In this section, we show a representative application of σ-convex functions. We will establish some new integral inequalities of Hermite–Hadamard type via σ-convex functions.
Let \(\mathcal{I}=[a,b]\) unless otherwise specified and σ be a continuous differentiable and strictly monotonic function in its domain. We denote by \(\mathbb{R}^{+}\) the set of positive real numbers.
Theorem 3.1
Suppose that
\(f:\mathcal{I}\rightarrow \mathbb{R} \)
is an integrable
σ-convex function with respect to the function
σ, then we have the following inequalities:
$$\begin{aligned} f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)&\leq \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x) \sigma '(x)\, \mathrm{d}x \\ & \leq \frac{f(a)+f(b)}{2}. \end{aligned}$$
(3.1)
Proof
Since f is a σ-convex function, we have
$$\begin{aligned} f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(x)+{\sigma }(y)}{2} \biggr) \biggr)\leq \frac{f(x)+f(y)}{2}. \end{aligned}$$
Substituting \(x=\sigma ^{-1}((1-t)\sigma (a)+t\sigma (b))\) and \(y=\sigma ^{-1}(t\sigma (a)+(1-t)\sigma (b))\) in the above inequality, we have
$$\begin{aligned} & f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{f(\sigma ^{-1}((1-t)\sigma (a)+t\sigma (b)))+f(\sigma ^{-1}(t\sigma (a)+(1-t)\sigma (b)))}{2}. \end{aligned}$$
(3.2)
Integrating both sides of (3.2) with respect to t on \([0,1]\), we get
$$\begin{aligned} f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)\leq \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x) \sigma '(x)\, \mathrm{d}x. \end{aligned}$$
(3.3)
Similarly, in light of the assumption in Theorem 3.1 that f is the σ-convex function, we have
$$\begin{aligned} f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\leq (1-t)f(a)+tf(b). \end{aligned}$$
Integrating both sides of the above inequality with respect to t on \([0,1]\), we obtain
$$\begin{aligned} \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)\sigma '(x) \, \mathrm{d}x\leq \frac{f(a)+f(b)}{2}. \end{aligned}$$
(3.4)
Combining (3.3) and (3.4) completes the proof of Theorem 3.1. □
Theorem 3.2
Let
\(f:\mathcal{I}\rightarrow \mathbb{R}^{+} \)
be an integrable
σ-convex function with respect to the function
σ. Then
$$\begin{aligned} &\frac{2f(a)}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl(\frac{{\sigma }(b)-{\sigma }(x)}{{\sigma }(b)-{\sigma }(a)} \biggr)f(x){\sigma '(x)\,\mathrm{d}x} \\ &\qquad {}+ \frac{2f(b)}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl(\frac{{\sigma }(x)-{\sigma }(a)}{{\sigma }(b)-{\sigma }(a)} \biggr)f(x){\sigma '(x)\,\mathrm{d}x} \\ &\quad \leq \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}}f^{2}(x){\sigma '(x) \,\mathrm{d}x}+\frac{f^{2}(a)+f(a)f(b)+f^{2}(b)}{3} \\ &\quad \leq \frac{2[f^{2}(a)+f(a)f(b)+f^{2}(b)]}{3}. \end{aligned}$$
(3.5)
Proof
Using the arithmetic-geometric means inequality gives
$$\begin{aligned} &2f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \bigl((1-t)f(a)+tf(b)\bigr) \\ &\quad \leq \bigl(f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{ \sigma }(b) \bigr) \bigr) \bigr)^{2}+ \bigl((1-t)f(a)+tf(b) \bigr)^{2} \\ &\quad = \bigl(f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \bigr)^{2}+(1-t)^{2}f^{2}(a)+t^{2}f^{2}(b)+2t(1-t)f(a)f(b). \end{aligned}$$
Integrating both sides of the above inequality with respect to t on \([0,1]\), we obtain
$$\begin{aligned} &2f(a) \int _{0}^{1}(1-t)f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \\ &\qquad {}+2f(b) \int _{0}^{1}tf \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \\ &\quad \leq \int _{0}^{1}f^{2} \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b) \bigr) \bigr)\,\mathrm{d}t +f^{2}(a) \int _{0}^{1}(1-t)^{2} \,\mathrm{d}t +f^{2}(b) \int _{0}^{1}t^{2}\,\mathrm{d}t \\ &\qquad{} +2f(a)f(b) \int _{0}^{1}t(1-t)\,\mathrm{d}t . \end{aligned}$$
(3.6)
By making the change of variable, inequality (3.6) can be rewritten as
$$\begin{aligned} &2f(a)\cdot \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl(\frac{{\sigma }(b)-{\sigma }(x)}{{\sigma }(b)-{\sigma }(a)} \biggr)f(x){\sigma '(x)\,\mathrm{d}x} \\ &\qquad {}+ 2f(b)\cdot \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl(\frac{{\sigma }(x)-{\sigma }(a)}{{\sigma }(b)-{\sigma }(a)} \biggr)f(x){\sigma '(x)\,\mathrm{d}x} \\ &\quad \leq \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}}f^{2}(x){\sigma '(x) \,\mathrm{d}x}+\frac{f^{2}(a)+f(a)f(b)+f^{2}(b)}{3}. \end{aligned}$$
(3.7)
On the other hand, since f is a σ-convex function, we have
$$\begin{aligned} f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \leq (1-t)f(a)+tf(b),\quad \forall t\in [0,1], \end{aligned}$$
therefore
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}}f^{2}(x){\sigma '(x) \,\mathrm{d}x}\\ &\quad= \int _{0}^{1}f^{2} \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \\ &\quad \leq \int _{0}^{1} \bigl[(1-t)f(a)+tf(b) \bigr]^{2}\,\mathrm{d}t=\frac{f ^{2}(a)+f(a)f(b)+f^{2}(b)}{3}. \end{aligned}$$
(3.8)
Combining (3.7) and (3.8) leads to the inequalities described in Theorem 3.2. □
Theorem 3.3
Let
\(f:\mathcal{I}\rightarrow \mathbb{R}^{+} \)
be an integrable
σ-convex function with respect to the function
σ. Then
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}}f(x){\sigma '(x) \, \mathrm{d}x} \\ &\quad \leq \frac{1}{2}f \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\qquad {}+\frac{1}{4({\sigma }(b)-{\sigma }(a))f ({\sigma }^{-1} (\frac{ {\sigma }(a)+{\sigma }(b)}{2} ) )}{ \int _{a}^{b}}f^{2}(x){\sigma '(x)\,\mathrm{d}x} \\ &\qquad {}+\frac{1}{24f ({\sigma }^{-1} ( \frac{{\sigma }(a)+{\sigma }(b)}{2} ) )} \bigl(f^{2}(a)+f^{2}(b)+4f(a)f(b) \bigr). \end{aligned}$$
(3.9)
Proof
Using the arithmetic-geometric means inequality and the σ convexity of f, it follows that
$$\begin{aligned} &f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \bigl[f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr)+f \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \bigr] \\ &\quad \leq f^{2} \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\qquad {}+\frac{1}{4} \bigl[f \bigl({\sigma }^{-1} \bigl((1-t){ \sigma }(a)+t{\sigma }(b) \bigr) \bigr)+f \bigl({\sigma }^{-1} \bigl(t{ \sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \bigr]^{2} \\ &\quad =f^{2} \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\qquad {}+\frac{1}{4} \bigl[ f^{2} \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b) \bigr) \bigr)+f^{2} \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t) { \sigma }(b) \bigr) \bigr) \\ &\qquad {} +2f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr)f \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t){ \sigma }(b) \bigr) \bigr) \bigr] \\ &\quad \leq f^{2} \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\qquad {}+\frac{1}{4} \bigl[ f^{2} \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b) \bigr) \bigr)+f^{2} \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t) { \sigma }(b) \bigr) \bigr) \\ &\qquad {} +2 \bigl((1-t){f}(a)+t{f}(b) \bigr) \bigl(t{f}(a)+(1-t){f}(b) \bigr) \bigr]. \end{aligned}$$
Integrating both sides of the above inequality with respect to t on \([0,1]\), we obtain
$$\begin{aligned} &f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \biggl[ \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \\ &\qquad {} + \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t) {\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \biggr] \\ &\quad \leq f^{2} \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \int _{0}^{1}\,\mathrm{d}t \\ &\qquad {}+ \frac{1}{4} \biggl[ \int _{0}^{1}f^{2} \bigl({\sigma }^{-1} \bigl((1-t) {\sigma }(a)+t{\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \\ &\qquad {} + \int _{0}^{1}f^{2} \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t) {\sigma }(b) \bigr) \bigr)\,\mathrm{d}t \\ &\qquad{} +2\bigl(f^{2}(a)+f^{2}(b)\bigr) \int _{0}^{1}t(1-t)\,\mathrm{d}t+2f(a)f(b) \int _{0}^{1}\bigl(t^{2}+(1-t)^{2} \bigr)\,\mathrm{d}t \biggr]. \end{aligned}$$
Performing the change of variable, we get
$$\begin{aligned} &f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \frac{2}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}}f(x){\sigma '(x) \, \mathrm{d}x} \\ &\quad \leq f^{2} \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr) \\ &\qquad {}+\frac{1}{2({\sigma }(b)-{\sigma }(a))}{ \int _{a}^{b}}f^{2}(x){\sigma '(x) \,\mathrm{d}x}\\ &\qquad {}+\frac{f^{2}(a)+f^{2}(b)+4f(a)f(b)}{12}. \end{aligned}$$
After a simple computation, one can transform the above inequality to the required inequality of Theorem 3.3. □
Theorem 3.4
Suppose that
\(f,h:\mathcal{I}\rightarrow \mathbb{R}^{+}\)
are two similarly ordered integrable
σ-convex functions with respect to the function
σ, then
hf
is also the
σ-convex function with respect to the function
σ.
Proof
Since \(f,h:\mathcal{I}\rightarrow \mathbb{R}^{+}\) are two similarly ordered integrable σ-convex functions, for \(\forall x,y \in \mathcal{I}\), \(t\in [0,1]\), we have
$$\begin{aligned} &f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(x)+t{\sigma }(y)\bigr) \bigr)h \bigl( {\sigma }^{-1}\bigl((1-t){\sigma }(x)+t{\sigma }(y)\bigr) \bigr) \\ &\quad \leq \bigl[(1-t)f(x)+tf(y)\bigr] \bigl[(1-t)h(x)+th(y)\bigr] \\ &\quad = (1-t)^{2}f(x)h(x)+t(1-t)\bigl[f(x)h(y)+f(y)h(x)\bigr] +t^{2}f(y)h(y) \\ &\quad = (1-t)f(x)h(x)+tf(y)h(y)+(1-t)^{2}f(x)h(x) \\ &\qquad{} +t(1-t)\bigl[f(x)h(y)+f(y)h(x)\bigr]+t^{2}f(y)h(y)-(1-t)f(x)h(x)-tf(y)h(y) \\ &\quad =(1-t)f(x)h(x)+tf(y)h(y)-t(1-t)\bigl[\bigl(f(x)-f(y)\bigr) \bigl(h(x)-h(y)\bigr)\bigr] \\ &\quad \leq (1-t)f(x)h(x)+tf(y)h(y). \end{aligned}$$
Thus
$$\begin{aligned} &f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(x)+t{\sigma }(y) \bigr) \bigr)h \bigl( {\sigma }^{-1}\bigl((1-t){\sigma }(x)+t{\sigma }(y) \bigr) \bigr) \\ &\quad \leq (1-t)f(x)h(x)+tf(y)h(y). \end{aligned}$$
(3.10)
Using the definition of σ-convex function (Definition 2.2), we conclude that hf is the σ-convex function with respect to the function σ. The proof of Theorem 3.4 is complete. □
Theorem 3.5
Let
\(f,h:\mathcal{I} \rightarrow \mathbb{R}^{+} \)
be integrable
σ-convex functions with respect to the function
σ. Then
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma }'(x)\, \mathrm{d}x \\ &\quad \leq \frac{1}{3} \mathbb{M}_{2}(a,b)+\frac{1}{6} \mathbb{N}_{2}(a,b) \\ &\quad \leq \frac{1}{6} \bigl[ \bigl[ \mathbb{M}_{1}(a,b) \bigr]^{2}+ \bigl[ \mathbb{N}_{1}(a,b) \bigr]^{2}- \bigl[f(a)f(b)+h(a)h(b) \bigr] \bigr], \end{aligned}$$
where
$$\begin{aligned} &\mathbb{M}_{1}(a,b) =f(a)+f(b) , \\ &\mathbb{N}_{1}(a,b) =h(a)+h(b) , \\ &\mathbb{M}_{2}(a,b) =f(a)h(a)+f(b)h(b) , \\ &\mathbb{N}_{2}(a,b) = f(a)h(b)+f(b)h(a) . \end{aligned}$$
Proof
Since \(f,h:\mathcal{I} \rightarrow \mathbb{R}^{+} \) are integrable σ-convex functions, we have
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma '(x)\, \mathrm{d}x} \\ &\quad = \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl[(1-t){\sigma }(a)+t{\sigma }(b)\bigr] \bigr)h \bigl({\sigma }^{-1} \bigl[(1-t){\sigma }(a)+t{\sigma }(b)\bigr] \bigr) \,\mathrm{d} t \\ &\quad \leq \int _{0}^{1}\bigl[(1-t)f(a)+tf(b)\bigr] \bigl[(1-t)h(a)+th(b)\bigr]\,\mathrm{d}t \\ &\quad =f(a)h(a) \int _{0}^{1}(1-t)^{2}\,\mathrm{d}t + \bigl[f(a)h(b)+f(b)h(a)\bigr] \int _{0}^{1}t(1-t)\,\mathrm{d}t +f(b)h(b) \int _{0}^{1}t^{2}\,\mathrm{d}t \\ &\quad =\frac{1}{3}\bigl[f(a)h(a)+f(b)h(b)\bigr]+\frac{1}{6} \bigl[f(a)h(b)+f(b)h(a)\bigr] \\ &\quad =\frac{1}{3} \mathbb{M}_{2}(a,b)+\frac{1}{6} \mathbb{N}_{2}(a,b). \end{aligned}$$
On the other hand, we have
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma '(x)\, \mathrm{d}x} \\ &\quad \leq \int _{0}^{1}\bigl[(1-t)f(a)+tf(b)\bigr] \bigl[(1-t)h(a)+th(b)\bigr]\,\mathrm{d}t \\ &\quad \leq \int _{0}^{1} \biggl[\frac{ [(1-t)f(a)+tf(b) ]^{2}+ [(1-t)h(a)+th(b) ]^{2}}{2} \biggr]\, \mathrm{d}t \\ &\quad =\frac{1}{2} \int _{0}^{1} \bigl[(1-t)^{2} \bigl[f^{2}(a)+{h}^{2}(a)\bigr]+t^{2}\bigl[f ^{2}(b)+{h}^{2}(b)\bigr] \\ &\qquad {}+2t(1-t)\bigl[f(a)f(b)+h(a)h(b)\bigr] \bigr]\,\mathrm{d}t \\ &\quad =\frac{1}{6} \bigl[ \bigl[f(a)+f(b) \bigr]^{2}+ \bigl[h(a)+h(b) \bigr]^{2}- \bigl[f(a)f(b)+h(a)h(b) \bigr] \bigr] \\ &\quad =\frac{1}{6} \bigl[ \bigl[ \mathbb{M}_{1}(a,b) \bigr]^{2}+ \bigl[ \mathbb{N}_{1}(a,b) \bigr]^{2}- \bigl[f(a)f(b)+h(a)h(b) \bigr] \bigr]. \end{aligned}$$
This completes the proof of Theorem 3.5. □
Theorem 3.6
Let
\(f,h: \mathcal{I} \rightarrow \mathbb{R}^{+} \)
be two similarly ordered integrable
σ-convex functions with respect to the function
σ. Then
$$\begin{aligned} \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma }'(x) \,\mathrm{d} x\leq \frac{1}{2}\mathbb{M}_{2}(a,b), \end{aligned}$$
(3.11)
where
\(\mathbb{M}_{2}(a,b)=f(a)h(a)+f(b)h(b)\).
Proof
Using inequality (3.10) gives
$$\begin{aligned} &f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr)h \bigl( {\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \\ &\quad \leq (1-t)f(a)h(a)+tf(b)h(b). \end{aligned}$$
Integrating both sides of the above inequality with respect to t on \([0,1]\) yields the inequality asserted by Theorem 3.6. □
Theorem 3.7
Let
\(f,h:\mathcal{I} \rightarrow \mathbb{R}^{+} \)
be two integrable
σ-convex functions with respect to the function
σ. Then
$$\begin{aligned} &2f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad {}-\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma }'(x)\, \mathrm{d}x \leq \frac{1}{6} \mathbb{M}_{2}(a,b)+ \frac{1}{3} \mathbb{N}_{2}(a,b), \end{aligned}$$
(3.12)
where
\(\mathbb{M}_{2}(a,b)=f(a)h(a)+f(b)h(b)\), \(\mathbb{N}_{2}(a,b)=f(a)h(b)+f(b)h(a)\).
Proof
Since f and h are both integrable σ-convex functions, by the same way as in the proof of Theorem 3.1, we deduce that
$$\begin{aligned} &\begin{aligned} & f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{f ({\sigma }^{-1}((1-t){\sigma }(a)+t{\sigma }(b)) )+ f ({\sigma }^{-1}(t{\sigma }(a)+(1-t){\sigma }(b)) )}{2}, \end{aligned} \\ &\begin{aligned} & h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{h ({\sigma }^{-1}((1-t){\sigma }(a)+t{\sigma }(b)) )+h ({\sigma }^{-1}(t{\sigma }(a)+(1-t){\sigma }(b)) )}{2}. \end{aligned} \end{aligned}$$
Hence, we have
$$\begin{aligned} &f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{1}{4} \bigl[f \bigl({\sigma }^{-1}\bigl[(1-t){ \sigma }(a)+t{\sigma }(b)\bigr] \bigr)+f \bigl({\sigma }^{-1}\bigl[t{ \sigma }(a)+(1-t){\sigma }(b)\bigr] \bigr) \bigr] \\ &\qquad{} \times \bigl[h \bigl({\sigma }^{-1}\bigl[(1-t){\sigma }(a)+t{ \sigma }(b)\bigr] \bigr)+h \bigl({\sigma }^{-1}\bigl[t{\sigma }(a)+(1-t){ \sigma }(b)\bigr] \bigr) \bigr] \\ &\quad =\frac{1}{4} \bigl[f \bigl({\sigma }^{-1}\bigl((1-t){ \sigma }(a)+t{\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1}\bigl((1-t){ \sigma }(a)+t{\sigma }(b)\bigr) \bigr) \\ &\qquad {}+f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \\ &\qquad {} + f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) \\ &\qquad {} +f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \bigr] \\ &\quad \leq \frac{1}{4} \bigl[ f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr) \\ &\qquad {} +f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \\ &\qquad {}+ \bigl((1-t)f(a)+tf(b) \bigr) \bigl(th(a)+(1-t)h(b) \bigr) \\ &\qquad {} + \bigl(tf(a)+(1-t)f(b) \bigr) \bigl((1-t)h(a)+th(b) \bigr) \bigr]. \end{aligned}$$
Integrating both sides of the above inequality with respect to t on \([0,1]\) gives
$$\begin{aligned} &f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{1}{4} \biggl[ \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad {} + \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t) {\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad {} +2\bigl(f(a)h(a)+f(b)h(b)\bigr) \int _{0}^{1}t(1-t)\,\mathrm{d}t \\ &\qquad {} +\bigl(f(a)h(b)+f(b)h(a)\bigr) \int _{0}^{1}\bigl(t^{2}+(1-t)^{2} \bigr) \,\mathrm{d}t \biggr] \\ &\quad =\frac{1}{4} \biggl[ \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr) h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad {} + \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t) {\sigma }(b)\bigr) \bigr) h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad {}+\frac{1}{3}\bigl(f(a)h(a)+f(b)h(b)\bigr)+\frac{2}{3} \bigl(f(a)h(b)+f(b)h(a)\bigr) \biggr] \\ &\quad = \frac{1}{2} \biggl[\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma '(x)\, \mathrm{d}x} +\frac{1}{6} \mathbb{M}_{2}(a,b)+ \frac{1}{3}\mathbb{N}_{2}(a,b) \biggr]. \end{aligned}$$
Theorem 3.7 is proved. □
Theorem 3.8
Let
\(f,h:\mathcal{I} \rightarrow \mathbb{R}^{+} \)
be two integrable
σ-convex functions with respect to the function
σ. Then
$$ f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \leq \frac{\mathbb{M}_{2}(a,b)+\mathbb{N}_{2}(a,b)}{4}, $$
(3.13)
where
\(\mathbb{M}_{2}(a,b)=f(a)h(a)+f(b)h(b)\), \(\mathbb{N}_{2}(a,b)=f(a)h(b)+f(b)h(a)\).
Proof
From Theorem 3.6 and Theorem 3.7, one can obtain the required result. □
Theorem 3.9
Let
\(f,h:\mathcal{I} \rightarrow \mathbb{R}^{+} \)
be two integrable
σ-convex functions with respect to the strictly increasing function
σ. Then
$$\begin{aligned} &{ \int _{a}^{b}} { \int _{a}^{b}} \int _{0}^{1} f \bigl({\sigma }^{-1}\bigl[(1-t){\sigma }(x)+t{\sigma }(y)\bigr] \bigr)h \bigl({\sigma } ^{-1}\bigl[(1-t)h(x)+th(y)\bigr] \bigr){\sigma '(x)\sigma '(y)\,\mathrm{d}t \,\mathrm{d}y\,\mathrm{d}x} \\ &\quad \leq \frac{2({\sigma }(b)-{\sigma }(a))}{3}{ \int _{a}^{b}}f(x)h(x) {\sigma '(x)\, \mathrm{d}x} +\frac{({\sigma }(b)-{\sigma }(a))^{2}}{12} \bigl[\mathbb{M}_{2}(a,b)+ \mathbb{N}_{2}(a,b) \bigr], \end{aligned}$$
(3.14)
where
\(\mathbb{M}_{2}(a,b)=f(a)h(a)+f(b)h(b)\), \(\mathbb{N}_{2}(a,b)=f(a)h(b)+f(b)h(a)\).
Proof
Since f and h are σ-convex functions, then for \(\forall x,y \in \mathcal{I}\), \(t\in [0,1]\) we have
$$\begin{aligned} & f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(x)+t{\sigma }(y)\bigr) \bigr) h \bigl( {\sigma }^{-1}\bigl((1-t)\sigma (x)+t\sigma (y)\bigr) \bigr) \\ &\quad \leq \bigl[(1-t)f(x)+tf(y) \bigr] \bigl[(1-t)h(x)+th(y) \bigr] \\ &\quad = (1-t)^{2}f(x)h(x)+t(1-t) \bigl[f(x)h(y)+h(x)f(y) \bigr]+t^{2}f( y)h(y). \end{aligned}$$
Integrating both sides of the above inequality with respect to t on \([0, 1]\), we have
$$\begin{aligned} & \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(x)+t{\sigma }(y)\bigr) \bigr) h \bigl({\sigma }^{-1} \bigl((1-t)\sigma (x)+t\sigma (y)\bigr) \bigr) \,\mathrm{d}t \\ &\quad \leq f(x)h(x) \int _{0}^{1}(1-t)^{2}\,\mathrm{d}t +f(y)h(y) \int _{0}^{1}t ^{2}\,\mathrm{d}t \\ &\qquad{} + \bigl[f(x)h(y)+h(x)f(y) \bigr] \int _{0}^{1}t(1-t)\,\mathrm{d}t \\ &\quad = \frac{1}{3} \bigl[f(x)h(x)+f(y)h(y) \bigr]+\frac{1}{6} \bigl[f(x)h(y)+h(x)f(y) \bigr]. \end{aligned}$$
Again, integrating both sides of the above inequality over the plane domain \(\{(x,y): x\in [a,b], y\in [a,b]\}\) and then using the right-hand side of the Hermite–Hadamard inequality (3.1), we deduce that
$$\begin{aligned} &{ \int _{a}^{b}} { \int _{a}^{b}} \int _{0}^{1} f \bigl({\sigma }^{-1} \bigl((1-t) {\sigma }(x)+t{\sigma }(y)\bigr) \bigr)h \bigl({\sigma }^{-1} \bigl((1-t)\sigma (x)+t \sigma (y)\bigr) \bigr)\sigma '(x)\sigma '(y)\,\mathrm{d}t\,\mathrm{d}y{\,\mathrm{d}x} \\ &\quad \leq \frac{2({\sigma }(b)-{\sigma }(a))}{3}{ \int _{a}^{b}}f(x)h(x) {\sigma '(x)\, \mathrm{d}x}+\frac{1}{6}{ \int _{a}^{b}}h (y){\sigma }'(y) \, \mathrm{d}y{ \int _{a}^{b}}f(x){\sigma '(x)\, \mathrm{d}x} \\ &\qquad{} +\frac{1}{6}{ \int _{a}^{b}}f (y){\sigma }'(y)\, \mathrm{d}y{ \int _{a}^{b}}h(x){\sigma '(x)\, \mathrm{d}x} \\ &\quad \leq \frac{2({\sigma }(b)-{\sigma }(a))}{3}{ \int _{a}^{b}}f(x)h(x) {\sigma '(x)\, \mathrm{d}x} \\ &\qquad {}+\frac{({\sigma }(b)-{\sigma }(a))^{2}}{6} \biggl[ \biggl(\frac{h(a)+h(b)}{2} \biggr) \biggl( \frac{f(a)+f(b)}{2} \biggr) \\ &\qquad {} + \biggl(\frac{f(a)+f(b)}{2} \biggr) \biggl( \frac{h(a)+h(b)}{2} \biggr) \biggr] \\ &\quad = \frac{2({\sigma }(b)-{\sigma }(a))}{3}{ \int _{a}^{b}}f(x)h(x){\sigma '(x) \, \mathrm{d}x} +\frac{({\sigma }(b)-{\sigma }(a))^{2}}{12} \bigl[\bigl(f(a)+f(b)\bigr) \bigl(h(a)+h(b)\bigr) \bigr] \\ &\quad = \frac{2({\sigma }(b)-{\sigma }(a))}{3}{ \int _{a}^{b}}f(x)h(x){\sigma '(x) \, \mathrm{d}x} +\frac{({\sigma }(b)-{\sigma }(a))^{2}}{12} \bigl[ \mathbb{M}_{2}(a,b)+ \mathbb{N}_{2}(a,b) \bigr], \end{aligned}$$
which is the required result. The proof of Theorem 3.9 is complete. □
Theorem 3.10
Let
\(f,h:\mathcal{I} \rightarrow \mathbb{R} \)
be two integrable
σ-convex functions with respect to the function
σ. Then
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl(\frac{{\sigma }(b)-{\sigma }(x)}{{\sigma }(b)-{\sigma }(a)} \biggr) \bigl(f(a)h(x)+h(a)f(x) \bigr){\sigma '(x)\,\mathrm{d}x} \\ &\qquad{} +\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl(\frac{{\sigma }(x)-{\sigma }(a)}{{\sigma }(b)-{\sigma }(a)} \biggr) \bigl(f(b)h(x)+h(b)f(x) \bigr){\sigma '(x)\,\mathrm{d}x} \\ &\quad \leq \frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} f(x)h(x) {\sigma '(x)\, \mathrm{d}x}+\frac{ \mathbb{M}_{2}(a,b)}{3}+\frac{ \mathbb{N}_{2}(a,b)}{6}, \end{aligned}$$
(3.15)
where
\(\mathbb{M}_{2}(a,b)=f(a)h(a)+f(b)h(b)\), \(\mathbb{N}_{2}(a,b)=f(a)h(b)+f(b)h(a)\).
Proof
Since f and h are σ-convex functions, then for \(\forall t\in [0,1]\) we have
$$\begin{aligned} &f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr) \leq (1-t)f(a)+tf(b), \\ &h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\leq (1-t)h(a)+th(b). \end{aligned}$$
Utilizing the rearrangement inequality, we obtain
$$\begin{aligned} &f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr) \bigl((1-t)h(a)+th(b)\bigr) \\ &\qquad {} +h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \bigl((1-t)f(a)+tf(b)\bigr) \\ &\quad \leq \bigl((1-t)f(a)+tf(b)\bigr) \bigl((1-t)h(a)+th(b)\bigr) \\ &\qquad {} +f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr). \end{aligned}$$
Integrating both sides of the above inequality over the interval \([0,1]\), we find
$$\begin{aligned} &h(a) \int _{0}^{1}(1-t)f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t +h(b) \int _{0}^{1}t f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad {} +f(a) \int _{0}^{1}(1-t)h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad {}+f(b) \int _{0}^{1}t h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\quad \leq f(a)h(a) \int _{0}^{1}(1-t)^{2}\,\mathrm{d}t +f(b)h(b) \int _{0}^{1}t ^{2}\,\mathrm{d}t + \bigl(f(a)h(b)+f(b)h(a)\bigr) \int _{0}^{1}t(1-t)\,\mathrm{d}t \\ &\qquad {} + \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t. \end{aligned}$$
Applying the change of variable to the integrations, we obtain the required result of Theorem 3.10. □
Theorem 3.11
Let
\(f,h:\mathcal{I} \rightarrow \mathbb{R}^{+} \)
be two integrable
σ-convex functions with respect to the function
σ. Then
$$\begin{aligned} &\frac{1}{{\sigma }(b)-{\sigma }(a)}{ \int _{a}^{b}} \biggl[f \biggl({\sigma } ^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) h(x) + h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) f(x) \biggr]\sigma '(x)\,\mathrm{d}x \\ &\quad \leq \frac{1}{2({\sigma }(b)-{\sigma }(a))}{ \int _{a}^{b}} f(x)h(x) {\sigma '(x)\, \mathrm{d}x}\\ &\qquad {}+f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\qquad {} +\frac{1}{12} \mathbb{M}_{2}(a,b)+\frac{1}{6} \mathbb{N}_{2}(a,b), \end{aligned}$$
where
\(\mathbb{M}_{2}(a,b)=f(a)h(a)+f(b)h(b)\), \(\mathbb{N}_{2}(a,b)=f(a)h(b)+f(b)h(a)\).
Proof
Note that f and h are integrable σ-convex functions. Taking \(t=\frac{1}{2}\) in (2.1) and letting \(x={\sigma }^{-1}((1-t) {\sigma }(a)+t{\sigma }(b))\), \(y={\sigma }^{-1}(t{\sigma }(a)+(1-t) {\sigma }(b))\), we have
$$\begin{aligned} & f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{f ({\sigma }^{-1}((1-t){\sigma }(a)+t{\sigma }(b)) )+ f ({\sigma }^{-1}(t{\sigma }(a)+(1-t){\sigma }(b)) )}{2}, \\ & h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{h ({\sigma }^{-1}((1-t){\sigma }(a)+t{\sigma }(b)) )+h ({\sigma }^{-1}(t{\sigma }(a)+(1-t){\sigma }(b)) )}{2}. \end{aligned}$$
Utilizing the rearrangement inequality, we obtain
$$\begin{aligned} &\frac{1}{2}f \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \bigl[h \bigl({\sigma } ^{-1}\bigl((1-t){\sigma }(a)+t{ \sigma }(b)\bigr) \bigr)+h \bigl({\sigma }^{-1}\bigl(t {\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) \bigr] \\ &\qquad {} + \frac{1}{2}h \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr) \bigl[f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr)\\ &\qquad {}+f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) \bigr] \\ &\quad \leq \frac{1}{4} \bigl[f \bigl({\sigma }^{-1}\bigl((1-t){ \sigma }(a)+t{\sigma }(b)\bigr) \bigr)+f \bigl({\sigma }^{-1}\bigl(t{ \sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) \bigr] \\ &\qquad{} \times \bigl[h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{ \sigma }(b)\bigr) \bigr)+h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){ \sigma }(b)\bigr) \bigr) \bigr] \\ &\qquad {}+f \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \\ &\quad = \frac{1}{4} \bigl[ f \bigl({\sigma }^{-1}\bigl((1-t){ \sigma }(a)+t{\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1}\bigl((1-t){ \sigma }(a)+t{\sigma }(b)\bigr) \bigr) \\ &\qquad{}+ f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \\ &\qquad{}+ f \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \\ &\qquad{}+ f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b) \bigr) \bigr) \bigr] \\ &\qquad{} +f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr) \\ &\quad \leq \frac{1}{4} \bigl[ f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr)h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr) \\ &\qquad{} + f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr)h \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b) \bigr) \bigr) \\ &\qquad{}+ \bigl((1-t)f(a)+tf(b)\bigr) \bigl(th(a)+(1-t)h(b)\bigr)\\ &\qquad {}+ \bigl(tf(a)+(1-t)f(b)\bigr) \bigl((1-t)h(a)+th(b)\bigr) \bigr] \\ &\qquad{} +f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr). \end{aligned}$$
Integrating the first and last expressions among the above inequalities over the interval \([0,1]\), we have
$$\begin{aligned} &\frac{1}{2}f \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr) \int _{0}^{1} \bigl[h \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\\ &\qquad {}+h \bigl( {\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) \bigr]\, \mathrm{d}t \\ &\qquad{} + \frac{1}{2}h \biggl({\sigma }^{-1} \biggl( \frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr) \int _{0}^{1} \bigl[f \bigl({\sigma }^{-1} \bigl((1-t) {\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\\ &\qquad {}+f \bigl({\sigma }^{-1}\bigl(t{\sigma }(a)+(1-t) {\sigma }(b)\bigr) \bigr) \bigr]\, \mathrm{d}t \\ &\quad \leq \frac{1}{4} \biggl[ \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl((1-t){\sigma }(a)+t {\sigma }(b)\bigr) \bigr) h \bigl({\sigma }^{-1}\bigl((1-t){\sigma }(a)+t{\sigma }(b)\bigr) \bigr)\,\mathrm{d}t \\ &\qquad{}+ \int _{0}^{1}f \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) h \bigl({\sigma }^{-1} \bigl(t{\sigma }(a)+(1-t){\sigma }(b)\bigr) \bigr) \,\mathrm{d}t \\ &\qquad{} + \frac{1}{3} \bigl(f(a)h(a)+f(b)h(b) \bigr)+ \frac{2}{3} \bigl(f(a)h(b)+f(b)h(a) \bigr) \biggr] \\ &\qquad{}+ f \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+{\sigma }(b)}{2} \biggr) \biggr)h \biggl({\sigma }^{-1} \biggl(\frac{{\sigma }(a)+ {\sigma }(b)}{2} \biggr) \biggr). \end{aligned}$$
From the above inequality, by simple computation and arrangement, we obtain the inequality asserted by Theorem 3.11. □