---
abstract: 'In this paper, using generalized k-fractional integral operator (in terms of the Gauss hypergeometric function), we establish new results on generalized k-fractional integral inequalities by considering the extended Chebyshev functional in case of synchronous function and some other inequalities.'
---

On Chebyshev type Inequalities using Generalized k-Fractional Integral Operator

$Vaijanath  \, L. Chinchane $

Department of Mathematics,\
Deogiri Institute of Engineering and Management\
Studies Aurangabad-431005, INDIA\
chinchane85@gmail.com

**Keywords :** Chebyshev inequality, generalized k-fractional integral.\
**Mathematics Subject Classification :** 26D10, 26A33.

Introduction 
=============

#### 

In recent years, many authors have worked on fractional integral inequalities by using different fractional integral operator such as Riemann-Liouville, Hadamard, Saigo and Erdelyi-Kober, see [@A; @BA; @BE; @C1; @C2; @C3; @C4; @C5; @D1; @KA; @P1; @YI]. In [@KI2] S. Kilinc and H. Yildirim establish new generalized k-fractional integral inequalities involving Gauss hypergeometric function related to Chebyshev functional. In [@C2; @D2] authors gave the following fractional integral inequalities, using the Hadamard and Riemann-Liouville fractional integral for extended Chebyshev functional.

Let $f$ and $g$ be two synchronous function on $[0,\infty[$, and $r,p,q:[0,\infty)\rightarrow [0,\infty)$. Then for all $t>0$, $\alpha>0$, we have $$\begin{split}
  &2_{H}D_{1,t}^{-\alpha}r(t) \left[_{H}D_{1,t}^{-\alpha}p(t) _{H}D_{1,t}^{-\alpha}(qfg)(t)+
  _{H}D_{1,t}^{-\alpha}q(t)_{H}D_{1,t}^{-\alpha}(pfg)(t)\right]+\\
  &2 _{H}D_{1,t}^{-\alpha}p(t)_{H}D_{1,t}^{-\alpha}q(t)_{H}D_{1,t}^{-\alpha}(rfg)(t)\geq\\
   &_{H}D_{1,t}^{-\alpha}r(t) \left[_{H}D_{1,t}^{-\alpha}(pf)(t)_{H}D_{1,t}^{-\alpha}(qg)(t)+_{H}D_{1,t}^{-\alpha}(qf)(t)_{H}D_{1,t}^{-\alpha}(pg)(t)\right]+\\
   &_{H}D_{1,t}^{-\alpha}p(t)\left[_{H}D_{1,t}^{-\alpha}(rf)(t)_{H}D_{1,t}^{-\alpha}(qg)(t)+_{H}D_{1,t}^{-\alpha}(qf)(t)_{H}D_{1,t}^{-\alpha}(rg)(t)\right]+\\
   &_{H}D_{1,t}^{-\alpha}q(t)\left[_{H}D_{1,t}^{-\alpha}(rf)(t)_{H}D_{1,t}^{-\alpha}(pg)(t)+_{H}D_{1,t}^{-\alpha}(pf)(t)_{H}D_{1,t}^{-\alpha}(rg)(t)\right]
  \end{split}$$

Let $f$ and $g$ be two synchronous function on $[0,\infty[$, and $r,p,q:[0,\infty)\rightarrow [0,\infty)$. Then for all $t>0$, $\alpha>0$, we have: $$\begin{split}
  &_{H}D_{1,t}^{-\alpha}r(t)\times\\
  & \left[_{H}D_{1,t}^{-\alpha}q(t) _{H}D_{1,t}^{-\beta}(pfg)(t)+2
  _{H}D_{1,t}^{-\alpha}p(t)_{H}D_{1,t}^{-\beta}(qfg)(t)+_{H}D_{1,t}^{-\beta}q(t)_{H}D_{1,t}^{-\alpha}(pfg)(t)\right]\\
  &+\left[_{H}D_{1,t}^{-\alpha}p(t)_{H}D_{1,t}^{-\beta}q(t)+_{H}D_{1,t}^{-\beta}p(t)_{H}D_{1,t}^{-\alpha}q(t)\right]_{H}D_{1,t}^{-\alpha}(rfg)(t)\geq\\
   &_{H}D_{1,t}^{-\alpha}r(t) \left[_{H}D_{1,t}^{-\alpha}(pf)(t)_{H}D_{1,t}^{-\beta}(qg)(t)+_{H}D_{1,t}^{-\beta}(qf)(t)_{H}D_{1,t}^{-\alpha}(pg)(t)\right]+\\
   &_{H}D_{1,t}^{-\alpha}p(t)\left[_{H}D_{1,t}^{-\alpha}(rf)(t)_{H}D_{1,t}^{-\beta}(qg)(t)+_{H}D_{1,t}^{-\beta}(qf)(t)_{H}D_{1,t}^{-\alpha}(rg)(t)\right]+\\
   &_{H}D_{1,t}^{-\alpha}q(t)\left[_{H}D_{1,t}^{-\alpha}(rf)(t)_{H}D_{1,t}^{-\beta}(pg)(t)+_{H}D_{1,t}^{-\beta}(pf)(t)_{H}D_{1,t}^{-\alpha}(rg)(t)\right].
  \end{split}$$

#### 

The main objective of this paper is to establish some Chebyshev type inequalities and some other inequalities using generalized k-fractional integral operator. The paper has been organized as follows. In Section 2, we define basic definitions related to generalized k-fractional integral operator. In section 3, we obtain Chebyshev type inequalities using generalized k-fractional. In Section 4 , we prove some inequalities for positive continuous functions.

 Preliminaries
==============

#### 

In this section, we present some definitions which will be used later discussion.

Two function $f$ and $g$ are said to synchronous (asynchronous) on $[a,b],$ if $$\left((f(u)-f(v))(g(u)-g(v))\right)\geq (\leq)0,$$ for all $ u, v \in [0,\infty)$.

[@KI2; @YI] The function $f(x)$, for all $x>0$ is said to be in the $L_{p,k}[0,\infty),$ if $$L_{p,k}[0,\infty)=\left\{f: \|f\|_{L_{p,k}[0,\infty)}=\left(\int_{0}^{\infty}|f(x)|^{p}x^{k}dx\right)^{\frac{1}{p}} < \infty \, \, 1 \leq p < \infty \, k \geq 0\right\},$$

[@KI2; @SAO; @YI] Let $f \in L_{1,k}[0,\infty),$. The generalized Riemann-Liouville fractional integral $I^{\alpha,k}f(x)$ of order $\alpha, k \geq 0$ is defined by $$I^{\alpha,k}f(x)= \frac{(k+1)^{1-\alpha}}{\Gamma (\alpha)}\int_{0}^{x}(x^{k+1}-t^{k+1})^{\alpha-1}t^{k} f(t)dt.$$

[@KI2; @YI] Let $k\geq0,\alpha>0 \mu >-1$ and $\beta, \eta \in R $. The generalized k-fractional integral $I^{\alpha,\beta,\eta,\mu}_{t,k}$ (in terms of the Gauss hypergeometric function)of order $\alpha$ for real-valued continuous function $f(t)$ is defined by $$\begin{split}
I^{\alpha,\beta,\eta,\mu}_{t,k}[f(t)]&
= \frac{(k+1)^{\mu+\beta+1}t^{(k+1)(-\alpha-\beta-2\mu)}}{\Gamma (\alpha)}\int_{0}^{t}\tau^{(k+1)\mu}(t^{k+1}-\tau^{k+1})^{\alpha-1}
\times \\
& _{2}F_{1} (\alpha+ \beta+\mu, -\eta; \alpha; 1-(\frac{\tau}{t})^{k+1})\tau^{k} f(\tau)d\tau.
\end{split}$$

where, the function $_{2}F_{1}(-)$ in the right-hand side of (2.4) is the Gaussian hypergeometric function defined by $$_{2}F_{1} (a, b; c; t)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{t^{n}}{n!},$$ and $(a)_{n}$ is the Pochhammer symbol\
$$(a)_{n}=a(a+1)...(a+n-1)=\frac{\Gamma(a+n)}{\Gamma(a)}, \,\,\,(a)_{0}=1.$$ Consider the function $$\begin{split}
F(t,\tau)&= \frac{(k+1)^{\mu+\beta+1}t^{(k+1)(-\alpha-\beta-2\mu)}}{\Gamma (\alpha)}\tau^{(k+1)\mu}\\
&(t^{k+1}-\tau^{k+1})^{\alpha-1} \times _{2}F_{1} (\alpha+ \beta+\mu, -\eta; \alpha; 1-(\frac{\tau}{t})^{k+1})\\
&=\sum_{n=0}^{\infty}\frac{(\alpha+\beta+\mu)_{n}(-n)_{n}}{\Gamma(\alpha+n)n!}t^{(k+1)(-\alpha-\beta-2\mu-\eta)}\tau^{(k+1)\mu}(t^{k+1}-\tau^{k+1})^{\alpha-1+n}(k+1)^{\mu+\beta+1}\\
&=\frac{\tau^{(k+1)\mu}(t^{k+1}-\tau^{k+1})^{\alpha-1}(k+1)^{\mu+\beta+1}}{t^{k+1}(\alpha+\beta+2\mu)\Gamma(\alpha)}+\\
&\frac{\tau^{(k+1)\mu}(t^{k+1}-\tau^{k+1})^{\alpha}(k+1)^{\mu+\beta+1}(\alpha+\beta+\mu)(-n)}{t^{k+1}(\alpha+\beta+2\mu+1)\Gamma(\alpha+1)}+\\
&\frac{\tau^{(k+1)\mu}(t^{k+1}-\tau^{k+1})^{\alpha+1}(k+1)^{\mu+\beta+1}(\alpha+\beta+\mu)(\alpha+\beta+\mu+1)(-n)(-n+1)}{t^{k+1}(\alpha+\beta+2\mu+1)\Gamma(\alpha+2)2!}+...
\end{split}$$ It is clear that $F(t,\tau)$ is positive because for all $\tau \in (0, t)$ , $(t>0)$ since each term of the (2.6) is positive.

Fractional Integral Inequalities for Extended Chebyshev Functional
==================================================================

In this section, we establish some Chebyshev type fractional integral inequalities by using the generalized k-fractional integral (in terms of the Gauss hypergeometric function) operator. The following lemma is used for the our main result.

Let $f$ and $g$ be two synchronous function on $[0,\infty[,$ and $x,y:[0,\infty)\rightarrow$ $[0,\infty)$ be two nonnegative functions. Then for all $k \geq 0,$ $t>0$, $\alpha > max\{0,-\beta-\mu\}$, $\beta < 1,$ $\mu >-1,$ $\beta -1< \eta <0,$ we have, $$\begin{split}
  &I^{\alpha,\beta,\eta,\mu}_{t,k}x(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(yfg)(t)+ I^{\alpha,\beta,\eta,\mu}_{t,k}y(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)\geq \\
  &I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(yg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(yf)(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$

**Proof**: Since $f$ and $g$ are synchronous on $[0,\infty[$ for all $\tau \geq 0$, $\rho\geq 0$, we have $$(f(\tau)-f(\rho)) (g(\tau)-g(\rho))\geq 0.$$ From (3.2), $$f(\tau)g(\tau)+f(\rho)g(\rho)\geq f(\tau)g(\rho)+f(\rho)g(\tau).$$ Now, multiplying both side of (3.3) by $ \tau^{k}x(\tau)F(t,\tau)$, $\tau \in (0,t)$, $t>0$. Then the integrating resulting identity with respect to $\tau$ from $0$ to $t$, we obtain by definition (2.4) $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)+f(\rho)g(\rho) I^{\alpha,\beta,\eta,\mu}_{t,k}(x)(t)\\
&I^{\alpha,\beta,\eta,\mu}_{t,k}(yg)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)+f(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$ Now, multiplying both side of (3.4) by $ \rho^{k}y(\rho)F(t,\rho)$, $\rho \in (0,t)$, $t>0$, where $F(t,\rho)$ defined in view of (2.6). Then the integrating resulting identity with respect to $\rho$ from $0$ to $t$, we obtain by definition (2.4) $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}y(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)+ I^{\alpha,\beta,\eta,\mu}_{t,k}(yfg)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(x)(t)\\
&\geq g(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(yf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$ This complete the proof of (3.1)\
Now, we gave our main result here.

Let $f$ and $g$ be two synchronous function on $[0,\infty[$, and $r,p,q:[0,\infty)\rightarrow [0,\infty)$. Then for all $k \geq 0,$ $t>0$, $\alpha > max\{0,-\beta-\mu\}$, $\beta < 1,$ $\mu >-1,$ $\beta -1< \eta <0,$ we have, $$\begin{split}
  &2I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) \left[I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qfg)(t)+
  I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)\right]+\\
  &2 I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\geq\\
   &I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) \left[I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t)\right]+\\
   &I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right]+\\
   &I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right]
  \end{split}$$

**Proof**: To prove above theorem, putting $x=p, \ y=q$, and using lemma 3.1, we get $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qfg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)\geq \\ &I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t).
\end{split}$$ Now, multiplying both side by (3.7) $I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)$, we have $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}p(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(qfg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)\right]\geq \\ &I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t)\right],
\end{split}$$ putting $x=r, y=q$, and using lemma 3.1, we get $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(qfg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\geq \\
&I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t),
\end{split}$$ multiplying both side by (3.9) $I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)$, we have $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(qfg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t) \right]\geq\\
&I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(qg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right].
\end{split}$$ With the same arguments as before, we can write $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\right]\geq\\
&I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right].
\end{split}$$ Adding the inequalities (3.8), (3.10) and (3.11), we get required inequality (3.6).\
Here, we give the lemma which is useful to prove our second main result.

Let $f$ and $g$ be two synchronous function on $[0,\infty[$. and $x,y:[0,\infty[\rightarrow$ $[0,\infty[$. Then for all $k \geq 0,$ $t>0$, $\alpha > max\{0,-\beta-\mu\}$,$\gamma> max\{0,-\delta-\upsilon\}$ $\beta,\delta < 1,$ $\upsilon,\mu >-1,$ $\beta -1< \eta <0,$ $\delta-1<\zeta <0,$ we have, $$\begin{split}
  &I^{\alpha,\beta,\eta,\mu}_{t,k}x(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfg)(t)+ I^{\gamma,\delta,\zeta,\upsilon}_{t,k}y(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)\geq \\
  &I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yf)(t) I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$

**Proof**: Now multiplying both side of (3.4) by $$\begin{split}
&\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}y(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
\end{split}$$ which remains positive in view of the condition stated in (3.12), $\rho \in (0,t)$, $t>0$, we obtain $$\begin{split}
&\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}y(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)\\
&+\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}y(\rho)f(\rho)g(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
I^{\alpha,\beta,\eta,\mu}_{t,k}x(t)\geq \\
&\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}y(\rho)g(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}y(\rho)f(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t),
\end{split}$$ then integrating (3.14) over (0,t), we obtain $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}y(t)+
I^{\alpha,\beta,\eta,\mu}_{t,k}(x)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfg)(t)\\
 &\geq I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}yg(t)
 +I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}yf(t),
\end{split}$$ this ends the proof of inequality (3.12).

Let $f$ and $g$ be two synchronous function on $[0,\infty[$, and $r,p,q:[0,\infty)\rightarrow [0,\infty)$. Then for all $t>0$, $\alpha>0$, we have: $$\begin{split}
  &I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)\times\\
  & \left[I^{\alpha,\beta,\eta,\mu}_{t,k}q(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(pfg)(t)+2
  I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qfg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)\right]\\
  &+\left[I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}q(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}p(t)I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\right]I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\geq\\
   &I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) \left[I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t)\right]+\\
   &I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right]+\\
   &I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(pg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right].
  \end{split}$$

**Proof**: To prove above theorem, putting $x=p, \ y=q$, and using lemma 3.3 we get $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}p(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qfg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)\geq \\ &I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t).
\end{split}$$ Now, multiplying both side by (3.17) $I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)$, we have $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}p(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qfg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pfg)(t)\right]\geq \\ &I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(pf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(pg)(t)\right],
\end{split}$$ putting $x=r, \ y=q$, and using lemma 3.3, we get $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qfg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\geq \\
&I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t),
\end{split}$$ multiplying both side by (3.19) $I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)$, we have $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}r(t) I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qfg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}q(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\geq \right]\\
&I^{\alpha,\beta,\eta,\mu}_{t,k}p(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(qf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right].
\end{split}$$ With the same argument as before, we obtain $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}r(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(pfg)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}p(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rfg)(t)\right]\geq \\
&I^{\alpha,\beta,\eta,\mu}_{t,k}q(t)\left[I^{\alpha,\beta,\eta,\mu}_{t,k}(rf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(pg)(t)+(pf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(rg)(t)\right].
\end{split}$$ Adding the inequalities (3.18), (3.20) and (3.21), we follows the inequality (3.16).

If $ f,g,r,p \  and\  q $ satisfies the following condition,

1.  The function f and g is asynchronous on $[0,\infty)$.

2.  The function r,p,q are negative on $[0,\infty)$.

3.  Two of the function r,p,q are positive and the third is negative on $[0,\infty)$.

then the inequality 3.6 and 3.16 are reversed.

Other fractional integral inequalities
======================================

In this section, we proved some fractional integral inequalities for positive and continuous functions which as follows:

Suppose that $f$, $g$ and $h$ be three positive and continuous functions on $[0,\infty[$, such that $$(f(\tau)-f(\rho))(g(\tau)-g(\tau))(h(\tau)+h(\rho))\geq 0; \  \tau, \rho \in(0,t)\ \  t>0,$$ and $x$ be a nonnegative function on $[0,\infty)$. Then for all $k \geq 0,$ $t>0$, $\alpha > max\{0,-\beta-\mu\}$,$\gamma> max\{0,-\delta-\upsilon\}$ $\beta,\delta < 1,$ $\upsilon,\mu >-1,$ $\beta -1< \eta <0,$ $\delta-1<\zeta <0,$ we have, $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(x)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xfgh)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xfg)(t)\\
&+I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xh)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xfgh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(x)(t)\\
& \geq I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xgh)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xfh)(t)\\
&+I^{\alpha,\beta,\eta,\mu}_{t,k}(xgh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xf)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xfh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xg)(t).
\end{split}$$

**Proof**: Since $f$, $g$ and $h$ be three positive and continuous functions on $[0,\infty[$ by (4.1), we can write $$\begin{split}
&f(\tau)g(\tau)h(\tau)+f(\rho)g(\rho)h(\rho)+f(\tau)g(\tau)h(\rho)+f(\rho)g(\rho)h(\tau)\\
&\geq f(\tau)g(\rho)h(\tau)+f(\tau)g(\rho)h(\rho)+f(\rho)g(\tau)h(\tau)+f(\rho)g(\tau)h(\rho).
\end{split}$$ Now, multiplying both side of (4.3) by $ \tau^{k}x(\tau)F(t,\tau)$, $\tau \in (0,t)$, $t>0$. Then the integrating resulting identity with respect to $\tau$ from $0$ to $t$, we obtain by definition (2.4) $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(xfgh)(t)+f(\rho)g(\rho)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}x(t)+g(\tau)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+f(\rho)g(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xh)(t)\geq g(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xfh)(t)+g(\rho)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+f(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xgh)(t)+f(\rho)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$

Now multiplying both side of (4.4) by $$\begin{split}
&\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}x(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
\end{split}$$ which remains positive in view of the condition stated in (4.2), $\rho \in (0,t)$, $t>0$ and integrating resulting identity with respective $\rho $ from $0$ to $t$, we obtain $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(xfgh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}x(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xfgh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}x(t)\\
&+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xgf)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xfg)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xh)(t)\\
&\geq I^{\gamma,\delta,\zeta,\upsilon}_{t,k}xg(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xfh)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xgh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xgh)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(xfh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).\end{split}$$ which implies the proof inequality 4.2.\
Here, we give another inequality which is as follows.

Let $f$, $g$ and $h$ be three positive and continuous functions on $[0,\infty[$, which satisfying the condition (4.1) and $x$ and $y$ be two nonnegative functions on $[0,\infty)$. Then for all $k \geq 0,$ $t>0$, $\alpha > max\{0,-\beta-\mu\}$,$\gamma> max\{0,-\delta-\upsilon\}$ $\beta,\delta < 1,$ $\upsilon,\mu >-1,$ $\beta -1< \eta <0,$ $\delta-1<\zeta <0,$ we have, $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(x)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfgh)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfg)(t)\\
&+I^{\alpha,\beta,\eta,\mu}_{t,k}(xfg)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yh)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xfgh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}y(t)\\
& \geq I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(ygh)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfh)(t)\\
&+I^{\alpha,\beta,\eta,\mu}_{t,k}(xgh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yf)(t)+I^{\alpha,\beta,\eta,\mu}_{t,k}(xfh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yg)(t).
\end{split}$$

**Proof**: Multiplying both side of (4.3) by $ \tau^{k}x(\tau)F(t,\tau)$, $\tau \in (0,t)$, $t>0$, where $F(t,\tau)$ defined by (2.6). Then the integrating resulting identity with respect to $\tau$ from $0$ to $t$, we obtain by definition (2.4) $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(xfgh)(t)+f(\rho)g(\rho)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}x(t)+g(\tau)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+f(\rho)g(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xh)(t)\geq g(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xfh)(t)+g(\rho)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+f(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xgh)(t)+f(\rho)h(\rho)I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$

Now multiplying both side of (4.8) by $$\begin{split}
&\frac{(k+1)^{\upsilon+\delta+1}t^{(k+1)(-\delta-\gamma-2\upsilon)}}{\Gamma (\gamma)}\rho^{(k+1)\upsilon}y(\rho)\\
&(t^{k+1}-\rho^{k+1})^{\gamma-1} \times _{2}F_{1} (\gamma+ \delta+\upsilon, -\zeta; \gamma; 1-(\frac{\rho}{t})^{k+1})\rho^{k}
\end{split}$$ which remains positive in view of the condition stated in (4.7), $\rho \in (0,t)$, $t>0$ and integrating resulting identity with respective $\rho $ from $0$ to $t$, we obtain $$\begin{split}
&I^{\alpha,\beta,\eta,\mu}_{t,k}(xfgh)(t)I^{\gamma,\delta,\zeta,\upsilon}_{t,k}y(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfgh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}x(t)\\
&+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xgf)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfg)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xh)(t)\\
&\geq I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yg)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xfh)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(ygh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xf)(t)\\
&+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yf)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xgh)(t)+I^{\gamma,\delta,\zeta,\upsilon}_{t,k}(yfh)(t)I^{\alpha,\beta,\eta,\mu}_{t,k}(xg)(t).
\end{split}$$ which implies the proof inequality 4.7.

[00]{} G. A. Anastassiou, *Fractional Differentiation Inequalities,* Springer Publishing Company, Incorporated, New York, NY, 2009. D.Baleanu, J.A.T.Machado and C.J.Luo, *Fractional Dynamic and Control,* Springer, 2012, pp.159-171. S. Belarbi and Z. Dahmani, *On some new fractional integral inequality,* J. Inequal. Pure and Appl. Math., 10(3)(2009), Art.86, 5 pp. V. L. Chinchane and D. B. Pachpatte, *A note on some integral inequalities via Hadamard integral,* J. Fractional Calculus Appl. 4(11)(2013), 1-5. V. L. Chinchane and D. B. Pachpatte, *On some integral inequalities using Hadamard fractional integral,* Malaya J. Math. 1(1)(2012), 62-66. V.L. Chinchane and D. B. Pachpatte, *Some new integral inequalities using Hadamard fractional integral operator,* Adv. Inequal. Appl. 2014, 2014:12. V.L. Chinchane and D. B. Pachpatte, *New fractional inequalities via Hadamard fractional integral,* Internat. J. Functional Analyisis, Operator Theory and Application, 5, 3(2013), 165-176. V.L. Chinchane and D. B. Pachpatte, *On some new Gr$\ddot{u}$ss-type inequality using Hadamard fractional integral operator,* J. Fractional Calculus Appl. Vol. 5(3S) No. 12, pp. 1-10. Z. Dahmani, *The Riemann-Liouville operator to genarate some new inequalities,* Int. J. Nonlinear Sci. volume 12(2011), No.4, pp.452-455. Z. Dahmani, *Some results associate with fractional integrals involving the extended Chebyshev,* Acta Univ. Apulensis Math. Inform. 27(2011), pp.217-224. S.L.Kalla and A. Rao, *On Gr$\ddot{u}$ss type inequality for hypergeometric fractional integrals,* Matematiche (Catania) 66(1)(2011), 57-64. Amersterdam, 2006. S.Kilinc and H.Yildirim, *Generalized fractional integral inequalities involving Hypergeometic operators,* International Journal of Pure and Applied Mathematics, 101(1), 2015, 71-82. S. D. Purohit and R. K. Raina, *Chebyshev type inequalities for the Saigo fractional integral and their q- analogues,* J. Math. Inequal., 7(2),(2013), 239-249. S.G.Somko, A.A.Kilbas and O.I.Marichev, *Fractional Integral and Derivative Theory and Application,* Gordon and Breach, Yverdon, Switzerland, 1993. H. Yildirim and Z. Kirtay, *Ostrowski inequality for generalized fractional integral and related equalities,*Malaya J. Mat. 2(3), (2014), 322-329.
