Hermite–Hadamard-type inequalities for functions whose derivatives are η-convex via fractional integrals

Abstract

In the present research, we develop some integral inequalities of Hermite–Hadamard type for differentiable η-convex functions. Moreover, our results include several new and known results as particular cases.

1 Introduction

Throughout this paper, let I be an interval in $\mathbb{R}$. Also consider $\eta: A\times A \rightarrow B$ for appropriate $A, B \subseteq\mathbb{R}$.

Let $f:I \subseteq\mathbb{R} \rightarrow\mathbb{R}$ be a convex function, and let $a_{1}$, ${a_{2}} \in I$ with $a_{1}< {a_{2}}$. The following double inequality

$$ f \biggl(\frac{a_{1} + {a_{2}}}{2} \biggr) \leq\frac{1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)\,dx \leq\frac{f(a_{1}) + f({a_{2}})}{2} $$

(1)

is known in the literature as the Hadamard inequality for convex functions. Fejer [1] gave a generalization of (1) as follows. If $f:[a_{1}, {a_{2}}] \rightarrow\mathbb{R}$ is a convex function and $g:[a_{1}, {a_{2}}] \rightarrow\mathbb{R}$ is nonnegative, integrable, and symmetric about $\frac{a_{1} + {a_{2}}}{2}$, then

$$ f \biggl(\frac{a_{1} + {a_{2}}}{2} \biggr) \int_{a_{1}}^{a_{2}} g(x)\,dx \leq \int_{a_{1}}^{a_{2}} f(x)g(x)\,dx \leq\frac{f(a_{1}) + f({a_{2}})}{2} \int _{a_{1}}^{a_{2}} g(x)\,dx. $$

(2)

Since the Hermite–Hadamard inequality and fractional integrals have a wide range of applications, many researchers extend their studies to Hermite–Hadamard-type inequalities involving fractional integrals.

In 2015, Iscan [2] obtained Hermite–Hadamard–Fejér-type inequalities for convex functions via fractional integrals. In 2017, Farid and Tariq [3] developed fractional integral inequalities for m-convex functions. Also, Farid and Abbas [4] established Hermite–Hadamard–Fejér-type inequalities for p-convex functions via generalized fractional integrals. For recent generalizations, we refer to [5,6,7], and [8].

Xi and Qi [9], Ozdemir et al. [10], and Sarikaya et al. [5] established Hermite–Hadamard-type inequalities for convex functions. Gordji et al. [11] introduced an important generalization of convexity known as η-convexity.

Definition 1.1

([11])

A function $f: I \rightarrow\mathbb{R}$ is called η-convex if

$$ f\bigl(\alpha x + (1 - \alpha)y\bigr) \leq f(y) + \alpha\eta \bigl(f(x) , f(y)\bigr) $$

(3)

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

Theorem 1.1

([10])

Let $f:I\subseteq[0, \infty) \rightarrow\mathbb{R}$ be a differentiable mapping on the interior $I^{\circ}$ of I such that $f'' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}$, ${a_{2}} \in I$ with ${a_{1}}< {a_{2}}$. If $|f|$ is convex on $[{a_{1}}, {a_{2}}]$, then

$$\begin{aligned} & \biggl\vert f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{192} \biggl\{ \bigl\vert f''({a_{1}}) \bigr\vert + 6 \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert + \bigl\vert f''({a_{2}}) \bigr\vert \biggr\} . \end{aligned}$$

(4)

Theorem 1.2

([10])

Let $f:I\subseteq[0, \infty) \rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$ such that $f'' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}$, ${a_{2}} \in I$ with ${a_{1}}< {a_{2}}$. If $|f''|^{q}$ for $q \geq1$ is convex on $[{a_{1}}, {a_{2}}]$, then

$$\begin{aligned} & \biggl\vert f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{48} \biggl(\frac{3}{4} \biggr)^{\frac {1}{q}} \biggl\{ \biggl(\frac{ \vert f''({a_{1}}) \vert ^{q}}{3} + \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} \biggr)^{\frac{1}{q}} \\ &\qquad {}+ \biggl( \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} + \frac{ \vert f''({a_{2}}) \vert ^{q}}{3} \biggr)^{\frac{1}{q}}\biggr\} . \end{aligned}$$

(5)

Lemma 1.1

([9])

Let $f:I\subseteq\mathbb{R} \rightarrow\mathbb{R}$ be a differentiable function on $I^{\circ}$ such that $f' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}$, ${a_{2}} \in I$ with ${a_{1}}< {a_{2}}$. If α, $\beta\in\mathbb{R}$, then

$$\begin{aligned} &\frac{\alpha f({a_{1}}) + \beta f({a_{2}})}{2} + \frac{2 - \alpha- \beta}{2} f\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int _{a_{1}}^{a_{2}} f(x)\,dx \\ &\quad = \frac{{a_{2}} - {a_{1}}}{4} \int_{0}^{1} \biggl[(1 - \alpha- t) f' \biggl(t{a_{1}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2} \biggr) \\ &\qquad {}+ (\beta- t)f' \biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 -t){a_{2}} \biggr) \biggr]\,dt. \end{aligned}$$

(6)

Lemma 1.2

([9])

For $s> 0$ and $0 \leq\epsilon\leq1$, we have

$$\begin{aligned} \begin{aligned} & \int_{0}^{1} |\epsilon- t|^{s} \,dt = \frac{\epsilon^{s + 1} + (1 - \epsilon)^{s + 1}}{s + 1}, \\ & \int_{0}^{1} t|\epsilon- t|^{s} \,dt = \frac{\epsilon^{s + 2} + (s + 1 + \epsilon)(1 - \epsilon)^{s + 1}}{(s + 1)(s + 2)}. \end{aligned} \end{aligned}$$

(7)

The paper is organized as follows. In Sect. 2, we establish Hermite–Hadamard- and Fejer-type inequalities for η-convex functions. In the last section, we derive Fractional integral inequalities for η-convex functions.

2 Hermite–Hadamard- and Fejer-type inequalities

Theorem 2.1

Let $f:I\subseteq\mathbb{R} \rightarrow\mathbb{R}$ be an η-convex function with $f \in L^{1}[a_{1}, {a_{2}}]$, where $a_{1}$, ${a_{2}} \in I$ with $a_{1}< {a_{2}}$,Then

$$\begin{aligned} &f \biggl(\frac{a_{1} + {a_{2}}}{2} \biggr) - \frac{1}{2({a_{2}} - a_{1})} \int_{a_{1}}^{a_{2}} \eta\bigl(f(a_{1} + {a_{2}} - x) , f(x)\bigr) \,dx \\ &\quad \leq\frac{1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)\,dx \leq f({a_{2}}) + \frac{1}{2}\eta\bigl(f(a_{1}) , f({a_{2}}) \bigr). \end{aligned}$$

(8)

Proof

According to (3), with $x = ta_{1} + (1 - t){a_{2}}$, $y = (1 - t)a_{1} + t{a_{2}}$, and $\alpha= \frac{1}{2}$, where $t \in[0, 1]$, we find that

$$ f \biggl(\frac{a_{1} + {a_{2}}}{2} \biggr) \leq f\bigl((1 - t)a_{1} + t{a_{2}}\bigr) + \frac{1}{2} \eta\bigl(f\bigl(ta_{1} + (1 - t){a_{2}}\bigr) , f\bigl((1 - t)a_{1} + t{a_{2}}\bigr)\bigr). $$

Thus by integrating we obtain

$$\begin{aligned} f \biggl(\frac{a_{1} + {a_{2}}}{2} \biggr) &\leq \int_{0}^{1}f\bigl((1 - t)a_{1} + t{a_{2}}\bigr)\,dt \\ &\quad {}+ \frac{1}{2} \int_{0}^{1}\eta\bigl(f\bigl(ta_{1} + (1 - t){a_{2}}\bigr) , f\bigl((1 - t)a_{1} + t{a_{2}} \bigr)\bigr)\,dt \\ &\leq\frac{1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)\,dx + \frac {1}{2({a_{2}} - a_{1})} \int_{a_{1}}^{a_{2}}\eta\bigl(f(a_{1} + {a_{2}} - x) , f(x)\bigr)\,dx, \end{aligned}$$

so that

$$ f \biggl(\frac{a_{1} + {a_{2}}}{2} \biggr) - \frac{1}{2({a_{2}} - a_{1})} \int_{a_{1}}^{a_{2}}\eta\bigl(f(a_{1} + {a_{2}} - x) , f(x)\bigr)\,dx \leq\frac {1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)\,dx, $$

(9)

and the first inequality is proved. Taking $x = a_{1}$ and $y = {a_{2}}$ in (3), we get

$$ f\bigl(\alpha a_{1} + (1 - \alpha)a_{2}\bigr) \leq f(a_{2}) + \alpha\eta\bigl(f(a_{1}), f(a_{2}) \bigr). $$

Integrating this inequality with respect to α over $[0, 1]$, we get

$$ \frac{1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)\,dx \leq f({a_{2}}) + \frac {1}{2} \eta\bigl(f(a_{1}) , f({a_{2}})\bigr). $$

(10)

Clearly, (9) and (10) yield (8). □

Remark 2.1

Taking $\eta(x , y) = x - y$, we reduce (8) to inequality (1).

Theorem 2.2

Let f and g be nonnegative η-convex functions with $fg \in L^{1}[a_{1}, {a_{2}}]$, where $a_{1}, {a_{2}} \in I$, $a_{1}< {a_{2}}$. Then

$$ \frac{1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)g(x)\,dx \leq M'(a_{1} , {a_{2}}), $$

(11)

where

$$\begin{aligned} M'(a_{1} , {a_{2}}) &= f({a_{2}})g({a_{2}}) + \frac{1}{2}f({a_{2}})\eta\bigl(g(a_{1}) , g({a_{2}})\bigr) + \frac{1}{2}g({a_{2}})\eta \bigl(f(a_{1}) , f({a_{2}})\bigr) \\ &\quad {}+ \frac{1}{3}\eta\bigl(f(a_{1}) , f({a_{2}}) \bigr) \eta\bigl(g(a_{1}) , g({a_{2}})\bigr). \end{aligned}$$

Proof

Since f and g are η-convex functions, we have

$$\begin{aligned}& f\bigl(ta_{1} + (1 - t){a_{2}}\bigr) \leq f({a_{2}}) + t \eta\bigl(f(a_{1}) , f({a_{2}}) \bigr), \\& g\bigl(ta_{1} + (1 - t){a_{2}}\bigr) \leq g({a_{2}}) + t \eta\bigl(g(a_{1}) , g({a_{2}}) \bigr) \end{aligned}$$

for all $t \in[0, 1]$. Since f and g are nonnegative, we have

$$\begin{aligned}& f\bigl(ta_{1} + (1 - t){a_{2}}\bigr) g\bigl(ta_{1} + (1 - t){a_{2}}\bigr) \\& \quad \leq f({a_{2}})g({a_{2}}) + tf({a_{2}})\eta \bigl(g(a_{1}) , g({a_{2}})\bigr) \\& \qquad {} + tg({a_{2}})\eta\bigl(f(a_{1}) , f({a_{2}})\bigr) + t^{2} \eta\bigl(f(a_{1}) , f({a_{2}})\bigr) \eta\bigl(g(a_{1}) , g({a_{2}}) \bigr). \end{aligned}$$

Integrating both sides of the inequality over $[0, 1]$, we obtain

$$\begin{aligned}& \int_{0}^{1} f\bigl(ta_{1} + (1 - t){a_{2}}\bigr) g\bigl(ta_{1} + (1 - t){a_{2}} \bigr) \,dt \\& \quad \leq f({a_{2}})g({a_{2}}) + \frac{1}{2}f({a_{2}}) \eta\bigl(g(a_{1}) , g({a_{2}})\bigr) + \frac{1}{2}g({a_{2}}) \eta\bigl(f(a_{1}) , f({a_{2}})\bigr) \\& \qquad {} + \frac{1}{3}\eta\bigl(f(a_{1}) , f({a_{2}})\bigr) \eta\bigl(g(a_{1}) , g({a_{2}}) \bigr). \end{aligned}$$

Then

$$ \frac{1}{{a_{2}} - a_{1}} \int_{a_{1}}^{a_{2}} f(x)g(x)\,dx \leq M'(a_{1} , {a_{2}}). $$

 □

Remark 2.2

By taking $\eta(x , y) = x - y$ inequality (11) becomes inequality (1.4) in [5].

Theorem 2.3

Let f be an η-convex function with $f \in L^{1}[a_{1}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$, $a_{1}< {a_{2}}$, and let $g: [a_{1}, {a_{2}}] \rightarrow\mathbb{R}$ be nonnegative, integrable, and symmetric about $\frac{(a_{1} + {a_{2}})}{2}$. Then

$$ \int_{a_{1}}^{a_{2}} f(y)g(y)\,dy \leq\biggl[f({a_{2}}) + \frac{1}{2}\eta\bigl(f(a_{1}), f({a_{2}})\bigr) \biggr] \int_{a_{1}}^{a_{2}} g(y)\,dy. $$

(12)

Proof

Since, f is an η-convex function and g is nonnegative, integrable. and symmetric about $\frac{(a_{1} + {a_{2}})}{2}$, we find that

$$\begin{aligned} \int_{a_{1}}^{a_{2}} f(y)g(y)\,dy & = \frac{1}{2} \biggl[ \int_{a_{1}}^{a_{2}} f(y)g(y)\,dy + \int_{a_{1}}^{a_{2}} f(a_{1} + {a_{2}} - y)g(a_{1} + {a_{2}} - y)\,dy \biggr] \\ &=\frac{1}{2} \int_{a_{1}}^{a_{2}} \bigl[ \bigl(f(y) + f(a_{1} + {a_{2}} - y) \bigr)g(y)\,dy \bigr] \\ &=\frac{1}{2} \int_{a_{1}}^{a_{2}} \biggl[f \biggl(\frac{{a_{2}} - y}{{a_{2}} - a_{1}} a_{1} + \frac{y - a_{1}}{{a_{2}} - a_{1}} {a_{2}} \biggr) \\ &\quad {}+ f \biggl( \frac {y - a_{1}}{{a_{2}} - a_{1}} a_{1} + \frac{{a_{2}} - y}{{a_{2}} - a_{1}} {a_{2}} \biggr) \biggr]g(y)\,dy \\ &\leq\frac{1}{2} \int_{a_{1}}^{a_{2}} \biggl[ \biggl(f({a_{2}}) + \frac {{a_{2}} - y}{{a_{2}} - a_{1}}\eta\bigl(f(a_{1}), f({a_{2}})\bigr) \biggr) \\ &\quad {}+ \biggl(f({a_{2}}) + \frac{y - a_{1}}{{a_{2}} - a_{1}}\eta \bigl(f(a_{1}), f({a_{2}})\bigr) \biggr) \biggr]g(y)\,dy \\ &\leq\biggl[f({a_{2}}) + \frac{1}{2}\eta\bigl(f(a_{1}), f({a_{2}})\bigr)\biggr] \int_{a_{1}}^{a_{2}} g(y)\,dy. \end{aligned}$$

 □

Remark 2.3

If we choose $\eta(x, y) = x - y$ and $g(x) = 1$, then (12) reduces to the second inequality in (1), and if we take $\eta (x, y) = x - y$, then (12) reduces to the second inequality in (2).

3 Fractional integral inequalities

Theorem 3.1

Let $f:I \rightarrow\mathbb{R}$, $I\subseteq\mathbb{R}$ be a differentiable mapping on $I^{0}$ with $f' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}$, ${a_{2}} \in I$, ${a_{1}}< {a_{2}}$. If $|f' (x)|^{q}$ for $q \geq1$ is η-convex on $[{a_{1}}, {a_{2}}]$ and $0 \leq\alpha$, $\beta\leq1$, then

$$\begin{aligned}& \biggl\vert \frac{\alpha f({a_{1}}) + \beta f({a_{2}})}{2} +\frac{2 - \alpha - \beta}{2} f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{8} \biggl(\frac{1}{6} \biggr)^{\frac {1}{q}}\bigl\{ \bigl(1 - 2\alpha+ 2\alpha^{2}\bigr)^{1 - \frac{1}{q}} \bigl[ \bigl(6 - 12\alpha+ 12\alpha^{2} \bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl(4 - 9\alpha+ 12\alpha^{2} - 2\alpha^{3} \bigr) \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr]^{\frac{1}{q}} \\& \qquad {} + \bigl(1 - 2\beta+ 2\beta^{2} \bigr)^{1 - \frac{1}{q}} \bigl[ \bigl(6 - 12\beta+ 12\beta^{2} \bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl(2 - 3\beta+ 2\beta^{3} \bigr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(13)

Proof

For $q> 1$, by Lemma 1.1, the η-convexity of $|f' (x)|^{q}$ on $[{a_{1}}, {a_{2}}]$, and the Hölder integral inequality, we have

$$\begin{aligned}& \biggl\vert \frac{\alpha f({a_{1}}) + \beta f({a_{2}})}{2} + \frac{2 - \alpha - \beta}{2} f\biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl[ \int_{0}^{1} \vert 1 - \alpha- t \vert \biggl\vert f' \biggl(t{a_{1}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert \,dt \\& \qquad {} + \int_{0}^{1} \vert \beta- t \vert \biggl\vert f' \biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 -t){a_{2}} \biggr) \biggr\vert \,dt\biggr] \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl\{ \biggl( \int_{0}^{1} \vert 1 - \alpha- t \vert \,dt \biggr)^{1 - \frac{1}{q}} \biggl[ \int_{0}^{1} \vert 1 - \alpha- t \vert \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \biggl(\frac{1 + t}{2}\biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \,dt \biggr]^{\frac{1}{q}} +\biggl( \int_{0}^{1} \vert \beta- t \vert \,dt \biggr)^{1 - \frac {1}{q}} \\& \qquad {} \times\biggl[ \int_{0}^{1} \vert \beta- t \vert \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl( \frac{t}{2}\biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert \bigr)\biggr) \,dt\biggr]^{\frac {1}{q}}\biggr\} . \end{aligned}$$

(14)

Using Lemma 1.2, by a direct calculation we get

$$\begin{aligned}& \int_{0}^{1} \vert 1 - \alpha- t \vert \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac{1 + t}{2} \biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \,dt \\& \quad = \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \frac{1}{2} \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \int_{0}^{1} \vert 1 - \alpha- t \vert \,dt \\& \qquad {} + \frac{1}{2} \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \int_{0}^{1} t \vert 1 - \alpha- t \vert \,dt \\& \quad = \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \frac{1}{2} \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \biggl(\frac{1}{2} - \alpha+ \alpha^{2} \biggr) \\& \qquad {} + \frac{1}{12} \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigl[(1 - \alpha)^{3} + \alpha^{2}(3 - \alpha)\bigr] \\& \quad = \frac{1}{2} \bigl(1 - 2 \alpha+ 2 \alpha^{2}\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \frac {1}{12} \bigl(4 - 9\alpha+ 12\alpha^{2} - 2 \alpha^{3}\bigr) \\& \qquad {} \times\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \end{aligned}$$

and

$$\begin{aligned}& \int_{0}^{1} \vert \beta- t \vert \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl( \frac{t}{2}\biggr) \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \,dt \\& \quad = \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \int_{0}^{1} \vert \beta- t \vert \,dt + \frac{1}{2} \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \int_{0}^{1} t \vert \beta- t \vert \,dt \\& \quad = \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \biggl(\frac{1}{2} - \beta- \beta^{2} \biggr) + \frac {1}{12} \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigl(\beta^{3} + (2 + \beta) (1 - \beta)^{2}\bigr) \\& \quad =\frac{1}{2} \bigl(1 - 2\beta+ 2\beta^{2}\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \frac{1}{12} \bigl(2 - 3\beta+ 2\beta^{3}\bigr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr). \end{aligned}$$

Substituting these two inequalities into inequality (14) and using Lemma 1.2 result in inequality (13) for $q> 1$.

For $q = 1$, from Lemmas 1.1 and 1.2 it follows that

$$\begin{aligned}& \biggl\vert \frac{\alpha f({a_{1}}) + \beta f({a_{2}})}{2} + \frac{2 - \alpha - \beta}{2} f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl\{ \int_{0}^{1} \vert 1 - \alpha- t \vert \biggl( \bigl\vert f'({a_{2}}) \bigr\vert + \biggl( \frac{1 + t}{2} \biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert , \bigl\vert f'({a_{2}}) \bigr\vert \bigr) \biggr)\,dt \\& \qquad {} + \int_{0}^{1} \vert \beta- t \vert \biggl( \bigl\vert f'({a_{2}}) \bigr\vert + \frac{t}{2} \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert , \bigl\vert f'({a_{2}}) \bigr\vert \bigr) \biggr)\,dt\biggr\} \\& \quad = \frac{{a_{2}} - {a_{1}}}{48} \bigl\{ \bigl(6 - 12\alpha+ 12\alpha ^{2} \bigr) \bigl\vert f'({a_{2}}) \bigr\vert + \bigl(4 - 9 \alpha+ 12\alpha^{2} - 2\alpha ^{3} \bigr) \\& \qquad {} \times\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert , \bigl\vert f'({a_{2}}) \bigr\vert \bigr) + \bigl(6 - 12\beta+ 12\beta ^{2} \bigr) \bigl\vert f'({a_{2}}) \bigr\vert \\& \qquad {} + \bigl(2 - 3\beta+ 2\beta^{3} \bigr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert , \bigl\vert f'({a_{2}}) \bigr\vert \bigr)\bigr\} . \end{aligned}$$

(15)

 □

Remark 3.1

If we take $\eta(x , y) = x - y$, then inequality (13) reduces to inequality (3.1) in [9].

Taking $\alpha= \beta$ in Theorem 3.1, we derive the following corollary.

Corollary 3.1

Let $f:I \rightarrow\mathbb{R}$, $I\subseteq\mathbb{R}$ be a differentiable mapping on $I^{0}$ with $f'\in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$, ${a_{1}}< {a_{2}}$. If $|f' (x)|^{q}$ for $q \geq1$ is η-convex on $[{a_{1}}, {a_{2}}]$ and $0 \leq\alpha\leq1$, then

$$\begin{aligned}& \biggl\vert \frac{\alpha}{2}\bigl[f({a_{1}}) + f({a_{2}})\bigr] + (1 - \alpha) f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int _{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{8} \biggl(\frac{1}{6} \biggr)^{\frac {1}{q}}\bigl(1 - 2\alpha+ 2\alpha^{2}\bigr)^{1 - \frac{1}{q}} \bigl[ \bigl(6 - 12\alpha+ 12\alpha^{2}\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl(4 - 9\alpha+ 12\alpha^{2} - 2\alpha^{3} \bigr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}} \\& \qquad {} + \bigl[\bigl(6 - 12\alpha+ 12\alpha^{2}\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \bigl(2 - 3 \alpha +2\alpha^{3}\bigr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}}. \end{aligned}$$

(16)

Remark 3.2

If we take $\eta(x , y) = x - y$, then inequality (16) reduces to inequality (3.5) in [9].

By choosing $\alpha= \beta= \frac{1}{2}, \frac{1}{3}$, respectively, in Theorem 3.1 we can deduce the following inequalities.

Corollary 3.2

Let $f:I \rightarrow\mathbb{R}$, $I\subseteq\mathbb{R}$, be a differentiable mapping on $I^{0}$ with $f' \in L^{1}[{a_{1}} , {a_{2}}]$, where ${a_{1}}, {a_{2}} \in$ I, ${a_{1}}<{a_{2}}$. If $|f' (x)|^{q}$ for $q \geq1$ is η-convex on $[{a_{1}} , {a_{2}}]$ and $0 \leq\alpha, \beta\leq1$, then

$$\begin{aligned} \begin{aligned} & \biggl\vert \frac{1}{2} \biggl[ \frac{f({a_{1}}) + f({a_{2}})}{2} + f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr] - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{{a_{2}} - {a_{1}}}{16} \biggl(\frac{1}{12} \biggr)^{\frac {1}{q}} \bigl\{ \bigl[12 \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + 9\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}} \\ &\qquad {} + \bigl[12 \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + 3\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac {1}{q}}\bigr\} , \\ & \biggl\vert \frac{1}{6} \biggl[f({a_{1}}) + f({a_{2}}) + 4f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr] - \frac{1}{{a_{2}} - {a_{1}}} \int _{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{5({a_{2}} - {a_{1}})}{72} \biggl(\frac{1}{90} \biggr)^{\frac {1}{q}}\bigl\{ \bigl[90 \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + 61\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}} \\ &\qquad {} + \bigl[90 \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + 29\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac {1}{q}}\bigr\} . \end{aligned} \end{aligned}$$

(17)

Setting $q = 1$ in Corollary 3.2, we have the following:

Corollary 3.3

Let $f:I \rightarrow\mathbb{R}$, $I\subseteq\mathbb{R}$, be a differentiable mapping on $I^{0}$ with $f' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$, ${a_{1}}<{a_{2}}$. If $|f' (x)|$ is η-convex on $[{a_{1}}, {a_{2}}]$, then

$$\begin{aligned} \begin{aligned} & \biggl\vert \frac{1}{2} \biggl[ \frac{f({a_{1}}) + f({a_{2}})}{2} + f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr)\biggr] - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{{a_{2}} - {a_{1}}}{16} \bigl[2 \bigl\vert f'({a_{2}}) \bigr\vert + \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert , \bigl\vert f'({a_{2}}) \bigr\vert \bigr) \bigr], \\ & \biggl\vert \frac{1}{6} \biggl[f({a_{1}}) + f({a_{2}}) + 4f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr)\biggr] - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{5({a_{2}} - {a_{1}})}{72} \bigl[2 \bigl\vert f'({a_{2}}) \bigr\vert + \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert , \bigl\vert f'({a_{2}}) \bigr\vert \bigr) \bigr]. \end{aligned} \end{aligned}$$

(18)

Remark 3.3

If we take $\eta(x , y) = x - y$, then inequalities (17) and (18) reduce to inequalities (3.6) and (3.7) in [9].

Theorem 3.2

Let $f:I \rightarrow\mathbb{R}$, $I\subseteq\mathbb{R}$, be a differentiable mapping on $I^{0}$ with $f' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in$ I, ${a_{1}}<{a_{2}}$. If $|f' (x)|^{q}$ for $q \geq1$ is η-convex on $[{a_{1}}, {a_{2}}]$ and $0 \leq\alpha, \beta\leq1$, then

$$\begin{aligned}& \biggl\vert \frac{\alpha f({a_{1}}) + \beta f({a_{2}})}{2} + \frac{2 - \alpha- \beta}{2} f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac {1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl[\frac{1}{2(q + 1)(q + 2)} \biggr]^{\frac{1}{q}} \\& \qquad {} \times\bigl\{ \bigl[\bigl(\bigl[2(q + 2) (1 - \alpha)^{q + 1}+ 2(q + 2) \alpha^{q + 1}\bigr]\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl[(q + 3 - \alpha) (1 - \alpha)^{q + 1} + (2q + 4 - \alpha)\alpha ^{q + 1}\bigr] \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}} \\& \qquad {} + \bigl[\bigl(2(q + 2) (1 - \beta)^{q + 1} + 2(q + 2) \beta^{q + 1}\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl(\beta^{q + 2} + (q + 1 + \beta) (1 - \beta)^{q + 1}\bigr) \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}}\bigr\} . \end{aligned}$$

(19)

Proof

For $q> 1$, by the η-convexity of $|f'(x)|^{q}$ on $[{a_{1}}, {a_{2}}]$ and Hölder’s integral inequality it follows that

$$\begin{aligned}& \biggl\vert \frac{\alpha f({a_{1}}) + \beta f({a_{2}})}{2} + \frac{2 - \alpha - \beta}{2} f\biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl[ \int_{0}^{1} \vert 1 - \alpha- t \vert \biggl\vert f' \biggl(t{a_{1}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert \,dt \\& \qquad {} + \int_{0}^{1} \vert \beta- t \vert \biggl\vert f' \biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 -t){a_{2}} \biggr) \biggr\vert \,dt\biggr] \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl\{ \biggl( \int_{0}^{1} \,dt\biggr)^{1 - \frac{1}{q}} \biggl[ \int_{0}^{1} \vert 1 - \alpha- t \vert ^{q} \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac{1 + t}{2}\biggr) \\& \qquad {} \times\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\biggr) \,dt \biggr]^{\frac {1}{q}}+ \biggl( \int_{0}^{1} \,dt\biggr)^{1 - \frac{1}{q}} \biggl[ \int_{0}^{1} \vert \beta- t \vert ^{q} \\& \qquad {} \times\biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac{t}{2}\biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert \bigr)\biggr) \,dt \biggr]^{\frac{1}{q}}\biggr\} \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl\{ \biggl[ \int_{0}^{1} \vert 1 - \alpha- t \vert ^{q} \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac{1 + t}{2}\biggr) \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\biggr)\,dt \biggr]^{\frac{1}{q}} \\& \qquad {} + \biggl[ \int_{0}^{1} \vert \beta- t \vert ^{q} \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac {t}{2}\biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert \bigr)\biggr)\,dt \biggr]^{\frac {1}{q}}\biggr\} . \end{aligned}$$

(20)

By Lemma 1.2 we have

$$\begin{aligned}& \int_{0}^{1} \vert 1 - \alpha- t \vert ^{q} \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac{1 + t}{2} \biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr)\,dt \\& \quad = \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \frac{1}{2}\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \int_{0}^{1} \vert 1 - \alpha- t \vert ^{q} \,dt \\& \qquad {} + \frac{1}{2}\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \int_{0}^{1} t \vert 1 - \alpha- t \vert ^{q} \,dt \\& \quad = \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \frac{1}{2}\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr) \biggl(\frac{(1 - \alpha)^{q + 1} + \alpha^{q + 1}}{q + 1} \biggr) \\& \qquad {} + \frac{1}{2}\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggl( \frac{(1 - \alpha)^{q +2} + (q + 2 - \alpha) \alpha^{q + 1}}{(q + 1)(q + 2)} \biggr) \\& \quad = \frac{1}{2(q + 1)(q + 2)} \bigl[2(q + 2) (1 - \alpha)^{q + 1} + 2(q + 2) \alpha^{q + 1} \bigr] \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl[2(q + 2) (1 - \alpha)^{q + 1} + (q + 2) \alpha^{q + 1} + (1 - \alpha)^{q +2} + (q + 2 - \alpha) \alpha^{q + 1} \bigr] \\& \qquad {} \times\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \\& \quad = \frac{1}{2(q + 1)(q + 2)} \bigl[2(q + 2) (1 - \alpha)^{q + 1} + 2(q + 2) \alpha^{q + 1} \bigr] \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl[(q + 3 - \alpha) (1 - \alpha)^{q +1} + (2q + 4 - \alpha )\alpha^{q + 1} \bigr]\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \end{aligned}$$

and

$$\begin{aligned}& \int_{0}^{1} \vert \beta- t \vert ^{q} \biggl( \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \biggl(\frac {t}{2} \biggr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \biggr)\,dt \\& \quad = \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \int_{0}^{1} \vert \beta- t \vert ^{q} \,dt + \frac{1}{2}\eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \int_{0}^{1} t \vert \beta- t \vert ^{q} \,dt \\& \quad = \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \biggl(\frac{\beta^{q + 1} + (1 - \beta)^{q + 1}}{q + 1} \biggr) + \frac{1}{2}\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \\& \qquad {} \times \biggl(\frac{\beta^{q + 2} + (q + 1 + \beta)(1 - \beta)^{q + 1}}{(q + 1)(q + 2)} \biggr) \\& \quad = \frac{1}{2(q + 1)(q + 2)} \bigl\{ \bigl[2(q + 2) (1 - \beta)^{q + 1} + 2(q + 2)\beta^{q + 1} \bigr] \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl[\beta^{q + 2}+ (q + 1 + \beta) (1 - \beta)^{q + 1} \bigr] \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr\} . \end{aligned}$$

Substituting the last two equalities into inequality (20) yields inequality (19) for $q> 1$.

For $q = 1$, the proof is the same as that of (15), and the theorem is proved. □

Remark 3.4

If we take $\eta(x , y) = x - y$, then inequality (19) reduces to inequality (3.8) in [9].

Similarly to corollaries of Theorem 3.1, we can obtain the following corollaries of Theorem 3.2.

Corollary 3.4

Let $f:I\subseteq\mathbb{R} \rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$ with $f' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$, ${a_{1}}< {a_{2}}$. If $|f' (x)|^{q}$ for $q \geq1$ is η-convex on $[{a_{1}}, {a_{2}}]$ and $0 \leq\alpha\leq 1$, then

$$\begin{aligned}& \biggl\vert \frac{\alpha}{2}\bigl[f({a_{1}}) + f({a_{2}})\bigr] + (1 - \alpha) f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} - {a_{1}}} \int _{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl[\frac{1}{2(q + 1)(q + 2)} \biggr]^{\frac{1}{q}} \\& \qquad {} \times\bigl\{ \bigl[ \bigl(2(q + 2) (1 - \alpha)^{q + 1} + 2(q + 2) \alpha^{q + 1} \bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl((q + 3 - \alpha) (1 - \alpha)^{q + 1} + (2q + 4 - \alpha )\alpha^{q + 1}\bigr) \eta\bigl( \bigl\vert f'(a) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr]^{\frac {1}{q}} \\& \qquad {} +\bigl[ \bigl(2(q + 2) (1 - \alpha)^{q + 1} + 2(q + 2) \alpha^{q + 1} \bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\& \qquad {} + \bigl(\alpha^{q + 2} + (q + 1 + \alpha) (1 - \alpha)^{q + 1} \bigr) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr)\bigr]^{\frac {1}{q}}\bigr\} . \end{aligned}$$

(21)

Remark 3.5

If we take $\eta(x , y) = x - y$, then inequality (21) reduces to inequality (3.11) in [9].

Corollary 3.5

Let $f:I\subseteq\mathbb{R} \rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$ with $f' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$, ${a_{1}}< {a_{2}}$. If $|f' (x)|^{q}$ for $q \geq1$ is η-convex on $[{a_{1}}, {a_{2}}]$ and $0 \leq\alpha$, $\beta\leq1$, then

$$\begin{aligned} \begin{aligned} & \biggl\vert \frac{1}{2} \biggl[\frac{f({a_{1}}) + f({a_{2}})}{2} + f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr] - \frac{1}{{a_{2}} - {a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{{a_{2}} - {a_{1}}}{8} \biggl[\frac{1}{4(q + 1)(q + 2)} \biggr]^{\frac{1}{q}} \\ &\qquad {} \times\bigl\{ \bigl[ \bigl((4q + 8) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + (3q + 6) \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr) \bigr]^{\frac{1}{q}} \\ &\qquad {} + \bigl[ \bigl((4q + 8) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + (q + 2) \eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr) \bigr]^{\frac{1}{q}}\bigr\} , \\ & \biggl\vert \frac{1}{6} \biggl[f({a_{1}}) + f({a_{2}}) + 4f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr] - \frac{1}{{a_{2}} - {a_{1}}} \int _{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\ &\quad \leq\frac{{a_{2}} - {a_{1}}}{12} \biggl[\frac{1}{18(q + 1)(q + 2)} \biggr]^{\frac{1}{q}} \bigl\{ \bigl[ \bigl((3q + 6) (2)^{q + 2} + 6(q + 2) \bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \\ &\qquad {} + \bigl((3q + 8) (2)^{q + 1} + (6q + 11)\eta\bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q}\bigr) \bigr)\bigr]^{\frac{1}{q}}+\bigl[\bigl((3q + 6) (2)^{q + 2} \\ &\qquad {} + 6(q + 2)\bigr) \bigl\vert f'({a_{2}}) \bigr\vert ^{q} + \bigl(1 + (3q + 4) (2)^{q + 1} \bigr) \eta \bigl( \bigl\vert f'({a_{1}}) \bigr\vert ^{q} , \bigl\vert f'({a_{2}}) \bigr\vert ^{q} \bigr)\bigr]^{\frac{1}{q}}\bigr\} . \end{aligned} \end{aligned}$$

(22)

Remark 3.6

If we take $\eta(x , y) = x - y$, then inequality (22) reduces to inequality (3.12) in [9] respectively.

If we take $q = 1$ in Corollary 3.5, then we get Corollary 3.3.

To prove our next results, we consider the following lemma proved in [10].

Lemma 3.1

Let $f:I \rightarrow\mathbb{R}$, $I\subseteq\mathbb{R}$ be a differentiable mapping on $I^{0}$ with $f'' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in$ I and ${a_{1}}<{a_{2}}$. Then

$$\begin{aligned}& \frac{1}{{a_{2}} -{a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx - f \biggl(\frac {{a_{1}} + {a_{2}}}{2} \biggr) \\& \quad = \frac{({a_{2}} - {a_{1}})^{2}}{16} \biggl[ \int_{0}^{1} t^{2}f'' \biggl(t\frac {{a_{1}} + {a_{2}}}{2} + (1 - t){a_{1}} \biggr)\,dt \\& \qquad {} + \int_{0}^{1} (t - 1)^{2} f'' \biggl(t{a_{2}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2} \biggr)\,dt \biggr]. \end{aligned}$$

(23)

Theorem 3.3

Let $f : I \subset[0, \infty) \rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$ with $f'' \in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$ and ${a_{1}}< {a_{2}}$. If $|f''|$ is η-convex on $[{a_{1}}, {a_{2}}]$, then

$$\begin{aligned}& \biggl\vert f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} -{a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16} \biggl[\frac{1}{3} \biggl( \bigl\vert f''({a_{1}}) \bigr\vert + \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert \biggr) \\& \qquad {} +\frac{1}{4} \biggl(\eta \biggl( \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert , \bigl\vert f''({a_{1}}) \bigr\vert \biggr) + \frac{1}{3} \eta \biggl( \bigl\vert f''({a_{2}}) \bigr\vert , \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert \biggr) \biggr) \biggr]. \end{aligned}$$

(24)

Proof

From Lemma 3.1 we have

$$\begin{aligned}& \biggl\vert f\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) - \frac{1}{{a_{2}} -{a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl[ \int_{0}^{1} t^{2} \biggl\vert f'' \biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 - t){a_{1}}\biggr) \biggr\vert \,dt \\& \qquad {} + \int_{0}^{1} (t - 1)^{2} \biggl\vert f''\biggl(t{a_{2}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert \,dt\biggr] \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl[ \int_{0}^{1} t^{2} \biggl( \bigl\vert f''({a_{1}}) \bigr\vert + t\eta\biggl( \biggl\vert f''\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert , \bigl\vert f''({a_{1}}) \bigr\vert \biggr)\biggr)\,dt \biggr] \\& \qquad {} + \frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl[ \int_{0}^{1}(t - 1)^{2}\biggl( \biggl\vert f''\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert \\& \qquad {}+ t\eta\biggl( \bigl\vert f''({a_{2}}) \bigr\vert , \biggl\vert f''\biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert \biggr)\biggr)\,dt\biggr] \\& \quad = \frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl[\frac{1}{3} \bigl\vert f''({a_{1}}) \bigr\vert + \frac {1}{3} \biggl\vert f''\biggl( \frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert + \frac {1}{4}\eta\biggl( \biggl\vert f''\biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert , \bigl\vert f''({a_{1}}) \bigr\vert \biggr) \\& \qquad {} + \frac{1}{12}\eta\biggl( \bigl\vert f''({a_{2}}) \bigr\vert , \biggl\vert f''\biggl( \frac {{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert \biggr)\biggr] \\& \quad = \frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl[\frac{1}{3} \biggl( \bigl\vert f''({a_{1}}) \bigr\vert + \biggl\vert f''\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert \biggr) \\& \qquad {} +\frac{1}{4}\biggl(\eta\biggl( \biggl\vert f''\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert , \bigl\vert f''({a_{1}}) \bigr\vert \biggr)+\frac{1}{3} \eta \biggl( \bigl\vert f''({a_{2}}) \bigr\vert , \biggl\vert f''\biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert \biggr)\biggr)\biggr]. \end{aligned}$$

This proves inequality (24). □

Remark 3.7

If we take $\eta(x, y) = x - y$, then inequality (24) reduces to inequality (4).

Theorem 3.4

Let $f : I \subset[0, \infty) \rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$ with $f''\in L^{1}[{a_{1}}, {a_{2}}]$, where ${a_{1}}, {a_{2}} \in I$ and ${a_{1}}< {a_{2}}$. If $|f''|^{q}$ for $q \geq1$ with $\frac{1}{p} + \frac{1}{q}= 1$ is η-convex on $[{a_{1}}, {a_{2}}]$, then

$$\begin{aligned}& \biggl\vert f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} -{a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16} \biggl(\frac{1}{3} \biggr)^{\frac {1}{p}} \biggl[ \biggl(\frac{1}{3} \bigl\vert f''({a_{1}}) \bigr\vert ^{q} + \frac{1}{4}\eta \biggl( \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q}, \bigl\vert f''({a_{1}}) \bigr\vert ^{q} \biggr) \biggr)^{\frac{1}{q}} \\& \qquad {} + \biggl(\frac{1}{3} \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} + \frac{1}{12}\eta \biggl( \bigl\vert f''({a_{2}}) \bigr\vert ^{q} , \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} \biggr) \biggr)^{\frac{1}{q}}\biggr]. \end{aligned}$$

(25)

Proof

Suppose that $p \geq1$. From Lemma 3.1, using the power mean inequality, we have

$$\begin{aligned}& \biggl\vert f\biggl(\frac{{a_{1}} + {a_{2}}}{2}\biggr) - \frac{1}{{a_{2}} -{a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl[ \int_{0}^{1} t^{2} \biggl\vert f'' \biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 - t){a_{1}}\biggr) \biggr\vert \,dt \\& \qquad {} + \int_{0}^{1} (t - 1)^{2} \biggl\vert f''\biggl(t{a_{2}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2}\biggr) \biggr\vert \,dt\biggr] \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl( \int_{0}^{1} t^{2} \,dt \biggr)^{\frac{1}{p}}\biggl( \int_{0}^{1} t^{2} \biggl\vert f''\biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 - t){a_{1}}\biggr) \biggr\vert ^{q} \,dt\biggr)^{\frac {1}{q}} \\& \qquad {} + \frac{({a_{2}} - {a_{1}})^{2}}{16}\biggl( \int_{0}^{1} (t - 1)^{2} \,dt \biggr)^{\frac{1}{p}}\biggl( \int_{0}^{1}(t - 1)^{2} \biggl\vert f''\biggl(t{a_{2}} + (1 - t)\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} \,dt\biggr)^{\frac {1}{q}}. \end{aligned}$$

Because $|f''|^{q}$ is η-convex, we have

$$\begin{aligned}& \int_{0}^{1} t^{2} \biggl\vert f'' \biggl(t\frac{{a_{1}} + {a_{2}}}{2} + (1 - t){a_{1}} \biggr) \biggr\vert ^{q} \,dt \\& \quad \leq\frac{1}{3} \bigl\vert f''({a_{1}}) \bigr\vert ^{q} + \frac{1}{4} \biggl(\eta \biggl( \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q}, \bigl\vert f''({a_{1}}) \bigr\vert ^{q} \biggr) \biggr) \end{aligned}$$

and

$$\begin{aligned}& \int_{0}^{1}(t - 1)^{2} \biggl\vert f'' \biggl(t{a_{2}} + (1 - t) \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} \,dt \\& \quad \leq\frac{1}{3} \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} + \frac{1}{12}\eta \biggl( \bigl\vert f''({a_{2}}) \bigr\vert ^{q} , \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} \biggr). \end{aligned}$$

Therefore we have

$$\begin{aligned}& \biggl\vert f \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) - \frac{1}{{a_{2}} -{a_{1}}} \int_{a_{1}}^{a_{2}} f(x)\,dx \biggr\vert \\& \quad \leq\frac{({a_{2}} - {a_{1}})^{2}}{16} \biggl(\frac{1}{3} \biggr)^{\frac {1}{p}} \biggl\{ \biggl(\frac{1}{3} \bigl\vert f''({a_{1}}) \bigr\vert ^{q} + \frac{1}{4}\eta \biggl( \biggl\vert f'' \biggl(\frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q}, \bigl\vert f''({a_{1}}) \bigr\vert ^{q} \biggr) \biggr)^{\frac{1}{q}} \\& \qquad {}+ \frac{1}{3} \biggl\vert f'' \biggl( \frac{{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} + \frac{1}{12}\eta \biggl( \bigl\vert f''({a_{2}}) \bigr\vert ^{q} , \biggl\vert f'' \biggl( \frac {{a_{1}} + {a_{2}}}{2} \biggr) \biggr\vert ^{q} \biggr)^{\frac{1}{q}} \biggr\} . \end{aligned}$$

 □

Remark 3.8

If we take $\eta(x, y) = x - y$, then inequality (25) reduces to inequality (5).

4 Application to means

For two positive numbers $a_{1} > 0$ and $a_{2} > 0$, define

$$ \begin{aligned} &A(a_{1}, a_{2}) = \frac{a_{1} + a_{2}}{2}, \qquad G(a_{1}, a_{2}) = \sqrt{a_{1}a_{2}}, \qquad H(a_{1}, a_{2}) = \frac{2a_{1}a_{2}}{a_{1} + a_{2}}, \\ &L(a_{1}, a_{2}) = \textstyle\begin{cases} [\frac{a_{2}^{s+1} - a_{1}^{s+1}}{(s + 1)(a_{2} - a_{1})} ]^{\frac {1}{s}}, & a_{1} \neq a_{2}, \\ a_{1}, & a_{1} = a_{2}, \end{cases}\displaystyle \\ &I(a_{1}, a_{2}) = \textstyle\begin{cases} \frac{1}{e} (\frac{a_{2}^{a_{2}}}{a_{1}^{a_{1}}} )^{\frac{1}{a_{2} - a_{1}}}, & a_{1} \neq a_{2}, \\ a_{1}, & a_{1} = a_{2}, \end{cases}\displaystyle \\ &H_{w,s}(a_{1}, a_{2}) = \textstyle\begin{cases} [\frac{a_{1}^{s} + w(a_{1}a_{2})^{\frac{s}{2}} + a_{2}^{s}}{w + 2} ]^{\frac{1}{s}}, & s\neq0, \\ \sqrt{a_{1}a_{2}}, & s = 0, \end{cases}\displaystyle \end{aligned} $$

(26)

for $0 \leq w < \infty$. These means are respectively called the arithmetic, geometric, harmonic, generalized logarithmic, identric, and Heronian means of two positive numbers $a_{1}$ and $a_{2}$.

Applying Theorems 3.1 and 3.2 to $f(x) = x^{s}$ for $s \neq0$ and $x > 0$ results in the following inequalities for means.

Theorem 4.1

Let $a_{1} > 0$, $a_{2} > 0$, $a_{1} \neq a_{2}$, $q\geq1$, and either $s> 1$ and $(s -1)q \geq1$ or $s < 0$. Then

$$\begin{aligned}& \biggl\vert A\bigl(\alpha a_{1}^{s}, \beta a_{2}^{s}\bigr) + \frac{2 - \alpha- \beta}{2} A^{s}(a_{1}, a_{2}) - L^{s}(a_{1}, a_{2}) \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{8} \biggl(\frac{1}{6} \biggr)^{\frac {1}{q}}\bigl\{ \bigl(1 - 2\alpha+ 2\alpha^{2}\bigr)^{1 - \frac{1}{q}} \bigl[ \bigl(6 - 12\alpha+ 12\alpha^{2} \bigr) \bigl\vert sa_{2}^{s - 1} \bigr\vert ^{q} \\& \qquad {} + \bigl(4 - 9\alpha+ 12\alpha^{2} - 2\alpha^{3} \bigr) \eta\bigl( \bigl\vert sa_{1}^{s - 1} \bigr\vert ^{q} , \bigl\vert sa_{2}^{s - 1} \bigr\vert ^{q}\bigr) \bigr]^{\frac{1}{q}} \\& \qquad {} + \bigl(1 - 2\beta+ 2\beta^{2} \bigr)^{1 - \frac{1}{q}} \bigl[ \bigl(6 - 12\beta+ 12\beta^{2} \bigr) \bigl\vert sa_{2}^{s - 1} \bigr\vert ^{q} \\& \qquad {} + \bigl(2 - 3\beta+ 2\beta^{3} \bigr) \eta\bigl( \bigl\vert sa_{1}^{s - 1} \bigr\vert ^{q} , \bigl\vert sa_{2}^{s - 1} \bigr\vert ^{q}\bigr) \bigr]^{\frac{1}{q}} \bigr\} . \end{aligned}$$

(27)

Theorem 4.2

Let $a_{1} > 0$, $a_{2} > 0$, $a_{1} \neq a_{2}$, $q\geq1$, and either $s> 1$ and $(s -1)q \geq1$ or $s < 0$. Then

$$\begin{aligned}& \biggl\vert A\bigl(\alpha a_{1}^{s}, \beta a_{2}^{s}\bigr) + \frac{2 - \alpha- \beta}{2} A^{s}(a_{1}, a_{2}) - L^{s}(a_{1}, a_{2}) \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl[\frac{1}{2(q + 1)(q + 2)} \biggr]^{\frac{1}{q}} \\& \qquad {} \times\bigl\{ \bigl[\bigl(\bigl[2(q + 2) (1 - \alpha)^{q + 1}+ 2(q + 2) \alpha^{q + 1}\bigr]\bigr) \bigl\vert sa_{2}^{s-1} \bigr\vert ^{q} \\& \qquad {} + \bigl[(q + 3 - \alpha) (1 - \alpha)^{q + 1} + (2q + 4 - \alpha)\alpha ^{q + 1}\bigr] \eta\bigl( \bigl\vert sa_{1}^{s-1} \bigr\vert ^{q} , \bigl\vert sa_{2}^{s-1} \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}} \\& \qquad {} + \bigl[\bigl(2(q + 2) (1 - \beta)^{q + 1} + 2(q + 2) \beta^{q + 1}\bigr) \bigl\vert sa_{2}^{s-1} \bigr\vert ^{q} \\& \qquad {} + \bigl(\beta^{q + 2} + (q + 1 + \beta) (1 - \beta)^{q + 1}\bigr) \eta \bigl( \bigl\vert sa_{1}^{s-1} \bigr\vert ^{q} , \bigl\vert sa_{2}^{s-1} \bigr\vert ^{q}\bigr)\bigr]^{\frac{1}{q}}\bigr\} . \end{aligned}$$

(28)

Taking $f(x) = \ln x$ for $x>0$ in Theorems 3.1 and 3.2 results in the following inequalities for means.

Theorem 4.3

For $a_{1} > 0$, $a_{2} > 0$, $a_{1} \neq a_{2}$ and $q\geq1$, we have

$$\begin{aligned}& \biggl\vert \frac{\ln G^{2}(a_{1}^{\alpha}, a_{2}^{\beta})}{2} + \frac{2 - \alpha - \beta}{2}\ln A(a_{1}, a_{2}) - \ln I(a_{1}, a_{2}) \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{8} \biggl(\frac{1}{6} \biggr)^{\frac {1}{q}}\biggl\{ \bigl(1 - 2\alpha+ 2\alpha^{2}\bigr)^{1 - \frac{1}{q}} \biggl[ \bigl(6 - 12\alpha+ 12\alpha^{2} \bigr) \biggl( \frac{1}{a_{2}} \biggr)^{q} \\& \qquad {} + \bigl(4 - 9\alpha+ 12\alpha^{2} - 2\alpha^{3} \bigr) \eta \biggl( \biggl(\frac{1}{a_{1}} \biggr)^{q} , \biggl( \frac{1}{a_{2}} \biggr)^{q} \biggr) \biggr]^{\frac{1}{q}} \\& \qquad {} + \bigl(1 - 2\beta+ 2\beta^{2} \bigr)^{1 - \frac{1}{q}} \biggl[ \bigl(6 - 12\beta+ 12\beta^{2} \bigr) \biggl(\frac{1}{a_{2}} \biggr)^{q} \\& \qquad {} + \bigl(2 - 3\beta+ 2\beta^{3} \bigr) \eta \biggl( \biggl( \frac {1}{a_{1}} \biggr)^{q} , \biggl(\frac{1}{a_{2}} \biggr)^{q} \biggr) \biggr]^{\frac{1}{q}} \biggr\} . \end{aligned}$$

(29)

Theorem 4.4

For $a_{1} > 0$, $a_{2} > 0$, $a_{1} \neq a_{2}$ and $q\geq1$, we have

$$\begin{aligned}& \biggl\vert \frac{\ln G^{2}(a_{1}^{\alpha}, a_{2}^{\beta})}{2} + \frac{2 - \alpha- \beta}{2}\ln A(a_{1}, a_{2}) - \ln I(a_{1}, a_{2}) \biggr\vert \\& \quad \leq\frac{{a_{2}} - {a_{1}}}{4} \biggl[\frac{1}{2(q + 1)(q + 2)} \biggr]^{\frac{1}{q}} \\& \qquad {} \times\biggl\{ \biggl[\bigl(\bigl[2(q + 2) (1 - \alpha)^{q + 1}+ 2(q + 2) \alpha^{q + 1}\bigr]\bigr) \biggl(\frac{1}{a_{2}} \biggr)^{q} \\& \qquad {} + \bigl[(q + 3 - \alpha) (1 - \alpha)^{q + 1} + (2q + 4 - \alpha)\alpha ^{q + 1}\bigr] \eta \biggl( \biggl(\frac{1}{a_{1}} \biggr)^{q}, \biggl(\frac {1}{a_{2}} \biggr)^{q} \biggr) \biggr]^{\frac{1}{q}} \\& \qquad {} + \biggl[\bigl(2(q + 2) (1 - \beta)^{q + 1} + 2(q + 2) \beta^{q + 1}\bigr) \biggl(\frac{1}{a_{2}} \biggr)^{q} \\& \qquad {} + \bigl(\beta^{q + 2} + (q + 1 + \beta) (1 - \beta)^{q + 1}\bigr) \eta \biggl( \biggl(\frac{1}{a_{1}} \biggr)^{q}, \biggl(\frac{1}{a_{2}} \biggr)^{q} \biggr) \biggr]^{\frac{1}{q}}\biggr\} . \end{aligned}$$

(30)

Finally, we can establish an inequality for the Heronian mean as follows.

Theorem 4.5

For $a_{2}> a_{1}> 0$, $a_{1} \neq a_{2}$, $w \geq0$, and $s \geq4$ or $0 \neq s<1$, we have

$$\begin{aligned}& \biggl\vert \frac{H^{s}_{w, s} (a_{1}, a_{2})}{H(a_{1}^{s}, a_{2}^{s})} + H^{\frac {s}{2} + 1}_{w, (\frac{s}{2} + 1)} \biggl( \frac{a_{2}}{a_{1}} + \frac {a_{1}}{a_{2}}, 1 \biggr) - H^{s}_{w, s} \biggl(\frac{L(a_{1}^{2}, a_{2}^{2})}{G^{2}(a_{1}, a_{2})}, 1 \biggr) \biggr\vert \\& \quad \leq \frac{(a_{2} - a_{1})A(a_{1}, a_{2})}{8G^{2}(a_{1}, a_{2})}\biggl[\frac {2|s|}{w + 2} \biggl(G^{2(s-1)} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) + \frac{w}{2}G^{s - \frac{1}{2}} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) \biggr) \\& \qquad {} + \eta \biggl(\frac{|s|}{w + 2} \biggl(G^{2(s-1)} \biggl(a_{1}, \frac {1}{a_{2}} \biggr) + \frac{w}{2}G^{s - \frac{1}{2}} \biggl(a_{1}, \frac {1}{a_{2}} \biggr) \biggr), \\& \qquad \frac{|s|}{w + 2} \biggl(G^{2(s-1)} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) + \frac{w}{2}G^{s - \frac{1}{2}} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) \biggr) \biggr)\biggr]. \end{aligned}$$

(31)

Proof

Let $f(x) = \frac{x^{s} + wx^{\frac{s}{2}} + 1}{w + 2}$ for $x>0$ and $s \notin(1, 4)$. Then

$$ f'(x) = \frac{s}{w + 2} \biggl(x^{s - 1} + \frac{w}{2}x^{\frac{s}{2} - 1} \biggr). $$

(32)

By Corollary 3.3 it follows that

$$\begin{aligned}& \biggl\vert \frac{1}{2}\biggl[\frac{f (\frac{a_{2}}{a_{1}} ) + f (\frac{a_{1}}{a_{2}} )}{2} + f \biggl( \frac{\frac{a_{2}}{a_{1}} + \frac{a_{1}}{a_{2}} }{2} \biggr)\biggr] - \frac{1}{\frac{a_{2}}{a_{1}} - \frac{a_{1}}{a_{2}}} \int_{\frac{a_{1}}{a_{2}}}^{\frac{a_{2}}{a_{1}}}f(x)\,dx \biggr\vert \\& \quad = \biggl\vert \frac{1}{2}\biggl\{ \frac{1}{2}\biggl[ \frac{a_{2}^{s} + w(a_{1}a_{2})^{\frac{s}{2}} + a_{1}^{s}}{a_{1}^{s}(w + 2)} + \frac{a_{1}^{s} + w(a_{1}a_{2})^{\frac{s}{2}} + a_{2}^{s}}{a_{2}^{s}(w + 2)}\biggr] \\& \qquad {} + \frac{ (\frac{a_{2}}{a_{1}} + \frac{a_{1}}{a_{2}} )^{s} + w (\frac{a_{2}}{a_{1}} + \frac{a_{1}}{a_{2}} )^{\frac{s}{2}} + 1}{w + 2}\biggr\} \\& \qquad {} - \frac{1}{w + 2}\biggl[\frac{ (\frac{a_{2}}{a_{1}} )^{s + 1} - (\frac{a_{1}}{a_{2}} )^{s + 1}}{(s + 1) (\frac {a_{2}}{a_{1}} - \frac{a_{1}}{a_{2}} )} + w \frac{ (\frac {a_{2}}{a_{1}} )^{\frac{s}{2} + 1} - (\frac{a_{1}}{a_{2}} )^{ \frac{s}{2} + 1}}{ (\frac{s}{2} + 1 ) (\frac {a_{2}}{a_{1}} - \frac{a_{1}}{a_{2}} )} + 1\biggr] \biggr\vert \\& \quad = \biggl\vert \frac{H^{s}_{w, s} (a_{1}, a_{2})}{H(a_{1}^{s}, a_{2}^{s})} + H^{\frac {s}{2} + 1}_{w, (\frac{s}{2} + 1)} \biggl(\frac{a_{2}}{a_{1}} + \frac {a_{1}}{a_{2}}, 1 \biggr) - H^{s}_{w, s} \biggl(\frac{L(a_{1}^{2}, a_{2}^{2})}{G^{2}(a_{1}, a_{2})}, 1 \biggr) \biggr\vert . \end{aligned}$$

(33)

On the other hand, we have

$$\begin{aligned}& \frac{\frac{a_{2}}{a_{1}} - \frac{a_{1}}{a_{2}}}{16} \biggl[2 \biggl\vert f' \biggl( \frac{a_{2}}{a_{1}} \biggr) \biggr\vert + \eta \biggl( \biggl\vert f' \biggl(\frac{a_{1}}{a_{2}} \biggr) \biggr\vert , \biggl\vert f' \biggl(\frac {a_{2}}{a_{1}} \biggr) \biggr\vert \biggr) \biggr] \\& \quad = \frac{a_{2}^{2} - a_{1}^{2}}{16a_{1}a_{2}} \biggl[2 \biggl\vert \frac{s}{w + 2} \biggl( \biggl(\frac{a_{2}}{a_{1}} \biggr)^{s - 1} + \frac{w}{2} \biggl( \frac {a_{2}}{a_{1}} \biggr)^{\frac{s}{2} - 1} \biggr) \biggr\vert \\& \qquad {} + \eta \biggl( \biggl\vert \frac{s}{w + 2} \biggl( \biggl( \frac {a_{1}}{a_{2}} \biggr)^{s - 1} + \frac{w}{2} \biggl( \frac{a_{1}}{a_{2}} \biggr)^{\frac{s}{2} - 1} \biggr) \biggr\vert , \biggl\vert \frac{s}{w + 2} \biggl( \biggl(\frac{a_{2}}{a_{1}} \biggr)^{s - 1} + \frac{w}{2} \biggl(\frac {a_{2}}{a_{1}} \biggr)^{\frac{s}{2} - 1} \biggr) \biggr\vert \biggr) \biggr] \\& \quad = \frac{(a_{2} - a_{1})A(a_{1}, a_{2})}{8G^{2}(a_{1}, a_{2})} \biggl[\frac{2|s|}{w + 2} \biggl(G^{2(s-1)} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) + \frac {w}{2}G^{s - \frac{1}{2}} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) \biggr) \\& \qquad {} + \eta \biggl(\frac{|s|}{w + 2} \biggl(G^{2(s-1)} \biggl(a_{1}, \frac {1}{a_{2}} \biggr) + \frac{w}{2}G^{s - \frac{1}{2}} \biggl(a_{1}, \frac {1}{a_{2}} \biggr) \biggr), \\& \qquad {}\frac{|s|}{w + 2} \biggl(G^{2(s-1)} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) + \frac{w}{2}G^{s - \frac{1}{2}} \biggl(a_{2}, \frac{1}{a_{1}} \biggr) \biggr) \biggr) \biggr]. \end{aligned}$$

(34)

Obviously (33) and (34) yield (31). □