## Appendix B Global Stability Result Proof
We employ the following Lyapunov function:  
\[+(d_{2}+k_{2})s_{1}\left(1-\dfrac{x_{1}}{x}\dfrac{x+v_{1}}{x_{1}+ v_{1}}+\dfrac{v}{v_{1}}\dfrac{x+v_{1}}{x+v}\right)\]4 \[+(d_{2}+k_{2})s_{1}\left(1-\dfrac{x_{1}}{x}\dfrac{x+v_{1}}{x_{1}+ v_{1}}+\dfrac{v}{v_{1}}\dfrac{x+v_{1}}{x+v}\right)\]5 \[+(d_{2}+k_{2})s_{1}\left(1-\dfrac{x_{1}}{x}\dfrac{x+v_{1}}{x_{1}+ v_{1}}+\dfrac{v}{v_{1}}\dfrac{x+v_{1}}{x+v}\right)\]6  
We have  
\[+(d_{2}+k_{2})s_{1}\left(1-\dfrac{x_{1}}{x}\dfrac{x+v_{1}}{x_{1}+ v_{1}}+\dfrac{v}{v_{1}}\dfrac{x+v_{1}}{x+v}\right)\]7 \[+(d_{2}+k_{2})s_{1}\left(1-\dfrac{x_{1}}{x}\dfrac{x+v_{1}}{x_{1}+ v_{1}}+\dfrac{v}{v_{1}}\dfrac{x+v_{1}}{x+v}\right)\]8  
Since,  
\[\left\{ \begin{array}{l}\lambda=d_{1}x_{2}+(d_{2}+k_{2})s_{2},\\ k_{1}x_{2}v_{2}=(x_{2}+v_{2})(d_{2}+k_{2})s_{2},\\ \frac{s_{2}}{y_{2}}=\frac{d_{3}}{k_{2}}+\frac{pz_{2}}{k_{2}},\ \frac{y_{2}}{v_{2}}= \frac{d_{4}}{a},\ \frac{s_{2}}{v_{2}}=\frac{d_{3}d_{4}}{ak_{2}}+\frac{d_{4}pz_{2}}{ak_{2}}, \end{array} \right.\]  
we have  
\[\hat{\mathcal{L}}(x,s,y,v,z)= \lambda-d_{1}x-\frac{x_{2}}{x}\frac{x+v_{2}}{x_{2}+v_{2}}(\lambda -d_{1}x)+((d_{2}+k_{2})s_{2})\,\frac{v}{v_{2}}\,\frac{x_{2}+v_{2}}{x+v}-(d_{2 }+k_{2})s\] \[-\frac{s_{2}}{s}\left(\frac{k_{1}xv}{x+v}-(d_{2}+k_{2})s\right)+ \frac{d_{2}+k_{2}}{k_{2}}(k_{2}s-d_{3}y-pyz)\] \[-\frac{d_{2}+k_{2}}{k_{2}}\frac{y_{2}}{y}(k_{2}s-d_{3}y-pyz)+ \frac{d_{3}(d_{2}+k_{2})+(d_{2}+k_{2})pz_{2}}{ak_{2}}\] \[\times(ay-d_{4}v)-\frac{d_{3}(d_{2}+k_{2})+(d_{2}+k_{2})pz_{2}}{ak _{2}}\frac{v_{2}}{v}(ay-d_{4}v)\] \[+\frac{p}{c}\frac{d_{2}+k_{2}}{k_{2}}(cyz-bz)-\frac{pz_{2}}{cz} \frac{d_{2}+k_{2}}{k_{2}}(cyz-bz).\]  
On the other hand, we have  
\[\left\{\begin{array}{l}\lambda-d_{1}x=d_{1}x_{2}+d_{2}+k_{2})s_{2}-d_{1}x, \\ \lambda-d_{1}x-(d_{2}+k_{2})s_{2}\frac{x+v_{2}}{k_{1}xv_{2}}\left(\lambda-d_{1 }x-\frac{k_{1}xv}{x+v}\right)=d_{1}x_{2}(1-\frac{x}{x_{2}}-\frac{x_{2}}{x} \frac{x+v_{2}}{x_{2}+v_{2}}  
\[\left\{ \begin{array}{l}\dfrac{(d_{2}+k_{2})pz_{2}}{ak_{2}}(ay-d_{4}v)-\dfrac{(d_ {2}+k_{2})pz_{2}}{ak_{2}}\dfrac{v_{2}}{v}(ay-d_{4}v)=\dfrac{(d_{2}+k_{2})pz_{2}y }{k_{2}}\\ \quad-\dfrac{(d_{2}+k_{2})pz_{2}y_{2}}{k_{2}}\dfrac{v}{v_{2}}-\dfrac{(d_{2}+k_{ 2})pz_{2}y}{k_{2}}\dfrac{v_{2}}{v}+\dfrac{(d_{2}+k_{2})pz_{2}y_{2}}{k_{2}},\\ \dfrac{(d_{2}+k_{2})pz_{2}}{ak_{2}}(ay-d_{4}v)-\dfrac{(d_{2}+k_{2})pz_{2}}{ak_{2 }}\dfrac{v_{2}}{v}(ay-d_{4}v)-\dfrac{z_{2}}{z}\dfrac{p}{c}\dfrac{d_{2}+k_{2}}{ k_{2}}\left(xyz-bz\right)\\ \quad+\dfrac{(d_{2}+k_{2})}{k_{2}}pz_{2}y_{2}\dfrac{v}{v_{2}}-\dfrac{(d_{2}+k_ {2})}{k_{2}}pz_{2}y_{2}+\dfrac{(d_{2}+k_{2})}{k_{2}}pz_{2}y_{2}\dfrac{v_{2}}{v }-\dfrac{(d_{2}+k_{2})}{k_{2}}pz_{2}y_{2}=0.\end{array} \right.\]  
All of these imply  
\[ \begin{array}{l}\dot{L}=-\dfrac{d_{1}v_{2}}{x(x_{2}+v_{2})}(x-x_{2})^{2}\\ \quad-(d_{2}+k_{2})s_{2}\left(\dfrac{x(v-v_{2})^{2}}{v_{2}(x+v_{2})(x+v)}\right) \\ \quad+(d_{2}+k_{2})s_{2}\left(5-\dfrac{x_{2}}{x}\dfrac{x+v_{2}}{x_{2}+v_{2}}- \dfrac{s_{2}}{s}\dfrac{xv}{x_{2}v_{2}}\dfrac{x_{2}+v_{2}}{x+v}-\dfrac{sy_{2}}{ s_{2}y}-\dfrac{yv_{2}}{y_{2}v}-\dfrac{x+v}{x+v_{2}}\right).\end{array} \]  
Again, since the arithmetic mean is greater than or equal to the geometric mean, it follows that  
\[5-\dfrac{x_{2}}{x}\dfrac{x+v_{2}}{x_{2}+v_{2}}-\dfrac{s_{2}}{s}\dfrac{xv}{x_{ 2}v_{2}}\dfrac{x_{2}+v_{2}}{x+v}-\dfrac{sy_{2}}{s_{2}y}-\dfrac{yv_{2}}{y_{2}v }-\dfrac{x+v}{x+v_{2}}\leq 0,\]  
which means that \(\dot{L}\leq 0\), and the equality hods when \(x=x_{2}\), \(s=s_{2}\), \(y=y_{2}\), \(v=v_{2}\) and \(z=z_{2}\). By the LaSalle invariance principle, the endemic point \(E_{2}\) is globally stable.