , β$ for some $n ≥ 1$. For $n = 1$, we take the sum in (5) to be zero.
We consider Green’s functions $G i , i = 1 , … , 5$ defined on $α , β × α , β$ by
and
All five Green’s functions are continuous, symmetric and convex with respect to both the variables s and t.
It can be easily shown by integrating by parts that every function $φ : α , β → R$, $φ ∈ C 2 α , β$ can be represented in the following five forms. If $x ∈ α , β$, then
$φ x = φ α + x − α φ ′ β + ∫ α β G 1 x , s φ ″ s d s ,$
$φ x = φ β + x − β φ ′ α + ∫ α β G 2 x , s φ ″ s d s ,$
$φ x = φ β − β − α φ ′ β + x − α φ ′ α + ∫ α β G 3 x , s φ ″ s d s ,$
$φ x = φ α + β − α φ ′ α − β − x φ ′ β + ∫ α β G 4 x , s φ ″ s d s ,$
$φ x = β − x β − α φ α + x − α β − α φ β + ∫ α β G 5 x , s φ ″ s d s .$
By easy calculation (see [11,12]), using representations (12)–(16), we can obtain the identity
$φ α + φ β − ∫ a b φ g x d λ x ∫ a b d λ x − φ α + β − ∫ a b g x d λ x ∫ a b d λ x = ∫ α β G i α , s + G i β , s − ∫ a b G i g x , s d λ x ∫ a b d λ x − G i α + β − ∫ a b g x d λ x ∫ a b d λ x , s φ ″ s d s ,$
for all $i = 1 , … , 5$. In the rest of the paper, for the sake of simplicity, let us denote
$x ¯ = ∫ a b g x d λ x ∫ a b d λ x , y ¯ = ∫ a b φ g x d λ x ∫ a b d λ x , z ¯ i s = ∫ a b G i g x , s d λ x ∫ a b d λ x ,$
$G i s = G i α , s + G i β , s − z ¯ i s − G i α + β − x ¯ , s , i = 1 , … , 5 .$
Since all five Green’s functions are continuous and convex, by Theorem 1, $G i s ≥ 0$ for all $i = 1 , … , 5$.

## 2. Main Results

We start with two identities which are very useful in obtaining generalizations of inequality (4) to the class of n-convex functions.
Lemma 1.
Let the functions g and λ be as in Theorem 1, and let the function K be defined by (6). Then, for every function $φ : α , β → R$ such that $φ n − 1$ is absolutely continuous on $α , β$ for some $n ≥ 1$, the identity
$φ α + φ β − y ¯ − φ α + β − x ¯ = 1 β − α ∑ k = 2 n − 1 n − k k ! φ k − 1 β α − β k − ∫ a b g x − β k d λ x ∫ a b d λ x − α − x ¯ k − 1 β − α ∑ k = 2 n − 1 n − k k ! φ k − 1 α β − α k − ∫ a b g x − α k d λ x ∫ a b d λ x − β − x ¯ k + 1 n − 1 ! β − α ∫ α β α − t n − 1 K t , α + β − t n − 1 K t , β − ∫ a b g x − t n − 1 K t , g x d λ x ∫ a b d λ x − α + β − x ¯ − t n − 1 K t , α + β − x ¯ φ n t d t$
holds.
Proof.
Using (5) in the left hand of equality (19), we obtain
$φ α + φ β − y ¯ − φ α + β − x ¯ = − ∑ k = 1 n − 1 n − k k ! α − β k − 1 φ k − 1 β + 1 n − 1 ! β − α ∫ α β α − t n − 1 K t , α φ n t d t − ∑ k = 1 n − 1 n − k k ! β − α k − 1 φ k − 1 α + 1 n − 1 ! β − α ∫ α β β − t n − 1 K t , β φ n t d t + 1 β − α ∑ k = 1 n − 1 n − k k ! φ k − 1 α ∫ a b g x − α k d λ x − φ k − 1 β ∫ a b g x − β k d λ x ∫ a b d λ x − 1 n − 1 ! β − α ∫ α β ∫ a b g x − t n − 1 K t , g x d λ x ∫ a b d λ x φ n t d t + 1 β − α ∑ k = 1 n − 1 n − k k ! φ k − 1 α β − x ¯ k − φ k − 1 β α − x ¯ k − 1 n − 1 ! β − α ∫ α β α + β − x ¯ − t n − 1 K t , α + β − x ¯ φ n t d t .$
By regrouping and calculating all the first members in the above sums, we obtain (19). □
Lemma 2.
Let the functions g and λ be as in Theorem 1, and let the functions $G i , i = 1 , … , 5$, and K be defined by (7)(11) and (6), respectively. Then, for every function $φ : α , β → R$ such that $φ n − 1$ is absolutely continuous on $α , β$ for some $n ≥ 3$, the identity
$φ α + φ β − y ¯ − φ α + β − x ¯ = 1 β − α ∑ k = 0 n − 3 n − k − 2 k ! ∫ α β G i s φ k + 1 β s − β k − φ k + 1 α s − α k d s + 1 n − 3 ! β − α ∫ α β φ n t ∫ α β G i s s − t n − 3 K t , s d s d t$
holds.
Proof.
From (5), we have
$φ ″ s = ∑ k = 0 n − 3 n − k − 2 k ! · − φ k + 1 α s − α k + φ k + 1 β s − β k β − α + 1 n − 3 ! β − α ∫ α β s − t n − 3 K t , s φ n t d t .$
Using (21) in (17) and applying Fubini’s theorem, we easily obtain (20). □
The following generalizations of inequality (4) for n-convex functions hold.
Theorem 2.
Let the functions g and λ be as in Theorem 1, and let the function K be defined by (6). Let $φ : α , β → R$ be a n-convex function such that $φ n − 1$ is absolutely continuous on $α , β$ for some $n ≥ 1$.
(i)
If
$α − t n − 1 K t , α + β − t n − 1 K t , β − ∫ a b g x − t n − 1 K t , g x d λ x ∫ a b d λ x − α + β − x ¯ − t n − 1 K t , α + β − x ¯ ≥ 0 , α ≤ t ≤ β ,$
then
$φ α + φ β − y ¯ − φ α + β − x ¯ ≥ 1 β − α ∑ k = 2 n − 1 n − k k ! φ k − 1 β α − β k − ∫ a b g x − β k d λ x ∫ a b d λ x − α − x ¯ k − 1 β − α ∑ k = 2 n − 1 n − k k ! φ k − 1 α β − α k − ∫ a b g x − α k d λ x ∫ a b d λ x − β − x ¯ k .$
(ii)
If n is even then (23) holds. Moreover, if $φ k − 1 α ≤ 0$ for $k = 2 , 3 , … , n − 1$, $φ k − 1 β ≥ 0$ for $k = 2 , 4 , … , n − 2$ and $φ k − 1 β ≤ 0$ for $k = 3 , 5 , … , n − 1$ then the right-hand side of (23) is nonnegative and (4) holds.
Proof.
$i$ Since $φ n − 1$ is absolutely continuous on $α , β$ and $φ$ is n-convex, $φ n$ exist almost everywhere on $α , β$ and $φ n ≥ 0$. This fact and assumption (22) show that the third member on the right-hand side of (19) is nonnegative, so inequality (23) immediately follows from identity (19).
$i i$ For a fixed $t ∈ α , β$ consider the function $ψ : α , β → R$ defined by
It is continuous for $n ≥ 2$ and, since
it is convex on $α , β$ for even n. Therefore, by Theorem 1,
$ψ α + ψ β − ∫ a b ψ g x d λ x ∫ a b d λ x − ψ α + β − x ¯ ≥ 0 ,$
i.e., inequality (22) holds and, by $i$, inequality (23) holds.
Further, the function $t ↦ t − α k$ is convex on $α , β$ for any k, and the function $t ↦ t − β k$ is convex (concave) on $α , β$ for even (odd) k. Hence, by Theorem 1,
$β − α k − ∫ a b g x − α k d λ x ∫ a b d λ x − β − x ¯ k ≥ 0 ,$
for all $k = 2 , … , n − 1$,
$α − β k − ∫ a b g x − β k d λ x ∫ a b d λ x − α − x ¯ k ≥ 0 ,$
for $k = 2 , 4 , … , n − 2$ and the reversed inequality (25) holds for $k = 3 , 5 , … , n − 1$. Hence, the second claim in $i i$ immediately follows. □
Remark 2.
Under the assumptions of Theorem 2, the right-hand side of (23) can be written in the form
$Ψ α + Ψ β − ∫ a b Ψ g x d λ x ∫ a b d λ x − Ψ α + β − x ¯$
where the function $Ψ : α , β → R$ is defined by
$Ψ t = 1 β − α ∑ k = 2 n − 1 n − k k ! φ k − 1 β t − β k − φ k − 1 α t − α k .$
Therefore, if n is even and the function $Ψ$ is convex on $α , β$ then, by Theorem 1, the right-hand side of (23) is nonnegative and (4) holds.
Theorem 3.
Let the functions g and λ be as in Theorem 1, and let the functions $G i , i = 1 , … , 5$, and K be defined by (7)(11) and (6), respectively. Let $φ : α , β → R$ be an n-convex function such that $φ n − 1$ is absolutely continuous on $α , β$ for some $n ≥ 3$. Then, for all $i = 1 , … , 5$:
(i)
If
$∫ α β G i s s − t n − 3 K t , s d s ≥ 0 , α ≤ t ≤ β$
then
$φ α + φ β − y ¯ − φ α + β − x ¯ ≥ 1 β − α ∑ k = 0 n − 3 n − k − 2 k ! ∫ α β G i s φ k + 1 β s − β k − φ k + 1 α s − α k d s$
holds.
(ii)
If n is even, then (27) holds. Moreover, if $φ k + 1 α ≤ 0$ for $k = 0 , 1 , … , n − 3$, $φ k − 1 β ≥ 0$ for $k = 0 , 2 , … , n − 4$ and $φ k − 1 β ≤ 0$ for $k = 1 , 3 , … , n − 3$ then the right-hand side of (27) is nonnegative and (4) holds.
Proof.
$i$ Similarly as in the proof of Theorem 2, we can assume that $φ$ is n-times differentiable and $φ n ≥ 0$. Hence, assumption (26) gives that the second member on the right-hand side of (20) is nonnegative, so inequality (27) immediately follows from identity (20).
$i i$ Since $G i s ≥ 0$, for $α ≤ s ≤ t ≤ β$ (26) holds when n is even, and the reversed (26) holds when n is odd, while for $α ≤ t ≤ s ≤ β$ (26) always holds. Therefore, $i i$ immediately follows from $i$. □
Remark 3.
Under the assumptions of Theorem 3, the right-hand side of (27) can be written in the form
$Ψ ^ i α + Ψ ^ i β − ∫ a b Ψ ^ i g x d λ x ∫ a b d λ x − Ψ ^ i α + β − x ¯ , i = 1 , … , 5$
where the functions $Ψ ^ i : α , β → R$, $i = 1 , … , 5$ are defined by
$Ψ ^ i t = 1 β − α ∑ k = 0 n − 3 n − k − 2 k ! ∫ α β G i t , s φ k + 1 β s − β k − φ k + 1 α s − α k d s .$
Therefore, if n is even and the functions $Ψ ^ i$, $i = 1 , … , 5$ are convex on $α , β$ then, by Theorem 1, the right-hand side of (27) is nonnegative for all $i = 1 , … , 5$ and (4) holds.

## 3. Related Results for $n$-Convex Functions at a Point

The class of n-convex functions at a point was introduced by Pečarić et al. in  as follows.
Let I be an interval in $R$, $d ∈ I ∘$ (interior of I), $n ∈ N$ and $P n x = x n$. A function $ϕ : I → R$ is said to be n-convex at point d if there exists a constant $C ϕ$ such that the function $Φ = ϕ − C ϕ n − 1 ! P n − 1$ is $n − 1$-concave on $I ∩ − ∞ , d$ and $n − 1$-convex on $I ∩ d , ∞$. It is shown that a function is n-convex on an interval if and only if it is n-convex at every point of its interior.
Additionally, the authors of  investigated the conditions which some linear functionals $A : C δ 1 , d → R$ and $B : C d , δ 2 → R$ have to fulfill so that the inequality $A ϕ ≤ B ϕ$ holds for any function $ϕ$ that is n-convex at point $d ∈ δ 1 , δ 2$. In this section, we give inequalities of this type for the linear functionals obtained as the differences of the left- and right-hand sides of inequalities from the previous section.
In the rest of this section, we take function $λ$ to be as in Theorem 1, functions $G i , i = 1 , … , 5$ and K defined by (7)–(11) and (6), respectively, and function $ϕ : α , β → R$ such that $ϕ n − 1$ is absolutely continuous on $α , β$. Let $δ 1 , δ 2 , d ∈ α , β$ be such that $δ 1 < d < δ 2$, and let $g 1 , g 2 : a , b → R$ be continuous and monotonic functions such that $g 1 a , b ⊂ δ 1 , d$, $g 2 a , b ⊂ d , δ 2$.
With $A ϕ / δ 1 , d$ and $B ϕ / d , δ 2$ we denote the difference of the left- and right-hand sides of inequality (23) for the function $ϕ$ on the intervals $δ 1 , d$ and $d , δ 2$, respectively, i.e.,
$A ϕ / δ 1 , d = ϕ δ 1 + ϕ d − ∫ a b ϕ g 1 x d λ x ∫ a b d λ x − ϕ δ 1 + d − x ¯ 1 − 1 d − δ 1 ∑ k = 2 n − 1 n − k k ! ϕ k − 1 d δ 1 − d k − ∫ a b g 1 x − d k d λ x ∫ a b d λ x − δ 1 − x ¯ 1 k + 1 d − δ 1 ∑ k = 2 n − 1 n − k k ! ϕ k − 1 δ 1 d − δ 1 k − ∫ a b g 1 x − δ 1 k d λ x ∫ a b d λ x − d − x ¯ 1 k ,$
where $x ¯ 1 = ∫ a b g 1 x d λ x ∫ a b d λ x ,$
$B ϕ / d , δ 2 = ϕ d + ϕ δ 2 − ∫ a b ϕ g 2 x d λ x ∫ a b d λ x − ϕ d + δ 2 − x ¯ 2 − 1 δ 2 − d ∑ k = 2 n − 1 n − k k ! ϕ k − 1 δ 2 d − δ 2 k − ∫ a b g 2 x − δ 2 k d λ x ∫ a b d λ x − d − x ¯ 2 k + 1 δ 2 − d ∑ k = 2 n − 1 n − k k ! ϕ k − 1 d δ 2 − d k − ∫ a b g 2 x − d k d λ x ∫ a b d λ x − δ 2 − x ¯ 2 k ,$
where $x ¯ 2 = ∫ a b g 2 x d λ x ∫ a b d λ x$.
For even $n ≥ 2$ from Theorem 2 $i i$ it follows that $B ϕ / d , δ 2 ≥ 0$ when $ϕ$ is n-convex, and $A ϕ / δ 1 , d ≤ 0$ when $ϕ$ is n-concave.
Similarly, for $i = 1 , … , 5$ with $A i ϕ / δ 1 , d$ and $B i ϕ / d , δ 2$ we denote the difference of the left- and right-hand sides of inequality (27) for the function $ϕ$ on the intervals $δ 1 , d$ and $d , δ 2$, respectively, i.e.,
$A i ϕ / δ 1 , d = ϕ δ 1 + ϕ d − ∫ a b ϕ g 1 x d λ x ∫ a b d λ x − ϕ δ 1 + d − x ¯ 1 − 1 d − δ 1 ∑ k = 0 n − 3 n − k − 2 k ! ∫ δ 1 d G i 1 s ϕ k + 1 d s − d k − ϕ k + 1 δ 1 s − δ 1 k d s ,$
where $x ¯ 1 = ∫ a b g 1 x d λ x ∫ a b d λ x$ and $G i 1 s = G i δ 1 , s + G i d , s − ∫ a b G i g 1 x , s d λ x ∫ a b d λ x − G i δ 1 + d − x ¯ 1 , s$,
$B i ϕ / d , δ 2 = ϕ d + ϕ δ 2 − ∫ a b ϕ g 2 x d λ x ∫ a b d λ x − ϕ d + δ 2 − x ¯ 2 − 1 δ 2 − d ∑ k = 0 n − 3 n − k − 2 k ! ∫ d δ 2 G i 2 s ϕ k + 1 δ 2 s − δ 2 k − ϕ k + 1 d s − d k d s ,$
where $x ¯ 2 = ∫ a b g 2 x d λ x ∫ a b d λ x$ and $G i 2 s = G i d , s + G i δ 2 , s − ∫ a b G i g 2 x , s d λ x ∫ a b d λ x − G i d + δ 2 − x ¯ 2 , s$.
For even $n ≥ 4$ and $i = 1 , … , 5$ from Theorem 3 $i i$ it follows that $B i ϕ / d , δ 2 ≥ 0$ when $ϕ$ is n-convex, and $A i ϕ / δ 1 , d ≤ 0$ when $ϕ$ is n-concave.
Theorem 4.
Under the above assumptions, let $n ∈ N$ be even and the function ϕ be $n + 1$-convex at point d.
(i)
If $n ≥ 2$, and
$∫ δ 1 d δ 1 − t n − 1 K t , δ 1 + d − t n − 1 K t , d − ∫ a b g 1 x − t n − 1 K t , g 1 x d λ x ∫ a b d λ x − δ 1 + d − x ¯ 1 − t n − 1 K t , δ 1 + d − x ¯ 1 d t = d − δ 1 δ 2 − d ∫ d δ 2 d − t n − 1 K t , d + δ 2 − t n − 1 K t , δ 2 − ∫ a b g 2 x − t n − 1 K t , g 2 x d λ x ∫ a b d λ x − d + δ 2 − x ¯ 2 − t n − 1 K t , d + δ 2 − x ¯ 2 d t$
then $A ϕ / δ 1 , d ≤ B ϕ / d , δ 2 .$
(ii)
If $n ≥ 4$, and
$∫ δ 1 d ∫ δ 1 d G i 1 s s − t n − 3 K t , s d s d t = d − δ 1 δ 2 − d ∫ d δ 2 ∫ d δ 2 G i 2 s s − t n − 3 K t , s d s d t$
then $A i ϕ / δ 1 , d ≤ B i ϕ / d , δ 2 , i = 1 , … , 5 .$
Proof.
$i$ Let $n ≥ 2$. Since $ϕ$ is $n + 1$-convex at point d, there exists $C ϕ$ such that the function $Φ = ϕ − C ϕ n ! P n$ is n-concave on $δ 1 , d$ and n-convex on $d , δ 2$. Therefore
$0 ≥ A Φ / δ 1 , d = A ϕ / δ 1 , d − C ϕ n ! A P n / δ 1 , d$
and
$0 ≤ B Φ / d , δ 2 = B ϕ / d , δ 2 − C ϕ n ! B P n / d , δ 2 .$
From identity (19) for the function $P n$ on the intervals $δ 1 , d$ and $d , δ 2$ we have
$A P n / δ 1 , d = n d − δ 1 ∫ δ 1 d δ 1 − t n − 1 K t , δ 1 + d − t n − 1 K t , d − ∫ a b g 1 x − t n − 1 K t , g 1 x d λ x ∫ a b d λ x − δ 1 + d − x ¯ 1 − t n − 1 K t , δ 1 + d − x ¯ 1 d t$
and
$B P n / d , δ 2 = n δ 2 − d ∫ d δ 2 d − t n − 1 K t , d + δ 2 − t n − 1 K t , δ 2 − ∫ a b g 2 x − t n − 1 K t , g 2 x d λ x ∫ a b d λ x − d + δ 2 − x ¯ 2 − t n − 1 K t , d + δ 2 − x ¯ 2 d t .$
Hence, assumption (28) is equivalent to $A P n / δ 1 , d = B P n / d , δ 2$ and from (30) and (31) it follows
$A ϕ / δ 1 , d ≤ C ϕ n ! A P n / δ 1 , d = C ϕ n ! B P n / d , δ 2 ≤ B ϕ / d , δ 2 .$
$i i$ Analogously as $i$ using identity (20). □
Remark 4.
Notice that from the proof of Theorem 4 it follows that it is also valid when the assumptions (28) and (29) are replaced with the weaker assumptions that $C ϕ B P n / d , δ 2 − A P n / δ 1 , d ≥ 0$ and $C ϕ B i P n / d , δ 2 − A i P n / δ 1 , d ≥ 0$ for $i = 1 , … , 5$, respectively.

## 4. Conclusions

In the papers [11,12], we generalized the Jensen–Mercer inequality to the class of n-convex functions by using the Taylor polynomial and Green’s functions. Here, we do so using Fink’s identity and use these generalizations to obtain some related results for n-convex functions at a point. In a similar way as in , we can obtain Grüss and Ostrowski type inequalities in terms of Čebyšev functionals. Using the main ideas from , we can also generate new families of n-exponentially convex and exponentially convex functions by considering the linear functionals defined above.

## Funding

This research was supported by the University of Split, FESB.

## Acknowledgments

We would like to thank the reviewers for their effort to read the paper thoroughly and give us very useful suggestions on how to improve it.

## Conflicts of Interest

The author declares no conflict of interest.

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Matković, A. Generalizations of the Jensen–Mercer Inequality via Fink’s Identity. Mathematics 2021, 9, 2406. https://doi.org/10.3390/math9192406

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Matković A. Generalizations of the Jensen–Mercer Inequality via Fink’s Identity. Mathematics. 2021; 9(19):2406. https://doi.org/10.3390/math9192406

Chicago/Turabian Style

Matković, Anita. 2021. "Generalizations of the Jensen–Mercer Inequality via Fink’s Identity" Mathematics 9, no. 19: 2406. https://doi.org/10.3390/math9192406

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