Since
f and
g are two integrable functions on
\([a,b]\) and satisfy the conditions (
3) and (
4), we define
$$\begin{aligned}& \begin{aligned}[b] B(x,y)&= \bigl[f(x)-f(y) \bigr] \bigl[g(x)-g(y) \bigr] \\ &=f(x)g(x)+f(y)g(y)-f(x)g(y)-f(y)g(x) \end{aligned} \end{aligned}$$
(7)
$$\begin{aligned}& \quad\Rightarrow \\& \begin{aligned}[b] &\frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int_{a}^{t} \int _{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha}{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac {\alpha}{k}-1} x^{r}y^{r}B(x,y)\,dx \,dy \\ &\quad=\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl(f(t)g(t) \bigr)-I_{a,k}^{\alpha ,r}f(t)I_{r,k}^{\alpha}g(t). \end{aligned} \end{aligned}$$
(8)
Now by using the value of
\(B(x,y)\) from (
7) to the left hand side of equality (
8) and then applying the Cauchy-Schwarz inequality, we have
$$\begin{aligned} & \biggl(\frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int _{a}^{t} \int_{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha }{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac{\alpha}{k}-1} x^{r}y^{r}B(x,y)\,dx \,dy \biggr)^{2} \\ &\quad\leq\frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int _{a}^{t} \int_{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha }{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac{\alpha}{k}-1} x^{r}y^{r} \bigl[f(x)-f(y) \bigr]^{2}\,dx \,dy \\ &\qquad{}\times \frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int _{a}^{t} \int_{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha }{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac{\alpha}{k}-1} x^{r}y^{r} \bigl[g(x)-g(y) \bigr]^{2}\,dx \,dy. \end{aligned}$$
(9)
Now since
$$\bigl[f(x)-f(y) \bigr]^{2}=f^{2}(x)+f^{2}(y)-2f(x)f(y), $$
one can easily prove that
$$\begin{aligned} &\frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int_{a}^{t} \int _{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha}{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac {\alpha}{k}-1} x^{r}y^{r} \bigl[f(x)-f(y) \bigr]^{2}\,dx \,dy \\ &\quad=\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r}f^{2}(t)- \bigl(I_{a,k}^{\alpha,r}f(t) \bigr)^{2}. \end{aligned}$$
(10)
Similarly,
$$\begin{aligned} &\frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int_{a}^{t} \int _{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha}{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac {\alpha}{k}-1} x^{r}y^{r} \bigl[g(x)-g(y) \bigr]^{2}\,dx \,dy \\ &\quad=\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r}g^{2}(t)- \bigl(I_{a,k}^{\alpha,r}g(t) \bigr)^{2}. \end{aligned}$$
(11)
Using equations (
10) and (
11) into (
9), we get
$$\begin{aligned} & \biggl(\frac{(r+1)^{2-\frac{2\alpha}{k}}}{2k^{2}\Gamma_{k}^{2}(\alpha)} \int _{a}^{t} \int_{a}^{t} \bigl(t^{r+1}-x^{r+1} \bigr)^{\frac{\alpha }{k}-1} \bigl(t^{r+1}-y^{r+1} \bigr)^{\frac{\alpha}{k}-1} x^{r}y^{r}B(x,y)\,dx \,dy \biggr)^{2} \\ &\quad\leq \biggl[\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha }{k}}}{\Gamma_{k}(\alpha+k)}I_{a,k}^{\alpha,r}f^{2}(t)- \bigl(I_{a,k}^{\alpha ,r}f(t) \bigr)^{2} \biggr] \\ &\qquad{}\times \biggl[\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha }{k}}}{\Gamma_{k}(\alpha+k)}I_{a,k}^{\alpha,r}g^{2}(t)- \bigl(I_{a,k}^{\alpha ,r}g(t) \bigr)^{2} \biggr]. \end{aligned}$$
(12)
Thus the equation (
8) together with the inequality (
12) implies that
$$\begin{aligned} & \biggl(\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl(f(t)g(t) \bigr)-I_{a,k}^{\alpha ,r}f(t)I_{a,k}^{\alpha,r}g(t) \biggr)^{2} \\ &\quad\leq \biggl[\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha }{k}}}{\Gamma_{k}(\alpha+k)}I_{a,k}^{\alpha,r}f^{2}(t)- \bigl(I_{a,k}^{\alpha ,r}f(t) \bigr)^{2} \biggr] \\ &\qquad{}\times \biggl[\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha }{k}}}{\Gamma_{k}(\alpha+k)}I_{a,k}^{\alpha,r}g^{2}(t)- \bigl(I_{a,k}^{\alpha ,r}g(t) \bigr)^{2} \biggr]. \end{aligned}$$
(13)
Now since
$$\bigl(\varphi_{2}(t)-f(t) \bigr) \bigl(f(t)-\varphi_{1}(t) \bigr)\geq0 $$
and
$$\bigl(\psi_{2}(t)-g(t) \bigr) \bigl(g(t)-\phi_{1}(t) \bigr) \geq0, $$
therefore,
$$\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl(\varphi_{2}(t)-f(t) \bigr) \bigl(f(t)-\varphi_{1}(t) \bigr)\geq0, \quad t\in[a,b] $$
and
$$\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl(\varphi_{2}(t)-f(t) \bigr) \bigl(f(t)-\varphi_{1}(t) \bigr)\geq0,\quad t\in[a,b]. $$
By Theorem
2.7, we have
$$\begin{aligned} &\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r}f^{2}(t)- \bigl[I_{a,k}^{\alpha,r}f(t) \bigr]^{2} \\ &\quad\leq \bigl(I_{a,k}^{\alpha,r}\varphi_{2}(t)-I_{a,k}^{\alpha ,r}f(t) \bigr) \bigl(I_{a,k}^{\alpha,r}f(t)-I_{a,k}^{\alpha,r} \varphi_{1}(t) \bigr) \\ &\qquad{}+\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl( \varphi_{1}(t)f(t) \bigr)-I_{a,k}^{\alpha ,r} \varphi_{1}(t)I_{a,k}^{\alpha,r}f(t) \\ &\qquad{}+\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl( \varphi_{2}(t)f(t) \bigr)-I_{a,k}^{\alpha ,r} \varphi_{2}(t)I_{a,k}^{\alpha,r}f(t) \\ &\qquad{}-\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl( \varphi_{1}(t)\varphi _{2}(t) \bigr)+I_{a,k}^{\alpha,r} \varphi_{1}(t)I_{a,k}^{\alpha,r}\varphi_{2}(t) \\ &\quad=S_{k}^{r}(f,\varphi_{1}, \varphi_{2}). \end{aligned}$$
(14)
Similarly,
$$\begin{aligned} &\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r}g^{2}(t)- \bigl[I_{a,k}^{\alpha,r}g(t) \bigr]^{2} \\ &\quad\leq \bigl(I_{a,k}^{\alpha,r}\psi_{2}(t)-I_{a,k}^{\alpha ,r}g(t) \bigr) \bigl(I_{a,k}^{\alpha,r}g(t)-I_{a,k}^{\alpha,r} \psi_{1}(t) \bigr) \\ &\qquad{}+\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl( \psi_{1}(t)g(t) \bigr)-I_{a,k}^{\alpha,r}\psi _{1}(t)I_{a,k}^{\alpha,r}g(t) \\ &\qquad{}+\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl( \psi_{2}(t)g(t) \bigr)-I_{a,k}^{\alpha,r}\psi _{2}(t)I_{a,k}^{\alpha,r}g(t) \\ &\qquad{}-\frac{(r+1)^{\frac{-\alpha}{k}}t^{(r+1)\frac{\alpha}{k}}}{\Gamma _{k}(\alpha+k)}I_{a,k}^{\alpha,r} \bigl( \psi_{1}(t)\psi_{2}(t) \bigr)+I_{a,k}^{\alpha ,r} \psi_{1}(t)I_{a,k}^{\alpha,r}\psi_{2}(t) \\ &\quad=S_{k}^{r}(g,\psi_{1},\psi_{2}). \end{aligned}$$
(15)
Equations (
14) and (
15) together with inequality (
13) yield inequality (
6). □