Refinement Mappings Related to Hermite-Hadamard Type Inequalities for GA-Convex Function

Abstract

In this paper, we present some new refinement mappings associated with the Hermite–Hadamard type inequalities that are constructed for GA-convex mappings. Our investigation of the mappings leads to the discovery of several interesting features as well as the development of some inequalities for the Hermite–Hadamard type inequalities, which have already been established for GA-convex functions, as well as refining the relationship between the middle, rightmost, and leftmost elements of the function. Some applications to special means of positive real numbers are also given.

1. Introduction

A function $\aleph :I\subseteq \mathbb{R}\to \mathbb{R}$ is said to be convex on the interval I, if for all ${\vartheta }_1,{\vartheta }_2\in I$ and $\tau \in (0,1)$ it satisfies the following inequality:

$$\aleph (\tau {\vartheta }_1+(1-\tau ){\vartheta }_2)\le \tau \aleph \left({\vartheta }_1\right)+(1-\tau )\aleph \left({\vartheta }_2\right).$$

Convex functions play an important role in the field of integral inequalities. For convex functions, many equalities and inequalities have been established, but one of the most important ones is Hermite–Hadamard’s integral inequality, which is restated as follows [1,2]:

Let $\aleph :I⟶\mathbb{R},\varnothing \ne I\subseteq \mathbb{R},{\vartheta }_1,{\vartheta }_2\in I$ with ${\vartheta }_1<{\vartheta }_1$, be a convex function then

$$\aleph \left(\frac{{\vartheta }_1+{\vartheta }_2}{2}\right)\le \frac{1}{{\vartheta }_2-{\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph \left(x\right)dx\le \frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}.$$

(1)

The inequalities (1) hold in reversed direction if ℵ is concave. In recent years, a number of mathematicians have devoted their efforts to generalizing, refining, counterparting, and extending the Hermite–Hadamard inequality (1) for different classes of convex functions and mappings. These inequalities have many extensions and generalizations, see [3,4,5,6,7,8,9,10,11,12,13,14]. Dragomir defined the following mappings $\mathcal{X},\mathcal{Y}:[0,1]\to \mathbb{R}$ as follows

$$\mathcal{X}\left(\tau \right)=\frac{1}{{\vartheta }_2-{\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph \left(\tau x+(1-\tau )\left(\frac{{\vartheta }_1+{\vartheta }_2}{2}\right)\right)dx$$

and

$$\mathcal{Y}\left(\tau \right)=\frac{1}{{({\vartheta }_2-{\vartheta }_1)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph (\tau x+(1-\tau )y)dxdy,$$

where $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R}$ is a convex function and obtained some refinements between the middle and the left most terms in [15] for (1).

Theorem 1

([15]). Let $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R}$ be a convex function on $[{\vartheta }_1,{\vartheta }_2].$ Then

(i) 

$\mathcal{X}$ is convex convex on $[{\vartheta }_1,{\vartheta }_2]$.

(ii) 

The following hold:

$$\underset{\tau \in [0,1]}{\inf }\mathcal{X}\left(\tau \right)=\mathcal{X}\left(0\right)=\aleph \left(\frac{{\vartheta }_1+{\vartheta }_2}{2}\right)$$

and

$$\underset{\tau \in [0,1]}{\sup }\mathcal{X}\left(\tau \right)=\mathcal{X}\left(1\right)=\frac{1}{{\vartheta }_2-{\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph \left(x\right)dx.$$

(iii) 

$\mathcal{X}$ increases monotonically on $[0,1]$.

Theorem 2

([15]). Let $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R}$ be a convex function on $[{\vartheta }_1,{\vartheta }_2]$. Then

(i) 

$\mathcal{Y}\left(\tau +\frac{1}{2}\right)=\mathcal{Y}\left(\frac{1}{2}-\tau \right)$ for all $\tau \in \left[0,\frac{1}{2}\right].$

(ii) 

$\mathcal{Y}$ is convex convex on $[{\vartheta }_1,{\vartheta }_2]$.

(iii) 

The following hold:

$$\underset{\tau \in [0,1]}{\sup }\mathcal{Y}\left(\tau \right)=\mathcal{Y}\left(1\right)=\mathcal{Y}\left(0\right)=\frac{1}{{\vartheta }_2-{\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph \left(x\right)dx$$

and

$$\underset{\tau \in [0,1]}{\inf }\mathcal{Y}\left(\tau \right)=\mathcal{Y}\left(\frac{1}{2}\right)=\frac{1}{{({\vartheta }_2-{\vartheta }_1)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph \left(\frac{x+y}{2}\right)dxdy.$$

(iv) 

The following inequality is valid:

$$\aleph \left(\frac{{\vartheta }_1+{\vartheta }_2}{2}\right)\le \mathcal{Y}\left(\frac{1}{2}\right).$$

(v) 

$\mathcal{Y}$ increases monotonically on $\left[\frac{1}{2},1\right]$ and decreases monotonically on $\left[0,\frac{1}{2}\right]$.

(vi) 

We have the inequality $\mathcal{X}\left(\tau \right)\le \mathcal{Y}\left(\tau \right)$ for all $\tau \in [0,1]$.

Yang and Hong [16] provided an improvement between the middle and the right most term by defining the following mapping $P:[0,1]\to \mathbb{R}$

$$\begin{array}{cc}\hfill \mathcal{W}\left(\tau \right)& =\frac{1}{2({\vartheta }_2-{\vartheta }_1)}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\left[\aleph \left(\left(\frac{1+\tau }{2}\right){\vartheta }_2+\left(\frac{1-\tau }{2}\right)x\right)\right.\hfill \\ \hfill & \left.+\aleph \left(\left(\frac{1+\tau }{2}\right){\vartheta }_1+\left(\frac{1-\tau }{2}\right)x\right)\right]dx,\hfill \end{array}$$

where $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R}$ is a convex function.

Theorem 3

([16]). Let $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R}$ be a convex function on $[{\vartheta }_1,{\vartheta }_2]$. Then

(i) 

$\mathcal{W}$ is convex on $[{\vartheta }_1,{\vartheta }_2]$.

(ii) 

$\mathcal{W}$ increases monotonically on $[0,1]$.

(iii) 

The following hold:

$$\underset{\tau \in [0,1]}{\inf }\mathcal{W}\left(\tau \right)=\mathcal{W}\left(0\right)=\frac{1}{{\vartheta }_2-{\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\aleph \left(x\right)dx$$

and

$$\underset{\tau \in [0,1]}{\sup }\mathcal{W}\left(\tau \right)=\mathcal{W}\left(1\right)=\frac{\aleph ({\vartheta }_1+\aleph \left({\vartheta }_2\right)}{2}.$$

Definition 1

([17]). A function $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ is said to be GA-convex (geometric arithmetically convex) if

$$\aleph \left({\vartheta }_1^{\tau }{\vartheta }_2^{1-\tau }\right)\le \tau \aleph \left({\vartheta }_1\right)+(1-\tau )\aleph \left({\vartheta }_2\right)$$

(2)

for all ${\vartheta }_1,{\vartheta }_2\in I$ and $\tau \in [0,1].$

Since condition 2 can be written as

$$(\aleph \circ \exp )(\tau \ln {\vartheta }_1+(1-\tau )\ln {\vartheta }_2)\le \tau (\aleph \circ \exp )\left(\ln {\vartheta }_1\right)+(1-\tau )(\aleph \circ \exp )\left(\ln {\vartheta }_2\right),$$

we observe that $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ is GA-convex on I if and only if $\aleph \circ \exp$ is convex on $\ln I:=\{\ln x:x\in I\}.$ If $I=[{\vartheta }_1,{\vartheta }_2]$, then $\ln I=[\ln {\vartheta }_1,\ln {\vartheta }_2].$ By using this useful property, we easily say that if $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ is GA-convex on I and ${\vartheta }_1,{\vartheta }_2\in I$ with ${\vartheta }_1<{\vartheta }_2$, then

$$\begin{array}{c}\hfill \aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\le \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}(\aleph \circ \exp )\left(x\right)dx=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx\le \frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}.\end{array}$$

(3)

The foregoing results inspire us to create some mappings for GA-convex functions that are similar to the preceding mappings and to refine the relationship between the middle, rightmost, and leftmost elements of the function (3).

2. The Main Results

Suppose that $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ is GA-convex on I and ${\vartheta }_1,{\vartheta }_2\in I$, let $\mathcal{H}:[0,1]\to \mathbb{R}$ be defined by

$$\mathcal{H}\left(\tau \right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x^{\tau }{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-\tau }\right)dx.$$

(4)

We sate some important facts which relate GA-convex and convex functions and use them to prove the main results of this section.

Theorem 4

([3]). If $[{\vartheta }_1,{\vartheta }_2]\subset (0,\infty )$ and the function $\mathcal{G}:[\ln {\vartheta }_1,\ln {\vartheta }_2]\to \mathbb{R}$ is convex (concave) on $[\ln {\vartheta }_1,\ln {\vartheta }_2]$, then the function $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R},\aleph \left(\tau \right)=\mathcal{G}\left(\ln \tau \right)$ is GA-convex (concave) on $[{\vartheta }_1,{\vartheta }_2]$.

Remark 1.

It is obvious from Theorem 4 that if $\aleph :[{\vartheta }_1,{\vartheta }_2]\to \mathbb{R}$ is GA-convex on $[{\vartheta }_1,{\vartheta }_2]\subset (0,\infty )$, then $\aleph \circ \exp$ is convex on $[\ln {\vartheta }_1,\ln {\vartheta }_2]$. It follows that $\aleph \circ \exp$ has finite lateral derivatives on $\left(\ln {\vartheta }_1,\ln {\vartheta }_2\right)$ and by gradient inequality for convex functions we have

$$\aleph \circ \exp \left(x\right)-\aleph \circ \exp \left(y\right)(x-y)\ge \phi \left(\exp y\right)\exp \left(y\right),$$

where $\phi \left(\exp y\right)\in \left[{\aleph }_{-}^{{}^{\prime }}\left(\exp y\right),{\aleph }_{+}^{{}^{\prime }}\left(\exp y\right)\right]$ for any $x,y\in \left(\ln {\vartheta }_1,\ln {\vartheta }_2\right)$.

Theorem 5

([3] Jensen’s inequality for GA-convex functions). Let $\aleph :I\subset \left(0,\infty \right)\to \mathbb{R}$ be a GA-convex function and $\left[k,K\right]\subset I^{\circ }$. Assume also that $h:\mathsf{\Omega }\to \mathbb{R}$ is μ-measurable, satisfying the bounds

$$0<k\le h\left(\tau \right)\le K<\infty for\mu \text{-}a.e.\tau \in \mathsf{\Omega }$$

and $w\ge 0$ μ-a.e. on Ω with ${\int }_{\mathsf{\Omega }}wd\mu =1$. If $\phi \in \partial \aleph$ the subdifferential of ℵ and $\aleph \circ h$, $\ln h\in L_w\left(\mathsf{\Omega },\mu \right)$, then

$$\aleph \circ \exp \left({\int }_{\mathsf{\Omega }}w\ln hd\mu \right)\le {\int }_{\mathsf{\Omega }}\left(\aleph \circ h\right)wd\mu .$$

(5)

The following theorem holds:

Theorem 6.

A function $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ as above. Then

(i) 

$\mathcal{H}$ is GA-convex on $[0,1]$.

(ii) 

We have

$$\underset{\tau \in [0,1]}{\inf }\mathcal{H}\left(\tau \right)=\mathcal{H}\left(0\right)=\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)$$

and

$$\underset{\tau \in [0,1]}{\sup }\mathcal{H}\left(\tau \right)=\mathcal{H}\left(1\right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx.$$

(iii) 

$\mathcal{H}$ increases monotonically on $[0,1]$.

Proof. 

(i) In order to prove that $\mathcal{H}:[0,1]\to \mathbb{R}$ is GA-convex on $(0,1]$, where $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ is GA-convex on I, it is suffices to prove that the mapping $\mathcal{G}:[0,1]\to \mathbb{R}$ defined by

$$\mathcal{G}\left(\tau \right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{S}\circ \exp \left(\tau \ln x+\left(1-\tau \right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)dx$$

is convex on $[\ln {\vartheta }_1,\ln {\vartheta }_2]$ for convex function $\mathcal{S}:[\ln {\vartheta }_1,\ln {\vartheta }_2]\to \mathbb{R}.$ Let $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$ and ${\tau }_1,{\tau }_2\in (0,1]$. Then

$$\begin{array}{cc}\hfill & \mathcal{G}\circ \exp \left(\alpha \ln {\tau }_1+\beta \ln {\tau }_2\right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{S}\circ \exp \left(\left(\alpha {\tau }_1+\beta {\tau }_2\right)\ln x\right.\hfill \\ \hfill & \left.+\left(\alpha +\beta -\left(\alpha {\tau }_1+\beta {\tau }_2\right)\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)dx\hfill \\ \hfill =& \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{S}\circ \exp \left(\alpha \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\right.\hfill \\ \hfill & \left.+\beta \left({\tau }_2\ln x+\left(1-{\tau }_2\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\right)dx\hfill \\ \hfill \le & \alpha \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{S}\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)dx\hfill \\ \hfill & +\beta \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{S}\circ \exp \left({\tau }_2\ln x+\left(1-{\tau }_2\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)dx\hfill \\ \hfill =& \alpha \mathcal{G}\left({\tau }_1\right)+\beta \mathcal{G}\left({\tau }_2\right),\hfill \end{array}$$

which shows that $\mathcal{H}$ is GA-convex on $(0,1]$.

(ii) By Jensen’s integral inequality, we have

$$\begin{array}{cc}\hfill \mathcal{H}\left(\tau \right)& =\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x^{\tau }{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-\tau }\right)dx\hfill \\ \hfill & ={\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)x}\aleph \left(x^{\tau }{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-\tau }\right)dx\hfill \\ \hfill & \ge \aleph \circ \exp \left({\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}\ln \left(x^{\tau }{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-\tau }\right)\right)dx=\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right).\hfill \end{array}$$

Now, using the GA-convexity of ℵ we get

$$\begin{array}{c}\mathcal{H}\left(\tau \right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x^{\tau }{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-\tau }\right)dx\hfill \\ \hfill \le \tau \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x\right)dx+\left(1-\tau \right)\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right).\end{array}$$

(6)

Since the mapping

$$\mathcal{P}\left(\tau \right):=\tau \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x\right)dx+\left(1-\tau \right)\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right),$$

increases monotonically on $[0,1]$. Hence it is proved that

$$\begin{array}{c}\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\le \mathcal{H}\left(\tau \right)\le \tau \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x\right)dx+\left(1-\tau \right)\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\hfill \\ \hfill \le \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx.\end{array}$$

(7)

Thus

$$\underset{\tau \in [0,1]}{\inf }\mathcal{H}\left(\tau \right)=\mathcal{H}\left(0\right)=\aleph \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)$$

and

$$\underset{\tau \in [0,1]}{\sup }\mathcal{H}\left(\tau \right)=\mathcal{H}\left(1\right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx.$$

(iii) To prove that $\mathcal{H}$ increases monotonically on $[0,1]$ is equivalent to prove that $\mathcal{G}$ increases monotonically on $[0,1]$. Since $\mathcal{G}$ is convex on $[0,1]$, so for ${\tau }_1,{\tau }_2\in (0,1)$, we get

$$\begin{array}{cc}\hfill & \frac{\mathcal{G}\left({\tau }_2\right)-\mathcal{G}\left({\tau }_1\right)}{{\tau }_2-{\tau }_1}\ge {\mathcal{G}}_{+}^{\prime }\left({\tau }_1\right)\hfill \\ \hfill & =\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\mathcal{S}}_{+}^{\prime }\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\hfill \\ \hfill & \times \left(x^{{\tau }_1}{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-{\tau }_1}\right)\ln \left(\frac{x}{\sqrt{{\vartheta }_1{\vartheta }_2}}\right)dx.\hfill \end{array}$$

The convexity of $\mathcal{G}$ on $\left[\ln {\vartheta }_1,\ln {\vartheta }_2\right]$ yields

$$\begin{array}{cc}\hfill & \mathcal{S}\circ \exp \left(\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)-\mathcal{S}\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\hfill \\ \hfill & \ge {\tau }_1{\mathcal{S}}_{+}^{\prime }\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\hfill \\ \hfill & \times \left(x^{{\tau }_1}{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-{\tau }_1}\right)\ln \left(\frac{\sqrt{{\vartheta }_1{\vartheta }_2}}{x}\right),\hfill \end{array}$$

for every $x\in \left[\ln {\vartheta }_1,\ln {\vartheta }_2\right]$. Thus

$$\begin{array}{cc}\hfill & \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\mathcal{S}}_{+}^{\prime }\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\hfill \\ \hfill & \times \left(x^{{\tau }_1}{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-{\tau }_1}\right)\ln \left(\frac{x}{\sqrt{{\vartheta }_1{\vartheta }_2}}\right)dx\hfill \\ \hfill & \ge \frac{1}{{\tau }_1}\left[\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{S}\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)dx\right.\hfill \\ \hfill & \left.-\mathcal{S}\circ \exp \left(\ln \left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right)\right]\hfill \\ \hfill & =\frac{1}{{\tau }_1}\left[\mathcal{G}\left({\tau }_1\right)-\mathcal{S}\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right]\ge 0.\hfill \end{array}$$

This proves that $\mathcal{G}\left({\tau }_2\right)-\mathcal{G}\left({\tau }_1\right)\ge 0$ for $1\ge {\tau }_2\ge {\tau }_1$. Hence $\mathcal{G}$ is monotonically increasing on $[0,1]$ which implies that $\mathcal{H}$ is also monotonically increasing on $[0,1]$. □

Now, we shall define the second mapping in connection with Hadamard’s inequalities. Let $\aleph :[{\vartheta }_1,{\vartheta }_2]\subseteq (0,\infty )\to \mathbb{R}$ be a GA-convex function on $[{\vartheta }_1,{\vartheta }_2].$ Put

$$\mathcal{F}:[0,1]\to \mathbb{R},\mathcal{F}\left(\tau \right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\tau }{y}^{1-\tau }\right)dxdy.$$

The following theorem holds:

Theorem 7.

Let $\aleph :[{\vartheta }_1,{\vartheta }_2]\subseteq (0,\infty )\to \mathbb{R}$ be as above. Then

(i) 

$\mathcal{F}(\tau +\frac{1}{2})=\mathcal{F}(\frac{1}{2}-\tau )$ for all τ in $[0,\frac{1}{2}]$.

(ii) 

$\mathcal{F}$ is GA-convex on $[0,1]$.

(iii) 

We have

$$\underset{\tau \in [0,1]}{\sup }\mathcal{F}\left(\tau \right)=\mathcal{F}\left(0\right)=\mathcal{F}\left(1\right)=\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x\right)dx$$

and

$$\underset{\tau \in [0,1]}{\inf }\mathcal{F}\left(\tau \right)=\mathcal{F}\left(\frac{1}{2}\right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(\sqrt{xy}\right)dxdy.$$

(iv) 

The following inequality is valid:

$$\aleph \left(\sqrt{xy}\right)\le \mathcal{F}\left(\frac{1}{2}\right).$$

(v) 

$\mathcal{F}$ decreases monotonically on $[0,\frac{1}{2}]$ and increases monotonically on $\left[\frac{1}{2},1\right]$.

(vi) 

We have the inequality $\mathcal{H}\left(\tau \right)\le \mathcal{F}\left(\tau \right)$ for all $\tau \in [0,1].$

Proof. 

(i) Let $\tau \in [0,\frac{1}{2}]$. We have

$$\begin{array}{c}\mathcal{F}\left(\tau +\frac{1}{2}\right)=\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\tau +\frac{1}{2}}{y}^{\frac{1}{2}-\tau }\right)dxdy\hfill \\ \hfill =\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\frac{1}{2}-\tau }{y}^{\tau +\frac{1}{2}}\right)dxdy=\mathcal{F}\left(\frac{1}{2}-\tau \right).\end{array}$$

(8)

(ii) The argument is similar to that in the proof of Theorem 6 (i) and we omit it.

(iii) Since $\aleph :[{\vartheta }_1,{\vartheta }_2]\subset (0,\infty )\to \mathbb{R}$ is GA-convex function on $[{\vartheta }_1,{\vartheta }_2]$, for all $x,y\in [{\vartheta }_1,{\vartheta }_2]$ and $\tau \in [0,1]$, we obtain

$$\begin{array}{c}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\tau }{y}^{1-\tau }\right)dxdy\hfill \\ \le {\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}[\tau \aleph \left(x\right)+(1-\tau )\aleph \left(y\right)]dxdy\\ \hfill =\tau {\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x\right)+(1-\tau ){\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(y\right)dxdy.\end{array}$$

(9)

Integrating this inequality over $[{\vartheta }_1,{\vartheta }_2]\times [{\vartheta }_1,{\vartheta }_2]$ we get

$$\begin{array}{c}\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\tau }{y}^{1-\tau }\right)dxdy\hfill \\ \le {\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}[\tau \aleph \left(x\right)+(1-\tau )\aleph \left(y\right)]dxdy\\ \hfill =\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx.\end{array}$$

(10)

Thus, $\mathcal{F}\left(\tau \right)\le \mathcal{F}\left(0\right)=\mathcal{F}\left(1\right)$ for all $\tau \in [0,1]$.

Since ℵ is convex on $[{\vartheta }_1,{\vartheta }_2]$, for all $\tau \in [0,1]$ and $x,y$ in $[{\vartheta }_1,{\vartheta }_2]$, we have

$$\frac{1}{2}[\aleph \left(x^{\tau }{y}^{1-\tau }\right)+\aleph \left(x^{1-\tau }{y}^{\tau }\right)]\ge \aleph \left(\sqrt{xy}\right).$$

Integrating this inequality in $[{\vartheta }_1,{\vartheta }_2]\times [{\vartheta }_1,{\vartheta }_2]$ we deduce

$$\begin{array}{c}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(\sqrt{xy}\right)dxdy\hfill \\ \le \frac{1}{2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}[\aleph \left(x^{\tau }{y}^{1-\tau }\right)+\aleph \left(x^{1-\tau }{y}^{\tau }\right)]dxdy\\ \hfill ={\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\tau }{y}^{1-\tau }\right)dxdy,\end{array}$$

(11)

which implies that $\mathcal{F}\left(\frac{1}{2}\right)\le \mathcal{F}\left(\tau \right)$ for all $\tau$ in $[0,1]$ and the statement is proven.

(iv) By Jensen’s integral inequality we have

$$\begin{array}{c}\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\aleph \left(x^{\tau }{y}^{1-\tau }\right)dxdy\hfill \\ \hfill \ge \aleph \circ \exp \left(\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\ln \left(x^{\tau }{y}^{1-\tau }\right)dxdy\right).\end{array}$$

(12)

Since a simple computation shows that

$$\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{xy}\ln \left(x^{\tau }{y}^{1-\tau }\right)dxdy=\sqrt{xy}.$$

(13)

(v) In order to prove that ℵ increases monotonically on $[\frac{1}{2},1]$ and decreases monotoniically on $[0,\frac{1}{2}]$ it is suffices to prove that $\mathcal{K}:[0,1]\to \mathbb{R}$ defined by

$$\begin{array}{cc}\hfill & \mathcal{K}\left(\tau \right)=\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{N}\circ \exp \left(\tau \ln x+\left(1-\tau \right)\ln y\right)dxdy\hfill \\ & \frac{\mathcal{K}\left({\tau }_2\right)-\mathcal{K}\left({\tau }_1\right)}{{\tau }_2-{\tau }_1}\ge {\mathcal{K}}_{+}^{\prime }\left({\tau }_1\right)\hfill \\ \hfill & =\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\mathcal{N}}_{+}^{\prime }\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln y\right)\hfill \\ \hfill & \times \left(x^{{\tau }_1}{y}^{1-{\tau }_1}\right)\ln \left(\frac{x}{y}\right)dxdy.\hfill \end{array}$$

The convexity of $\mathcal{K}$ on $\left[\ln {\vartheta }_1,\ln {\vartheta }_2\right]$ yields

$$\begin{array}{cc}\hfill & \mathcal{N}\circ \exp \left(\ln \sqrt{xy}\right)-\mathcal{N}\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln y\right)\hfill \\ \hfill & \ge \left(\frac{1}{2}-{\tau }_1\right){\mathcal{N}}_{+}^{\prime }\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln y\right)\ln \left(\frac{x}{y}\right)\left(x^{{\tau }_1}{y}^{1-{\tau }_1}\right),\hfill \end{array}$$

for every $x,y\in \left[\ln {\vartheta }_1,\ln {\vartheta }_2\right]$. Thus

$$\begin{array}{c}\frac{1}{{\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\mathcal{N}}_{+}^{\prime }\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln y\right)\hfill \\ \times \ln \left(\frac{x}{y}\right)\left(x^{{\tau }_1}{y}^{1-{\tau }_1}\right)dxdy\ge \frac{2}{\left(2{\tau }_1-1\right){\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}^2}\\ \times {\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{N}\circ \exp \left({\tau }_1\ln x+\left(1-{\tau }_1\right)\ln y\right)dxdy\\ \hfill -\mathcal{N}\circ \exp \left(\ln \sqrt{xy}\right)=\frac{2}{\left(2{\tau }_1-1\right)}\left[\mathcal{K}\left({\tau }_1\right)-\mathcal{N}\left(\sqrt{xy}\right)\right]\ge 0.\end{array}$$

This proves that $\mathcal{K}\left({\tau }_2\right)-\mathcal{K}\left({\tau }_1\right)\ge 0$ for $1\ge {\tau }_2\ge {\tau }_1$. Hence $\mathcal{K}$ is monotonically increasing on $[\frac{1}{2},1]$ which implies that $\mathcal{F}$ is also monotonically increasing on $[\frac{1}{2},1]$.

The fact that $\mathcal{F}$ increases monotonically on $[0,\frac{1}{2}]$ follows from the above conclusion using statement (i).

(vi) In this part we can easily prove that the following inequality

$$\mathcal{H}\left(\tau \right)\le \mathcal{F}\left(\tau \right)$$

for all $\tau \in [0,1]$ by using the concept of GA-Jensen’s integral inequality.

□

Theorem 8.

Let $\aleph :[{\vartheta }_1,{\vartheta }_2]\subset (0,\infty )\to \mathbb{R}$ is GA-convex function on $[{\vartheta }_1,{\vartheta }_2]$. Then

$$\mathcal{V}\left(\tau \right)=\frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\left[\aleph \left({\vartheta }_2^{\frac{1+\tau }{2}}{x}^{\frac{1-\tau }{2}}\right)+\aleph \left({\vartheta }_1^{\frac{1+\tau }{2}}{x}^{\frac{1-\tau }{2}}\right)\right]dx.$$

(i) 

$\mathcal{V}$ is GA-convex on $(0,1]$.

(ii) 

The following hold:

$$\underset{\tau \in [0,1]}{\inf }\mathcal{V}\left(\tau \right)=\mathcal{V}\left(0\right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx$$

and

$$\underset{\tau \in [0,1]}{\sup }\mathcal{V}\left(\tau \right)=\mathcal{V}\left(1\right)=\frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}.$$

(iii) 

$\mathcal{V}$ increases monotonically on $[0,1]$.

Proof. 

(i) In order to prove that $\mathcal{V}:[0,1]\to \mathbb{R}$ is GA-convex on $(0,1]$, where $\aleph :I\subseteq (0,\infty )\to \mathbb{R}$ is GA-convex on I, it is suffices to prove that the mapping $\mathcal{J}:[0,1]\to \mathbb{R}$ defined by

$$\begin{array}{c}\mathcal{J}\left(\tau \right)=\frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+\tau }{2}\right)\ln {\vartheta }_2+\left(\frac{1-\tau }{2}\right)\ln x\right)\right.\hfill \\ \hfill +\left.\mathcal{Q}\circ \exp \left(\left(\frac{1+\tau }{2}\right)\ln {\vartheta }_1+\left(\frac{1-\tau }{2}\right)\ln x\right)\right]dx\end{array}$$

(14)

is convex for convex function $\mathcal{Q}:[\ln {\vartheta }_1,\ln {\vartheta }_2]\to \mathbb{R}$ on $[\ln {\vartheta }_1,\ln {\vartheta }_2].$ Let $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$ and ${\tau }_1,{\tau }_2\in (0,1]$. Then

$$\begin{array}{c}\mathcal{J}\left(\alpha {\tau }_1+\beta {\tau }_2\right)=\frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+\alpha {\tau }_1+\beta {\tau }_2}{2}\right)\ln {\vartheta }_2\right.\right.\hfill \\ \left.+\left(\frac{1-\left(\alpha {\tau }_1+\beta {\tau }_2\right)}{2}\right)\ln x\right)+\mathcal{Q}\circ \exp \left(\left(\frac{1+\alpha {\tau }_1+\beta {\tau }_2}{2}\right)\ln {\vartheta }_1\right.\\ \left.\left.+\left(\frac{1-\left(\alpha {\tau }_1+\beta {\tau }_2\right)}{2}\right)\ln x\right)\right]dx=\frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}\\ \times {\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{\alpha +\beta +\alpha {\tau }_1+\beta {\tau }_2}{2}\right)\ln {\vartheta }_2\right.\right.\\ \left.+\left(\frac{\alpha +\beta -\left(\alpha {\tau }_1+\beta {\tau }_2\right)}{2}\right)\ln x\right)+\mathcal{Q}\circ \exp \left(\left(\frac{\alpha +\beta +\alpha {\tau }_1+\beta {\tau }_2}{2}\right)\ln {\vartheta }_1\right.\\ \left.\left.+\left(\frac{\alpha +\beta -\left(\alpha {\tau }_1+\beta {\tau }_2\right)}{2}\right)\ln x\right)\right]dx=\alpha \frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}\\ \hfill \times {\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\right.\end{array}$$

$$\begin{array}{c}+\left.\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_1+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\right]dx\hfill \\ +\beta \frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_2}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_2}{2}\right)\ln x\right)\right.\\ \hfill +\left.\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_2}{2}\right)\ln {\vartheta }_1+\left(\frac{1-{\tau }_2}{2}\right)\ln x\right)\right]dx=\alpha \mathcal{J}\left({\tau }_1\right)+\beta \mathcal{J}\left({\tau }_2\right),\end{array}$$

which shows that $\mathcal{V}$ is GA-convex on $(0,1]$.

(ii) Let ${\tau }_1,{\tau }_2\in (0,1]$

$$\begin{array}{cc}\hfill & \frac{\mathcal{J}\left({\tau }_2\right)-\mathcal{J}\left({\tau }_1\right)}{{\tau }_2-{\tau }_1}\ge {\mathcal{J}}_{+}^{\prime }\left({\tau }_1\right)=\frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}\hfill \\ \hfill & \times {\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\left({\vartheta }_2^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right)\ln \left({\displaystyle \frac{{\vartheta }_2}{x}}\right)\right.\hfill \\ \hfill & \left.+\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_1+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\left({\vartheta }_1^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right)\ln \left({\displaystyle \frac{{\vartheta }_1}{x}}\right)\right]dx.\hfill \end{array}$$

(15)

The convexity of $\mathcal{J}$ on $\left[\ln {\vartheta }_1,\ln {\vartheta }_2\right]$ yields

$$\begin{array}{c}\mathcal{Q}\circ \exp \left(\ln \sqrt{{\vartheta }_{2}x}\right)-\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\hfill \\ \hfill \ge \frac{{\tau }_1}{2}{\mathcal{Q}}_{+}^{\prime }\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\left(\ln \left(\frac{x}{{\vartheta }_2}\right)\right)\left({\vartheta }_2^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right).\end{array}$$

(16)

for every $x\in \left[\ln {\vartheta }_1,\ln {\vartheta }_2\right]$ and ${\tau }_1\in (0,1].$ Thus

$$\begin{array}{c}{\mathcal{Q}}_{+}^{\prime }\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\left(\ln \left(\frac{{\vartheta }_2}{x}\right)\right)\left({\vartheta }_2^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right)\hfill \\ \hfill \ge \frac{2}{{\tau }_1}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)-\mathcal{Q}\circ \exp \left(\ln \sqrt{{\vartheta }_{2}x}\right)\right].\end{array}$$

(17)

Similarly, we also get that

$$\begin{array}{c}{\mathcal{Q}}_{+}^{\prime }\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_1+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\left(\ln \left(\frac{{\vartheta }_1}{x}\right)\right)\left({\vartheta }_1^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right)\hfill \\ \hfill \ge \frac{2}{{\tau }_1}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_1+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)-\mathcal{Q}\circ \exp \left(\ln \sqrt{{\vartheta }_{1}x}\right)\right].\end{array}$$

(18)

Thus, we have

$$\begin{array}{c}\frac{1}{2\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[{\mathcal{Q}}_{+}^{\prime }\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\right.\hfill \\ \times \left(\ln \left(\frac{{\vartheta }_2}{x}\right)\right)\left({\vartheta }_2^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right)+{\mathcal{Q}}_{+}^{\prime }\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_1\right.\\ \hfill \left.\left.+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\left(\ln \left(\frac{{\vartheta }_1}{x}\right)\right)\left({\vartheta }_1^{\frac{1+{\tau }_1}{2}}{x}^{\frac{1-{\tau }_1}{2}}\right)\right]dx\end{array}$$

(19)

$$\begin{array}{c}\ge \frac{2}{{\tau }_1}\frac{1}{2\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\left[\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\right.\hfill \\ +\mathcal{Q}\circ \exp \left(\left(\frac{1+{\tau }_1}{2}\right)\ln {\vartheta }_2+\left(\frac{1-{\tau }_1}{2}\right)\ln x\right)\\ \left.-\mathcal{Q}\circ \exp \left(\ln \sqrt{{\vartheta }_{2}x}\right)-\mathcal{Q}\circ \exp \left(\ln \sqrt{{\vartheta }_{1}x}\right)\right]\\ =\frac{2}{{\tau }_1}\left[\mathcal{J}\left({\tau }_1\right)-\frac{1}{2\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{Q}\left(\sqrt{{\vartheta }_{2}x}\right)dx\right.\\ \hfill \left.-\frac{1}{2\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}{\int }_{\ln {\vartheta }_1}^{\ln {\vartheta }_2}\mathcal{Q}\left(\sqrt{{\vartheta }_{1}x}\right)dx\right]=\frac{2}{{\tau }_1}\left[\mathcal{J}\left({\tau }_1\right)-\mathcal{Q}\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)\right]\ge 0.\end{array}$$

This proves that $\mathcal{J}\left({\tau }_2\right)-\mathcal{J}\left({\tau }_1\right)\ge 0$ for $1\ge {\tau }_2\ge {\tau }_1$. Hence, $\mathcal{V}$ is monotonically increasing on $[0,1]$.

(iii) Since $\mathcal{V}\left(\tau \right)$ is monotonically increasing, we have

$$\begin{array}{c}\mathcal{V}\left(\tau \right)\ge \mathcal{V}\left(0\right)=\frac{1}{2\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\left[\aleph \left(\sqrt{{\vartheta }_{2}x}\right)+\aleph \left(\sqrt{{\vartheta }_{1}x}\right)\right]dx\hfill \\ \hfill =\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx.\end{array}$$

(20)

Using the GA-convexity of ℵ on $[{\vartheta }_1,{\vartheta }_2]$ and Hermite–Hadamard type inequalities for GA-convex functions, we get

$$\begin{array}{c}\mathcal{V}\left(\tau \right)=\frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\left[\aleph \left({\vartheta }_2^{\frac{1+\tau }{2}}{x}^{\frac{1-\tau }{2}}\right)+\aleph \left({\vartheta }_1^{\frac{1+\tau }{2}}{x}^{\frac{1-\tau }{2}}\right)\right]dx\hfill \\ \le \frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\left[\left(\frac{1+\tau }{2}\right)\aleph \left({\vartheta }_2\right)+\left(\frac{1-\tau }{2}\right)\aleph \left(x\right)\right.\\ \left.+\left(\frac{1+\tau }{2}\right)\aleph \left({\vartheta }_1\right)+\left(\frac{1-\tau }{2}\right)\aleph \left(x\right)\right]dx\\ =\left(\frac{1+\tau }{2}\right)\frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}+\left(\frac{1-\tau }{2}\right)\\ \hfill \times \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx\le \frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}.\end{array}$$

(21)

Thus

$$\begin{array}{c}\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx\le \mathcal{V}\left(\tau \right)\le \left(\frac{1+\tau }{2}\right)\frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}\hfill \\ \hfill +\left(\frac{1-\tau }{2}\right)\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx\le \frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}.\end{array}$$

(22)

It is proved that

$$\underset{\tau \in [0,1]}{\inf }\mathcal{V}\left(\tau \right)=\mathcal{V}\left(0\right)=\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx$$

and

$$\underset{\tau \in [0,1]}{\sup }\mathcal{V}\left(\tau \right)=\mathcal{V}\left(1\right)=\frac{\aleph \left({\vartheta }_1\right)+\aleph \left({\vartheta }_2\right)}{2}.$$

□

3. Application to Special Means

Let us recall the following special means of two nonnegative numbers ${\vartheta }_1,{\vartheta }_2$ with $>{\vartheta }_2>{\vartheta }_1$.

(1) The arithmetic mean

$$A({\vartheta }_1,{\vartheta }_2):=\frac{{\vartheta }_1+{\vartheta }_2}{2}.$$

(2) The geometric mean

$$G({\vartheta }_1,{\vartheta }_2):=\sqrt{{\vartheta }_1{\vartheta }_2}.$$

(3) The logarithmic mean

$$L({\vartheta }_1,{\vartheta }_2):=\frac{{\vartheta }_2-{\vartheta }_1}{\ln {\vartheta }_2-\ln {\vartheta }_1}.$$

(4) The p-logarithmic mean

$$L_p({\vartheta }_1,{\vartheta }_2):={\left(\frac{{\vartheta }_2^{p+1}-{\vartheta }_1^{p+1}}{\left(p+1\right)\left({\vartheta }_2-{\vartheta }_1\right)}\right)}^{\frac{1}{p}},p\in \mathbb{R}\setminus \{-1,0\}.$$

Let $p\ne 0$ and let $g\left(\tau \right)=\exp \left(p\tau \right)$, $\tau \in \mathbb{R}$. Then the function $\aleph :\left(0,\infty \right)\to \mathbb{R}$, $\aleph \left(\tau \right)=g\left(\ln \tau \right)=\exp \left(p\ln \tau \right)={\tau }^p$ is a GA-convex function on $\left(0,\infty \right)$. Let $0<{\vartheta }_1<{\vartheta }_2$, then

$$\begin{array}{cc}\hfill \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{\aleph \left(x\right)}{x}dx& =\frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}{x}^{p-1}dx\hfill \\ \hfill & =\frac{{\vartheta }_2^p-{\vartheta }_1^p}{p\left(\ln {\vartheta }_2-\ln {\vartheta }_1\right)}=L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right).\hfill \end{array}$$

Additionally,

$$\begin{array}{cc}\hfill & \frac{1}{\ln {\vartheta }_2-\ln {\vartheta }_1}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\aleph \left(x^{\tau }{\left(\sqrt{{\vartheta }_1{\vartheta }_2}\right)}^{1-\tau }\right)dx\hfill \\ \hfill & ={\vartheta }_1^{\frac{p}{2}(1-\tau )}{\vartheta }_2^{\frac{p}{2}(1-\tau )}{L}_{\frac{p}{\tau }-1}^{\frac{p}{\tau }-1}\left({\vartheta }_1,{\vartheta }_2\right).\hfill \end{array}$$

Thus, (7) takes the form

$$\begin{array}{c}\hfill G({\vartheta }_1,{\vartheta }_2)\le G\left({\vartheta }_1^{p(1-\tau )}{\vartheta }_2^{p(1-\tau )}\right)L_{\frac{p}{\tau }-1}^{\frac{p}{\tau }-1}\left({\vartheta }_1,{\vartheta }_2\right)\le \tau L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right)+\left(1-\tau \right)G({\vartheta }_1,{\vartheta }_2)\le L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right),\end{array}$$

(24)

where $G\left({\vartheta }_1,{\vartheta }_2\right)$ and $L_p\left({\vartheta }_1,{\vartheta }_2\right)$ are the geometric and the logarithmic means of ${\vartheta }_1$ and ${\vartheta }_1$, respectively.

If we choose $\tau =\frac{1}{2}$ in (23), we get

$$\begin{array}{c}\hfill 0\le G^{\frac{1}{2}}\left({\vartheta }_1^p,{\vartheta }_2^p\right)L_{2p-1}^{2p-1}\left({\vartheta }_1,{\vartheta }_2\right)-G\left({\vartheta }_1,{\vartheta }_2\right)\le \frac{1}{2}\left[L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right)-G\left({\vartheta }_1,{\vartheta }_2\right)\right]\le L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right)-G\left({\vartheta }_1,{\vartheta }_2\right).\end{array}$$

Moreover

$$\begin{array}{cc}\hfill & \frac{1}{2(\ln {\vartheta }_2-\ln {\vartheta }_1)}{\int }_{{\vartheta }_1}^{{\vartheta }_2}\frac{1}{x}\left[\aleph \left({\vartheta }_2^{\frac{1+\tau }{2}}{x}^{\frac{1-\tau }{2}}\right)+\aleph \left({\vartheta }_1^{\frac{1+\tau }{2}}{x}^{\frac{1-\tau }{2}}\right)\right]dx\hfill \\ \hfill & =A\left({\vartheta }_1^{p\left(\frac{1+\tau }{2}\right)},{\vartheta }_2^{p\left(\frac{1+\tau }{2}\right)}\right)L_{{}^{p\left(\frac{1-\tau }{2}\right)-1}}^{{}^{p\left(\frac{1-\tau }{2}\right)-1}}\left({\vartheta }_1,{\vartheta }_2\right).\hfill \end{array}$$

Thus, (22) takes the form

$$\begin{array}{c}{L}_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right)\le A\left({\vartheta }_1^{p\left(\frac{1+\tau }{2}\right)},{\vartheta }_2^{p\left(\frac{1+\tau }{2}\right)}\right)L_{{}^{p\left(\frac{1-\tau }{2}\right)-1}}^{{}^{p\left(\frac{1-\tau }{2}\right)-1}}\left({\vartheta }_1,{\vartheta }_2\right)\hfill \\ \hfill \le \left(\frac{1+\tau }{2}\right)A\left({\vartheta }_1^p,{\vartheta }_2^p\right)+\left(\frac{1-\tau }{2}\right)L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right)\le A\left({\vartheta }_1^p,{\vartheta }_2^p\right),\end{array}$$

(24)

where $A\left({\vartheta }_1^p,{\vartheta }_2^p\right)$ and $L\left({\vartheta }_1^p,{\vartheta }_2^p\right)$ are the arithmetic and the logarithmic means of ${\vartheta }_1^p$ and ${\vartheta }_2^p$, respectively.

If we choose $\tau =\frac{1}{2}$ in (24), we get

$$\begin{array}{c}0\le A\left({\vartheta }_1^{p\left(\frac{3}{4}\right)},{\vartheta }_2^{p\left(\frac{3}{4}\right)}\right)L_{{}^{p\left(\frac{1}{4}\right)-1}}^{{}^{p\left(\frac{1}{4}\right)-1}}\left({\vartheta }_1,{\vartheta }_2\right)-L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right)\hfill \\ \hfill \le \frac{3}{4}\left[A\left({\vartheta }_1^p,{\vartheta }_2^p\right)-L\left({\vartheta }_1^p,{\vartheta }_2^p\right)\right]\le A\left({\vartheta }_1^p,{\vartheta }_2^p\right)-L_{p-1}^{p-1}\left({\vartheta }_1,{\vartheta }_2\right).\end{array}$$

(25)

4. Conclusions

Hadamard’s inequalities are important in many different fields of science, including engineering, economics, astronomy, and mathematics. Hadamard’s inequality has been very important to many mathematicians because it can be used in many different types of pure and applied math. As a result, the aims of this article were to investigate certain refinement mappings for GA-convex functions that are similar to the preceding mappings and to refine the relationship between the middle, rightmost, and leftmost elements of the function.