In this section, we use fractional conformable integral operator to develop reverse Minkowski and Hermite–Hadamard integral inequalities. The reverse Minkowski fractional integral inequality is presented in the following theorems.
Theorem 2.1
Let
\(\beta ,\alpha >0\), \(\sigma \geq 1\), and let
Φ, Ψ
be two positive functions on
\([0,\infty )\)
such that, for all
\(x>0\), \({}^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x)<\infty \), \({}^{\beta }\mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x)<\infty \). If
\(0< m\leq \frac{\varPhi (t)}{\varPsi (t)}\leq M\), \(t\in [0,x]\), then the following inequality holds:
$$\begin{aligned} \bigl({}^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x) \bigr)^{\frac{1}{ \sigma }}+ \bigl({}^{\beta } \mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }} \leq \frac{1+M(m+2)}{(m+1)(M+1)} \bigl({}^{ \beta }\mathfrak{I}^{\alpha }( \varPhi +\varPsi )^{\sigma }(x) \bigr)^{\frac{1}{ \sigma }}. \end{aligned}$$
(7)
Proof
Using the condition \(\frac{\varPhi (t)}{\varPsi (t)}< M\), \(t\in [0,x]\), \(x>0\), we have
$$ (M+1)^{\sigma }\varPhi ^{\sigma }(t)\leq M^{\sigma } (\varPhi +\varPsi )^{\sigma }(t). $$
(8)
Multiplying both sides of (8) by \(\frac{1}{\varGamma (\beta )} (\frac{x^{\alpha }-t^{\alpha }}{\alpha } )^{\beta -1}t^{\alpha -1}\) and integrating the resultant inequality with respect to t from 0 to x, we have
$$\begin{aligned} \frac{(M+1)^{\sigma }}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{ \alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha -1}\varPhi ^{ \sigma }(t)\,dt &\leq \frac{M^{\sigma }}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha -1} (\varPhi +\varPsi )^{\sigma }(t)\,dt, \end{aligned}$$
which can be written as
$$ ^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x)\leq \frac{M^{\sigma }}{(M+1)^{\sigma }}{}^{\beta }\mathfrak{I}^{\alpha } (\varPhi + \varPsi )^{\sigma }(x). $$
Hence, it follows that
$$ \bigl({}^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x) \bigr)^{\frac{1}{ \sigma }}\leq \frac{M}{(M+1)} \bigl( {}^{\beta }\mathfrak{I}^{\alpha } (\varPhi +\varPsi )^{\sigma }(x) \bigr) ^{\frac{1}{\sigma }}. $$
(9)
Now, using the condition \(m\varPsi (t)\leq \varPhi (t)\), we have
$$ \biggl(1+\frac{1}{m} \biggr)\varPsi (t)\leq \frac{1}{m} \bigl(\varPhi (t)+\varPsi (t) \bigr), $$
which yields
$$ \biggl(1+\frac{1}{m} \biggr)^{\sigma }\varPsi ^{\sigma }(t)\leq \biggl(\frac{1}{m} \biggr)^{ \sigma } \bigl(\varPhi (t)+\varPsi (t) \bigr)^{\sigma }. $$
(10)
Multiplying both sides of (10) by \(\frac{1}{\varGamma (\beta )} (\frac{x^{\alpha }-t^{\alpha }}{\alpha } )^{\beta -1}t^{\alpha -1}\) and integrating the resultant inequality with respect to t from 0 to x, we get
$$ \bigl({}^{\beta }\mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{1}{ \sigma }}\leq \frac{1}{(m+1)} \bigl( {}^{\beta }\mathfrak{I}^{\alpha } (\varPhi +\varPsi )^{\sigma }(x) \bigr) ^{\frac{1}{\sigma }}. $$
(11)
Thus, adding inequalities (9) and (11) yields the desired inequality. □
Theorem 2.2
Let
\(\beta , \alpha >0\), \(\beta \in \mathbb{C}\), \(\sigma \geq 1\), and let
Φ, Ψ
be two positive functions on
\([0,\infty )\)
such that, for all
\(x>0\), \({}^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x)< \infty \), \({}^{\beta }\mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x)<\infty \). If
\(0< m\leq \frac{\varPhi (t)}{\varPsi (t)}\leq M\), \(t\in [0,x]\), then the following inequality holds:
$$\begin{aligned} & \bigl({}^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x) \bigr)^{\frac{2}{ \sigma }}+ \bigl({}^{\beta } \mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{2}{\sigma }} \\ &\quad \geq \biggl(\frac{(M+1)(m+1)}{M}-2 \biggr) \bigl({}^{\beta } \mathfrak{I} ^{\alpha }\varPhi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }} \bigl({}^{\beta } \mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }}. \end{aligned}$$
(12)
Proof
From the multiplication of inequalities (9) and (11), we have
$$\begin{aligned} \biggl(\frac{(M+1)(m+1)}{M} \biggr) \bigl({}^{\beta } \mathfrak{I}^{\alpha } \varPhi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }} \bigl({}^{\beta } \mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }} \leq \bigl( \bigl[{}^{\beta }\mathfrak{I}^{\alpha } \bigl(\varPhi (x)+ \varPsi (x) \bigr)^{\sigma } \bigr]^{\frac{1}{\sigma }} \bigr)^{2}. \end{aligned}$$
(13)
Now, applying the Minkowski inequality to the right-hand side of (13), we obtain
$$\begin{aligned} &\bigl( \bigl[{}^{\beta }\mathfrak{I}^{\alpha } \bigl(\varPhi (x)+ \varPsi (x) \bigr)^{\sigma } \bigr]^{\frac{1}{\sigma }} \bigr)^{2} \\ &\quad \leq \bigl[ \bigl({} ^{\beta }\mathfrak{I}^{\alpha } \varPhi ^{\sigma }(x) \bigr)^{\frac{1}{ \sigma }}+ \bigl({}^{\beta } \mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }} \bigr]^{2} \\ &\quad \leq \bigl({}^{\beta }\mathfrak{I}^{\alpha }\varPhi ^{\sigma }(x) \bigr)^{\frac{1}{ \sigma }}+ \bigl({}^{\beta } \mathfrak{I}^{\alpha }\varPsi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }}+2 \bigl({}^{\beta }\mathfrak{I}^{\alpha } \varPhi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }} \bigl({}^{\beta } \mathfrak{I}^{\alpha } \varPsi ^{\sigma }(x) \bigr)^{\frac{1}{\sigma }}. \end{aligned}$$
(14)
Thus, from inequalities (13) and (14), we get the desired inequality (12). □
Lemma 2.3
([2])
Let
f
be a concave function on
\([a,b]\), then the following inequalities hold:
$$\begin{aligned} f(a)+f(b)\leq f(b+a-x)+f(x)\leq 2f \biggl(\frac{a+b}{2} \biggr). \end{aligned}$$
(15)
Theorem 2.4
Let
\(\beta , \alpha >0\), \(\beta \in \mathbb{C}\), \(r, s>1\), and let
Φ
and
Ψ
be two positive functions on
\([0,\infty )\). If
\(\varPhi ^{r}\)
and
\(\varPsi ^{s}\)
are two concave functions on
\([0,\infty )\), then the following inequality holds:
$$\begin{aligned} & 2^{-r-s} \bigl(\varPhi (0)+\varPhi \bigl(x^{\alpha } \bigr) \bigr)^{r} \bigl(\varPsi (0)+\varPsi \bigl(x ^{\alpha } \bigr) \bigr)^{s} \bigl({}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha } \bigr) \bigr)^{2} \\ &\quad \leq {}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha } \varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) {}^{\beta } \mathfrak{I}^{\alpha } \bigl(x ^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr). \end{aligned}$$
(16)
Proof
Since the functions \(\varPhi ^{r}\) and \(\varPsi ^{s}\) are concave on \([0,\infty )\), therefore for any \(x>0\), \(\alpha >0\) and by Lemma 2.3, we have
$$\begin{aligned} \varPhi ^{r}(0)+\varPhi ^{r} \bigl(x^{\alpha } \bigr)\leq \varPhi ^{r} \bigl(x^{\alpha }-t^{\alpha } \bigr)+\varPhi ^{r} \bigl(t^{\alpha } \bigr)\leq 2\varPhi ^{r} \biggl(\frac{x^{\alpha }}{2} \biggr) \end{aligned}$$
(17)
and
$$\begin{aligned} \varPsi ^{s}(0)+\varPsi ^{s} \bigl(x^{\alpha } \bigr)\leq \varPsi ^{s} \bigl(x^{\alpha }-t^{\alpha } \bigr)+\varPhi ^{s} \bigl(t^{\alpha } \bigr)\leq 2\varPsi ^{s} \biggl(\frac{x^{\alpha }}{2} \biggr). \end{aligned}$$
(18)
Multiplying both sides of (17) and (18) by \(\frac{1}{\varGamma (\beta )} (\frac{x^{\alpha }-t^{\alpha }}{\alpha } )^{\beta -1}t^{\alpha \beta -1}\), \(t\in (0,x)\), and integrating the resultant inequalities from 0 to x, we get
$$\begin{aligned} &\frac{\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha })}{\varGamma (\beta )} \int _{0} ^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{ \alpha \beta -1} \,dt \\ &\quad \leq \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t ^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha \beta -1}\varPhi ^{r} \bigl(x^{ \alpha }-t^{\alpha } \bigr)\,dt \\ &\qquad {}+\frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{ \alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha \beta -1}\varPhi ^{r} \bigl(t^{ \alpha } \bigr)\,dt \\ &\quad \leq \frac{2\varPhi ^{r}(\frac{x^{\alpha }}{2})}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t ^{\alpha \beta -1}\,dt \end{aligned}$$
(19)
and
$$\begin{aligned} &\frac{\varPsi ^{s}(0)+\varPsi ^{s}(x^{\alpha })}{\varGamma (\beta )} \int _{0} ^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{ \alpha \beta -1} \,dt \\ &\quad \leq \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t ^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha \beta -1}\varPsi ^{s} \bigl(x^{ \alpha }-t^{\alpha } \bigr)\,dt \\ &\qquad {}+\frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{ \alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha \beta -1}\varPsi ^{s} \bigl(t^{ \alpha } \bigr)\,dt \\ &\quad \leq \frac{2\varPsi ^{s}(\frac{x^{\alpha }}{2})}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t ^{\alpha \beta -1}\,dt. \end{aligned}$$
(20)
Taking \(x^{\alpha }-t^{\alpha }=y^{\alpha }\), we have
$$\begin{aligned} \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha \beta -1}\varPhi ^{r} \bigl(x^{\alpha }-t ^{\alpha } \bigr)\,dt ={}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \end{aligned}$$
(21)
and
$$\begin{aligned} \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\alpha \beta -1}\varPsi ^{s} \bigl(x^{\alpha }-t ^{\alpha } \bigr)\,dt ={}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr). \end{aligned}$$
(22)
Thus the use of (19) and (21) yields
$$\begin{aligned} & \bigl(\varPhi ^{r}(0)+\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \bigl(^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha } \bigr) \bigr) \\ &\quad \leq 2 \bigl({}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta - \alpha }\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \bigr)\leq 2\varPhi ^{r} \biggl( \frac{x^{\alpha }}{2} \biggr) \bigl({}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{ \alpha \beta -\alpha } \bigr) \bigr). \end{aligned}$$
(23)
Similarly, the use of (20) and (22) yields
$$\begin{aligned} \bigl(\varPsi ^{s}(0)+\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) \bigl(^{\beta }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \beta -\alpha } \bigr) \bigr) &\leq 2 \bigl( {}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) \bigr) \\ & \leq 2\varPsi ^{s} \biggl(\frac{x^{\alpha }}{2} \biggr) \bigl( {}^{\beta }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \beta -\alpha } \bigr) \bigr). \end{aligned}$$
(24)
From inequalities (23) and (24), it follows that
$$\begin{aligned} &\bigl(\varPhi ^{r}(0)+\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \bigl(\varPsi ^{s}(0)+\varPsi ^{s} \bigl(x ^{\alpha } \bigr) \bigr) \bigl({}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha } \bigr) \bigr)^{2} \\ &\quad \leq 4 \bigl({}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha } \varPhi ^{s} \bigl(x^{\alpha } \bigr) \bigr) \bigr) \bigl( {}^{\beta }\mathfrak{I}^{\alpha } \bigl(x ^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) \bigr). \end{aligned}$$
(25)
Since Φ and Ψ are positive functions, therefore for any \(x>0\), \(\alpha >0\), \(r\geq 1\), and \(s\geq 1\), we have
$$ \biggl(\frac{\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha })}{2} \biggr)^{\frac{1}{r}} \geq 2^{-1} \bigl(\varPhi (0)+\varPhi \bigl(x^{\alpha } \bigr) \bigr) $$
(26)
and
$$ \biggl(\frac{\varPsi ^{s}(0)+\varPsi ^{s}(x^{\alpha })}{2} \biggr)^{\frac{1}{s}} \geq 2^{-1} \bigl(\varPsi (0)+\varPsi \bigl(x^{\alpha } \bigr) \bigr). $$
(27)
Hence, it follows that
$$ \biggl(\frac{\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha })}{2} \biggr) \bigl({}^{\beta } \mathfrak{I}^{\alpha } \bigl(t^{\alpha \beta -\alpha } \bigr) \bigr)\geq 2^{-r} \bigl(\varPhi (0)+\varPhi \bigl(x^{\alpha } \bigr) \bigr)^{r} \bigl({}^{\beta }\mathfrak{I} ^{\alpha } \bigl(t^{\alpha \beta -\alpha } \bigr) \bigr) $$
(28)
and
$$ \biggl(\frac{\varPsi ^{s}(0)+\varPsi ^{s}(x^{\alpha })}{2} \biggr) \bigl({}^{\beta } \mathfrak{I}^{\alpha } \bigl(t^{\alpha \beta -\alpha } \bigr) \bigr)\geq 2^{-s} \bigl(\varPsi (0)+\varPsi \bigl(x^{\alpha } \bigr) \bigr)^{s} \bigl({}^{\beta }\mathfrak{I} ^{\alpha } \bigl(t^{\alpha \beta -\alpha } \bigr) \bigr). $$
(29)
From inequalities (28) and (29), we obtain
$$\begin{aligned} &\frac{ (\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha }) ) (\varPsi ^{s}(0)+ \varPsi ^{s}(x^{\alpha }) )}{4} \bigl({}^{\beta } \mathfrak{I}^{\alpha } \bigl(t ^{\alpha \beta -\alpha } \bigr) \bigr)^{2} \\ &\quad \geq 2^{-r-s} \bigl(\varPhi (0)+\varPhi \bigl(x^{\alpha } \bigr) \bigr)^{r} \bigl(\varPsi (0)+ \varPsi \bigl(x^{\alpha } \bigr) \bigr)^{s} \bigl({}^{\beta }\mathfrak{I}^{\alpha } \bigl(t ^{\alpha \beta -\alpha } \bigr) \bigr)^{2}. \end{aligned}$$
(30)
Thus, by combining (25) and (30), we get the desired result. □
Theorem 2.5
Let
\(\beta , \mu , \alpha >0\), \(\beta , \mu \in \mathbb{C}\), \(r>1\), \(s>1\), and let
Φ, Ψ
be two positive functions on
\([0,\infty )\). If
\(\varPhi ^{r}\)
and
\(\varPsi ^{s}\)
are two concave functions on
\([0,\infty )\), then we have the following inequality:
$$\begin{aligned} &2^{2-r-s} \bigl(\varPhi (0)+\varPhi (x) \bigr)^{r} \bigl(\varPsi (0)+\varPsi (x) \bigr)^{s} \bigl({}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr) \bigr)^{2} \\ &\quad \leq \biggl[\frac{\varGamma (\mu )}{\varGamma (\beta )}{}^{\mu } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPhi ^{r} \bigl(x^{ \alpha } \bigr) \bigr)+{}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\mu \alpha - \alpha } \varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \biggr] \\ &\qquad {}\times \biggl[\frac{\varGamma (\mu )}{\varGamma (\beta )}{}^{\mu } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{ \alpha } \bigr) \bigr)+{}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\mu \alpha - \alpha } \varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) \biggr]. \end{aligned}$$
(31)
Proof
Multiplying both sides of inequalities (17) and (18) by \(\frac{1}{\varGamma (\beta )} (\frac{x^{\alpha }-t ^{\alpha }}{\alpha } )^{\beta -1}t^{\mu \alpha -1}\), \(t\in (0,x)\) and then integrating the resultant inequalities with respect to t from 0 to x, we have
$$\begin{aligned} &\frac{\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha })}{\varGamma (\beta )} \int _{0} ^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{ \mu \alpha -1}\,dt \\ &\quad \leq \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t ^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\mu \alpha -1} \varPhi ^{r} \bigl(x^{ \alpha }-t^{\alpha } \bigr)\,dt \\ &\qquad {}+\frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x ^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\mu \alpha -1} \varPhi ^{r} \bigl(t^{\alpha } \bigr)\,dt \\ &\quad \leq \frac{2\varPhi ^{r}(\frac{x^{\alpha }}{2})}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t ^{\mu \alpha -1}\,dt \end{aligned}$$
(32)
and
$$\begin{aligned} &\frac{\varPsi ^{s}(0)+\varPsi ^{s}(x^{\alpha })}{\varGamma (\beta )} \int _{0} ^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{ \mu \alpha -1}\,dt \\ &\quad \leq \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t ^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\mu \alpha -1} \varPsi ^{s} \bigl(x^{ \alpha }-t^{\alpha } \bigr)\,dt \\ &\qquad {}+\frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x ^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\mu \alpha -1} \varPsi ^{s} \bigl(t^{\alpha } \bigr)\,dt \\ &\quad \leq \frac{2\varPsi ^{s}(\frac{x^{\alpha }}{2})}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t ^{\mu \alpha -1}\,dt. \end{aligned}$$
(33)
Now, using \(x^{\alpha }-t^{\alpha }=y^{\alpha }\), we have
$$ \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\mu \alpha -1} \varPhi ^{r} \bigl(x^{\alpha }-t ^{\alpha } \bigr)\,dt =\frac{\varGamma (\mu )}{\varGamma (\beta )} {}^{\mu } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPhi ^{r} \bigl(x^{ \alpha } \bigr) \bigr) $$
(34)
and
$$ \frac{1}{\varGamma (\beta )} \int _{0}^{x} \biggl(\frac{x^{\alpha }-t^{\alpha }}{\alpha } \biggr)^{\beta -1}t^{\mu \alpha -1} \varPsi ^{s} \bigl(x^{\alpha }-t ^{\alpha } \bigr)\,dt =\frac{\varGamma (\mu )}{\varGamma (\beta )} {}^{\mu } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{ \alpha } \bigr) \bigr). $$
(35)
Thus, from (32) and (34), we can write
$$\begin{aligned} \bigl(\varPhi ^{r}(0)+\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr){}^{\beta }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr) &\leq \frac{\varGamma ( \mu )}{\varGamma (\beta )}{} ^{\mu }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) +{}^{\beta } \mathfrak{I} ^{\alpha } \bigl(x^{\alpha \mu -\alpha }\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \\ &\leq 2\varPhi ^{r} \biggl(\frac{x^{\alpha }}{2} \biggr) {}^{\beta }\mathfrak{I}^{ \alpha } \bigl(x^{\alpha \mu -\alpha } \bigr). \end{aligned}$$
(36)
Similarly, from inequalities (33) and (35), we obtain
$$\begin{aligned} \bigl(\varPsi ^{s}(0)+\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr){}^{\beta }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr) &\leq \frac{\varGamma ( \mu )}{\varGamma (\beta )}{} ^{\mu }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) +{}^{\beta } \mathfrak{I} ^{\alpha } \bigl(x^{\alpha \mu -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) \\ &\leq 2\varPsi ^{s} \biggl(\frac{x^{\alpha }}{2} \biggr) {}^{\beta }\mathfrak{I}^{ \alpha } \bigl(x^{\alpha \mu -\alpha } \bigr). \end{aligned}$$
(37)
From (36) and (37), it follows that
$$\begin{aligned} & \bigl(\varPhi ^{r}(0)+\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \bigl(\varPsi ^{s}(0)+\varPsi ^{s} \bigl(x ^{\alpha } \bigr) \bigr) \bigl({}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr) \bigr)^{2} \\ &\quad \leq \biggl[\frac{\varGamma (\mu )}{\varGamma (\beta )}{}^{\mu } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPhi ^{r} \bigl(x^{ \alpha } \bigr) \bigr) +{}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \mu -\alpha } \varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr) \biggr] \\ & \qquad {}\times\biggl[\frac{\varGamma (\mu )}{\varGamma (\beta )}{}^{\mu }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \beta -\alpha }\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr) + {}^{\beta }\mathfrak{I}^{\alpha } \bigl(x^{\alpha \mu -\alpha } \varPsi ^{s} \bigl(x ^{\alpha } \bigr) \bigr) \biggr]. \end{aligned}$$
(38)
Since Φ and Ψ are positive functions, therefore for any \(x>0\), \(\alpha >0\), \(r\geq 1\), \(s\geq 1\), we have
$$\begin{aligned} &\frac{\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha })}{2}{}^{\beta }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr)\geq 2^{-r} \bigl( \varPhi ^{r}(0)+ \varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr)^{r}{}^{\beta }\mathfrak{I}^{\alpha } \bigl(x ^{\alpha \mu -\alpha } \bigr) \end{aligned}$$
(39)
and
$$\begin{aligned} &\frac{\varPsi ^{s}(0)+\varPsi ^{s}(x^{\alpha })}{2}{}^{\beta }\mathfrak{I} ^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr)\geq 2^{-s} \bigl( \varPsi ^{s}(0)+ \varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr)^{s}{}^{\beta }\mathfrak{I}^{\alpha } \bigl(x ^{\alpha \mu -\alpha } \bigr). \end{aligned}$$
(40)
Thus from (39) and (40) it follows that
$$\begin{aligned} &\frac{ (\varPhi ^{r}(0)+\varPhi ^{r}(x^{\alpha }) ) (\varPsi ^{s}(0)+ \varPsi ^{s}(x^{\alpha }) )}{4} \bigl[ {}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr) \bigr]^{2} \\ &\quad \geq 2^{-r-s} \bigl(\varPhi ^{r}(0)+\varPhi ^{r} \bigl(x^{\alpha } \bigr) \bigr)^{r} \bigl( \varPsi ^{s}(0)+\varPsi ^{s} \bigl(x^{\alpha } \bigr) \bigr)^{s} \bigl[{}^{\beta } \mathfrak{I}^{\alpha } \bigl(x^{\alpha \mu -\alpha } \bigr) \bigr]^{2}. \end{aligned}$$
(41)
Combining inequalities (38) and (41), we get the desired proof. □
Remark 1
Letting \(\beta =\mu \) in Theorem 2.5, we obtain Theorem 2.4.