New general integral inequalities for quasi-geometrically convex functions via fractional integrals

Abstract

In this paper, the author introduces the concept of the quasi-geometrically convex functions, gives Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms and defines a new identity for fractional integrals. By using this identity, the author obtains new estimates on generalization of Hadamard et al. type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

MSC:26A33, 26A51, 26D15.

1 Introduction

Let a real function f be defined on some nonempty interval I of a real line ℝ. The function f is said to be convex on I if inequality

$$f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$$

(1)

holds for all $x,y\in I$ and $t\in [0,1]$.

We recall that the notion of quasi-convex function generalizes the notion of convex function. More exactly, a function $f:[a,b]\subset \mathbb{R}\to \mathbb{R}$ is said to be quasi-convex on $[a,b]$ if

$$f(tx+(1-t)y)\le max\{f(x),f(y)\}$$

for all $x,y\in [a,b]$ and $t\in [0,1]$. Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex (see [1]).

The following inequalities are well known in the literature as the Hermite-Hadamard inequality, the Ostrowski inequality and the Simpson inequality, respectively.

Theorem 1.1 Let $f:I\subseteq \mathbb{R}\to \mathbb{R}$ be a convex function defined on the interval I of real numbers, and let $a,b\in I$ with $a<b$. The following double inequality holds:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2}.$$

(2)

Theorem 1.2 Let $f:I\subseteq \mathbb{R}\to \mathbb{R}$ be a mapping differentiable in $I^{\circ }$, the interior of I, and let $a,b\in I^{\circ }$ with $a<b$. If $|f^{\prime }(x)|\le M$, $x\in [a,b]$, then the following inequality holds:

$$|f(x)-\frac{1}{b-a}{\int }_a^{b}f(t)dt|\le \frac{M}{b-a}\left[\frac{{(x-a)}^2+{(b-x)}^2}{2}\right]$$

for all $x\in [a,b]$.

Theorem 1.3 Let $f:[a,b]\to \mathbb{R}$ be a four times continuously differentiable mapping on $(a,b)$ and ${\parallel f^{(4)}\parallel }_{\mathrm{\infty }}={sup}_{x\in (a,b)}|f^{(4)}(x)|<\mathrm{\infty }$. Then the following inequality holds:

$$|\frac{1}{3}[\frac{f(a)+f(b)}{2}+2f\left(\frac{a+b}{2}\right)]-\frac{1}{b-a}{\int }_a^{b}f(x)dx|\le \frac{1}{2\text{,}880}{\parallel f^{(4)}\parallel }_{\mathrm{\infty }}{(b-a)}^4.$$

The following definitions are well known in the literature.

Definition 1.1 ([2, 3])

A function $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ is said to be GA-convex (geometric-arithmetically convex) if

$$f\left(x^t{y}^{1-t}\right)\le tf(x)+(1-t)f(y)$$

for all $x,y\in I$ and $t\in [0,1]$.

Definition 1.2 ([2, 3])

A function $f:I\subseteq (0,\mathrm{\infty })\to (0,\mathrm{\infty })$ is said to be GG-convex (called in [4] a geometrically convex function) if

$$f\left(x^t{y}^{1-t}\right)\le f{(x)}^{t}f{(y)}^{(1-t)}$$

for all $x,y\in I$ and $t\in [0,1]$.

We will now give definitions of the right-hand side and left-hand side Hadamard fractional integrals which are used throughout this paper.

Definition 1.3 Let $f\in L[a,b]$. The right-hand side and left-hand side Hadamard fractional integrals $J_{a^{+}}^{\alpha }f$ and $J_{b^{-}}^{\alpha }f$ of order $\alpha >0$ with $b>a\ge 0$ are defined by

$$J_{a+}^{\alpha }f(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^x{(ln\frac{x}{t})}^{\alpha -1}f(t)\frac{dt}{t},a<x<b$$

and

$$J_{b-}^{\alpha }f(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_x^b{(ln\frac{t}{x})}^{\alpha -1}f(t)\frac{dt}{t},a<x<b,$$

respectively, where $\mathrm{\Gamma }(\alpha )$ is the Gamma function defined by $\mathrm{\Gamma }(\alpha )={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\alpha -1}dt$ (see [5]).

In recent years, many authors have studied error estimations for Hermite-Hadamard, Ostrowski and Simpson inequalities; for refinements, counterparts, generalization see [4, 6–20].

In this paper, the concept of the quasi-geometrically convex function is introduced, Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms are established, and a new identity for Hadamard fractional integrals is defined. By using this identity, author obtains a generalization of Hadamard, Ostrowski and Simpson type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals.

2 Main results

Let $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function on $I^{\circ }$, the interior of I, throughout this section we will take

$$\begin{array}{r}{I}_f(x,\lambda ,\alpha ,a,b)\\ =(1-\lambda )[{ln}^{\alpha }\frac{x}{a}+{ln}^{\alpha }\frac{b}{x}]f(x)+\lambda [f(a){ln}^{\alpha }\frac{x}{a}+f(b){ln}^{\alpha }\frac{b}{x}]\\ -\mathrm{\Gamma }(\alpha +1)[J_{x-}^{\alpha }f(a)+J_{x+}^{\alpha }f(b)],\end{array}$$

where $a,b\in I$ with $a<b$, $x\in [a,b]$, $\lambda \in [0,1]$, $\alpha >0$ and Γ is the Euler Gamma function.

Definition 2.1 A function $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ is said to be quasi-geometrically convex on I if

$$f\left(x^t{y}^{1-t}\right)\le sup\{f(x),f(y)\},$$

for any $x,y\in I$ and $t\in [0,1]$.

Remark 2.1 Clearly, any GA-convex and geometrically convex functions are quasi-geometrically convex functions. Furthermore, there exist quasi-geometrically convex functions which are neither GA-convex nor geometrically convex. In that context, we point out an elementary example. The function $f:(0,4]\to \mathbb{R}$,

$$f(x)=\{\begin{array}{ll}1,& x\in (0,1],\\ {(x-2)}^2,& x\in [1,4]\end{array}$$

is neither GA-convex nor geometrically convex on $(0,4]$, but it is a quasi-geometrically convex function on $(0,4]$.

Proposition 2.1 If $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ is convex and nondecreasing, then it is quasi-geometrically convex on I.

Proof This follows from

$$\begin{array}{rcl}f\left(x^t{y}^{1-t}\right)& \le & f(tx+(1-t)y)\\ \le & tf(x)+(1-t)f(y)\le sup\{f(x),f(y)\},\end{array}$$

for all $x,y\in I$ and $t\in [0,1]$. □

Proposition 2.2 If $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ is quasi-convex and nondecreasing, then it is quasi-geometrically convex on I. If $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ is quasi-geometrically convex and nonincreasing, then it is quasi-convex on I.

Proof These conclusions follows from

$$f\left(x^t{y}^{1-t}\right)\le f(tx+(1-t)y)\le sup\{f(x),f(y)\}$$

and

$$f(tx+(1-t)y)\le f\left(x^t{y}^{1-t}\right)\le sup\{f(x),f(y)\}$$

for all $x,y\in I$ and $t\in [0,1]$, respectively. □

Hermite-Hadamard’s inequalities can be represented for GA-convex functions in fractional integral forms as follows.

Theorem 2.1 Let $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ be a function such that $f\in L[a,b]$, where $a,b\in I$ with $a<b$. If f is a GA-convex function on $[a,b]$, then the following inequalities for fractional integrals hold:

$$f(\sqrt{ab})\le \frac{\mathrm{\Gamma }(\alpha +1)}{2{(ln\frac{b}{a})}^{\alpha }}\{J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)\}\le \frac{f(a)+f(b)}{2}$$

(3)

with $\alpha >0$.

Proof Since f is a GA-convex function on $[a,b]$, we have for all $x,y\in [a,b]$ (with $t=1/2$ in inequality (1)),

$$f(\sqrt{xy})\le \frac{f(x)+f(y)}{2}.$$

Choosing $x=a^t{b}^{1-t}$, $y=b^t{a}^{1-t}$, we get

$$f(\sqrt{ab})\le \frac{f(a^t{b}^{1-t})+f(b^t{a}^{1-t})}{2}.$$

(4)

Multiplying both sides of (4) by $t^{\alpha -1}$, then integrating the resulting inequality with respect to t over $[0,1]$, we obtain

$$\begin{array}{rcl}f(\sqrt{ab})& \le & \frac{\alpha }{2}\{{\int }_0^{1}f\left(a^t{b}^{1-t}\right)dt+{\int }_0^{1}f\left(b^t{a}^{1-t}\right)dt\}\\ =& \frac{\alpha }{2}\{{\int }_a^b{\left(\frac{lnb-lnu}{lnb-lna}\right)}^{\alpha -1}f(u)\frac{du}{uln\frac{b}{a}}+{\int }_a^b{\left(\frac{lnu-lna}{lnb-lna}\right)}^{\alpha -1}f(u)\frac{du}{uln\frac{b}{a}}\}\\ =& \frac{\alpha \mathrm{\Gamma }(\alpha )}{2{(ln\frac{b}{a})}^{\alpha }}\{J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)\}\\ =& \frac{\mathrm{\Gamma }(\alpha +1)}{2{(ln\frac{b}{a})}^{\alpha }}\{J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)\},\end{array}$$

and the first inequality is proved.

For the proof of the second inequality in (3), we first note that if f is a convex function, then for $t\in [0,1]$, it yields

$$f\left(a^t{b}^{1-t}\right)\le tf(a)+(1-t)f(b)$$

and

$$f\left(b^t{a}^{1-t}\right)\le tf(b)+(1-t)f(a).$$

By adding these inequalities, we have

$$f\left(a^t{b}^{1-t}\right)+f\left(b^t{a}^{1-t}\right)\le f(a)+f(b).$$

(5)

Then multiplying both sides of (5) by $t^{\alpha -1}$, and integrating the resulting inequality with respect to t over $[0,1]$, we obtain

$${\int }_0^{1}f\left(a^t{b}^{1-t}\right)t^{\alpha -1}dt+{\int }_0^{1}f\left(b^t{a}^{1-t}\right)t^{\alpha -1}dt\le [f(a)+f(b)]{\int }_0^1{t}^{\alpha -1}dt,$$

i.e.,

$$\frac{\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}\{J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)\}\le f(a)+f(b).$$

The proof is completed. □

In order to prove our main results, we need the following identity.

Lemma 2.1 Let $f:I\subseteq (0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function on $I^{\circ }$ such that $f^{\prime }\in L[a,b]$, where $a,b\in I$ with $a<b$. Then for all $x\in [a,b]$, $\lambda \in [0,1]$ and $\alpha >0$, we have:

$$\begin{array}{rcl}{I}_f(x,\lambda ,\alpha ,a,b)& =& a{(ln\frac{x}{a})}^{\alpha +1}{\int }_0^1(t^{\alpha }-\lambda ){\left(\frac{x}{a}\right)}^t{f}^{\prime }\left(x^t{a}^{1-t}\right)dt\\ -b{(ln\frac{b}{x})}^{\alpha +1}{\int }_0^1(t^{\alpha }-\lambda ){\left(\frac{x}{b}\right)}^t{f}^{\prime }\left(x^t{b}^{1-t}\right)dt.\end{array}$$

(6)

Proof By integration by parts and twice changing the variable, for $x\ne a$, we can state that

$$\begin{array}{r}aln\frac{x}{a}{\int }_0^1(t^{\alpha }-\lambda ){\left(\frac{x}{a}\right)}^t{f}^{\prime }\left(x^t{a}^{1-t}\right)dt\\ ={\int }_0^1(t^{\alpha }-\lambda )df\left(x^t{a}^{1-t}\right)\\ =(t^{\alpha }-\lambda )f\left(x^t{a}^{1-t}\right){|}_0^1-\frac{\alpha }{{(ln\frac{x}{a})}^{\alpha }}{\int }_a^x{(ln\frac{u}{a})}^{\alpha -1}\frac{f(u)}{u}du\\ =(1-\lambda )f(x)+\lambda f(a)-\frac{\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{x}{a})}^{\alpha }}{J}_{x-}^{\alpha }f(a),\end{array}$$

(7)

and for $x\ne b$, similarly, we get

$$\begin{array}{r}-bln\frac{b}{x}{\int }_0^1(t^{\alpha }-\lambda ){\left(\frac{x}{b}\right)}^t{f}^{\prime }\left(x^t{b}^{1-t}\right)dt\\ ={\int }_0^1(t^{\alpha }-\lambda )df\left(x^t{b}^{1-t}\right)\\ =(t^{\alpha }-\lambda )f\left(x^t{b}^{1-t}\right){|}_0^1-\frac{\alpha }{{(ln\frac{b}{x})}^{\alpha }}{\int }_x^b{(ln\frac{b}{u})}^{\alpha -1}\frac{f(u)}{u}du\\ =(1-\lambda )f(x)+\lambda f(b)-\frac{\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{x})}^{\alpha }}{J}_{x+}^{\alpha }f(b).\end{array}$$

(8)

Multiplying both sides of (7) and (8) by ${(ln\frac{x}{a})}^{\alpha }$ and ${(ln\frac{b}{x})}^{\alpha }$, respectively, and adding the resulting identities, we obtain the desired result. For $x=a$ and $x=b$, the identities

$$I_f(a,\lambda ,\alpha ;a,b)=b{(ln\frac{b}{a})}^{\alpha +1}{\int }_0^1(t^{\alpha }-\lambda ){\left(\frac{a}{b}\right)}^t{f}^{\prime }\left(a^t{b}^{1-t}\right)dt,$$

and

$$I_f(b,\lambda ,\alpha ;a,b)=a{(ln\frac{b}{a})}^{\alpha +1}{\int }_0^1(t^{\alpha }-\lambda ){\left(\frac{b}{a}\right)}^t{f}^{\prime }\left(b^t{a}^{1-t}\right),$$

can be proved respectively easily by performing an integration by parts in the integrals from the right-hand side and changing the variable. □

Theorem 2.2 Let $f:I\subset (0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function on $I^{\circ }$ such that $f^{\prime }\in L[a,b]$, where $a,b\in I^{\circ }$ with $a<b$. If ${|f^{\prime }|}^q$ is quasi-geometrically convex on $[a,b]$ for some fixed $q\ge 1$, $x\in [a,b]$, $\lambda \in [0,1]$ and $\alpha >0$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|I_f(x,\lambda ,\alpha ,a,b)|\\ \le A_1^{1-\frac{1}{q}}(\alpha ,\lambda )\{a{(ln\frac{x}{a})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{B}_1^{\frac{1}{q}}(x,\alpha ,\lambda ,q)\\ +b{(ln\frac{b}{x})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{B}_2^{\frac{1}{q}}(x,\alpha ,\lambda ,q)\},\end{array}$$

(9)

where

$$\begin{array}{c}{A}_1(\alpha ,\lambda )=\frac{2\alpha {\lambda }^{1+\frac{1}{\alpha }}+1}{\alpha +1}-\lambda ,\hfill \\ B_1(x,\alpha ,\lambda ,q)={\int }_0^1|t^{\alpha }-\lambda |{\left(\frac{x}{a}\right)}^{qt}dt,\hfill \\ B_2(x,\alpha ,\lambda ,q)={\int }_0^1|t^{\alpha }-\lambda |{\left(\frac{x}{b}\right)}^{qt}dt.\hfill \end{array}$$

Proof Since $|f^{\prime }{|}^q$ is quasi-geometrically convex on $[a,b]$, for all $t\in [0,1]$,

$$|f^{\prime }\left(x^t{a}^{1-t}\right){|}^q\le sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\}$$

and

$$|f^{\prime }\left(x^t{b}^{1-t}\right){|}^q\le sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\}.$$

Hence, using Lemma 2.1 and power mean inequality, we get

$$\begin{array}{r}|I_f(x,\lambda ,\alpha ,a,b)|\\ \le a{(ln\frac{x}{a})}^{\alpha +1}{\left({\int }_0^1\right|t^{\alpha }-\lambda \left|dt\right)}^{1-\frac{1}{q}}{\left({\int }_0^1\right|t^{\alpha }-\lambda |{\left(\frac{x}{a}\right)}^{qt}sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\}dt)}^{\frac{1}{q}}\\ +b{(ln\frac{b}{x})}^{\alpha +1}{\left({\int }_0^1\right|t^{\alpha }-\lambda \left|dt\right)}^{1-\frac{1}{q}}{\left({\int }_0^1\right|t^{\alpha }-\lambda |{\left(\frac{x}{b}\right)}^{qt}sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\}dt)}^{\frac{1}{q}},\\ |I_f(x,\lambda ,\alpha ,a,b)|\le {\left({\int }_0^1\right|t^{\alpha }-\lambda \left|dt\right)}^{1-\frac{1}{q}}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|\le }\times \{a{(ln\frac{x}{a})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{\left({\int }_0^1\right|t^{\alpha }-\lambda \left|{\left(\frac{x}{a}\right)}^{qt}dt\right)}^{\frac{1}{q}}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|\le }+b{(ln\frac{b}{x})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{\left({\int }_0^1\right|t^{\alpha }-\lambda \left|{\left(\frac{x}{b}\right)}^{qt}dt\right)}^{\frac{1}{q}}\}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|}\le A_1^{1-\frac{1}{q}}(\alpha ,\lambda )\{a{(ln\frac{x}{a})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{B}_1^{\frac{1}{q}}(x,\alpha ,\lambda ,q)\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|\le }+b{(ln\frac{b}{x})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{B}_2^{\frac{1}{q}}(x,\alpha ,\lambda ,q)\},\end{array}$$

which completes the proof. □

Corollary 2.1 Under the assumptions of Theorem  2.2 with $q=1$, inequality (9) reduces to the following inequality:

$$\begin{array}{rcl}|I_f(x,\lambda ,\alpha ,a,b)|& \le & \{a{(ln\frac{x}{a})}^{\alpha +1}{B}_1(x,\alpha ,\lambda ,1)sup\left\{\right|f^{\prime }(x)|,|f^{\prime }(a)\left|\right\}\\ +b{(ln\frac{b}{x})}^{\alpha +1}{B}_2(x,\alpha ,\lambda ,1)sup\left\{\right|f^{\prime }(x)|,|f^{\prime }(b)\left|\right\}\}.\end{array}$$

Corollary 2.2 Under the assumptions of Theorem  2.2 with $\alpha =1$, inequality (9) reduces to the following inequality:

$$\begin{array}{r}{(ln\frac{b}{a})}^{-1}|I_f(x,\lambda ,\alpha ,a,b)|\\ \le |(1-\lambda )f(x)+\lambda \left[\frac{f(a)ln\frac{x}{a}+f(b)ln\frac{b}{x}}{ln\frac{b}{a}}\right]-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le {(ln\frac{b}{a})}^{-1}{\left(\frac{2{\lambda }^2-2\lambda +1}{2}\right)}^{1-\frac{1}{q}}\{a{(ln\frac{x}{a})}^2{B}_1^{\frac{1}{q}}(x,1,\lambda ,q){(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}\\ +b{(ln\frac{b}{x})}^2{B}_2^{\frac{1}{q}}(x,1,\lambda ,q){(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}\},\end{array}$$

where

$$\begin{array}{c}{B}_1(x,1,\lambda ,q)=h_{\lambda }\left({\left(\frac{x}{a}\right)}^q\right),B_2(x,1,\lambda ,q)=h_{\lambda }\left({\left(\frac{x}{b}\right)}^q\right),\hfill \\ h(u,\lambda )=\frac{2u^{\lambda }-u-1}{{(lnu)}^2}+\frac{(1-\lambda )u-\lambda }{lnu},u\in (0,\mathrm{\infty })\mathrm{\setminus }\{1\},\hfill \end{array}$$

(10)

specially for $x=\sqrt{ab}$, we get

$$\begin{array}{r}|(1-\lambda )f(\sqrt{ab})+\lambda \left(\frac{f(a)+f(b)}{2}\right)-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{ln\frac{b}{a}}{4}{\left(\frac{2{\lambda }^2-2\lambda +1}{2}\right)}^{1-\frac{1}{q}}\{a{h}^{\frac{1}{q}}({\left(\frac{b}{a}\right)}^{\frac{q}{2}},\lambda ){(sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}\\ +b{h}^{\frac{1}{q}}({\left(\frac{a}{b}\right)}^{\frac{q}{2}},\lambda ){(sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}\}.\end{array}$$

(11)

Corollary 2.3 In Theorem  2.2,

• 1. If we take $x=\sqrt{ab}$, $\lambda =\frac{1}{3}$, then we get the following Simpson-type inequality for fractional integrals:

  $$\begin{array}{r}|\frac{1}{6}[f(a)+4f(\sqrt{ab})+f(b)]-\frac{2^{\alpha -1}\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{ln\frac{b}{a}}{4}{A}_1^{1-\frac{1}{q}}(\alpha ,\frac{1}{3})\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{B}_1^{\frac{1}{q}}(\sqrt{ab},\alpha ,\frac{1}{3},q)\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{B}_2^{\frac{1}{q}}(\sqrt{ab},\alpha ,\frac{1}{3},q)\},\end{array}$$

specially for $\alpha =1$, we get

$$\begin{array}{r}|\frac{1}{6}[f(a)+4f(\sqrt{ab})+f(b)]-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{ln\frac{b}{a}}{4}{\left(\frac{5}{18}\right)}^{1-\frac{1}{q}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab})|,|f^{\prime }(a)\left|\right\}]}^{\frac{1}{q}}{h}^{\frac{1}{q}}({\left(\frac{b}{a}\right)}^{\frac{q}{2}},\frac{1}{3})\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{h}^{\frac{1}{q}}({\left(\frac{a}{b}\right)}^{\frac{q}{2}},\frac{1}{3})\},\end{array}$$

where h is defined as in (10).

Remark 2.2

• 1. If we take $x=\sqrt{ab}$, $\lambda =0$, then we get the following midpoint- type inequality for fractional integrals:

  $$\begin{array}{r}|f(\sqrt{ab})-\frac{2^{\alpha -1}\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{ln\frac{b}{a}}{4}{\left(\frac{1}{\alpha +1}\right)}^{1-\frac{1}{q}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{B}_1^{\frac{1}{q}}(\sqrt{ab},1,0,q)\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{B}_2^{\frac{1}{q}}(\sqrt{ab},1,0,q)\},\end{array}$$

specially for $\alpha =1$, we get

$$\begin{array}{r}|f(\sqrt{ab})-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{2^{\frac{1}{q}}ln\frac{b}{a}}{8}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{h}^{\frac{1}{q}}({\left(\frac{b}{a}\right)}^{\frac{q}{2}},0)\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{h}^{\frac{1}{q}}({\left(\frac{a}{b}\right)}^{\frac{q}{2}},0)\},\end{array}$$

where h is defined as in (10).

• 2. If we take $x=\sqrt{ab}$, $\lambda =1$, then we get the following trapezoid-type inequality for fractional integrals:

  $$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{2^{\alpha -1}\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{ln\frac{b}{a}}{4}{\left(\frac{1}{\alpha +1}\right)}^{1-\frac{1}{q}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{B}_1^{\frac{1}{q}}(\sqrt{ab},\alpha ,1,q)\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{B}_2^{\frac{1}{q}}(\sqrt{ab},\alpha ,1,q)\},\end{array}$$

specially for $\alpha =1$, we get

$$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{2^{\frac{1}{q}}ln\frac{b}{a}}{8}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{h}^{\frac{1}{q}}({\left(\frac{b}{a}\right)}^{\frac{q}{2}},1)\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{h}^{\frac{1}{q}}({\left(\frac{a}{b}\right)}^{\frac{q}{2}},1)\},\end{array}$$

where h is defined as in (10).

Corollary 2.4 Let the assumptions of Theorem  2.2 hold. If $|f^{\prime }(x)|\le M$ for all $x\in [a,b]$ and $\lambda =0$, then we get the following Ostrowski-type inequality for fractional integrals from inequality (9):

$$\begin{array}{r}|[{(ln\frac{x}{a})}^{\alpha }+{(ln\frac{b}{x})}^{\alpha }]f(x)-\mathrm{\Gamma }(\alpha +1)[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{M}{{(\alpha +1)}^{1-\frac{1}{q}}}[a{(ln\frac{x}{a})}^{\alpha +1}{B}_1^{\frac{1}{q}}(x,\alpha ,0,q)+b{(ln\frac{b}{x})}^{\alpha +1}{B}_2^{\frac{1}{q}}(x,\alpha ,0,q)].\end{array}$$

Theorem 2.3 Let $f:I\subset (0,\mathrm{\infty })\to \mathbb{R}$ be a differentiable function on $I^{\circ }$ such that $f^{\prime }\in L[a,b]$, where $a,b\in I^{\circ }$ with $a<b$. If ${|f^{\prime }|}^q$ is quasi-geometrically convex on $[a,b]$ for some fixed $q>1$, $x\in [a,b]$, $\lambda \in [0,1]$ and $\alpha >0$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|I_f(x,\lambda ,\alpha ,a,b)|\\ \le A_2^{\frac{1}{p}}(\alpha ,\lambda ,p)\{a{(ln\frac{x}{a})}^{\alpha +\frac{1}{p}}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{\left(\frac{x^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{(ln\frac{b}{x})}^{\alpha +\frac{1}{p}}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{\left(\frac{b^q-x^q}{q}\right)}^{\frac{1}{q}}\},\end{array}$$

(12)

where

$$\begin{array}{r}{A}_2(\alpha ,\lambda ,p)\\ =\{\begin{array}{ll}\frac{1}{\alpha p+1},& \lambda =0,\\ \frac{{\lambda }^{\frac{\alpha p+1}{\alpha }}}{\alpha }\{\beta (\frac{1}{\alpha },p+1)+\frac{{(1-\lambda )}^{p+1}}{p+1}\\ {\times }_2{F}_1(\frac{1}{\alpha }+p+1,p+1,p+2;1-\lambda )\},& 0<\lambda <1,\\ \frac{1}{\alpha }\beta (p+1,\frac{1}{\alpha }),& \lambda =1,\end{array}\end{array}$$

${}_{2}F_1$ is hypergeometric function defined by

$${}_{2}F_1(a,b;c;z)=\frac{1}{\beta (b,c-b)}{\int }_0^1{t}^{b-1}{(1-t)}^{c-b-1}{(1-zt)}^{-a}dt,c>b>0,|z|<1(\mathit{\text{see }}\text{[21]}),$$

and $\frac{1}{p}+\frac{1}{q}=1$.

Proof Using Lemma 2.1, the Hölder inequality and quasi-geometrical convexity of ${|f^{\prime }|}^q$, we get

$$\begin{array}{r}|I_f(x,\lambda ,\alpha ,a,b)|\\ \le a{(ln\frac{x}{a})}^{\alpha +1}{\left({\int }_0^1\right|t^{\alpha }-\lambda {|}^{p}dt)}^{\frac{1}{p}}{({\int }_0^1{\left(\frac{x}{a}\right)}^{qt}sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\}dt)}^{\frac{1}{q}}\\ +b{(ln\frac{b}{x})}^{\alpha +1}{\left({\int }_0^1\right|t^{\alpha }-\lambda {|}^{p}dt)}^{\frac{1}{p}}{({\int }_0^1{\left(\frac{x}{b}\right)}^{qt}sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\}dt)}^{\frac{1}{q}},\\ |I_f(x,\lambda ,\alpha ,a,b)|\le {\left({\int }_0^1\right|t^{\alpha }-\lambda {|}^{p}dt)}^{\frac{1}{p}}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|\le }\times \{a{(ln\frac{x}{a})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{\left({\int }_0^1{\left(\frac{x}{a}\right)}^{qt}dt\right)}^{\frac{1}{q}}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|\le }+b{(ln\frac{b}{x})}^{\alpha +1}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{\left({\int }_0^1{\left(\frac{x}{b}\right)}^{qt}dt\right)}^{\frac{1}{q}}\}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|}\le A_2^{\frac{1}{p}}(\alpha ,\lambda ,p)\{a{(ln\frac{x}{a})}^{\alpha +1-\frac{1}{q}}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{\left(\frac{x^q-a^q}{q}\right)}^{\frac{1}{q}}\\ \phantom{|I_f(x,\lambda ,\alpha ,a,b)|\le }+b{(ln\frac{b}{x})}^{\alpha +1-\frac{1}{q}}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{\left(\frac{b^q-x^q}{q}\right)}^{\frac{1}{q}}\},\end{array}$$

here, it is seen by a simple computation that

$$\begin{array}{rcl}{A}_2(\alpha ,\lambda ,p)& =& {\int }_0^1|t^{\alpha }-\lambda {|}^{p}dt\\ =& \{\begin{array}{ll}\frac{1}{\alpha p+1},& \lambda =0,\\ \frac{{\lambda }^{\frac{\alpha p+1}{\alpha }}}{\alpha }\{\beta (\frac{1}{\alpha },p+1)+\frac{{(1-\lambda )}^{p+1}}{p+1}\\ {\times }_2{F}_1(\frac{1}{\alpha }+p+1,p+1,2+p;1-\lambda )\},& 0<\lambda <1,\\ \frac{1}{\alpha }\beta (p+1,\frac{1}{\alpha }),& \lambda =1.\end{array}\end{array}$$

Hence, the proof is completed. □

Corollary 2.5 Under the assumptions of Theorem  2.3 with $\alpha =1$, inequality (12) reduces to the following inequality:

$$\begin{array}{r}|(1-\lambda )f(x)+\lambda \left[\frac{f(a)ln\frac{x}{a}+f(b)ln\frac{b}{x}}{ln\frac{b}{a}}\right]-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le {(ln\frac{b}{a})}^{-1}{\left(\frac{{\lambda }^{p+1}+{(1-\lambda )}^{p+1}}{p+1}\right)}^{\frac{1}{p}}\\ \times \{a{(ln\frac{x}{a})}^{1+\frac{1}{p}}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{\left(\frac{x^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{(ln\frac{b}{x})}^{1+\frac{1}{p}}{(sup\left\{\right|f^{\prime }(x){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{\left(\frac{b^q-x^q}{q}\right)}^{\frac{1}{q}}\},\end{array}$$

specially for $x=\sqrt{ab}$, we get

$$\begin{array}{r}|(1-\lambda )f(\sqrt{ab})+\lambda \left(\frac{f(a)+f(b)}{2}\right)-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}({\lambda }^{p+1}+{(1-\lambda )}^{p+1})}{2(p+1)}\right)}^{\frac{1}{p}}\{a{(sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\})}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{(sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\})}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\}.\end{array}$$

(13)

Corollary 2.6 In Theorem  2.3,

• 1. If we take $x=\sqrt{ab}$, $\lambda =\frac{1}{3}$, then we get the following Simpson-type inequality for fractional integrals:

  $$\begin{array}{r}|\frac{1}{6}[f(a)+4f(\sqrt{ab})+f(b)]-\frac{2^{\alpha -1}\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}(1+2^{p+1})}{3^{p+1}(p+1)2}\right)}^{\frac{1}{p}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\},\end{array}$$

specially for $\alpha =1$, we get

$$\begin{array}{r}|\frac{1}{6}[f(a)+4f(\sqrt{ab})+f(b)]-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}(1+2^{p+1})}{3^{p+1}(p+1)2}\right)}^{\frac{1}{p}}\\ \times \{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\}.\end{array}$$

Remark 2.3

• 1. If we take $x=\sqrt{ab}$, $\lambda =0$, then we get the following midpoint- type inequality for fractional integrals:

  $$\begin{array}{r}|f(\sqrt{ab})-\frac{2^{\alpha -1}\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}}{2(\alpha p+1)}\right)}^{\frac{1}{p}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\},\end{array}$$

specially for $\alpha =1$, we get

$$\begin{array}{r}|f(\sqrt{ab})-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}}{2(p+1)}\right)}^{\frac{1}{p}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\}.\end{array}$$

• 2. If we take $x=\sqrt{ab}$, $\lambda =1$, then we get the following trapezoid-type inequality for fractional integrals $\frac{1}{\alpha }\beta (p+1,\frac{1}{\alpha })$:

  $$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{2^{\alpha -1}\mathrm{\Gamma }(\alpha +1)}{{(ln\frac{b}{a})}^{\alpha }}[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}\beta (p+1,\frac{1}{\alpha })}{2\alpha }\right)}^{\frac{1}{p}}\\ \times \{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\},\end{array}$$

specially for $\alpha =1$, we get

$$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{1}{ln\frac{b}{a}}{\int }_a^b\frac{f(u)}{u}du|\\ \le \frac{1}{2}{\left(\frac{ln\frac{b}{a}}{2(p+1)}\right)}^{\frac{1}{p}}\{a{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(a){|}^q\}]}^{\frac{1}{q}}{\left(\frac{{\sqrt{ab}}^q-a^q}{q}\right)}^{\frac{1}{q}}\\ +b{[sup\left\{\right|f^{\prime }(\sqrt{ab}){|}^q,|f^{\prime }(b){|}^q\}]}^{\frac{1}{q}}{\left(\frac{b^q-{\sqrt{ab}}^q}{q}\right)}^{\frac{1}{q}}\}.\end{array}$$

Corollary 2.7 Let the assumptions of Theorem  2.3 hold. If $|f^{\prime }(x)|\le M$ for all $x\in [a,b]$ and $\lambda =0$, then we get the following Ostrowski-type inequality for fractional integrals from inequality (12):

$$\begin{array}{r}|[{(ln\frac{x}{a})}^{\alpha }+{(ln\frac{b}{x})}^{\alpha }]f(x)-\mathrm{\Gamma }(\alpha +1)[J_{\sqrt{ab}-}^{\alpha }f(a)+J_{\sqrt{ab}+}^{\alpha }f(b)]|\\ \le \frac{M}{{(\alpha p+1)}^{\frac{1}{p}}}[a{(ln\frac{x}{a})}^{\alpha +\frac{1}{p}}{\left(\frac{x^q-a^q}{q}\right)}^{\frac{1}{q}}+b{(ln\frac{b}{x})}^{\alpha +\frac{1}{p}}{\left(\frac{b^q-x^q}{q}\right)}^{\frac{1}{q}}].\end{array}$$

3 Application to special means

Let us recall the following special means of two nonnegative numbers a, b with $b>a$:

• 1. The arithmetic mean

  $$A=A(a,b):=\frac{a+b}{2}.$$

• 2. The geometric mean

  $$G=G(a,b):=\sqrt{ab}.$$

• 3. The logarithmic mean

  $$L=L(a,b):=\frac{b-a}{lnb-lna}.$$

• 4. The p-logarithmic mean

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

Proposition 3.1 For $b>a>0$, $n>0$ and $q\ge 1$, we have

$$\begin{array}{r}|(1-\lambda )G^{n+1}(a,b)+\lambda A(a^{n+1},b^{n+1})-(n+1)L(a,b)L_n^n(a,b)|\\ \le \frac{(n+1)ln\frac{b}{a}}{4}{\left(\frac{2{\lambda }^2-2\lambda +1}{2}\right)}^{1-\frac{1}{q}}\{a{G}^n(a,b)h^{\frac{1}{q}}({\left(\frac{b}{a}\right)}^{\frac{q}{2}},\lambda )\\ +b^{n+1}{h}^{\frac{1}{q}}({\left(\frac{a}{b}\right)}^{\frac{q}{2}},\lambda )\},\end{array}$$

where h is defined as in (10).

Proof Let $f(x)=\frac{x^{n+1}}{n+1}$, $x>0$, $n>0$ and $q\ge 1$. Then the function $|f^{\prime }(x){|}^q=x^{nq}$ is quasi-geometrically convex on $(0,\mathrm{\infty })$. Thus, by inequality (11), Proposition 3.1 is proved. □

Proposition 3.2 For $b>a>0$, $n>0$ and $q>1$, we have

$$\begin{array}{r}|(1-\lambda )G^{n+1}(a,b)+\lambda A(a^{n+1},b^{n+1})-(n+1)L(a,b)L_n^n(a,b)|\\ \le \frac{n+1}{2}{\left(\frac{ln\frac{b}{a}({\lambda }^{p+1}+{(1-\lambda )}^{p+1})}{2(p+1)}\right)}^{\frac{1}{p}}\{a{G}^n(a,b){\left(\frac{G^q(a,b)-a^q}{q}\right)}^{\frac{1}{q}}\\ +b^{n+1}{\left(\frac{b^q-G^q(a,b)}{q}\right)}^{\frac{1}{q}}\}.\end{array}$$

Proof Let $f(x)=\frac{x^{n+1}}{n+1}$, $x>0$, $n>0$ and $q>1$. Then the function $|f^{\prime }(x){|}^q=x^{nq}$ is quasi-geometrically convex on $(0,\mathrm{\infty })$. Thus, by inequality (13), Proposition 3.2 is proved. □