Theorem 3.1
Let
\(\nu >0\)
and
\(p>1,q>1,\frac{1}{p}+\frac{1}{q}=1\). Let
\(u, v \in C_{\nu }[a,b]\)
be two positive functions in
\([0,\infty [\)
such that
\({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t)<\infty \)
and
\({}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}v(t)<\infty \)
for all
\(t>a\). If
\(0<\alpha \leq \frac{u(t)}{v(t)}\leq \theta \)
for some
\(\alpha ,\theta \in \mathbb{R}_{+}^{*}\)
and all
\(t\in [a,b]\), then
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \bigr)^{\frac{1}{p}} \bigl(^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}v(t) \bigr)^{\frac{1}{q}}\leq \biggl(\frac{ \theta }{\alpha } \biggr)^{\frac{1}{pq}} \bigl[^{AB}{{}_{a}\mathcal{I}_{t} ^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]. $$
(3.1)
Proof
Using the condition \(\frac{u(t)}{v(t)}\leq \theta \), we get
$$ u^{\frac{1}{q}}\leq \theta ^{\frac{1}{q}} v^{\frac{1}{q}}. $$
(3.2)
Multiplying (3.2) by \(u^{\frac{1}{p}}\) and using the condition \(\frac{1}{p}+\frac{1}{q}=1\), we have
$$ u\leq \theta ^{\frac{1}{q}} u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.3)
Now let us use (3.3) twice. First, multiplying by \(\frac{1- \nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u\leq \theta ^{\frac{1}{q}} \frac{1- \nu }{\mathbb{B}(\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.4)
Second, multiplying by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )}\), we obtain
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u\leq \theta ^{\frac{1}{q}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.5)
Integrating both sides of Eq. (3.5) from 0 to t, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)\,ds \leq \theta ^{\frac{1}{q}} \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds. $$
(3.6)
Now, by adding Eq. (3.4) and Eq. (3.6) we find
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}u(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)\,ds \leq{} & \theta ^{ \frac{1}{q}} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )}u(t)^{\frac{1}{p}} v(t)^{ \frac{1}{q}} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \leq \theta ^{\frac{1}{q}} \bigl[^{AB} {{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]. $$
(3.7)
Taking the \(\frac{1}{p}\)th power of both sides of (3.7), we have
$$ \bigl[^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \bigr]^{\frac{1}{p}}\leq \theta ^{\frac{1}{pq}} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u ^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]^{\frac{1}{p}}. $$
(3.8)
Now, by the condition \(\alpha \leq \frac{u(t)}{v(t)}\) we have
$$ v^{\frac{1}{p}}\leq \alpha ^{\frac{-1}{p}} u^{\frac{1}{p}}. $$
(3.9)
Multiplying Eq. (3.9) by \(v^{\frac{1}{q}}\), we get
$$ v\leq \alpha ^{\frac{-1}{p}} u^{\frac{1}{p}}v^{\frac{1}{q}}. $$
(3.10)
Now let us use (3.10) twice. First, multiplying by \(\frac{1- \nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}v\leq \alpha ^{\frac{-1}{p}} \frac{1- \nu }{\mathbb{B}(\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.11)
Second, multiplying by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )}\), we obtain
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v\leq \alpha ^{\frac{-1}{p}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )}u^{\frac{1}{p}} v^{\frac{1}{q}}. $$
(3.12)
Integrating both sides of Eq. (3.12) from 0 to t, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v(s)\,ds \leq \alpha ^{\frac{-1}{p}} \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds. $$
(3.13)
Now, by adding Eq. (3.11) and Eq. (3.13) we find
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}v(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v(s)\,ds \leq {}& \alpha ^{ \frac{-1}{p}} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )}u(t)^{\frac{1}{p}} v(t)^{ \frac{1}{q}} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}u(s)^{\frac{1}{p}} v(s)^{\frac{1}{q}}\,ds \biggr]. \end{aligned}$$
(3.14)
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(t) \leq \alpha ^{\frac{-1}{p}} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{ \frac{1}{q}}(t) \bigr) \bigr]. $$
(3.15)
Taking the \(\frac{1}{q}\)th power of both sides of (3.15), we have
$$ \bigl[^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(s) \bigr]^{\frac{1}{q}}\leq \alpha ^{\frac{-1}{pq}} \bigl[^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u ^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]^{\frac{1}{q}}. $$
(3.16)
Finally, multiplying Eq. (3.8) and Eq. (3.16), we obtain
$$ \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t) \bigr)^{\frac{1}{p}} \bigl(^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}v(t) \bigr)^{\frac{1}{q}}\leq \biggl(\frac{ \theta }{\alpha } \biggr)^{\frac{1}{pq}} \bigl[^{AB}{{}_{a}\mathcal{I}_{t} ^{\nu }} \bigl(u^{\frac{1}{p}}(t)v^{\frac{1}{q}}(t) \bigr) \bigr]. $$
(3.17)
□
Theorem 3.2
Let
\(\nu >0\)
and
\(p>1,q>1,\frac{1}{p}+\frac{1}{q}=1\). Let
\(u, v \in C_{\nu }[a,b]\)
be two positive functions in
\([0,\infty [\)
such that
\({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t)<\infty\), \({}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{q}(t)<\infty \), \({}^{AB}{{}_{a}\mathcal{I}_{t} ^{\nu }}v^{p}(t)<\infty \), and
\({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v ^{q}(t)<\infty \)
for all
\(t>a\). If
\(0<\alpha \leq \frac{u(t)}{v(t)} \leq \theta \)
for some
\(\alpha ,\theta \in \mathbb{R}_{+}^{*}\)
and all
\(t\in [a,b]\), then
$$ {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t) \bigr)\leq \mathcal{A}^{*} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{p}(t)+v^{p}(t) \bigr)+ \mathcal{B}^{*}_{m} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u^{q}(t)+v ^{q}(t) \bigr), $$
(3.18)
where
$$ \mathcal{A}^{*}=\frac{2^{p-1}\theta ^{p}}{p(\theta +1)^{p}},\qquad \mathcal{B}^{*}_{m}= \frac{2^{q-1}}{q(1+\alpha )^{q}.} $$
Proof
Using the condition \(\frac{u(t)}{v(t)}\leq \theta \), we obtain
$$ u(t)\leq \biggl(\frac{\theta (u(t)+v(t))}{1+\theta }\biggr). $$
(3.19)
Taking the pth power of both sides of Eq. (2.2), we have
$$ u^{p}(t)\leq \biggl(\frac{\theta }{\theta +1} \biggr)^{p}\bigl(u(t)+v(t)\bigr)^{p}. $$
(3.20)
Multiplying both sides of (3.20) by \(\frac{1-\nu }{\mathbb{B}( \nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u^{p}(t)\leq \biggl( \frac{\theta }{\theta +1}\biggr)^{p}\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t) \bigr)^{p}. $$
(3.21)
Also, replacing t by s in Eq. (3.20) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u^{p}(s) \leq \biggl( \frac{\theta }{\theta +1}\biggr)^{p}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}. $$
(3.22)
Integrating both sides of Eq. (3.21) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u ^{p}(s) \,ds\leq \biggl(\frac{\theta }{\theta +1}\biggr)^{p} \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds. $$
(3.23)
Adding (3.21) and (3.23), we obtain
$$\begin{aligned} &\frac{1-\nu }{\mathbb{B}(\nu )}u^{p}(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u^{p}(s) \,ds\\ &\quad\leq \biggl(\frac{\theta }{ \theta +1}\biggr)^{p} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )} \bigl(u(t)+v(t)\bigr)^{p}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u^{p}(t) \leq \biggl(\frac{\theta }{\theta +1}\biggr)^{p} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}\bigl(u(t)+v(t)\bigr)^{p}. $$
(3.24)
Multiplying (2.7) by the constant \(\frac{1}{p}\), we find
$$ \frac{1}{p} \bigl(^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)\leq \frac{1}{p} \biggl( \frac{\theta }{\theta +1} \biggr)^{p} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p} \bigr]. $$
(3.25)
On the other hand, by using the condition \(0<\alpha \leq \frac{u(t)}{v(t)}\) we directly get
$$ v^{q}(t)\leq \frac{1}{(1+\alpha )^{q}}\bigl(u(t)+v(t) \bigr)^{q}. $$
(3.26)
Multiplying (3.26) by \(\frac{1-\nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}v^{q}(t)\leq \frac{1}{(1+\alpha )^{q}} \frac{1- \nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{q}. $$
(3.27)
Also, replacing t by s in Eq. (3.26) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v^{q}(s) \leq \frac{1}{(1+\alpha )^{q}} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}( \nu )\varGamma (\nu )}\bigl(u(s)+v(s)\bigr)^{q}. $$
(3.28)
Integrating both sides of Eq. (3.28) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v ^{q}(s) \,ds\leq \frac{1}{(1+\alpha )^{q}} \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{q}\,ds. $$
(3.29)
Adding (3.27) and (3.29), we obtain
$$\begin{aligned} \frac{1-\nu }{\mathbb{B}(\nu )}v^{q}(t)+ \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}v^{q}(s) \,ds \leq{} & \frac{1}{(1+ \alpha )^{q}} \biggl[\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t) \bigr)^{q} \\ &{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}\bigl(u(s)+v(s) \bigr)^{q}\,ds \biggr]. \end{aligned}$$
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{q}(t) \leq \frac{1}{(1+\alpha )^{q}} {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q}. $$
(3.30)
Multiplying (2.14) by\(\frac{1}{q}\), we have
$$ \frac{1}{q} \bigl(^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}v^{q}(t) \bigr)\leq \frac{1}{q} \frac{1}{(1+\alpha )^{q}} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t} ^{\nu }}\bigl(u(t)+v(t)\bigr)^{q} \bigr]. $$
(3.31)
By means of Eqs. (3.25) and (3.31) we get
$$\begin{aligned} &\frac{1}{p} \bigl(^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{p}(t) \bigr)+ \frac{1}{q} \bigl(^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v^{q}(t) \bigr) \\ &\quad\leq \frac{1}{p} \biggl(\frac{\theta }{\theta +1} \biggr)^{p} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p} \bigr]+\frac{1}{q} \frac{1}{(1+ \alpha )^{q}} \bigl[ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q} \bigr]. \end{aligned}$$
(3.32)
To complete our proof, we have to use Young’s inequality
$$ u(t)v(t)\leq \frac{u^{p}(t)}{p}+\frac{v^{q}(t)}{q}. $$
(3.33)
Multiplying (3.33) by \(\frac{1-\nu }{\mathbb{B}(\nu )}\), we get
$$ \frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t)\leq \frac{1-\nu }{\mathbb{B}( \nu )} \biggl( \frac{u^{p}(t)}{p}+\frac{v^{q}(t)}{q} \biggr). $$
(3.34)
Also, replacing t by s in Eq. (3.33) and multiplying both sides by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s) \leq \frac{\nu (t-s)^{\nu -1}}{p \mathbb{B}(\nu )\varGamma (\nu )}u^{p}(s)+ \frac{ \nu (t-s)^{\nu -1}}{q \mathbb{B}(\nu )\varGamma (\nu )}v^{q}(s). $$
(3.35)
Integrating both sides of Eq. (3.35) with respect to s, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s)\,ds \leq \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{p \mathbb{B}(\nu )\varGamma ( \nu )}u^{p}(s) \,ds+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{q \mathbb{B}( \nu )\varGamma (\nu )}v^{q}(s) \,ds. $$
(3.36)
Adding (3.34) and (3.36), we obtain
$$\begin{aligned} & \frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t) + \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s)\,ds \\ &\quad\leq \frac{1-\nu }{ \mathbb{B}(\nu )} \biggl(\frac{u^{p}(t)}{p}+\frac{v^{q}(t)}{q} \biggr) \\ &\qquad{}+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{p \mathbb{B}(\nu )\varGamma ( \nu )}u^{p}(s) \,ds+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{q \mathbb{B}( \nu )\varGamma (\nu )}v^{q}(s) \,ds. \end{aligned}$$
(3.37)
This implies
$$ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t)v(t) \leq \frac{1}{p} {}^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }}u^{p}(t)+ \frac{1}{q} {}^{AB}{{}_{a}\mathcal{I} _{t}^{\nu }}v^{q}(t). $$
(3.38)
Using (3.32) and (3.38), we have
$$\begin{aligned} &{} ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}u(t)v(t) \\ &\quad \leq \frac{1}{p} \biggl(\frac{ \theta }{\theta +1} \biggr)^{p} \bigl[ ^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p} \bigr]+\frac{1}{q} \frac{1}{(1+\alpha )^{q}} \bigl[ ^{AB}{{}_{a} \mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q} \bigr]. \end{aligned}$$
(3.39)
Using the inequality
$$ (u+v)^{r}\leq 2^{r-1}\bigl(u^{r}+v^{r} \bigr),\quad u,v\geq 0,r>1, $$
(3.40)
with \(r=p\) and multiplying (3.40) by the constant \(\frac{1-\nu }{ \mathbb{B}(\nu )}\), we find
$$ \frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p}\leq 2^{p-1}\frac{1- \nu }{\mathbb{B}(\nu )}\bigl(u(t)^{p}+v(t)^{p} \bigr). $$
(3.41)
Then multiplying Eq. (3.40) with \(r=p \) by \(\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$ \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}(u+v)^{p} \leq 2^{p-1} \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u ^{p}+v^{p}\bigr). $$
(3.42)
Integrating Eq. (3.42) from a to t, we have
$$ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds \leq 2^{p-1} \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu ) \varGamma (\nu )} \bigl(u^{p}(s)+v^{p}(s)\bigr)\,ds. $$
(3.43)
Adding Eq. (3.41) and Eq. (3.43), we obtain
$$\begin{aligned} &\frac{1-\nu }{\mathbb{B}(\nu )}\bigl(u(t)+v(t)\bigr)^{p}+ \int _{a}^{t}\frac{ \nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\bigl(u(s)+v(s) \bigr)^{p}\,ds \\ &\quad \leq2^{p-1} \biggl(\frac{1-\nu }{\mathbb{B}(\nu )} \bigl(u(t)^{p}+v(t)^{p}\bigr)+ \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )} \bigl(u^{p}(s)+v^{p}(s)\bigr)\,ds \biggr). \end{aligned}$$
(3.44)
This implies
$$ {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{p}\leq 2^{p-1} {}^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}\bigl(u^{p}(t)+v^{p}(t) \bigr). $$
(3.45)
Repeating the same process with \(r=q\), we get
$$ {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t)\bigr)^{q}\leq 2^{q-1} {}^{AB} {{}_{a}\mathcal{I}_{t}^{\nu }}\bigl(u^{q}(t)+v^{q}(t) \bigr). $$
(3.46)
Substituting by (3.45) and (3.46) into Eq. (3.39), the proof completed. □
Theorem 3.3
Let
\(\nu >0\), and let
\(u, v \in C_{\nu }[a,b]\)
be two positive functions in
\([0,\infty [\)
such that
\({}^{AB}{{}_{a}\mathcal{I}_{t}^{ \nu }}u(t)<\infty \)
and
\({}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }}v(t)<\infty \)
for all
\(t>a\), If
\(0<\alpha \leq \frac{u(t)}{v(t)}\leq \theta \)
for some
\(\alpha ,\theta \in \mathbb{R}_{+}^{*}\)
and all
\(t\in [a,b]\), then
$$ \frac{1}{\theta } {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)v(t) \bigr)\leq {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)+v(t) \bigr)^{2} \leq \frac{1}{\alpha } {}^{AB}{{}_{a}\mathcal{I}_{t}^{\nu }} \bigl(u(t)v(t) \bigr), $$
(3.47)
Proof
Using the condition
$$ 0< \alpha \leq \frac{u(t)}{v(t)}\leq \theta , $$
(3.48)
we conclude that
$$\begin{aligned} & (1+\alpha )v(t)\leq \bigl(u(t)+v(t)\bigr)\leq (\theta +1)v(t), \end{aligned}$$
(3.49)
$$\begin{aligned} & \frac{\theta +1}{\theta }u(t)\leq \bigl(u(t)+v(t)\bigr)\leq \frac{1+\alpha }{ \alpha }u(t). \end{aligned}$$
(3.50)
By (3.49) and (3.50) we obtain
$$\begin{aligned} \frac{1}{\theta }u(t)v(t)\leq \frac{(u(t)+v(t))^{2}}{(1+\alpha )( \theta +1)} \leq \frac{1}{\alpha }u(t)v(t). \end{aligned}$$
(3.51)
Multiplying (3.51) by \(\frac{1-\nu }{\mathbb{B}(\nu )} \) and then by \(\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\), we get
$$\begin{aligned} & \frac{1}{\theta }\frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t)\leq \frac{1- \nu }{\mathbb{B}(\nu )}\frac{(u(t)+v(t))^{2}}{(1+\alpha )(\theta +1)} \leq \frac{1}{\alpha }\frac{1-\nu }{\mathbb{B}(\nu )}u(t)v(t), \end{aligned}$$
(3.52)
$$\begin{aligned} & \frac{1}{\theta }\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}u(t)v(t)\leq \frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma ( \nu )}\frac{(u(t)+v(t))^{2}}{(1+\alpha )(\theta +1)} \leq \frac{1}{ \alpha }\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}u(t)v(t). \end{aligned}$$
(3.53)
Integrating Eq. (3.53) from 0 to t with respect to s, we have
$$\begin{aligned} \frac{1}{\theta } \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{\mathbb{B}( \nu )\varGamma (\nu )}u(s)v(s)\,ds &\leq \int _{a}^{t}\frac{\nu (t-s)^{ \nu -1}}{\mathbb{B}(\nu )\varGamma (\nu )}\frac{(u(s)+v(s))^{2}}{(1+ \alpha )(\theta +1)} \,ds \\ &\leq \frac{1}{\alpha } \int _{a}^{t}\frac{\nu (t-s)^{\nu -1}}{ \mathbb{B}(\nu )\varGamma (\nu )}u(s)v(s)\,ds. \end{aligned}$$
(3.54)
Adding Eqs. (3.52) and Eq. (3.54), we obtain the required inequality. □