## 3 Main Results
: _If \(p=1\) and \(q\to 1\), then (19) reduces to:_  
\[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]8 \[\quad-\tfrac{1}{(b-a)^{2}}\int_{a}^{pb+(1-p)a}\int_{a}^{pb+(1-p)a} f(tx+(1-t)y)\,_{a}d_{p,q}x_{a}d_{p,q}y \tag{17}\]9  
_which readily appeared in [25]._  
**Theorem 9**. : _Let \(f:[a,b]\to\mathbb{R}\) be a \((p,q)\)-differentiable convex continuous function, which is defined at the point \(\frac{qa+pb}{p+q}\in(a,b)\) and \(0<q<p\leq 1\). Then, the following inequalities:_  
\[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]0 \[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]1  
_are valid for all \(t\in[0,1]\)._  
Proof. : Since \(f\) is convex on \([a,b]\) and using Theorem 3, we have:  
\[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]2 \[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]3 \[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]4  
\[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]5  
for all \(t\in[0,1]\). On the other hand, since \(f\) is the \((p,q)\)-differentiable convex on \([a,b]\), we have:  
\[\leq t\left[\tfrac{p^{2}f(a)+pqf(pb+(1-p)a)}{p+q}-\tfrac{p}{b-a} \int_{a}^{pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right],\]6Taking the double \((p,q)\)-integration on both sides of the above inequality on \([a,b]\), we obtain:  
\[ \begin{array}{ll}\frac{1}{p(b-a)}&\int_{a}^{pb+(1-p)a}f\left(tx+(1-t)\frac{qa+pb }{p+q}\right)\,_{a}d_{p,q}x-\frac{1}{p(b-a)}\int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x\\ &\geq\frac{(1-t)}{p(b-a)}\int_{a}^{pb+(1-p)a}\left(\frac{qa+pb}{p+q}-x\right) \,_{a}D_{p,q}f(x)_{a}d_{p,q}x.\end{array}  \tag{21}\]  
Since,  
\[ \begin{array}{ll}\int_{a}^{pb+(1-p)a}&\left(\frac{qa+pb}{p+q}-x \right)\,_{a}D_{p,q}f(x)_{a}d_{p,q}x\\ &=\int_{a}^{pb+(1-p)a}f(qx+(1-q)a)_{a}d_{p,q}x-(b-a)\frac{pf(a)+qf(pb+(1-p)a)}{ p+q}.\end{array}  \tag{22}\]  
This completes the proof. Corollary 3: _Let \(f:[a,b]\to\mathbb{R}\) be a \((p,q)\)-differentiable convex continuous function and \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{array}{ll}0&\leq\frac{1}{b-a}\int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x- \frac{2}{b-a}\int_{\frac{(p+2p)(b-a)+(p+2a)a+pb}{2(p+q)}}^{\frac{(p^{2}+pb)(b- a)+(p+2a)a+pb}{2(1+q)}}f(x)\,_{a}d_{p,q}x\\ &\leq\frac{1}{2}\left[\frac{pf(a)+qf(pb+(1-p)a)}{p+q}-\frac{1}{b-a}\int_{a}^{ pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right].\end{array}  \tag{23}\]  
Theorem 10: _Let \(f:[a,b]\to\mathbb{R}\) be a \((p,q)\)-differentiable convex continuous function, which is defined at the point \(\frac{pa+qb}{p+q}\in(a,b)\) and \(0<q<p\leq 1\). Then, the following inequalities:_  
\[ \begin{array}{ll}(1-t)&\frac{p(p-q)(b-a)}{p+q}f^{\prime}\left(\frac{pa+qb}{ p+q}\right)\\ &\leq\frac{1}{b-a}\int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x-\frac{1}{b-a}\int_{a} ^{pb+(1-p)a}f\left(tx+(1-t)\frac{pa+qb}{p+q}\right)\,_{a}d_{p,q}x\\ &\leq(1-t)\left[\frac{qf(a)+pf(pb+(1-p)a)}{p+q}-\frac{1}{b-a}\int_{a}^{pb+(1- p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right]\end{array}  \tag{24}\]  
_are valid for all \(t\in[0,1]\)._  
Proof: The proof of this theorem follows a similar procedure as Theorem 9 by using Theorem 4. Corollary 4: _Let \(f:[a,b]\to\mathbb{R}\) be a \((p,q)\)-differentiable convex continuous function and \(0<q<p\leq 1\). Then, we have:_  
\[ \begin{array}{ll}&\frac{p(p-q)(b-a)}{2(p+q)}f^{\prime}\left(\frac{pa+qb}{p+q }\right)\\ &\leq\frac{1}{b-a}\int_{a}^{pb+(1-p)a}f(x)\,_{a}d_{p,q}x-\frac{2}{b-a}\int_{ \frac{2p+q(a+b)}{2(p+q)}}^{\frac{(p^{2}+pq)(b-a)+(2p+q)a+pb}{2(1+q)}}f(x)\,_{a} d_{p,q}x\\ &\leq\frac{1}{2}\left[\frac{qf(a)+pf(pb+(1-p)a)}{p+q}-\frac{1}{b-a}\int_{a}^{ pb+(1-p)a}f(qx+(1-q)a)\,_{a}d_{p,q}x\right].\end{array}