## 3 Main Results
\begin{split}\frac{1}{(pb-pa)^{2}}&\int_{a}^{pb+(1 -p)a}\int_{a}^{pb+(1-p)a}f\left(\frac{px+qy}{p+q}\right)_{a}d_{q}x_{a}d_{q}y\\ &=g\left(\frac{p}{p+q}\right)\leq\frac{1}{p}\int_{0}^{p}g(t)\,_{a }d_{p,q}t\\ &=\frac{1}{p(pb-pa)^{2}}\int_{0}^{p}\int_{a}^{pb+(1-p)a}\int_{a}^{ pb+(1-p)a}f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y_{a}d_{p,q}t\\ &\leq\frac{qg(0)+pg(1)}{p(p+q)}=\frac{1}{p(pb-pa)}\int_{a}^{pb+( 1-p)a}f(x)\,_{a}d_{p,q}x.\end{split} \]  
This completes the proof. **Remark 3**. : _If \(p=1\) and \(q\to 1\), then (16) reduces to:_  
\[ \begin{split}\frac{1}{(b-a)^{2}}\int_{a}^{b}\int_{a}^{b}f\left( \frac{x+y}{2}\right)dxdy&\leq\frac{1}{(b-a)^{2}}\int_{0}^{1}\int_{a }^{b}\int_{a}^{b}f(tx+(1-t)y)dxdydt\\ &\leq\frac{1}{b-a}\int_{a}^{b}f(x)dx,\end{split} \]  
_which readily appeared in [25]._  
**Theorem 8**. : _Let \(f:[a,b]\to\mathbb{R}\) be a \((p,q)\)-differentiable convex continuous function and \(0<q<p\leq 1\), then the following inequalities:_  
\[0 \leq\tfrac{p}{b-a}\int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x\] \[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\] \[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]  
_are valid for all \(t\in[0,1]\)._  
Proof. : Since \(f\) is convex on \(J\), it follows that:  
\[f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)\]  
for all \(x,y\in[a,b]\) and \(t\in[0,1]\). Taking double \((p,q)\)-integration on both sides of the above inequality on \([a,pb+(1-p)a]\times[a,pb+(1-p)a]\), we obtain:  
\[\int_{a}^{pb+(1-p)a} \int_{a}^{pb+(1-p)a}f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y\] \[\leq\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}[tf(x)+(1-t)f(y)]\,_{a }d_{p,q}x_{a}d_{p,q}y\] \[=p(b-a)\int_{a}^{pb+(1-p)a}f(x)_{a}d_{p,q}x.\]  
On the other hand, since \(f\) is \((p,q)\)-differentiable convex on \([a,b]\) and \(f^{\prime}\geq\,_{a}D_{p,q}f\), we have:  
\[f(tx+(1-t)y)-f(y)\geq t(x-y)\,_{a}D_{p,q}f(y)\]  
for all \(x,y\in[a,b]\) and \(t\in[0,1]\). Taking the double \((p,q)\)-integration on both sides of the above inequality on \([a,pb+(1-p)a]\times[a,pb+(1-p)a]\), we obtain:  
\[\int_{a}^{pb+(1-p)a} \int_{a}^{pb+(1-p)a}f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y-(pb-pa) \int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x\] \[\geq t\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a}(x-y)_{a}D_{p,q}f(y) \,_{a}d_{p,q}x_{a}d_{p,q}y. \tag{18}\]  
Since,  
\[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]0 \[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]1  
Substituting the above inequality in (18), we have:  
\[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]2 \[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]3  
for all \(t\in[0,1]\), which completes the proof. **Remark 4**. : _If \(p=1\) and \(q\to 1\), then (17) reduces to:_  
\[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]4 \[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]5  
_which readily appeared in [25; 26]._  
**Corollary 2**. : _Let \(f:[a,b]\to\mathbb{R}\) be a \((p,q)\)-differentiable convex continuous function and \(0<q<p\leq 1\). Then, we have:_  
\[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]6 \[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]7  
**Remark 5**.