Hermite–Hadamard and Jensen-Type Inequalities for Harmonical (h₁, h₂)-Godunova–Levin Interval-Valued Functions

Abstract

There is no doubt that convex and non-convex functions have a significant impact on optimization. Due to its behavior, convexity also plays a crucial role in the discussion of inequalities. The principles of convexity and symmetry go hand-in-hand. With a growing connection between the two in recent years, we can learn from one and apply it to the other. There have been significant studies on the generalization of Godunova–Levin interval-valued functions in the last few decades, as it has tremendous applications in both pure and applied mathematics. In this paper, we introduce the notion of interval- valued harmonical (h₁, h₂)-Godunova–Levin functions. Using the new concept, we establish a new interval Hermite–Hadamard and Jensen-type inequalities that generalize the ones that exist in the literature. Additionally, we provide some examples to prove the validity of our main results.

1. Introduction

As introduced in Moore’s celebrated book, interval analysis is one of the key methods used in numerical analysis [1]. As a result of its success in the last 50 years, various fields have benefited from its use, such as aeroelasticity [2], differential equations for the interval [3], automatic error analysis [4], neural network output optimization [5], computer graphics [6], and so on. Please refer to the following for a deeper understanding and applications, see e.g., [7,8,9,10,11].

Mathematics and other scientific fields are highly influenced by inequalities. Many types of inequalities exist, but those involving Jensen, Ostrowski, Hermite–Hadamard, and Minkowski hold particular significance among them. Chalco-Cano et al. recently extended a number of these inequalities to interval-valued functions.; see, e.g., [12,13,14,15]. According to several scholars, function convexity is based on inequalities. Many scholars studied the Hermite–Hadamard inequality for convex functions due to its importance. Here is the classical H-H inequality.

$$\vartheta \left(\frac{e+f}{2}\right)\le \frac{1}{f-e}{\int }_e^f\vartheta \left(\alpha \right)d\alpha \le \frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}.$$

There are different classes of convexity that can be represented by this inequality, see e.g., [16,17,18,19,20]. The concept of harmonic convexity was also introduced in 2014, along with a few H-H inequalities for this type of function [21]. In 2015, Noor et al. introduced harmonic h-convex functions and some related Hermite–Hadamard inequalities; see e.g., [22]. In combination with interval analysis, Zhao Dafang et al. and Ruonan Liu recently extended H-H inequality by extending it to interval h-convex functions [23], interval harmonic h-convex functions [24], interval $(h_1,h_2)$-convex functions [25], and interval harmonical $(h_1,h_2)$-convex functions [26]. Ohud Almutairi and Adem Kiliman used the definition of the h-Godunova–Levin function and established the following inequality [27]. Moreover, for the fuzzy interval- valued functions, Costa [14] presents a fuzzy Jensen-type integral inequality, and Hongxin Bai et al. developed Jensen-type inequality for $(h_1,h_2)$ interval-nonconvex functions [28].

Our research is primarily motivated by Ohud Almutairi [27], Hongxin Bai et al. [28], and Ruonan Liu [26]. We begin by initiating the notion of interval-valued harmonical $(h_1,h_2)$-Godunova–Levin functions. Then, we construct some new H-H and Jensen-type inequalities for the above-said generalization.

As a final note, the rest of the paper is structured as follows. Preliminaries and mathematical backgrounds are provided in Section 2. Section 3 describes the problem and our main findings. The conclusion is provided in Section 4.

2. Mathematical Backgrounds and Preliminaries

To begin, let us review some definitions, properties, and notations that will be used throughout the article; see [23]. Let us say that I denote the pack of all intervals of a set of real numbers R; [e] $\in I$ is defined as follows.

$\left[e\right]=[\underline{e},\overline{e}]=\{x\in R|\underline{e}\le x\le \overline{e}\}$, $\underline{e},\overline{e}\in R$, where real interval $\left[e\right]$ is closed and bounded subset of R. The interval $\left[e\right]$ is said to have degenerated when $\underline{e}$ and $\overline{e}$ are both equal. We call $\left[e\right]$ positive when $\underline{e}>0$ or negative when $\overline{e}<0$. We denote the pack of all intervals of the set of real numbers by $R_I$ of R and use $R_I^{+}$ and $R_I^{-}$ to define the pack of all positive and negative intervals, respectively. The inclusion “$\subseteq$” is defined as follows.

$$\left[e\right]\subseteq \left[f\right]⟺[\underline{e},\overline{e}]\subseteq [\underline{\vartheta },\overline{\vartheta }]⟺\underline{f}\le \underline{e},\overline{e}\le \overline{f}.$$

For any random real number $\upsilon$ and $\left[e\right]$, interval $\upsilon \left[e\right]$ is given as follows.

$$\upsilon .[\underline{e},\overline{e}]=\left\{\begin{array}{c}[\upsilon \underline{e},\upsilon \overline{e}],if \upsilon >0\hfill \\ \left\{0\right\},if \upsilon =0\hfill \\ [\upsilon \overline{e},\upsilon \underline{e}],if \upsilon <0\hfill \end{array}\right.$$

For $\left[e\right]=[\underline{e},\overline{e}]$ and $\left[f\right]=[\underline{f},\overline{f}]$, algebraic operations are defined as follows:

$$\left[e\right]+\left[f\right]=[\underline{e}+\underline{f},\overline{e}+\overline{f}],$$

$$\left[e\right]-\left[f\right]=[\underline{e}-\underline{f},\overline{e}-\overline{f}],$$

$$\left[e\right]·\left[f\right]=[min\{\underline{ef},\underline{e}\overline{f},\overline{e}\underline{f},\overline{ef}\},max\{\underline{ef},\underline{e}\overline{f},\overline{ef},\overline{e}\underline{f}\}],$$

$$\left[e\right]/\left[f\right]=[min\{\underline{e}/\underline{f},\underline{e}/\overline{f},\overline{e}/\underline{f},\overline{e}/\overline{f}\},max\{\underline{e}/\underline{f},\underline{e}/\overline{f},\overline{e}/\overline{f},\overline{e}/\overline{f}\}],$$

where

$$0\notin [\underline{e},\overline{e}].$$

For intervals $[\underline{e},\overline{e}],[\underline{f},\overline{f}]$, the Hausdorff–Pompeiu distance is defined as follows; see, e.g., [29].

$$d([\underline{e},\overline{e}],[\underline{f},\overline{f}])=max\left\{\right|\underline{e}-\underline{f}|,|\overline{e}-\overline{f}\left|\right\}.$$

It is generally known that the entire metric space $(R_I,d)$ is complete.

Definition 1

([30]). Let $\vartheta :[e,f]\to R_I$ be such that $\vartheta \left(\upsilon \right)=[\underline{\vartheta }\left(\upsilon \right),\overline{\vartheta }\left(\upsilon \right)]$ for each $\upsilon \in [e,f]$ and $\underline{\vartheta },\overline{\vartheta }$ are Riemann integrable over interval $[e,f]$. Then, we say that our function ϑ is Riemann integrable over interval $[e,f]$ and denoted as

$${\int }_e^f\vartheta \left(\upsilon \right)d\upsilon =\left[{\int }_e^f\underline{\vartheta }\left(\upsilon \right)d\upsilon ,{\int }_e^f\overline{\vartheta }\left(\upsilon \right)d\upsilon \right].$$

Definition 2

([21]). A set $S\subset R-\left\{0\right\}$ is known as harmonical convex set if

$$\frac{ef}{\upsilon e+(1-\upsilon )f}\in S$$

$\forall e,f\in S$ and $\upsilon \in [0,1]$.

Definition 3

([27]). A positive function $\vartheta :S\to R$ is said to be a Godunova–Levin if

$$\vartheta (\upsilon e+(1-\upsilon )f)\le \frac{\vartheta \left(e\right)}{\upsilon }+\frac{\vartheta \left(f\right)}{(1-\upsilon )}$$

$\forall e,f\in S$ and $\upsilon \in (0,1)$.

Definition 4

([21]). A function $\vartheta :S\to R$ is known as harmonically convex function, if

$$\vartheta \left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)\le \upsilon \vartheta \left(e\right)+(1-\upsilon )\vartheta \left(f\right)$$

$\forall e,f\in S$ and $\upsilon \in [0,1]$.

Definition 5

([22]). Let $h:[0,1]\subseteq S\to R$ be a non-negative function and S be harmonically convex set with $h\ne 0$. We state that $\vartheta :S\to R$ is known as harmonical h-convex function, if

$$\vartheta \left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)\le h\left(\upsilon \right)\vartheta \left(e\right)+h(1-\upsilon )\vartheta \left(f\right)$$

$\forall e,f\in S$ and $\upsilon \in [0,1]$.

Definition 6

([27]). Let $h:(0,1)\subseteq S\to R$ be a non-negative function. We state that $\vartheta :S\to R$ is known as h-Godunova–Levin function, if

$$\vartheta (\upsilon e+(1-\upsilon )f)\le \frac{\vartheta \left(e\right)}{h\left(\upsilon \right)}+\frac{\vartheta \left(f\right)}{h(1-\upsilon )}$$

for all $e,f\in S$ and $\upsilon \in (0,1)$.

Definition 7

([31]). Let $h:(0,1)\subseteq S\to R$ be a non-negative function. We state that $\vartheta :S\to R$ is known as harmonical h-Godunova–Levin function, if

$$\vartheta \left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)\le \frac{\vartheta \left(e\right)}{h\left(\upsilon \right)}+\frac{\vartheta \left(f\right)}{h(1-\upsilon )}$$

for all $e,f\in S$ and $\upsilon \in (0,1)$.

Definition 8

([32]). Let $h_1,h_2:(0,1)\subseteq S\to R$ be a non-negative functions. We state $\vartheta :S\to R$ is known as harmonical $(h_1,h_2)$-Godunova–Levin function, if

$$\vartheta \left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)\le \frac{\vartheta \left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\vartheta \left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)}$$

for all $e,f\in S$ and $\upsilon \in (0,1)$.

Remark 1.

• If $h_1\left(\upsilon \right)=h_2\left(\upsilon \right)=1$, then Definition 8 reduces to a harmonical P-convex function [31].
• If $h_1\left(\upsilon \right)=\frac{1}{\upsilon }$ and $h_2\left(\upsilon \right)=1$, then Definition 8 reduces to a harmonical convex function [31].
• If $h_1\left(\upsilon \right)=\frac{1}{h\left(\upsilon \right)}$ and $h_2\left(\upsilon \right)=1$, then Definition 8 reduces to a harmonical h-convex function [22].
• If $h_1\left(\upsilon \right)={\left(\upsilon \right)}^s$ and $h_2\left(\upsilon \right)=1$, then Definition 8 reduces to a harmonical s-Godunova–Levin function [22].

3. Main Results

Preliminaries were closed by introducing the new harmonic interval-valued $(h_1,h_2)$ Godunova–Levin functions. This idea is influenced by An et al. [25].

Definition 9.

Suppose $h_1,h_2:(0,1)\subseteq S\to R$ is a non-negative functions such that $h_1,h_2\ne 0$ and let S be a harmonical Godunova–Levin set. We state that $\vartheta :S\to R_I^{+}$ is known as a harmonical $(h_1,h_2)$-Godunova–Levin(GL) interval-valued convex function, if

$$\frac{\vartheta \left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\vartheta \left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)}\subseteq \vartheta \left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)$$

(1)

$\forall e,f\in S$ and $\upsilon \in (0,1)$. If the above inclusion is reveresed in Definition 9, then function ϑ is known as harmonical $(h_1,h_2)$-GL concave interval-valued functions. The spaces of all harmonical $(h_1,h_2)$-GL convex and $(h_1,h_2)$-GL concave interval-valued functions are denoted by $SGHX((\frac{1}{h_1},\frac{1}{h_2}),S,{R_I}^{+})$ and $SGHV((\frac{1}{h_1},\frac{1}{h_2}),S,{R_I}^{+})$, respectively.

Proposition 1.

Let $\vartheta :[e,f]\to {R_I}^{+}$ be harmonical interval- valued $(h_1,h_2)$-GL function such that $\vartheta \left(\upsilon \right)=[\underline{\vartheta }\left(\upsilon \right),\overline{\vartheta }\left(\upsilon \right)]$. Then, if $\vartheta \in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ iff $\underline{\chi }\in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ and if $\overline{\vartheta }\in SGHV((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+}).$

Proof. 

Let $\vartheta$ be $(h_1,h_2)$-GL convex interval valued function. Assume $x,y\in [e,f]$, $\upsilon \in (0,1)$. Then, we have the following.

$$\frac{\vartheta \left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\vartheta \left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)}\subseteq \vartheta \left(\frac{ef}{\kappa e+(1-\upsilon )f}\right)$$

That is, the following is obtained.

$$\left[\frac{\underline{\vartheta }\left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\underline{\vartheta }\left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)},\frac{\overline{\vartheta }\left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\overline{\vartheta }\left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)}\right]\subseteq \vartheta \left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)$$

It follows that we have

$$\frac{\underline{\vartheta }\left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\underline{\vartheta }\left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)}\ge \underline{\vartheta }\left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)$$

and

$$\frac{\overline{\vartheta }\left(e\right)}{h_1\left(\upsilon \right)h_2(1-\upsilon )}+\frac{\overline{\vartheta }\left(f\right)}{h_1(1-\upsilon )h_2\left(\upsilon \right)}\le \overline{\vartheta }\left(\frac{ef}{\upsilon e+(1-\upsilon )f}\right)$$

This shows that $\underline{\vartheta }\in SGX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ and $\overline{\vartheta }\in SGV((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+}).$ Conversely, suppose that $\underline{\vartheta }\in SGX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ and $\overline{\vartheta }\in SGV((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$. Based on the above definition and set inclusion, we obtain the following: $\vartheta \in SGX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+}).$ The proof has now been completed. □

Proposition 2.

Let $\vartheta :[e,f]\to {R_I}^{+}$ be harmonical interval valued $(h_1,h_2)$-GL function such that $\vartheta \left(\upsilon \right)=[\underline{\vartheta }\left(\upsilon \right),\overline{\vartheta }\left(\upsilon \right)]$. Then, if $\vartheta \in SGHV((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ iff $\underline{\vartheta }\in SGHV((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ and if $\overline{\vartheta }\in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+}).$ This is similar to Proposition 1.

3.1. Hermite–Hadamard Inequalities

In this section, we prove some inequalities of the Hermite-Hadamard type for harmonically ($h_1,h_2$)-Godunova-Levin interval-valued functions.

Theorem 1.

Throughout the entire process, we can follow $H(x,y)=h_1\left(x\right)h_2\left(y\right)$ $\forall x,y\in (0,1)$. Let $\vartheta :[e,f]\to {R_I}^{+},h_1,h_2:(0,1)\to R^{+}$ be a continuous functions. If $\vartheta \in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],$ ${R_I}^{+})$ and $\vartheta \in I{R}_{[}e,f]$, then the following is the case.

$$\frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{2}\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq [\vartheta \left(e\right)+\vartheta \left(f\right)]{\int }_0^1\frac{dx}{H(x,1-x)}.$$

(2)

Proof. 

We begin by assuming that $\vartheta \in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$, then

$$\frac{\vartheta \left(p\right)}{\left[H(\frac{1}{2},\frac{1}{2})\right]}+\frac{\vartheta \left(q\right)}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\subseteq \vartheta \left(\frac{2pq}{p+q}\right)$$

where

$$p=\frac{ef}{xe+(1-x)f}$$

and

$$q=\frac{ef}{(1-x)e+xf}$$

Then, we obtain the following.

$$\frac{1}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\left[\vartheta (\frac{ef}{xe+(1-x)f}\right)+\vartheta (\frac{ef}{(1-x)e+xf}\left)\right]\subseteq \vartheta \left(\frac{2ef}{e+f}\right)$$

(3)

Multiplying both sides by the following:

$$H\left(\frac{1}{2},\frac{1}{2}\right)$$

we obtain

$$\left[\vartheta (\frac{ef}{xe+(1-x)f}\right)+\vartheta (\frac{ef}{(1-x)e+xf}\left)\right]\subseteq H\left(\frac{1}{2},\frac{1}{2}\right)\vartheta \left(\frac{2ef}{e+f}\right)$$

(4)

Integrating both sides of the above inequality over $(0,1)$, we have

$$\begin{gathered} {\int }_0^1\left[\vartheta (\frac{ef}{xe+(1-x)f}\right)+\vartheta (\frac{ef}{(1-x)e+xf}\left)\right]dx \\ \subseteq H\left(\frac{1}{2},\frac{1}{2}\right){\int }_0^1\vartheta \left(\frac{2ef}{e+f}\right)dx \end{gathered}$$

Therefore,

$$\begin{gathered} {\int }_0^1\underline{\vartheta }(\frac{ef}{xe+(1-x)f})dx+{\int }_0^1\underline{\vartheta }(\frac{ef}{(1-x)e+xf})dx\ge \\ H\left(\frac{1}{2},\frac{1}{2}\right){\int }_0^1\underline{\vartheta }\left(\frac{2ef}{e+f}\right)dx \end{gathered}$$

and

$$\begin{gathered} {\int }_0^1\overline{\vartheta }(\frac{ef}{xe+(1-x)f})dx+{\int }_0^1\overline{\vartheta }(\frac{ef}{(1-x)e+xf})dx\le \\ H\left(\frac{1}{2},\frac{1}{2}\right){\int }_0^1\overline{\vartheta }\left(\frac{2ef}{e+f}\right)dx \end{gathered}$$

It follows that we have the following.

$$\begin{gathered} \frac{2ef}{f-e}{\int }_e^f\frac{\underline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \ge H\left(\frac{1}{2},\frac{1}{2}\right){\int }_0^1\underline{\vartheta }\left(\frac{2ef}{e+f}\right)dx \\ =H\left(\frac{1}{2},\frac{1}{2}\right)\underline{\vartheta }\left(\frac{2ef}{e+f}\right) \end{gathered}$$

Similarly, we have the following.

$$\begin{gathered} \frac{2ef}{f-e}{\int }_e^f\frac{\overline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \le H\left(\frac{1}{2},\frac{1}{2}\right){\int }_0^1\overline{\vartheta }\left(\frac{2ef}{e+f}\right)dx \\ =H\left(\frac{1}{2},\frac{1}{2}\right)\overline{\vartheta }\left(\frac{2ef}{e+f}\right) \end{gathered}$$

This implies the following.

$$\left[H\left(\frac{1}{2},\frac{1}{2}\right)\right]\left[\underline{\vartheta }\left(\frac{2ef}{e+f}\right),\overline{\vartheta }\left(\frac{2ef}{e+f}\right)\right]\supseteq \frac{2ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon$$

Divide both sides by $\frac{1}{2}$; then, the first inclusion of Theorem 2 is proved.

$$\frac{\left[H\left(\frac{1}{2},\frac{1}{2}\right)\right]}{2}\left[\underline{\vartheta }\left(\frac{2ef}{e+f}\right),\overline{\vartheta }\left(\frac{2ef}{e+f}\right)\right]\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon$$

(5)

Based on our hypothesis, we can obtain the following.

$$\frac{\vartheta \left(e\right)}{h_1\left(x\right)h_2(1-x)}+\frac{\vartheta \left(f\right)}{h_1(1-x)h_2\left(x\right)}\subseteq \vartheta \left(\frac{ef}{xe+(1-x)f}\right)$$

$$\frac{\vartheta \left(e\right)}{h_1(1-x)h_2\left(x\right)}+\frac{\vartheta \left(f\right)}{h_1\left(x\right)h_2(1-x)}\subseteq \vartheta \left(\frac{ef}{(1-x)e+xf}\right)$$

Add these two and integrate over $(0,1)$; we then have the following.

$$\begin{gathered} \left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]{\int }_0^1\frac{1}{h_1\left(x\right)h_2(1-x)}dx+\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]{\int }_0^1\frac{1}{h_1(1-x)h_2\left(x\right)}dx \\ \subseteq {\int }_0^1\left[\vartheta \left(\frac{ef}{xe+(1-x)f}\right)+\vartheta \left(\frac{ef}{(1-x)e+xf}\right)\right]dx \end{gathered}$$

Since, at $x=\frac{1}{2}$, both integrals ${\int }_0^1\frac{1}{h_1\left(x\right)h_2(1-x)}dx={\int }_0^1\frac{1}{h_1(1-x)h_2\left(x\right)}dx$ are equal, the following is implied.

$$2\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]{\int }_0^1\frac{1}{H(x,1-x)}dx\subseteq \frac{2ef}{f-e}{\int }_e^f\frac{\vartheta \left(\kappa \right)}{{\kappa }^2}d\kappa$$

Dividing by two, we obtain the following.

$$\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]{\int }_0^1\frac{1}{H(x,1-x)}dx\subseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\kappa \right)}{{\kappa }^2}d\kappa$$

(6)

Now, by combining (5) and (6), we obtain the required result.

$$\begin{gathered} \frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{2}\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\kappa \right)d\kappa }{{\kappa }^2}\supseteq \\ [\vartheta \left(e\right)+\vartheta \left(f\right)]{\int }_0^1\frac{dx}{H(x,1-x)}. \end{gathered}$$

□

Remark 2.

Theorem 1 reduces to harmonical P-functions for interval-valued if $h_1\left(x\right)=h_2\left(x\right)=1$:

$$\frac{1}{2}\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)d\upsilon }{{\upsilon }^2}\supseteq [\vartheta \left(e\right)+\vartheta \left(f\right)].$$

Theorem 1 reduces to harmonical h-Godunova–Levin interval-valued functions if $H(x,y)=h\left(x\right)$:

$$\frac{\left[h\left(\frac{1}{2}\right)\right]}{2}\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq [\vartheta \left(e\right)+\vartheta \left(f\right)]{\int }_0^1\frac{dx}{h\left(x\right)}.$$

Theorem 1 reduces to harmonical interval-valued h-convex functions if $H(x,y)=\frac{1}{h\left(x\right)}$:

$$\frac{1}{2\left[h\left(\frac{1}{2}\right)\right]}\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq [\vartheta \left(e\right)+\vartheta \left(f\right)]{\int }_0^{1}h\left(x\right)dx.$$

Theorem 1 reduces to harmonic $(h_1,h_2)$-convex interval-valued functions if $H(x,y)=\frac{1}{H(x,1-x)}$:

$$\frac{1}{2\left[H(\frac{1}{2},\frac{1}{2})\right]}\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq [\vartheta \left(e\right)+\vartheta \left(f\right)]{\int }_0^{1}H(x,1-x)dx.$$

Example 1.

Let $h_1\left(x\right)=\frac{1}{x}$ and $h_2\left(x\right)=1$ where $x\in (0,1)$, $[e,f]=[\frac{1}{2},1]$ and $\vartheta :[e,f]\to {R_I}^{+}$ is defined by $\vartheta \left(\upsilon \right)=[{\upsilon }^2,5-e^{\upsilon }].$ Then, the following is the case.

$$\begin{gathered} \frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{2}f\left(\frac{2ef}{e+f}\right)=\vartheta \left(\frac{2}{3}\right)=\left[\frac{4}{9},5-e^{\frac{2}{3}}\right], \\ \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon =\left[{\int }_{\frac{1}{2}}^{1}d\upsilon ,{\int }_{\frac{1}{2}}^1\frac{(5-e^{\upsilon })}{{\upsilon }^2}d\upsilon \right] \\ =\left[\frac{1}{2},2.979941375566026\right] \\ \left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]{\int }_{\frac{1}{2}}^1\frac{dx}{H(x,1-x)}=\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]{\int }_0^{1}xdx \\ =\left[\frac{5}{8},\frac{\left(10-\sqrt{e}-e\right)}{2}\right] \end{gathered}$$

Thus, we obtain the following.

$$\left[\frac{4}{9},5-e^{\frac{2}{3}}\right]\supseteq \left[\frac{1}{2},2.979941375566026\right]\supseteq \left[\frac{5}{8},\frac{\left(10-\sqrt{e}-e\right)}{2}\right]$$

Consequently, the above theorem is verified.

Theorem 2.

Let $\vartheta :[e,f]\to {R_I}^{+},h_1,h_2:(0,1)\to R^{+}$ be a continuous function. If $\vartheta \in SGHX((\frac{1}{h},[e,f],{R_I}^{+})$ and $\vartheta \in I{R}_{[}e,f]$, then we have

$$\begin{gathered} \frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{4}f\left(\frac{2ef}{e+f}\right)\supseteq {△}_1\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq {△}_2 \\ \supseteq \left\{\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]\left[\frac{1}{2}+\frac{1}{H(\frac{1}{2},\frac{1}{2})}\right]\right\}{\int }_0^1\frac{dx}{H(x,1-x)}. \end{gathered}$$

where

$$\begin{gathered} {△}_1=\frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\left[\vartheta \left(\frac{4ef}{e+3f}\right)+\vartheta \left(\frac{4ef}{f+3e}\right)\right], \\ {△}_2=\left[\vartheta \left(\frac{2ef}{e+f}\right)+\left(\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}\right)\right]{\int }_0^1\frac{dx}{H(x,1-x)}. \end{gathered}$$

Proof. 

We begin by assuming that $\vartheta \in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[s,r],{R_I}^{+})$ and $\vartheta \in I{R}_{[}e,f]$; for $[e,\frac{2ef}{e+f}],$ we have

$$\frac{\vartheta \left(\frac{e\frac{2ef}{e+f}}{xe+(1-x)\frac{2ef}{e+f}}\right)}{\left[H(\frac{1}{2},\frac{1}{2})\right]}+\frac{\vartheta \left(\frac{e\frac{2ef}{e+f}}{(1-x)e+x\frac{2ef}{e+f}}\right)}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\subseteq \vartheta \left(\frac{4ef}{e+3f}\right).$$

and we obtain the following.

$$\frac{1}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\left[\vartheta \left(\frac{e\frac{2ef}{e+f}}{xe+(1-x)\frac{2ef}{e+f}}\right)+\vartheta \left(\frac{e\frac{2ef}{e+f}}{(1-x)e+x\frac{2ef}{e+f}}\right)\right]\subseteq \vartheta \left(\frac{4ef}{e+3f}\right).$$

Integrating both sides of the above inequality over $(0,1)$, we have the following.

$$\begin{gathered} \frac{1}{\left[H(\frac{1}{2},\frac{1}{2})\right]}{\int }_0^1\left[\underline{\vartheta }\left(\frac{e\frac{2ef}{e+f}}{xe+(1-x)\frac{2ef}{e+f}}\right)dx+\underline{\vartheta }\left(\frac{e\frac{2ef}{e+f}}{(1-x)e+x\frac{2ef}{e+f}}\right)dx,\overline{\vartheta }\left(\frac{e\frac{2ef}{e+f}}{xe+(1-x)\frac{2ef}{e+f}}\right)dx+\overline{\vartheta }\left(\frac{e\frac{2ef}{e+f}}{(1-x)e+x\frac{2ef}{e+f}}\right)dx\right] \\ \subseteq \vartheta \left(\frac{4ef}{e+3f}\right). \end{gathered}$$

Then, the above inequality become the following.

$$=\frac{1}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\left[\frac{2ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\underline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon +\frac{2ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\underline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon ,\frac{2ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\overline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon +\frac{2ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\overline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right]$$

$$\subseteq f\left(\frac{4ef}{e+3f}\right),$$

$$=\frac{1}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\left[\frac{4ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\underline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon ,\frac{4ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\overline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right]\subseteq f\left(\frac{4ef}{e+3f}\right),$$

$$=\frac{4}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\left[\frac{ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\underline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon ,\frac{ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\overline{\vartheta }\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right]\subseteq \vartheta \left(\frac{4ef}{e+3f}\right),$$

$$=\frac{4}{\left[H(\frac{1}{2},\frac{1}{2})\right]}\left[\frac{ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right]\subseteq f\left(\frac{4ef}{e+3f}\right)$$

$$=\frac{ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \subseteq \frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\vartheta \left(\frac{4ef}{e+3f}\right)$$

(7)

Similarly, for interval $[\frac{2ef}{e+f},f],$ we have the following.

$$\frac{ef}{f-e}{\int }_{\frac{2ef}{e+f}}^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \subseteq \frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}f\left(\frac{4ef}{f+3e}\right)$$

(8)

Adding Equations (7) and (8), we have the following.

$$\begin{gathered} {△}_1=\frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\left[\vartheta \left(\frac{4ef}{3e+f}\right)+\vartheta \left(\frac{4ef}{e+3f}\right)\right] \\ \supseteq \left[\frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right] \\ =\frac{1}{2}\left[\frac{2ef}{f-e}{\int }_e^{\frac{2ef}{e+f}}\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon +\frac{2ef}{f-e}{\int }_{\frac{2ef}{e+f}}^f\frac{\vartheta \left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right] \\ \supseteq \frac{1}{2}\left[\left[\vartheta \left(e\right)+\vartheta \left(\frac{2ef}{e+f}\right)\right]{\int }_0^1\frac{dx}{H(x,1-x)}\right] \\ +\frac{1}{2}\left[\left[\vartheta \left(f\right)+\vartheta \left(\frac{2ef}{e+f}\right)\right]{\int }_0^1\frac{dx}{H(x,1-x)}\right] \\ =\frac{1}{2}\left[\left\{\vartheta \left(e\right)+\vartheta \left(f\right)+2\vartheta \left(\frac{2ef}{e+f}\right)\right\}{\int }_0^1\frac{dx}{H(x,1-x)}\right] \\ =\left[\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}+\vartheta \left(\frac{2ef}{e+f}\right)\right]{\int }_0^1\frac{dx}{H(x,1-x)}={△}_2 \end{gathered}$$

Now, the following is obtained.

$$\begin{gathered} \frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{4}\vartheta \left(\frac{2ef}{e+f}\right) \\ =\frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{4}\vartheta \left(\frac{2\frac{4ef}{e+3f}\frac{4ef}{f+3e}}{\frac{4ef}{e+3f}+\frac{4ef}{f+3e}}\right) \\ \supseteq \frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{4}\left[\frac{\vartheta \left(\frac{4ef}{e+3f}\right)}{H(\frac{1}{2},\frac{1}{2})}+\frac{\vartheta \left(\frac{4ef}{f+3e}\right)}{H(\frac{1}{2},\frac{1}{2})}\right] \\ =\frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{4H(\frac{1}{2},\frac{1}{2})}\left[\vartheta \left(\frac{4ef}{e+3f}\right)+\vartheta \left(\frac{4ef}{f+3e}\right)\right] \\ =\frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\left[\vartheta \left(\frac{4ef}{e+3f}\right)+\vartheta \left(\frac{4ef}{e+3f}\right)\right] \\ ={△}_1 \\ \supseteq \frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\left\{\frac{1}{H(\frac{1}{2},\frac{1}{2})}\left[\vartheta \left(e\right)+\vartheta \left(\frac{2ef}{e+f}\right)\right]+\frac{1}{H(\frac{1}{2},\frac{1}{2})}\left[\vartheta \left(f\right)+\vartheta \left(\frac{2ef}{e+f}\right)\right]\right\} \\ =\frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\left\{\frac{1}{H(\frac{1}{2},\frac{1}{2})}\left[\vartheta \left(e\right)+\vartheta \left(f\right)+2\vartheta \left(\frac{2ef}{e+f}\right)\right]\right\} \\ =\frac{1}{4}\left\{\vartheta \left(e\right)+\vartheta \left(f\right)+2\vartheta \left(\frac{2ef}{e+f}\right)\right\} \\ =\frac{1}{2}\left[\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}+\vartheta \left(\frac{2ef}{e+f}\right)\right] \\ \supseteq \left[\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}+\vartheta \left(\frac{2ef}{e+f}\right)\right]{\int }_0^1\frac{dx}{H(x,1-x)} \\ ={△}_2 \\ \supseteq \left[\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}+\frac{\vartheta \left(e\right)}{H(\frac{1}{2},\frac{1}{2})}+\frac{\vartheta \left(f\right)}{H(\frac{1}{2},\frac{1}{2})}\right]{\int }_0^1\frac{dx}{H(x,1-x)} \\ =\left[\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}+\frac{1}{H(\frac{1}{2},\frac{1}{2})}\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]\right]{\int }_0^1\frac{dx}{H(x,1-x)} \\ =\left\{\left[\vartheta \left(e\right)+\vartheta \left(f\right)\right]\left[\frac{1}{2}+\frac{1}{H(\frac{1}{2},\frac{1}{2})}\right]\right\}{\int }_0^1\frac{dx}{H(x,1-x)}. \end{gathered}$$

This complete the proof. □

Example 2.

Let $h_1\left(x\right)=\frac{1}{x}$ and $h_2\left(x\right)=1$ where $x\in (0,1)$, $[e,f]=[\frac{1}{2},1]$ and $\vartheta :[e,f]\to {R_I}^{+}$ are defined by $\vartheta \left(\upsilon \right)=[{\upsilon }^2,4-e^{\upsilon }].$

Consider the following.

$$\begin{gathered} \frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{4}\vartheta \left(\frac{2ef}{e+f}\right)=\vartheta \left(\frac{2}{3}\right)=[\frac{4}{9},4-e^{\frac{2}{3}}], \\ {△}_1=\frac{\left[H(\frac{1}{2},\frac{1}{2})\right]}{4}\left[\vartheta \left(\frac{4ef}{e+3f}\right)+\vartheta \left(\frac{4ef}{f+3e}\right)\right] \\ =\frac{1}{2}\left[[\frac{16}{49},4-e^{\frac{4}{7}}]+[\frac{16}{25},4-e^{\frac{4}{5}}]\right]=[\frac{592}{1225},\frac{8-e^{\frac{4}{7}}-e^{\frac{4}{5}}}{2}], \\ {△}_2=\left[f\left(\frac{2ef}{e+f}\right)+\left(\frac{\vartheta \left(e\right)+\vartheta \left(f\right)}{2}\right)\right]{\int }_0^1\frac{dx}{H(x,1-x)} \\ =\frac{1}{2}\left[\vartheta \left(\frac{2}{3}\right)+\left(\frac{\vartheta \left(\frac{1}{2}\right)+\vartheta \left(1\right)}{2}\right)\right]=\left[\frac{77}{144},\frac{-e-\sqrt{e}-2e^{\frac{2}{3}}+16}{4}\right] \end{gathered}$$

Thus, we obtain the following.

$$\begin{gathered} \left[\frac{4}{9},4-e^{\frac{2}{3}}\right]\supseteq [\frac{592}{1225},\frac{8-e^{\frac{4}{7}}-e^{\frac{4}{5}}}{2}]\supseteq \left[\frac{1}{2},1.979941375566026\right] \\ \supseteq \left[\frac{77}{144},\frac{-e-\sqrt{e}-2e^{\frac{2}{3}}+16}{4}\right] \end{gathered}$$

Consequently, the above theorem is verified.

Theorem 3.

Let $\vartheta :[e,f]\to {R_I}^{+},h_1,h_2:(0,1)\to R^{+}$ be a continuous functions. If $\vartheta \in SGHX((\frac{1}{h_1},\frac{1}{h_2}),$ $[e,f],{R_I}^{+})$ and $\vartheta \in I{R}_{[}e,f]$, then we have

$$\begin{gathered} \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq M(e,f){\int }_0^1\frac{1}{H^2(x,1-x)}dx \\ +N(e,f){\int }_0^1\frac{1}{H(x,x)H(1-x,1-x)}dx. \end{gathered}$$

where

$$\begin{gathered} M(e,f)=\vartheta \left(e\right)g\left(e\right)+\vartheta \left(f\right)g\left(f\right), \\ N(e,f)=\vartheta \left(e\right)g\left(f\right)+\vartheta \left(f\right)g\left(e\right). \end{gathered}$$

Proof. 

We assume that $\vartheta ,g\in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$; then, we have the following.

$$\frac{\vartheta \left(e\right)}{h_1\left(x\right)h_2(1-x)}+\frac{\vartheta \left(f\right)}{h_1(1-x)h_2\left(x\right)}\subseteq \vartheta \left(\frac{ef}{xe+(1-x)f}\right)$$

$$\frac{g\left(e\right)}{h_1\left(x\right)h_2(1-x)}+\frac{g\left(f\right)}{h_1(1-x)h_2\left(x\right)}\subseteq g\left(\frac{ef}{xe+(1-x)f}\right)$$

Then, the following is obtained.

$$\begin{gathered} \vartheta \left(\frac{ef}{xe+(1-x)f}\right)g\left(\frac{ef}{xe+(1-x)f}\right)\supseteq \\ \frac{\vartheta \left(e\right)g\left(e\right)}{H^2(x,1-x)}+\frac{\vartheta \left(e\right)g\left(f\right)+\vartheta \left(f\right)g\left(e\right)}{H(x,x)H(1-x,1-x)}+\frac{\vartheta \left(f\right)g\left(f\right)}{H^2(1-x,x)} \end{gathered}$$

Integrating both sides of the above inequality over $(0,1)$, we have the following.

$$\begin{gathered} {\int }_0^1\vartheta \left(\frac{ef}{xe+(1-x)f}\right)g\left(\frac{ef}{xe+(1-x)f}\right)dx \\ =[{\int }_0^1\underline{\vartheta }\left(\frac{ef}{xe+(1-x)f}\right)\underline{g}\left(\frac{ef}{xe+(1-x)f}\right)dx, \\ {\int }_0^1\overline{\vartheta }\left(\frac{ef}{xe+(1-x)f}\right)\overline{g}\left(\frac{ef}{xe+(1-x)f}\right)dx] \\ =\left[\frac{ef}{f-e}{\int }_e^f\frac{\underline{\vartheta }\left(\upsilon \right)\underline{g}\left(\upsilon \right)}{{\upsilon }^2}d\upsilon ,\frac{ef}{f-e}{\int }_e^f\frac{\overline{\vartheta }\left(\upsilon \right)\overline{g}\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right] \\ =\frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq {\int }_0^1\frac{\left[\vartheta \left(e\right)g\left(e\right)+\vartheta \left(f\right)g\left(f\right)\right]}{H^2(x,1-x)}dx+{\int }_0^1\frac{\left[\vartheta \left(e\right)g\left(f\right)+\vartheta \left(f\right)g\left(e\right)\right]}{H(x,x)H(1-x,1-x)}dx \end{gathered}$$

It follows that

$$\frac{ef}{f-e}{\int }_s^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \supseteq M(e,f){\int }_0^1\frac{dx}{H^2(x,1-x)}+N(e,f){\int }_0^1\frac{dx}{H(x,x)H(1-x,1-x)}.$$

The theorem is proved. □

Example 3.

Let $h_1\left(x\right)=\frac{1}{x}$ and $h_2\left(x\right)=1$ where $x\in (0,1)$, $[e,f]=[\frac{1}{2},1]$, and ϑ, $g:[e,f]\to {R_I}^{+}$ is defined as $\vartheta \left(\upsilon \right)=[{\upsilon }^2,4-e^{\upsilon }],g\left(\upsilon \right)=[\upsilon ,3-{\upsilon }^2].$

Then, the following is the case.

$$\begin{gathered} \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon =\left[{\int }_{\frac{1}{2}}^1\upsilon d\upsilon ,{\int }_{\frac{1}{2}}^1\frac{(4-e^{\upsilon })(3-{\upsilon }^2)}{{\upsilon }^2}d\upsilon \right] \\ =\left[\frac{3}{8},5.009384684456994\right] \\ M(e,f){\int }_0^1\frac{1}{H^2(x,1-x)}dx=M\left(\frac{1}{2},1\right){\int }_0^1{x}^{2}dx \\ =\left[\frac{9}{24},\frac{19}{3}-\frac{2\sqrt{e}}{3}-\frac{11e}{12}\right] \\ N(e,f){\int }_0^1\frac{1}{H(x,x)H(1-x,1-x)}dx=N\left(\frac{1}{2},1\right){\int }_0^1(x-x^2)dx \\ =\left[\frac{3}{24},\frac{19}{6}-\frac{11\sqrt{e}}{24}-\frac{e}{3}\right] \end{gathered}$$

It follows that

$$\begin{gathered} \left[\frac{3}{8},5.009384684456994\right]\supseteq \left[\frac{9}{24},\frac{19}{3}-\frac{2\sqrt{e}}{3}-\frac{11e}{12}\right] \\ +\left[\frac{3}{24},\frac{19}{6}-\frac{11\sqrt{e}}{24}-\frac{e}{3}\right]=\left[\frac{1}{2},\frac{19}{6}+\frac{-30e-27\sqrt{e}+152}{24}\right] \end{gathered}$$

Consequently, the above theorem is verified.

Theorem 4.

Let $\vartheta ,g:[e,f]\to {R_I}^{+},h_1,h_2:(0,1)\to R^{+}$ be a continuous functions. If $\vartheta ,g\in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ and $\vartheta ,g\in I{R}_{[}e,f]$, then we have

$$\begin{gathered} \frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{2}\vartheta \left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \\ +M(e,f){\int }_0^1\frac{1}{H(x,x)H(1-x,1-x)}dx+N(e,f){\int }_0^1\frac{1}{H^2(x,1-x)}dx. \end{gathered}$$

Proof. 

By hypothesis, one has the following.

$$\vartheta \left(\frac{2ef}{e+f}\right)\supseteq \frac{\vartheta \left(\frac{ef}{xe+(1-x)f}\right)}{H(\frac{1}{2},\frac{1}{2})}+\frac{\vartheta \left(\frac{ef}{xe+(1-x)f}\right)}{H(\frac{1}{2},\frac{1}{2})},$$

$$g\left(\frac{2ef}{e+f}\right)\supseteq \frac{g\left(\frac{ef}{xe+(1-x)f}\right)}{H(\frac{1}{2},\frac{1}{2})}+\frac{g\left(\frac{ef}{xe+(1-x)f}\right)}{H(\frac{1}{2},\frac{1}{2})}.$$

Then, the following is the case.

$$\vartheta \left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)$$

$$\supseteq \frac{1}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[\vartheta \left(\frac{ef}{xe+(1-x)f}\right)g\left(\frac{ef}{xe+(1-x)f}\right)+\vartheta \left(\frac{ef}{xf+(1-x)e}\right)g\left(\frac{ef}{xf+(1-x)e}\right)\right]$$

$$+\frac{1}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[\vartheta \left(\frac{ef}{xe+(1-x)f}\right)g\left(\frac{ef}{xf+(1-x)e}\right)+\vartheta \left(\frac{ef}{xf+(1-x)e}\right)g\left(\frac{ef}{xe+(1-x)f}\right)\right]$$

$$\supseteq \frac{1}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[\vartheta \left(\frac{ef}{xe+(1-x)f}\right)g\left(\frac{ef}{xe+(1-x)f}\right)+\vartheta \left(\frac{ef}{xf+(1-x)e}\right)g\left(\frac{ef}{xf+(1-x)e}\right)\right]$$

$$+\frac{1}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}[\left(\frac{\vartheta \left(e\right)}{H(x,1-x)}+\frac{\vartheta \left(f\right)}{H(1-x,x)}\right)\left(\frac{g\left(e\right)}{H(1-x,x)}+\frac{g\left(f\right)}{H(x,1-x)}\right)+$$

$$\left(\frac{\vartheta \left(e\right)}{H(1-x,x)}+\frac{\vartheta \left(f\right)}{H(x,1-x)}\right)\left(\frac{g\left(e\right)}{H(x,1-x)}+\frac{g\left(f\right)}{H(1-x,x)}\right)]$$

$$=\frac{1}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[\vartheta \left(\frac{ef}{xe+(1-x)f}\right)g\left(\frac{ef}{xe+(1-x)f}\right)+\vartheta \left(\frac{ef}{xf+(1-x)e}\right)g\left(\frac{ef}{xf+(1-x)e}\right)\right]$$

$$+\frac{1}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[\left(\frac{2}{H(x,x)H(1-x,1-x)}\right)M(e,f)+\left(\frac{1}{H^2(x,1-x)}+\frac{1}{H^2(1-x,x)}\right)N(e,f)\right].$$

Integrating both sides of the above inequality over $(0,1)$, we have the following.

$${\int }_0^1\vartheta \left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)dx=\left[{\int }_0^1\underline{\vartheta }\left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)dx,{\int }_0^1\overline{\vartheta }\left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)dx\right]$$

$$=\vartheta \left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)dx\supseteq \frac{2}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[\frac{2ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon \right]$$

$$+\frac{2}{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}\left[M(e,f){\int }_0^1\frac{dx}{H(x,x)H(1-x,1-x)}+N(e,f){\int }_0^1\frac{dx}{H^2(x,1-x)}\right].$$

Multiply both sides by $\frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{2}$ for the above equation, we obtain the following.

$$\frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{2}\vartheta \left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)\supseteq \frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon$$

$$+M(e,f){\int }_0^1\frac{dx}{H(x,x)H(1-x,1-x)}dx+N(e,f){\int }_0^1\frac{dx}{H^2(x,1-x)}.$$

This complete the proof. □

Example 4.

Let $h_1\left(x\right)=\frac{1}{x}$ and $h_2\left(x\right)=1$ where $x\in (0,1)$, $[e,f]=[\frac{1}{2},1]$, and $\vartheta ,$ $g:[e,f]\to {R_I}^{+}$ is defined as $\vartheta \left(\upsilon \right)=[{\upsilon }^2,4-e^{\upsilon }],g\left(\upsilon \right)=[\upsilon ,3-{\upsilon }^2].$

Then, we obtain the following.

$$\frac{{\left[H(\frac{1}{2},\frac{1}{2})\right]}^2}{2}\vartheta \left(\frac{2ef}{e+f}\right)g\left(\frac{2ef}{e+f}\right)=2\vartheta \left(\frac{2}{3}\right)g\left(\frac{2}{3}\right)$$

$$=\left[\frac{16}{27},\frac{46(4-e^{\frac{2}{3}})}{9}\right]$$

$$\frac{ef}{f-e}{\int }_e^f\frac{\vartheta \left(\upsilon \right)g\left(\upsilon \right)}{{\upsilon }^2}d\upsilon =\left[{\int }_{\frac{1}{2}}^1\upsilon d\upsilon ,{\int }_{\frac{1}{2}}^1\frac{(4-e^{\upsilon })(3-{\upsilon }^2)}{{\upsilon }^2}d\upsilon \right]$$

$$=\left[\frac{3}{8},5.009384684456994\right]$$

$$M(e,f){\int }_0^1\frac{1}{H(x,x)H(1-x,1-x)}dx=M\left(\frac{1}{2},1\right){\int }_0^1(x-x^2)dx$$

$$=\left[\frac{9}{48},\frac{19}{6}-\frac{\sqrt{e}}{3}-\frac{11e}{24}\right]$$

$$N(e,f){\int }_0^1\frac{1}{H^2(x,1-x)}dx=N\left(\frac{1}{2},1\right){\int }_0^1{x}^{2}dx$$

$$=\left[\frac{3}{12},\frac{19}{3}-\frac{11\sqrt{e}}{12}-\frac{2e}{3}\right]$$

It follows that

$$\left[\frac{16}{27},\frac{184-46e^{\frac{2}{3}}}{9}\right]\supseteq \left[\frac{3}{8},5.009384684456994\right]$$

$$+\left[\frac{9}{48},\frac{19}{6}-\frac{\sqrt{e}}{3}-\frac{11e}{24}\right]+\left[\frac{3}{12},\frac{19}{3}-\frac{11\sqrt{e}}{12}-\frac{2e}{3}\right]$$

$$=\left[\frac{13}{16},\frac{19}{6}+\frac{-27e-30\sqrt{e}+152}{24}+5.009384684456994\right]$$

Consequently, the above theorem is verified.

3.2. Jensen-Type Inequality

In this section, we prove Jensen-type inequality for harmonically $(h_1,h_2)$-Godunova-Levin interval-valued functions.

Theorem 5

([28,33]). Let $f_1,f_2,f_3\dots f_l\in R^{+}$ with $l\ge 2$. If $h_1,h_2$ is non-negative super multiplicative functions and ϑ be non-negative harmonic interval-valued $(h_1,h_2)$-Godunova–Levin function or $\vartheta \in SGHX((\frac{1}{h_1},\frac{1}{h_2}),[e,f],{R_I}^{+})$ with $j_1,j_2,j_3,\dots ,j_l\in I\subseteq {R_I}^{+}$. Then, one has the following:

$$\vartheta \left(\frac{1}{\frac{1}{F_l}{\sum }_{i=1}^l{f}_i{j}_i}\right)\supseteq \sum _{i=1}^l\left[\frac{\vartheta \left(j_i\right)}{H\left(\frac{f_i}{F_l},\frac{F_l-1}{F_l}\right)}\right]$$

(9)

where $F_l={\sum }_{i=1}^l$.

Proof. 

By mathematical induction, when $l=2$, then Equation (9) is true. Suppose that Equation (9) holds for $l-1$; then, we have the following.

$$\vartheta \left(\frac{1}{\frac{1}{F_l}{\sum }_{i=1}^l{f}_i{j}_i}\right)=\vartheta \left(\frac{1}{\frac{f_l}{F_l}{j}_k+{\sum }_{i=1}^{l-1}\frac{f_i}{F_l}{j}_i}\right)$$

$$\supseteq \frac{\vartheta \left(j_l\right)}{h_1\left(\frac{f_l}{F_l}\right)h_2\left(\frac{F_l-1}{F_l}\right)}+\frac{\vartheta \left({\sum }_{i=1}^{l-1}\frac{f_i}{F_l}{j}_i\right)}{h_1\left(\frac{F_l-1}{F_l}\right)h_2\left(\frac{f_l}{F_l}\right)}$$

$$\supseteq \frac{\vartheta \left(j_l\right)}{h_1\left(\frac{f_l}{F_l}\right)h_2\left(\frac{F_l-1}{F_l}\right)}+\sum _{i=1}^{l-1}\left[\frac{\vartheta \left(j_i\right)}{H\left(\frac{f_i}{F_l},\frac{F_l-2}{F_l-1}\right)}\right]\frac{1}{h_1\left(\frac{F_l-1}{F_l}\right)h_2\left(\frac{f_l}{F_l}\right)}$$

$$\supseteq \frac{\vartheta \left(j_l\right)}{h_1\left(\frac{f_l}{F_l}\right)h_2\left(\frac{F_l-1}{F_l}\right)}+\sum _{i=1}^{l-1}\left[\frac{\vartheta \left(j_i\right)}{H\left(\frac{f_i}{F_l},\frac{F_l-2}{F_l-1}\right)}\right]$$

$$\supseteq \sum _{i=1}^l\left[\frac{\vartheta \left(j_i\right)}{H\left(\frac{f_i}{F_l},\frac{F_l-1}{F_l}\right)}\right]$$

Therefore, the theorem holds by mathematical induction. □

4. Conclusions

In the literature, Jensen’s and Hermite–Hadamard inequalities are extensively discussed. Furthermore, the close relationship between convex inequalities and optimization opens up an entirely new research perspective. In this study, we introduced harmonical interval-valued $(h_1,h_2)$-Godunova–Levin function, and we established Jensen- and Hermite–Hadamard-type inequalities. To demonstrate the validity of our primary findings, we also covered several exceptional instances and included some examples. This study extends many existing results. In the future, we will try to investigate this idea by utilizing various fractional integral operators, including Riemann–Louisville, Katugampola, and generalized K-fractional operators.