Introduction {#Sec1}
============

During the last few decades, the theory of convex analysis has turned into one of the most interesting and useful fields to study a wide class of problems arising in pure and applied sciences. Innovative techniques and calculations have yielded different directions for the study of convex analysis. In recent years, various inequalities for convex functions and their variant forms have been developed using novel techniques; see \[[@CR1]--[@CR7]\].

The theory of convex functions is closely related to theory of inequalities. It is well known that a function is convex, if and only if it satisfies an integral inequality, which is known as the Hermite-Hadamard inequality; see \[[@CR8], [@CR9]\]. Such types of integral inequalities are useful in finding the upper and lower bounds. For recent developments and applications, see \[[@CR10]--[@CR14]\].

The convex sets and convex functions have been extended and generalized in different directions using innovative ideas to study different problems in a general and unified frame work; see \[[@CR15]--[@CR21]\]. One of the most recent significant generalizations of convex functions is the *φ*-convex function, introduced by Gordji *et al.* \[[@CR22]\]. These functions are non-convex functions. For recent developments, see \[[@CR16], [@CR22]--[@CR26]\] and the references therein.

The main purpose of this paper is to introduce a new class of generalized convex functions, which are called generalized geometrically convex functions. We establish some new results by using the basic inequalities. We derive new Hermite-Hadamard integral inequalities for the generalized geometrically convex functions. Several special cases are considered. Our results are a significant and important refinement of well-known results for inequalities.

Preliminaries {#Sec2}
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Let *I* be an interval in real line $\documentclass[12pt]{minimal}
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                \begin{document}$\mathbb{R}$\end{document}$. Let $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R} $\end{document}$ be a differentiable function on the interior $\documentclass[12pt]{minimal}
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                \begin{document}$I^{0} $\end{document}$ of *I* and let $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (\cdot ,\cdot ): \mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$\end{document}$ be a continuous bifunction. Throughout this paper, we will use the following notation: $$\documentclass[12pt]{minimal}
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                \begin{document}$$\mathbb{R} = (-\infty ,+\infty ), \qquad \mathbb{R}_{+} = (0,\infty ) \quad \mbox{and} \quad \mathbb{R}_{-} = (-\infty ,0). $$\end{document}$$

Definition 2.1 {#FPar1}
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\[[@CR22]\]

Let *I* be an interval in real line $\documentclass[12pt]{minimal}
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                \begin{document}$\mathbb{R}$\end{document}$. A function $\documentclass[12pt]{minimal}
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                \begin{document}$f:I=[x,y] \rightarrow \mathbb{R}$\end{document}$ is said to be generalized convex with respect to an arbitrary bifunction $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (\cdot ,\cdot ):\mathbb{R} \times \mathbb{R}\rightarrow \mathbb{R}$\end{document}$, if $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f \bigl(tx +(1-t)y \bigr) \leq f(y)+t\eta \bigl( f(x),f(y) \bigr), \quad \forall x,y\in I, t \in [0,1]. \end{aligned}$$ \end{document}$$

If $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (x,y) = x-y$\end{document}$, then the generalized convex function reduces to a convex function.

Every convex function is a generalized convex function, but the converse is not true; see, for example, Examples [2.2](#FPar2){ref-type="sec"} and [2.3](#FPar3){ref-type="sec"}.

Example 2.2 {#FPar2}
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\[[@CR12]\]

For a convex function *f*, we may find another function *η* other than the function $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (x,y) = x-y $\end{document}$ such that f is generalized convex. Consider $\documentclass[12pt]{minimal}
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                \begin{document}$f(x) = x^{2}$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (x,y) = 2x+y$\end{document}$. Then we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f \bigl(tx+(1-t)y \bigr) =& \bigl(tx+(1-t)y \bigr)^{2} \\ \leq & t x^{2}+ y^{2} +t(1-t)2xy \\ \leq & t x^{2}+ y^{2} +t(1-t) \bigl(x^{2}+y^{2} \bigr) \\ \leq & y^{2} +t \bigl(x^{2}+x^{2}+y^{2} \bigr) \\ =& y^{2} +t \bigl(2x^{2}+y^{2} \bigr) \\ =& f(y) +t\eta \bigl(f(x),f(y) \bigr), \quad \forall x,y \in \mathbb{R},t\in (0,1). \end{aligned}$$ \end{document}$$ Also the fact that $\documentclass[12pt]{minimal}
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                \begin{document}$x^{2} \leq y^{2} +(2x^{2} +y^{2})$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$y^{2} \leq y^{2}$\end{document}$, for all $\documentclass[12pt]{minimal}
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                \begin{document}$x,y \in \mathbb{R}$\end{document}$ shows the correctness of the inequality for $\documentclass[12pt]{minimal}
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                \begin{document}$t = 1$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$t = 0$\end{document}$, respectively. This means that f is a generalized convex function. Note that the function $\documentclass[12pt]{minimal}
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                \begin{document}$f (x) = x^{2}$\end{document}$ is generalized convex with respect to all $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (x,y) = ax+by$\end{document}$ with $\documentclass[12pt]{minimal}
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                \begin{document}$a\geq 1, b \geq -1$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$x,y \in \mathbb{R}$\end{document}$.

Example 2.3 {#FPar3}
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\[[@CR12]\]

Consider $\documentclass[12pt]{minimal}
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                \begin{document}$f:\mathbb{R} \rightarrow \mathbb{R}$\end{document}$ as $$\documentclass[12pt]{minimal}
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                \begin{document}$$ f(x ) = \textstyle\begin{cases} -x & \text{if } x\geq 0, \\ x & \text{if } x< 0, \end{cases} $$\end{document}$$ and define a bifunction $\documentclass[12pt]{minimal}
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                \begin{document}$\eta =-x-y$\end{document}$ for all $\documentclass[12pt]{minimal}
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                \begin{document}$x,y \in \mathbb{R}^{-}=(-\infty ,0)$\end{document}$. Then *f* is *η*-convex, but the converse is not true.

Definition 2.4 {#FPar4}
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\[[@CR6]\]

Let $\documentclass[12pt]{minimal}
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                \begin{document}$I\subset \mathbb{R}_{+}$\end{document}$. The set *I* is said to be a geometrically convex set, if $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} x^{t} y^{1-t} \in I, \quad \forall x, y\in I, t \in [0,1]. \end{aligned}$$ \end{document}$$

Definition 2.5 {#FPar5}
--------------

\[[@CR6]\]

A function $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R} $\end{document}$ is said to be geometrically convex on *I*, if $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f \bigl(y^{1-t}x^{t} \bigr)\leq (1-t)f(y) +t\bigl(f(x)\bigr), \quad \forall x, y\in I, t \in [0,1], \end{aligned}$$ \end{document}$$

where $\documentclass[12pt]{minimal}
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                \begin{document}$(y^{1-t}x^{t})$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$(1-t)f(y)+t(f(x))$\end{document}$ are the weighted geometric mean of two positive numbers *x* and *y* and the weighted arithmetic mean of $\documentclass[12pt]{minimal}
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                \begin{document}$f(x)$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$f(y)$\end{document}$, respectively.

We now introduce a new class of generalized convex functions on the geometrically convex set with respect to an arbitrary bifunction $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (\cdot,\cdot)$\end{document}$, which is called the generalized geometrically convex function.

Definition 2.6 {#FPar6}
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A function $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow\mathbb{R} $\end{document}$ is said to be generalized geometrically with respect to a bifunction $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (\cdot ,\cdot ):\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R}$\end{document}$, if $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f \bigl(y^{1-t}x^{t} \bigr)\leq (1-t)f(y)+t\bigl(f(y)\bigr)+ \eta \bigl(f(x),f(y) \bigr),\quad \forall x, y\in I, t\in [0,1]. \end{aligned}$$ \end{document}$$

If $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (x,y)=x-y$\end{document}$, then the generalized geometrically convex functions reduce to geometrically convex functions given in Definition [2.5](#FPar5){ref-type="sec"}.

If $\documentclass[12pt]{minimal}
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                \begin{document}$t=\frac{1}{2}$\end{document}$ in ([2.1](#Equ1){ref-type=""}), then $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f(\sqrt{xy})\leq f(y) +\frac{1}{2}\eta \bigl(f(x),f(y) \bigr),\quad \forall x, y\in I, t\in [0,1], \end{aligned}$$ \end{document}$$ which is called a generalized Jensen geometrically convex function.

We will use the following notations throughout this paper: arithmetic mean: $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} A(a,b)=\frac{a+b}{2}\quad \forall a, b\in \mathbb{R}_{+}, a\neq b, \end{aligned}$$ \end{document}$$logarithmic mean: $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} L(a,b) = \frac{b-a}{\log b-\log a}\quad \forall a, b\in \mathbb{R} _{+}, a \neq b, \end{aligned}$$ \end{document}$$generalized logarithmic mean: $$\documentclass[12pt]{minimal}
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                \begin{document}$$ \L (a,b ) = \textstyle\begin{cases} [\frac{b^{p+1}-a^{p+1}}{(p+1)(b-a)}]^{\frac{1}{p}} & \text{if } p \neq -1,0, \\ L(a,b) & \text{if } p=-1, \\ \frac{1}{e}(\frac{b^{b}}{a^{a}})^{\frac{1}{b-a}} & \text{if } p=0. \end{cases} $$\end{document}$$

Main results {#Sec3}
============

In this section, we derive some new Hermite-Hadamard type inequalities for generalized geometrically convex functions. We denote $\documentclass[12pt]{minimal}
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                \begin{document}$I=[a,b]$\end{document}$, unless otherwise specified.

Theorem 3.1 {#FPar7}
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*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
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                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
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                \begin{document}$a< b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} &f(\sqrt{ab})-\frac{1}{2(\ln b-\ln a)} \int_{a}^{b}\frac{1}{x} \biggl[ \eta \biggl(f \biggl(\frac{a b}{x} \biggr),f(x) \biggr) \biggr]\,\mathrm{d}x \\ &\quad \leq \int_{a}^{b}\frac{1}{x}f(x)\,\mathrm{d}x\leq \frac{f(a)+f(b)}{2}+ \frac{1}{4}\bigl(\eta \bigl(f(a),f(b) \bigr)+\eta\bigl(f(b),f(a) \bigr)\bigr). \end{aligned}$$ \end{document}$$

Proof {#FPar8}
-----

Let *f* be a generalized geometrically convex function on *I*. Then, $\documentclass[12pt]{minimal}
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                \begin{document}$\forall a,b\in I$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f \bigl(a^{t} b^{1-t} \bigr) \leq f(b)+t\eta \bigl(f(a),f(b) \bigr). \end{aligned}$$ \end{document}$$ Integrating ([3.1](#Equ3){ref-type=""}) over *t* on $\documentclass[12pt]{minimal}
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                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} \int_{0}^{1}f \bigl(a^{t} b^{1-t} \bigr)\,\mathrm{d}t \leq & \int_{0}^{1} \bigl[f(b)+t \eta \bigl(f(a),f(b) \bigr) \bigr]\,\mathrm{d}t \\ =& \biggl[f(b)+\frac{1}{2}\eta \bigl(f(a),f(b) \bigr) \biggr]. \end{aligned}$$ \end{document}$$ Thus $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr]\, \mathrm{d}x\leq \biggl[f(b)+ \frac{1}{2}\eta \bigl(f(a),f(b) \bigr) \biggr]. \end{aligned}$$ \end{document}$$ Using ([2.2](#Equ2){ref-type=""}) and taking $\documentclass[12pt]{minimal}
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                \begin{document}$x=(a^{t} b^{1-t})$\end{document}$ and $\documentclass[12pt]{minimal}
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                \begin{document}$y=(a^{1-t} b^{t})$\end{document}$, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} f(\sqrt{ab}) \leq \biggl[f \bigl( \bigl(a^{1-t} b^{t} \bigr) \bigr)+\frac{1}{2}\eta \bigl(f \bigl( \bigl(a^{t} b^{1-t} \bigr) \bigr),f \bigl( \bigl(a^{1-t} b^{t} \bigr) \bigr) \bigr) \biggr]. \end{aligned}$$ \end{document}$$ Integrating ([3.2](#Equ4){ref-type=""}) over *t* on $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} f(\sqrt{ab})= \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)+ \frac{1}{2}\eta \biggl(f \biggl(\frac{ab}{x} \biggr),f(x) \biggr)\biggr]\,\mathrm{d}x. \end{aligned}$$ \end{document}$$ Thus $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} f(\sqrt{ab})-\frac{1}{2(\ln b-\ln a)} \int_{a}^{b}\frac{1}{x}\eta \biggl(f \biggl( \frac{a b}{x} \biggr),f(x) \biggr)\,\mathrm{d}x\leq \frac{1}{\ln b-\ln a} \int_{a}^{b} \frac{1}{x} \bigl[f(x) \bigr]\, \mathrm{d}x. \end{aligned}$$ \end{document}$$ Since *f* is a generalized geometrically convex function, $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr) \leq f(b)+t\eta \bigl(f(a),f(b) \bigr), \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{1-t} b^{t} \bigr) \leq f(a)+t\eta \bigl(f(b),f(a) \bigr). \end{aligned}$$ \end{document}$$ Adding ([3.2](#Equ4){ref-type=""}) and ([3.3](#Equ5){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} f \bigl(a^{t} b^{1-t} \bigr)+f \bigl(a^{1-t} b^{t} \bigr) \leq f(a)+f(b)+t\eta \bigl(f(b),f(a) \bigr)+t \eta \bigl(f(a),f(b) \bigr). \end{aligned}$$ \end{document}$$ Integrating ([3.4](#Equ6){ref-type=""}) over *t* on $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \frac{2}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \, \mathrm{d}x\leq {f(a)+f(b)}+\frac{1}{2}\bigl(\eta \bigl(f(a),f(b) \bigr)+\eta\bigl(f(b),f(a) \bigr)\bigr). \end{aligned}$$ \end{document}$$ This completes the proof. □

Corollary 3.1 {#FPar9}
-------------

*If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$ *in Theorem * [3.1](#FPar7){ref-type="sec"}, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} f (\sqrt{ab} )\leq \int_{a}^{b}\frac{1}{x}f(x)\,\mathrm{d}x\leq \frac{f(a)+f(b)}{2}. \end{aligned}$$ \end{document}$$

Theorem 3.2 {#FPar10}
-----------

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a< b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x)g(x) \bigr] \,\mathrm{d}x \leq \frac{1}{3}M(a,b)+\frac{1}{6}N(a,b), \end{aligned}$$ \end{document}$$ *where* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& M(a,b)= \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr]+f(b)g(b) \bigr], \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& N(a,b)= \bigl[f(b) \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr]+g(b) \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigr]. \end{aligned}$$ \end{document}$$

Proof {#FPar11}
-----

Let *f*, *g* be generalized geometrically convex functions on *I*. Then, $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\forall a,b\in I$\end{document}$ and $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr) \leq (1-t)f(b)+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr], \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& g \bigl(a^{t} b^{1-t} \bigr) \leq (1-t)g(b)+t \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr]. \end{aligned}$$ \end{document}$$ From ([3.7](#Equ9){ref-type=""}) and ([3.8](#Equ10){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & f \bigl(a^{t} b^{1-t} \bigr)g \bigl(a^{t} b^{1-t} \bigr) \\ &\quad \leq (1-t)^{2}f(b)g(b)+t(1-t) \bigl[f(b) \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr]+g(b) \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigr] \\ &\qquad {}+t^{2} \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr] \bigr]. \end{aligned}$$ \end{document}$$ Integrating both sides of ([3.9](#Equ11){ref-type=""}) over *t* on $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \int_{0}^{1}f \bigl(a^{t} b^{1-t} \bigr)g \bigl(a^{t} b^{1-t} \bigr)\,\mathrm{d}t \\& \quad \leq f(b)g(b) \int_{0}^{1}(1-t)^{2}\,\mathrm{d}t \\& \qquad {}+ \bigl[f(b) \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr]+g(b) \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigr] \int_{0}^{1}t(1-t)\,\mathrm{d}t \\& \qquad{} + \bigl[ \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(b)+ \eta \bigl(g(a),g(b) \bigr) \bigr] \bigr] \int_{0}^{1}t^{2}\,\mathrm{d}t \\& \quad =\frac{1}{3}f(b)g(b)+\frac{1}{6} \bigl[f(b) \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr]+g(b) \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigr] \\& \qquad {}+ \frac{1}{3} \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(b)+\eta \bigl(g(a),g(b) \bigr) \bigr] \bigr] \\& \quad =\frac{1}{3}M(a,b)+\frac{1}{6}N(a,b). \end{aligned}$$ \end{document}$$ Thus $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x)g(x) \bigr] \,\mathrm{d}x \leq \frac{1}{3}M(a,b)+\frac{1}{6}N(a,b). \end{aligned}$$ \end{document}$$ This completes the proof. □

Theorem 3.3 {#FPar12}
-----------

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)g \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x\leq \frac{1}{6}M(a,b)+ \frac{1}{3}N(a,b), \end{aligned}$$ \end{document}$$ *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$M(a,b)$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$N(a,b)$\end{document}$ *are defined by* ([3.5](#Equ7){ref-type=""}) *and* ([3.6](#Equ8){ref-type=""}).

Proof {#FPar13}
-----

Let *f*, *g* be generalized geometrically convex functions on *I*. Then, $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\forall a,b\in I$\end{document}$ and $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr) \leq (1-t)f(b)+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr], \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& g \bigl(a^{1-t} b^{t} \bigr) \leq (1-t)g(a)+t \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]. \end{aligned}$$ \end{document}$$ From ([3.10](#Equ12){ref-type=""}) and ([3.11](#Equ13){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr)g \bigl(a^{1-t} b^{t} \bigr) \\& \quad \leq (1-t)^{2}f(b)g(a)+t(1-t) \bigl[f(b) \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \\& \quad \quad {}+g(a) \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigr] \\& \quad \quad {} +t^{2} \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]. \end{aligned}$$ \end{document}$$ Integrating both sides of ([3.12](#Equ14){ref-type=""}) over *t* on $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \int_{0}^{1}f \bigl(a^{t} b^{1-t} \bigr)g \bigl(a^{1-t} b^{t} \bigr)\,\mathrm{d}t \\& \quad {}\leq f(b)g(a) \int_{0}^{1}(1-t)^{2}\,\mathrm{d}t \\& \quad \quad {} + \bigl[f(b) \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]+g(a) \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigr] \int_{0}^{1}t(1-t)\,\mathrm{d}t \\& \quad\quad {} + \bigl[ \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+ \eta \bigl(g(b),g(a) \bigr) \bigr] \bigr] \int_{0}^{1}t^{2}\,\mathrm{d}t \\& \quad = \frac{1}{3}f(b)g(a)+\frac{1}{6} \bigl[f(b) \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]+g(a) \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr] \bigr] \\& \quad \quad {}+ \frac{1}{3} \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr] \\& \quad = \frac{1}{3}N(a,b)+\frac{1}{6}M(a,b). \end{aligned}$$ \end{document}$$ Thus $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)g \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x\leq \frac{1}{6}M(a,b)+ \frac{1}{3}N(a,b). \end{aligned}$$ \end{document}$$ This completes the proof. □

Corollary 3.2 {#FPar14}
-------------

*If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f = g$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (x,y)=x-y$\end{document}$ *in Theorem * [3.3](#FPar12){ref-type="sec"}, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)f \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x\leq \frac{2}{3}f(b)f(a)+ \frac{1}{6} \bigl[f^{2}(b)+f^{2}(a) \bigr]. \end{aligned}$$ \end{document}$$

Theorem 3.4 {#FPar15}
-----------

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)g \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x \\& \quad \leq \frac{1}{6} \bigl\{ \bigl[f^{2}(b)+g^{2}(a) \bigr] \\& \quad \quad {} + \bigl[ \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]^{2}+ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr]^{2} \bigr] \\& \quad \quad {} + \bigl[f(b) \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr]+g(a) \bigl[g(a)+ \eta \bigl(g(b),g(a) \bigr) \bigr] \bigr] \bigr\} . \end{aligned}$$ \end{document}$$

Proof {#FPar16}
-----

Let *f*, *g* be generalized geometrically convex functions on *I*. Then, $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\forall a,b\in I$\end{document}$ and $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr) \leq (1-t)f(b)+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr], \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& g \bigl(a^{1-t} b^{t} \bigr) \leq (1-t)g(a)+t \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]. \end{aligned}$$ \end{document}$$ Now, $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)g \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x \\& \quad = \int_{0}^{1} \bigl(f \bigl(a^{t} b^{1-t} \bigr) g \bigl(a^{1-t} b^{t} \bigr) \bigr) \, \mathrm{d}t \\& \quad \leq \frac{1}{2} \int_{0}^{1} \bigl[ \bigl[ f \bigl(a^{t} b^{1-t} \bigr) \bigr]^{2}+ \bigl[g \bigl(a^{1-t} b^{t} \bigr) \bigr]^{2} \bigr]\,\mathrm{d}t \\& \quad \leq \frac{1}{2} \biggl\{ \int_{0}^{1} \bigl[(1-t)f(b)+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigr]^{2} \\& \quad \quad {}+ \bigl[(1-t)g(a)+t \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]^{2} \biggr\} \,\mathrm{d}t \\& \quad = \frac{1}{2} \biggl\{ \bigl[f^{2}(b)+g^{2}(a) \bigr] \int_{0}^{1}(1-t)^{2} \,\mathrm{d}t \\& \quad \quad {}+ \bigl[ \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]^{2}+ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr]^{2} \bigr] \int_{0}^{1}t^{2}\,\mathrm{d}t \\& \quad \quad {}+ \bigl[ \bigl[f(b) \bigr] \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr]+ \bigl[g(a) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr] \int_{0}^{1}2t(1-t)\,\mathrm{d}t \biggr\} \\& \quad = \frac{1}{6} \bigl\{ \bigl[f^{2}(b)+g^{2}(a) \bigr] + \bigl[ \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]^{2}+ \bigl[f(b)+ \eta \bigl(f(a),f(b) \bigr) \bigr]^{2} \bigr] \\& \quad \quad {}+ \bigl[f(b) \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr]+g(a) \bigl[g(a)+ \eta \bigl(g(b),g(a) \bigr) \bigr] \bigr] \bigr\} , \end{aligned}$$ \end{document}$$ which is the required result. □

Corollary 3.3 {#FPar17}
-------------

*If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f=g$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$ *in Theorem * [3.4](#FPar15){ref-type="sec"}, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)f \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x \\& \quad \leq \frac{1}{6} \bigl\{ 2 \bigl[f^{2}(b)+f^{2}(a) \bigr] + \bigl[f(b)f(a)+f(a)f(b) \bigr] \bigr\} . \end{aligned}$$ \end{document}$$

Theorem 3.5 {#FPar18}
-----------

*Let* $\documentclass[12pt]{minimal}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)g \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x \\& \quad =\frac{1}{12} \bigl\{ \bigl[f(b)+g(a) \bigr]^{2}+ \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+ \eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]^{2} \\& \quad \quad {}+ \bigl(f(b)+g(a) \bigr) \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr\} . \end{aligned}$$ \end{document}$$

Proof {#FPar19}
-----

Let *f*, *g* be generalized geometrically convex functions on *I*. Then, $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\forall a,b \in I$\end{document}$ and $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr) \leq (1-t)f(b)+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr], \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& g \bigl(a^{1-t} b^{t} \bigr) \leq (1-t)g(a)+t \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr]. \end{aligned}$$ \end{document}$$ Now, $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)g \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x \\& \quad = \int_{0}^{1} \bigl(f \bigl(a^{t} b^{1-t} \bigr) g \bigl(a^{1-t} b^{t} \bigr) \bigr) \, \mathrm{d}t \\& \quad \leq \frac{1}{4} \int_{0}^{1} \bigl[ \bigl[ f \bigl(a^{t} b^{1-t} \bigr) \bigr]+ \bigl[g \bigl(a^{1-t} b ^{t} \bigr) \bigr] \bigr] ^{2}\,\mathrm{d}t \\& \quad \leq \frac{1}{4} \biggl\{ \int_{0}^{1} [(1-t)f(b)+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \\& \quad \quad {}+ \bigl[(1-t)g(a)+t \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]^{2} \biggr\} \,\mathrm{d}t \\& \quad = \frac{1}{4} \biggl\{ \int_{0}^{1} \bigl[(1-t) \bigl[f(b)+g(a) \bigr]+t \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+ \eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]^{2} \biggr\} \,\mathrm{d}t \\& \quad = \frac{1}{4} \biggl\{ \int_{0}^{1} [(1-t)^{2} \bigl[f(b)+g(a) \bigr]^{2}+t^{2} \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]^{2} \\& \quad \quad {}+2t(1-t) \bigl(f(b)+g(a) \bigr) \bigl[f(b)+\eta\bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \biggr\} \,\mathrm{d}t \\& \quad = \frac{1}{12} \bigl\{ \bigl[f(b)+g(a) \bigr]^{2}+ \bigl[ \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+ \eta \bigl(g(b),g(a) \bigr) \bigr] \bigr]^{2} \\& \quad \quad {}+ \bigl(f(b)+g(a) \bigr) \bigl[f(b)+\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(a)+\eta \bigl(g(b),g(a) \bigr) \bigr] \bigr\} , \end{aligned}$$ \end{document}$$ which is the required result. □

Corollary 3.4 {#FPar20}
-------------

*If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f=g$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$ *in Theorem * [3.5](#FPar18){ref-type="sec"}, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \biggl[f(x)f \biggl( \frac{a b}{x} \biggr) \biggr] \,\mathrm{d}x \\& \quad \leq \frac{1}{12} \bigl\{ \bigl[f(b)+f(a) \bigr]^{2}+ \bigl[f(a)f(b) \bigr]^{2}+ \bigl[f(a)f(b) \bigl(f(b)+f(a) \bigr) \bigr] \bigr\} . \end{aligned}$$ \end{document}$$

Theorem 3.6 {#FPar21}
-----------

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a< b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} &f(\sqrt{a b} ) g(\sqrt{ab} ) \\ &\quad \leq \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x)g(x) \bigr] \,\mathrm{d}x \\ &\quad \quad {}+\frac{1}{2(\ln b-\ln a)} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \biggl[ \eta \biggl(g \biggl(\frac{ab}{x} \biggr),g(x)\biggr) \biggr] \\ &\quad \quad {}+ \bigl[g(x) \bigr] \biggl[\eta \biggl(f \biggl(\frac{ab}{x}\biggr),f(x)\biggr) \biggr]\,\mathrm{d}x \\ &\quad \quad {}+\frac{1}{4(\ln b-\ln a)} \int_{a}^{b}\frac{1}{x} \biggl[\eta (f \biggl( \frac{ab}{x} \biggr),f(x) \biggr] \biggl[ \eta \biggl(g \biggl(\frac{ab}{x}\biggr),g(x)\biggr) \biggr]\,\mathrm{d}x. \end{aligned}$$ \end{document}$$

Proof {#FPar22}
-----

Let *f*, *g* be generalized geometrically convex functions on *I*. Then, $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\forall a,b \in I$\end{document}$ and $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f(\sqrt{ab} )\leq f \bigl(a^{1-t}b^{t} \bigr)+ \frac{1}{2}\eta \bigl(f \bigl(a^{t}b^{1-t} \bigr),f \bigl(a ^{1-t}b^{t} \bigr) \bigr), \end{aligned}$$ \end{document}$$ $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& g(\sqrt{ab} )\leq g \bigl(a^{1-t}b^{t} \bigr)+ \frac{1}{2}\eta \bigl(g \bigl(a^{t}b^{1-t} \bigr),g \bigl(a ^{1-t}b^{t} \bigr) \bigr). \end{aligned}$$ \end{document}$$ From ([3.17](#Equ19){ref-type=""}) and ([3.18](#Equ20){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} &f(\sqrt{a b} ) g(\sqrt{ab} ) \\ &\quad \leq \biggl[f \bigl(a^{1-t}b^{t} \bigr)+ \frac{1}{2}\eta \bigl(f \bigl(a^{t}b^{1-t} \bigr),f \bigl(a^{1-t}b ^{t} \bigr) \bigr) \biggr] \biggl[g \bigl(a^{1-t}b^{t} \bigr)+\frac{1}{2}\eta \bigl(g \bigl(a^{t}b^{1-t} \bigr),g \bigl(a^{1-t}b ^{t} \bigr) \bigr) \biggr] \\ &\quad =f \bigl(a^{1-t}b^{t} \bigr) \bigl[g \bigl(a^{1-t}b^{t} \bigr) \bigr]+\frac{1}{2} \bigl[ \bigl[f \bigl(a^{1-t}b^{t} \bigr) \bigr] \bigl[ \eta \bigl(g \bigl(a^{t}b^{1-t} \bigr),g \bigl(a^{1-t}b^{t} \bigr) \bigr) \bigr] \\ &\quad \quad {}+ \bigl[g \bigl(a^{1-t}b^{t} \bigr) \bigr] \bigl[\eta \bigl(f \bigl(a^{t}b^{1-t} \bigr),f \bigl(a^{1-t}b^{t} \bigr) \bigr) \bigr] \bigr] \\ &\quad \quad {} +\frac{1}{4} \bigl[ \bigl[\eta \bigl(g \bigl(a^{t}b^{1-t} \bigr),g \bigl(a^{1-t}b^{t} \bigr) \bigr) \bigr] \bigl[ \eta \bigl(f \bigl(a^{t}b^{1-t} \bigr),f \bigl(a^{1-t}b^{t} \bigr) \bigr) \bigr] \bigr]. \end{aligned}$$ \end{document}$$ Integrating ([3.19](#Equ21){ref-type=""}) over *t* on $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \int_{0}^{1}f(\sqrt{a b} ) g(\sqrt{ab} )\, \mathrm{d}t \\ &\quad \leq \int_{0}^{1}f \bigl(a^{1-t}b^{t} \bigr) \bigl[g \bigl(a^{1-t}b^{t} \bigr) \bigr]\,\mathrm{d}t \\ &\quad \quad {}+ \frac{1}{2}\int_{0}^{1} \bigl[ \bigl[f \bigl(a^{1-t}b^{t}\bigr) \bigr] \bigl[\eta \bigl(g \bigl(a^{t}b^{1-t} \bigr),g \bigl(a ^{1-t}b^{t} \bigr) \bigr) \bigr] \\ &\quad \quad {}+ \bigl[g \bigl(a^{1-t}b^{t} \bigr) \bigr] \bigl[\eta \bigl(f \bigl(a^{t}b^{1-t} \bigr),f \bigl(a^{1-t}b^{t} \bigr) \bigr) \bigr] \bigr]\,\mathrm{d}t \\ &\quad \quad {} +\frac{1}{4} \int_{0}^{1} \bigl[ \bigl[\eta \bigl(g \bigl(a^{t}b^{1-t} \bigr),g \bigl(a^{1-t}b ^{t} \bigr) \bigr) \bigr] \bigl[\eta \bigl(f \bigl(a^{t}b^{1-t} \bigr),f \bigl(a^{1-t}b^{t} \bigr) \bigr) \bigr] \bigr]\, \mathrm{d}t. \end{aligned}$$ \end{document}$$ Thus $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} &f(\sqrt{a b} ) g(\sqrt{ab} ) \\ &\quad \leq \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x)g(x) \bigr] \,\mathrm{d}x \\ &\quad \quad {}+\frac{1}{2(\ln b-\ln a)} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \biggl[ \eta \biggl(g \biggl(\frac{ab}{x} \biggr),g(x) \biggr)\biggr] \\ &\quad \quad {}+ \bigl[g(x) \bigr] \biggl[\eta \biggl(f \biggl(\frac{ab}{x}\biggr),f(x) \biggr)\biggr]\,\mathrm{d}x \\ &\quad \quad {}+\frac{1}{4( \ln b-\ln a)} \int_{a}^{b}\frac{1}{x} \biggl[\eta \biggl(f \biggl(\frac{ab}{x} \biggr),f(x) \biggr)\biggr] \biggl[ \eta \biggl(g \biggl(\frac{ab}{x}\biggr),g(x) \biggr)\biggr]\,\mathrm{d}x. \end{aligned}$$ \end{document}$$ This completes the proof. □

Corollary 3.5 {#FPar23}
-------------

*If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f=g$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$ *in Theorem * [3.6](#FPar21){ref-type="sec"}, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} f(\sqrt{a b} ) f(\sqrt{ab} )\leq \frac{1}{\ln b-\ln a} \int_{a} ^{b}\frac{1}{x} \bigl[f(x)f(x) \bigr] \,\mathrm{d}x. \end{aligned}$$ \end{document}$$

Theorem 3.7 {#FPar24}
-----------

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f,g:I=[a,b]\rightarrow (0,\infty )$\end{document}$ *be increasing and generalized geometrically convex functions on* *I* *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b \in I $\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} &\frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \biggl[g(b)+\frac{1}{2} \eta \bigl(g(a),g(b) \bigr) \biggr]\,\mathrm{d}x \\ &\quad \quad {}+\frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[g(x) \bigr] \biggl[f(b)+ \frac{1}{2}\eta \bigl(f(a),f(b) \bigr) \biggr]\,\mathrm{d}x \\ &\quad \leq \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \bigl[g(x) \bigr] \,\mathrm{d}x \\ &\quad \quad {}+ \biggl[f(b)+\frac{1}{2}\eta \bigl(f(a),f(b) \bigr) \biggr] \biggl[g(b)+\frac{1}{2}\eta \bigl(g(a),g(b) \bigr) \biggr]. \end{aligned}$$ \end{document}$$

Proof {#FPar25}
-----

Let *f* and *g* be generalized geometrically convex functions on *I*. Then $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\forall a,b\in I$\end{document}$ and $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$t \in [0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& f \bigl(a^{t} b^{1-t} \bigr) \leq \bigl[f(b)+t \eta \bigl(f(a),f(b) \bigr) \bigr], \end{aligned}$$ \end{document}$$ and $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned}& g \bigl(a^{t} b^{1-t} \bigr) \leq \bigl[g(b)+t \eta \bigl(g(a),g(b) \bigr) \bigr]. \end{aligned}$$ \end{document}$$ Using the inequality $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} (a-b) (c-d)\geq 0 \quad \forall a,b,c,d\in \mathbb{R},a < b , c< d, \end{aligned}$$ \end{document}$$ we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned}& \bigl[f \bigl(a^{t} b^{1-t} \bigr) \bigr] \bigl[g(b)+t\eta \bigl(g(a),g(b) \bigr) \bigr]+ \bigl[g \bigl(a^{t} b^{1-t} \bigr) \bigr] \bigl[f(b)+t\eta \bigl(f(a),f(b) \bigr) \bigr] \\& \quad \leq f \bigl(a^{t} b^{1-t} \bigr) g \bigl(a^{t} b^{1-t} \bigr)+ \bigl[f(b)+t\eta \bigl(f(a),f(b) \bigr) \bigr] \bigl[g(b)+t \eta \bigl(g(a),g(b) \bigr) \bigr]. \end{aligned}$$ \end{document}$$ Integrating the above inequality over *t* on $\documentclass[12pt]{minimal}
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                \begin{document}$[0,1]$\end{document}$, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned}& \int_{0}^{1} \bigl[f \bigl(a^{t} b^{1-t} \bigr) \bigr]\,\mathrm{d}t \int_{0}^{1} \bigl[g(b)+t\eta \bigl(g(a),g(b) \bigr) \bigr] \,\mathrm{d}t \\& \qquad {} + \int_{0}^{1} \bigl[g \bigl(a^{t} b^{1-t} \bigr) \bigr]\,\mathrm{d}t \int_{0}^{1} \bigl[f(b)+t\eta \bigl(f(a),f(b) \bigr) \bigr] \,\mathrm{d}t \\& \quad \leq \int_{0}^{1}f \bigl(a^{t} b^{1-t} \bigr)\,\mathrm{d}t \int_{0}^{1} g \bigl(a^{t} b ^{1-t} \bigr)\,\mathrm{d}t \\& \qquad {} + \int_{0}^{1} \bigl[f(b)+t\eta \bigl(f(a),f(b) \bigr) \bigr]\,\mathrm{d}t \int_{0}^{1} \bigl[g(b)+t \eta \bigl(g(a),g(b) \bigr) \bigr]\,\mathrm{d}t. \end{aligned}$$ \end{document}$$

Now after simple integration, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned}& \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \biggl[g(b)+\frac{1}{2} \eta \bigl(g(a),g(b) \bigr) \biggr]\,\mathrm{d}x \\& \quad \quad {}+\frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[g(x) \bigr] \biggl[f(b)+ \frac{1}{2}\eta \bigl(f(a),f(b) \bigr) \biggr]\,\mathrm{d}x \\& \quad \leq \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \bigl[g(x) \bigr] \,\mathrm{d}x \\& \quad \quad {}+ \biggl[f(b)+\frac{1}{2}\eta \bigl(f(a),f(b) \bigr) \biggr] \biggl[g(b)+\frac{1}{2}\eta \bigl(g(a),g(b) \bigr) \biggr]. \end{aligned}$$ \end{document}$$ This completes the proof. □

Corollary 3.6 {#FPar26}
-------------

*If* $\documentclass[12pt]{minimal}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f=g$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$ *in Theorem * [3.7](#FPar24){ref-type="sec"}, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsfonts} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned}& \frac{2}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \biggl[ \frac{f(b)+f(a)}{2} \biggr]\,\mathrm{d}x \\& \quad \leq \frac{1}{\ln b-\ln a} \int_{a}^{b}\frac{1}{x} \bigl[f(x) \bigr] \bigl[f(x) \bigr] \,\mathrm{d}x+ \bigl[f(b)+f(a) \bigr]. \end{aligned}$$ \end{document}$$

Integral inequalities {#Sec4}
=====================

In this section, we will use the following result to obtain our main results.

Lemma 4.1 {#FPar27}
---------

\[[@CR27]\]

*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R} $\end{document}$ *be a differentiable function on the interior* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsfonts} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0} $\end{document}$ *of* *I*, *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b \in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \begin{document}$f^{\prime }\in L[a,b]$\end{document}$. *Then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned}& b f(b)-a f(a)- \int_{a}^{b} f(x)\,\mathrm{d}x \\& \quad = \frac{(\ln b-\ln a)}{2} \biggl[ \int_{0}^{1} b^{1+t} a^{1-t}f' \bigl(b^{ \frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr)\,\mathrm{d}t+ \int_{0}^{1} b^{1-t} a ^{1+t}f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr)\,\mathrm{d}t \biggr]. \end{aligned}$$ \end{document}$$

Theorem 4.2 {#FPar28}
-----------

*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R}$\end{document}$ *be a differentiable function on the interior* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0} $\end{document}$ *of* *I*, *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f^{\prime }\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\vert f^{\prime} \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$q\geq 1$\end{document}$, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned}& \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\& \quad \leq\frac{(b-a)^{1-\frac{1}{q}}}{(2)^{\frac{1}{q}+1}} \bigl\{ b \bigl[ \bigl(2b-a-L(a,b) \bigr) \bigl\vert f^{\prime }(b) \bigr\vert ^{q} \\& \quad \quad {}+{ \bigl(L(a,b)-a \bigr)} \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \\& \quad \quad {}+a \bigl[ \bigl(b-2a+L(a,b) \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}+ \bigl({b-L(a,b)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{ \frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Proof {#FPar29}
-----

Using Lemma [4.1](#FPar27){ref-type="sec"} and Hölder's inequality, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad =\frac{ab(\ln b-\ln a)}{2} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert \,\mathrm{d}t + \int_{0}^{1} \biggl( \frac{a}{b} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert \,\mathrm{d}t \biggr] \\ &\quad \leq \frac{ab(\ln b-\ln a)}{2} \biggl\{ \biggl[ \int_{0}^{1} \biggl( \frac{b}{a} \biggr)^{t}\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \biggl( \frac{b}{a} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q} \,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad \quad {}+ \biggl[ \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{t}\,\mathrm{d}t \biggr]^{1- \frac{1}{q}} \biggl[ \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}} \biggr\} . \end{aligned}$$ \end{document}$$ Now consider $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I_{1} =& \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \\ \leq & \int_{0}^{1} \bigl\vert f'(b) \bigr\vert ^{q} \biggl(\frac{1+t}{2} \biggr)+ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \biggl( \frac{1-t}{2} \biggr)\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{t} \biggl(\frac{1+t}{2} \biggr)\,\mathrm{d}t + \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t} \biggl( \frac{1-t}{2} \biggr)\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \frac{2b-a-L(a,b)}{2a(\ln b-\ln a)}+ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}\frac{L(a,b)-a}{2a( \ln b-\ln a)}. \end{aligned}$$ \end{document}$$ Similarly $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document}$I_{2}$\end{document}$ becomes $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned} & \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \\ &\quad \leq \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}\frac{b-2a+L(a,b)}{2b(\ln b- \ln a)}+ \bigl\vert f'(b) \bigr\vert ^{q}\frac{b-L(a,b)}{2b(\ln b-\ln a)}. \end{aligned}$$ \end{document}$$ Also $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t}\,\mathrm{d}t \biggr]^{1-\frac{1}{q}}= \biggl( \frac{b-a}{a( \ln b-\ln a)} \biggr)^{1-\frac{1}{q}}. \end{aligned}$$ \end{document}$$ Combining ([4.1](#Equ25){ref-type=""}), ([4.2](#Equ26){ref-type=""}), ([4.3](#Equ27){ref-type=""}) and ([4.4](#Equ28){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(b-a)^{1-\frac{1}{q}}}{(2)^{\frac{1}{q}+1}} \bigl\{ b \bigl[ \bigl(2b-a-L(a,b) \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \\ &\quad \quad {}+{ \bigl(L(a,b)-a \bigr)} \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \\ &\quad \quad {}+a \bigl[ \bigl(b-2a+L(a,b) \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}+ \bigl({b-L(a,b)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{ \frac{1}{q}} \bigr\} , \end{aligned}$$ \end{document}$$ which is the required result. □

If $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$, then Theorem [4.2](#FPar28){ref-type="sec"} reduces to the following result.

Corollary 4.1 {#FPar30}
-------------

\[[@CR27]\]

*Let* $\documentclass[12pt]{minimal}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow\mathbb{R} $\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0}$\end{document}$ (*interior of* *I*), *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a, b \in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f'\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$q\geq 1$\end{document}$, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(b-a)^{1-\frac{1}{q}}}{(2)^{\frac{1}{q}+1}} \bigl\{ b \bigl[ \bigl(2b-a-L(a,b) \bigr) \bigl\vert f'(b) \bigr\vert ^{q}+{ \bigl(L(a,b)-a \bigr)} \bigl\vert f'(a) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \\ &\quad \quad {}+a \bigl[ \bigl(b-2a+L(a,b) \bigr) \bigl\vert f'(a) \bigr\vert ^{q}+ \bigl({b-L(a,b)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{ \frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Theorem 4.3 {#FPar31}
-----------

*Let* $\documentclass[12pt]{minimal}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty)\rightarrow \mathbb{R}$\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0}$\end{document}$ (*interior of* *I*), *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a, b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f'\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$q>1$\end{document}$, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)}{(2)^{1+\frac{1}{q}}} \bigl[L(a^{\frac{q}{q-1}},b ^{\frac{q}{q-1}} \bigr]^{1-\frac{1}{q}} \bigl\{ b \bigl[A \bigl(3 \bigl\vert f'(b) \bigr\vert ^{q}, \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \bigr) \bigr]^{\frac{1}{q}} \\ &\quad \quad {}+a \bigl[A \bigl(3 \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \bigr),\bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{ \frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Proof {#FPar32}
-----

Using Lemma [4.1](#FPar27){ref-type="sec"} and Hölder's inequality, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)-\frac{1}{b-a} \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad =\frac{ab(\ln b-\ln a)}{2} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert \,\mathrm{d}t + \int_{0}^{1} \biggl( \frac{a}{b} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert \,\mathrm{d}t \biggr] \\ &\quad \leq \frac{ab(\ln b-\ln a)}{2} \biggl\{ \biggl[ \int_{0}^{1} \biggl( \frac{b}{a} \biggr)^{\frac{qt}{q-1}}\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q} \, \mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad \quad {}+ \biggl[ \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{\frac{qt}{q-1}}\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q} \, \mathrm{d}t \biggr]^{\frac{1}{q}} \biggr\} . \end{aligned}$$ \end{document}$$ Now we consider $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \begin{document} $$\begin{aligned} I_{1} =& \int_{0}^{1} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q} \, \mathrm{d}t \\ \leq & \int_{0}^{1} \bigl\vert f'(b) \bigr\vert ^{q} \biggl(\frac{1+t}{2} \biggr)+ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \biggl( \frac{1-t}{2} \biggr)\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{1+t}{2} \biggr) \, \mathrm{d}t + \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{1-t}{2} \biggr)\, \mathrm{d}t \\ =&\frac{3}{4} \bigl\vert f'(b) \bigr\vert ^{q}+\frac{1}{4} \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q}. \end{aligned}$$ \end{document}$$ Similarly $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I_{2}$\end{document}$ becomes $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \int_{0}^{1} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q} \, \mathrm{d}t \leq \frac{3}{4} \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}+ \frac{1}{4} \bigl\vert f'(b) \bigr\vert ^{q}. \end{aligned}$$ \end{document}$$ Also $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{\frac{qt}{q-1}}\,\mathrm{d}t ]^{1- \frac{1}{q}}= \biggl(\frac{b^{\frac{q}{q-1}}-a^{\frac{q}{q-1}}}{( \frac{q}{q-1}) a^{\frac{q}{q-1}}(\ln b-\ln a)} \biggr)^{1-\frac{1}{q}}. \end{aligned}$$ \end{document}$$ Combining ([4.5](#Equ29){ref-type=""}), ([4.6](#Equ30){ref-type=""}), ([4.7](#Equ31){ref-type=""}) and ([4.8](#Equ32){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)}{(2)^{1+\frac{1}{q}}} \bigl[L\bigl(a^{\frac{q}{q-1}},b ^{\frac{q}{q-1}}\bigr) \bigr]^{1-\frac{1}{q}} \bigl\{ b \bigl[A \bigl(3 \bigl\vert f'(b) \bigr\vert ^{q}, \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \bigr) \bigr]^{\frac{1}{q}} \\ &\quad \quad {}+a \bigl[A\bigl(3 \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}\bigr), \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{ \frac{1}{q}} \bigr\} , \end{aligned}$$ \end{document}$$ which is the required result. □

If $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$, then Theorem [4.3](#FPar31){ref-type="sec"} reduces to the following result.

Corollary 4.2 {#FPar33}
-------------

\[[@CR27]\]

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow\mathbb{R} $\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0} $\end{document}$ (*interior of* *I*), *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a, b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f'\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$q>1$\end{document}$, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)}{(2)^{1+\frac{1}{q}}} \bigl[L\bigl(a^{\frac{q}{q-1}},b ^{\frac{q}{q-1}}\bigr)\bigr]^{1-\frac{1}{q}} \bigl\{ b \bigl[A \bigl(3 \bigl\vert f'(b) \bigr\vert ^{q}, \bigl\vert f'(a) \bigr\vert ^{q} \bigr) \bigr]^{\frac{1}{q}} \\ &\quad \quad {}+a \bigl[A \bigl(3 \bigl\vert f'(a) \bigr\vert ^{q} \bigr), \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Theorem 4.4 {#FPar34}
-----------

*Let* $\documentclass[12pt]{minimal}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R}$\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0} $\end{document}$ (*interior of* *I*), *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a,b \in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f' \in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$q\geq 1$\end{document}$, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)^{1-\frac{1}{q}}}{2(2q)^{\frac{1}{q}}} \bigl\{ b \bigl[ \bigl({L \bigl(a^{q},b^{q} \bigr)-a^{q}} \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \\ &\quad \quad {}+ \bigl(2b^{q}-a^{q}-L \bigl(a^{q},b^{q} \bigr) \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \\ &\quad \quad {}+a \bigl[ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \bigl(b^{q}-2a^{q}+L \bigl(a^{q},b^{q} \bigr) \bigr) \\ &\quad \quad {}+ \bigl({b^{q}-L \bigl(a^{q},b^{q} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Proof {#FPar35}
-----

Using Lemma [4.1](#FPar27){ref-type="sec"} and Hölder's inequality, we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad =\frac{ab(\ln b-\ln a)}{2} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert \,\mathrm{d}t + \int_{0}^{1} \biggl( \frac{a}{b} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert \,\mathrm{d}t \biggr] \\ &\quad \leq \frac{ab(\ln b-\ln a)}{2} \biggl\{ \biggl[ \int_{0}^{1} 1\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{qt} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad \quad {}+ \biggl[ \int_{0}^{1} 1\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int _{0}^{1} \biggl(\frac{a}{b} \biggr)^{qt} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q} \,\mathrm{d}t \biggr]^{\frac{1}{q}} \biggr\} . \end{aligned}$$ \end{document}$$ Now consider $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsfonts} 
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} I_{1} =& \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{qt} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \\ \leq & \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{qt} \biggl[ \bigl\vert f'(b) \bigr\vert ^{t} \biggl(\frac{1+t}{2} \biggr)+ \bigl\vert f'(b)+ \eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \biggl( \frac{1-t}{2} \biggr) \biggr]\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{qt} \biggl(\frac{1+t}{2} \biggr)\,\mathrm{d}t+ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{qt} \biggl( \frac{1-t}{2} \biggr)\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \frac{2b^{q}-a^{q}-L(a^{q},b^{q})}{2qa^{q}(\ln b-\ln a)}+ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q}\frac{L(a^{q},b^{q})-a^{q}}{2qa^{q}(\ln b- \ln a)}. \end{aligned}$$ \end{document}$$ Similarly $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I_{2}$\end{document}$ becomes $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
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                \usepackage{mathrsfs}
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                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{qt} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \\ &\quad \leq \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}\frac{b^{q}-2a^{q}+L(a^{q},b^{q})}{2qb ^{q}(\ln b-\ln a)}+ \bigl\vert f'(b) \bigr\vert ^{q} \frac{b^{q}-L(a^{q},b^{q})}{2qb^{q}(\ln b-\ln a)}. \end{aligned}$$ \end{document}$$ Combining ([4.9](#Equ33){ref-type=""}), ([4.10](#Equ34){ref-type=""}) and ([4.11](#Equ35){ref-type=""}), we have $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)^{1-\frac{1}{q}}}{2(2q)^{\frac{1}{q}}} \bigl\{ b \bigl[ \bigl({L \bigl(a^{q},b^{q} \bigr)-a^{q}} \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \\ &\quad \quad + \bigl({2b^{q}-a^{q}-L \bigl(a^{q},b^{q} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \\ &\quad \quad {} +a \bigl[ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \bigl({b^{q}-2a^{q}+L \bigl(a^{q},b^{q} \bigr)} \bigr)+ \bigl({b^{q}-L \bigl(a^{q},b^{q} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert \bigr]^{\frac{1}{q}} \bigr\} , \end{aligned}$$ \end{document}$$ which is the required result. □

If $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\eta (b,a)=b-a$\end{document}$, then Theorem [4.4](#FPar34){ref-type="sec"} reduces to the following result.

Corollary 4.3 {#FPar36}
-------------

\[[@CR27]\]

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow\mathbb{R} $\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$I^{0} $\end{document}$ (*interior of* *I*), *where* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a, b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$f'\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document}$q\geq 1$\end{document}$, *then* $$\documentclass[12pt]{minimal}
                \usepackage{amsmath}
                \usepackage{wasysym} 
                \usepackage{amsfonts} 
                \usepackage{amssymb} 
                \usepackage{amsbsy}
                \usepackage{mathrsfs}
                \usepackage{upgreek}
                \setlength{\oddsidemargin}{-69pt}
                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)^{1-\frac{1}{q}}}{2(2q)^{\frac{1}{q}}} \bigl\{ b \bigl[ \bigl({L \bigl(a^{q},b^{q} \bigr)-a^{q}} \bigr) \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {}+ \bigl({2b^{q}-a^{q}-L \bigl(a^{q},b^{q} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} +a \bigl[ \bigl\vert f'(a) \bigr\vert ^{q} \bigl( {b^{q}-2a^{q}+L \bigl(a^{q},b^{q} \bigr)} \bigr) \\ &\quad \quad {}+ \bigl({b^{q}-L \bigl(a^{q},b^{q} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Theorem 4.5 {#FPar37}
-----------

*Let* $\documentclass[12pt]{minimal}
                \usepackage{amsmath}
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                \usepackage{amsbsy}
                \usepackage{mathrsfs}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R}$\end{document}$ *be a differentiable function on* $\documentclass[12pt]{minimal}
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                \begin{document}$I^{0} $\end{document}$ (*interior of* *I*), *where* $\documentclass[12pt]{minimal}
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                \begin{document}$a, b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
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                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
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                \begin{document}$f'\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
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                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
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                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
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                \begin{document}$q>1$\end{document}$ *and* $\documentclass[12pt]{minimal}
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                \begin{document}$q>p>0$\end{document}$, *then* $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)}{2(2p)^{\frac{1}{q}}} \bigl[L\bigl(a^{\frac{q-p}{q-1}},b ^{\frac{q-p}{q-1}}\bigr) \bigr]^{1-\frac{1}{q}} \bigl\{ a^{1-\frac{p}{q}}b \bigl[ \bigl({L \bigl(a ^{p},b^{p} \bigr)-a^{p}} \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \\ &\quad \quad {}+ \bigl({2b^{p}-a^{p}-L \bigl(a^{p},b^{p} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \bigr\} \\ &\quad \quad {}+ \bigl\{ ab^{1-\frac{p}{q}} \bigl[ \bigl({b^{p}-2a^{p}+L \bigl(a^{p},b^{p} \bigr) \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q}} \\ &\quad \quad {}+ \bigl({b^{q}-L \bigl(a^{p},b^{p} \bigr) \bigr) \bigl\vert f'(b) \bigr\vert ^{q}} \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Proof {#FPar38}
-----

Using Lemma [4.1](#FPar27){ref-type="sec"} and Hölder's inequality, we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad =\frac{ab(\ln b-\ln a)}{2} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert \,\mathrm{d}t + \int_{0}^{1} \biggl( \frac{a}{b} \biggr)^{t} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert \,\mathrm{d}t \biggr] \\ &\quad \leq \frac{ab(\ln b-\ln a)}{2} \biggl\{ \biggl[ \int_{0}^{1} \biggl( \frac{b}{a} \biggr)^{\frac{(q-p)t}{q-1}}\,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{pt} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}} \\ &\quad\quad {}+ \biggl[ \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{\frac{(q-p)t}{q-1}} \,\mathrm{d}t \biggr]^{1-\frac{1}{q}} \biggl[ \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{pt} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \biggr]^{\frac{1}{q}} \biggr\} . \end{aligned}$$ \end{document}$$ Now consider $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} I_{1} =& \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{pt} \bigl\vert f' \bigl(b^{\frac{1+t}{2}}a^{\frac{1-t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \\ \leq & \int_{0}^{1} \biggl(\frac{b}{a}\biggr)^{pt} \biggl[ \bigl\vert f'(b) \bigr\vert ^{t} \biggl(\frac{1+t}{2} \biggr)+ \bigl\vert f'(b)+ \eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \biggl( \frac{1-t}{2} \biggr) \biggr]\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{pt} \biggl(\frac{1+t}{2} \biggr)\,\mathrm{d}t + \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q} \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{pt} \biggl( \frac{1-t}{2} \biggr)\,\mathrm{d}t \\ =& \bigl\vert f'(b) \bigr\vert ^{q} \frac{2b^{p}-a^{p}-L(a^{p},b^{p})}{2pa^{p}(\ln b-\ln a)}+ \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \frac{L(a^{p},b^{p})-a^{p}}{2pa^{p}(\ln b-\ln a)}. \end{aligned}$$ \end{document}$$ Similarly $\documentclass[12pt]{minimal}
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                \begin{document}$I_{2}$\end{document}$ becomes $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl[ \int_{0}^{1} \biggl(\frac{a}{b} \biggr)^{pt} \bigl\vert f' \bigl(b^{\frac{1-t}{2}}a^{\frac{1+t}{2}} \bigr) \bigr\vert ^{q}\,\mathrm{d}t \biggr] \\ &\quad \leq \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}\frac{b^{p}-2a^{p}+L(a^{p},b^{p})}{2pb ^{p}(\ln b-\ln a)}+ \bigl\vert f'(b) \bigr\vert ^{q} \frac{b^{p}-L(a^{p},b^{p})}{2pb^{p}(\ln b-\ln a)}. \end{aligned}$$ \end{document}$$ Also $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} \biggl[ \int_{0}^{1} \biggl(\frac{b}{a} \biggr)^{\frac{(q-p)t}{q-1}}\,\mathrm{d}t \biggr]^{1- \frac{1}{q}}= \biggl( \frac{b^{\frac{q-p}{q-1}}-a^{\frac{q-p}{q-1}}}{( \frac{q-p}{q-1}) a^{\frac{q-p}{q-1}}(\ln b-\ln a)} \biggr)^{1- \frac{1}{q}}. \end{aligned}$$ \end{document}$$ Combining ([4.12](#Equ36){ref-type=""}), ([4.13](#Equ37){ref-type=""}), ([4.14](#Equ38){ref-type=""}) and ([4.15](#Equ39){ref-type=""}), we have $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)}{2(2p)^{\frac{1}{q}}} \bigl[L\bigl(a^{\frac{q-p}{q-1}},b ^{\frac{q-p}{q-1}}\bigr) \bigr]^{1-\frac{1}{q}} \bigl\{ a^{1-\frac{p}{q}}b \bigl[ \bigl({L \bigl(a^{p},b^{p} \bigr)-a^{p}} \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b) \bigr) \bigr\vert ^{q} \\ &\quad \quad {}+({2b^{p}-a^{p}-L \bigl(a^{p},b^{p} \bigr)} \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \bigr\} \\ &\quad \quad {}+ \bigl\{ ab^{1-\frac{p}{q}} \bigl[ \bigl({b^{p}-2a^{p}+L \bigl(a^{p},b^{p} \bigr) \bigr) \bigl\vert f'(b)+\eta \bigl(f'(a),f'(b)\bigr) \bigr\vert ^{q}} \\ &\quad \quad {}+ \bigl(b^{q}-L \bigl(a^{p},b^{p} \bigr) \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \bigr\} , \end{aligned}$$ \end{document}$$ which is the required result. □

If $\documentclass[12pt]{minimal}
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                \begin{document}$\eta (b,a)=b-a$\end{document}$, then Theorem [4.5](#FPar37){ref-type="sec"} reduces to the following result.

Corollary 4.4 {#FPar39}
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\[[@CR27]\]

*Let* $\documentclass[12pt]{minimal}
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                \begin{document}$f:I\subset \mathbb{R}_{+}=(0,\infty )\rightarrow \mathbb{R} $\end{document}$ *be a differentiable function on the interior* $\documentclass[12pt]{minimal}
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                \begin{document}$I^{0} $\end{document}$ *of* *I*, *where* $\documentclass[12pt]{minimal}
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                \begin{document}$a, b\in I$\end{document}$ *with* $\documentclass[12pt]{minimal}
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                \begin{document}$a < b$\end{document}$ *and* $\documentclass[12pt]{minimal}
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                \begin{document}$f'\in L[a,b]$\end{document}$. *If* $\documentclass[12pt]{minimal}
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                \begin{document}$\vert f' \vert $\end{document}$ *is a generalized geometrically convex function on* $\documentclass[12pt]{minimal}
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                \begin{document}$[a,b]$\end{document}$ *for* $\documentclass[12pt]{minimal}
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                \begin{document}$q>1$\end{document}$ *and* $\documentclass[12pt]{minimal}
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                \begin{document}$q>p>0$\end{document}$, *then* $$\documentclass[12pt]{minimal}
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                \begin{document} $$\begin{aligned} & \biggl\vert b f(b)-a f(a)- \int_{a}^{b}f(x)\,\mathrm{d}x \biggr\vert \\ &\quad \leq \frac{(\ln b-\ln a)}{2(2p)^{\frac{1}{q}}} \bigl[L\bigl(a^{\frac{q-p}{q-1}},b ^{\frac{q-p}{q-1}}\bigr) \bigr]^{1-\frac{1}{q}} \bigl\{ a^{1-\frac{p}{q}}b \bigl[ \bigl({L \bigl(a ^{p},b^{p} \bigr)-a^{p}} \bigr) \bigl\vert f'(a) \bigr\vert ^{q} \\ &\quad \quad {}+ \bigl({2b^{p}-a^{p}-L \bigl(a^{p},b^{p} \bigr)} \bigr) \bigl\vert f'(b) \bigr\vert ^{q} \bigr]^{\frac{1}{q}} \bigr\} + \bigl\{ ab^{1-\frac{p}{q}} \bigl[ \bigl({b^{p}-2a^{p}+L \bigl(a^{p},b^{p} \bigr) \bigr) \bigl\vert f'(a) \bigr\vert ^{q}} \\ &\quad \quad {}+ \bigl({b^{q}-L \bigl(a^{p},b^{p} \bigr) \bigr) \bigl\vert f'(b) \bigr\vert ^{q}} \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$ \end{document}$$

Conclusion {#Sec5}
==========

In this paper, we have introduced and investigated a new class of generalized functions, called generalized geometrically convex functions. Some basic inequalities related to generalized geometrically convex functions have been derived. Several new inequalities of the Hermite-Hadamard type for generalized geometrically convex functions have been established. Several special cases are discussed, which can be deduced from our main results. The techniques and ideas of this paper may stimulate further research in this dynamic field. It is an interesting problem to consider these estimates in the setting of fractional calculus of meromorphic functions; see \[[@CR3], [@CR4], [@CR17], [@CR28], [@CR29]\] and the references therein.

**Competing interests**

The authors declare that they have no competing interests.

**Authors' contributions**

MAN, KIN and FS worked jointly. All the authors read and approved the final manuscript.

**Publisher's Note**

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

The authors would like to thank the Rector of the COMSATS Institute of Information Technology, Pakistan, for providing excellent research and academic environments. The authors are pleased to acknowledge the support of the Distinguished Scientist Fellowship Program (DSFP), from King Saud University, Riyadh, Saudi Arabia. The authors are grateful to the referees for their valuable and constructive comments.
