## 2 Preliminary Results
\(\left(1\right)\) For each \(\eta\in\left(0,1\right)\), there exists a \(T_{\eta}\geq t_{1}\) such that, for all \(t\geq T_{\eta}\),  
\[tz^{\prime}(\tau(t))\geq\eta\tau(t)z^{\prime}(t).\]  
\(\left(2\right)\) For each \(t\in\left[t_{1},\infty\right)\),  
\[z\left(t\right)\geq\frac{1}{2}\left(t-t_{1}\right)z^{\prime}\left(t\right). \tag{4}\]  
\(\left(3\right)\) For each \(t\in\left[t_{1},\infty\right)\),  
\[z^{\prime}\left(t\right)\geq\left(t-t_{1}\right)z^{\prime\prime}\left(t\right).\]  
Assume that \(r_{i}^{\prime}\left(t\right)>0\) for \(i=1,2\), \(x\) is a positive solution of (1) with corresponding function \(z\in\widetilde{\aleph}\) for all \(t\geq t_{1}\geq t_{0}\). Then, the following facts are verified:  
\[z\left(t\right)\geq\frac{1}{2}\left(t-t_{0}\right)^{1+1/\alpha_{1}}\left( \frac{\left(L_{1}z\left(t\right)\right)^{\prime}}{r_{1}\left(t\right)}\right)^ {1/\alpha_{1}}, \tag{5}\]  
\[\left(L_{2}z\left(t\right)\right)^{\prime}\leq-kq\left(t\right)\left(1-p\left( g\left(t\right)\right)\right)^{\beta}z^{\beta}\left(g\left(t\right)\right), \tag{6}\]  
\[z^{\prime}\left(t\right)\geq\left(\frac{R_{2}\left(t,t_{1}\right)}{r_{1}\left( t\right)}\right)^{1/\alpha_{1}}\left(L_{2}z\left(t\right)\right)^{1/\beta} \tag{7}\]  
and there exists a \(t_{2}\geq t_{1}\) such that  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]0  
for all \(t\geq t_{2}\). Let \(x\) be a positive solution of (1) with corresponding function \(z\in\widetilde{\aleph}\) for all \(t\geq t_{1}\geq t_{0}\). Then, \(z\left(t\right)>0\), \(L_{1}z\left(t\right)>0\) and \(L_{2}z\left(t\right)>0\) for \(t\geq t_{1}\). Since \(L_{3}z\left(t\right)\leq 0\) and \(r_{2}^{\prime}\left(t\right)>0\), we get  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]1  
Thus,  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]2 \[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]3  
and so  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]4It follows from (4) that  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]5  
Since \(z\left(t\right)>x\left(t\right)\) and \(z^{\prime}\left(t\right)>0\), we obtain \(x\left(t\right)\geq\left(1-p\left(t\right)\right)z\left(t\right),\) and hence  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]6  
Now, we have  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]7 \[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]8  
and so  
\[\frac{d}{dt}L_{1}z(t)<-K_{1}(r_{2}(t))^{-1/a_{2}},\]9  
for \(t\geq t_{2}\geq t_{1}\). By integrating the latter inequality from \(t_{2}\) to \(t\) and using \(\left(L_{2}z\left(t\right)\right)^{\prime}<0\), we get  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]0 \[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]1  
The proof is complete. **Lemma 5**. : _Assume that \(x\) is a positive solution of (1) with corresponding function \(z\in\aleph\) for all \(t\geq t_{1}\geq t_{0}\). Then:_  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]2  
_for \(u\leq v\), and_  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]3  
Proof. : Suppose that \(x\) is positive solution of (1). Then, there exists a \(t_{1}\geq t_{0}\) such that \(x\left(t\right)>0\), \(x\left(\tau\left(t\right)\right)>0\) and \(x\left(g\left(t\right)\right)>0\) for \(t\geq t_{1}\). From Lemma 1, we obtain  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]4  
it follows from the monotonicity of \(L_{2}z\left(t\right)\) that  
\[L_{1}z(t)<L_{1}z(t_{3})-K_{1}\int_{t_{3}}^{t}(r_{2}(s))^{-1/a_{2}}\mathrm{d}s.\]5  
for \(v\geq u\geq t_{1}\).