*Proof of Theorem 3.1*. It follows from (15) that one could choose enough small positive constant ε_{1} > 0 such that
$$\frac{{r}_{1L}}{{r}_{2M}}>max\left\{\frac{{a}_{1M}({d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{1})+{f}_{2M}({M}_{2}+{\epsilon}_{1}))}{{b}_{2L}},\frac{{c}_{1M}}{{c}_{2L}},\frac{{b}_{1M}}{{d}_{1L}{a}_{2L}}\right\}.$$(24)

(24) is equivalent to
$$\begin{array}{l}\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{1})+{f}_{2M}({M}_{2}+{\epsilon}_{1})}}<\frac{{r}_{1L}}{{r}_{2M}},\\ \frac{\frac{{b}_{1M}}{{d}_{1L}}}{{a}_{2L}}<\frac{{r}_{1L}}{{r}_{2M}},\frac{{c}_{1M}}{{c}_{2L}}<\frac{{r}_{1L}}{{r}_{2M}}.\end{array}$$(25)

Therefore, there exist two constants *α, β* such that
$$\begin{array}{l}\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{1})+{f}_{2M}({M}_{2}+{\epsilon}_{1})}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}},\\ \frac{\frac{{b}_{1M}}{{d}_{1L}}}{{a}_{2L}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}},\frac{{c}_{1M}}{{c}_{2L}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}}.\end{array}$$(26)

That is
$$\begin{array}{l}\alpha {c}_{1M}-\beta {c}_{2L}<0,\alpha {a}_{1M}-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{1})+{f}_{2M}({M}_{2}+{\epsilon}_{1})}<0,\\ \alpha \frac{{b}_{1M}}{{d}_{1L}}-\beta {a}_{2L}<0,-\alpha {r}_{1L}+\beta {r}_{2M}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}-{\delta}_{1}<0.\end{array}$$(27)

Let *x*(*t*)* =* (*x*_{1}(*t*)*, x*_{2}(*t*))^{T} be a solution of system (1) with *x*_{i}(0) > 0, *i* = 1,2. For above *ε*_{1} > 0, from Lemma 2.2 there exists *T*_{1} large enough such that
$${x}_{1}(t)<{M}_{1}+{\epsilon}_{1},{x}_{2}(t)<{M}_{2}+{\epsilon}_{1}\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{1}.$$(28)

From(l) we have
$$\frac{{\dot{x}}_{1}(t)}{{x}_{1}(t)}={r}_{1}(t)-{a}_{1}(t){x}_{1}(t)-\frac{{b}_{1}(t){x}_{2}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}-{c}_{1}(t){x}_{1}(t){x}_{2}(t),\frac{{\dot{x}}_{2}(t)}{{x}_{2}(t)}={r}_{2}(t)-\frac{{b}_{2}(t){x}_{1}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}-{a}_{2}(t){x}_{2}(t)-{c}_{2}(t){x}_{1}(t){x}_{2}(t).$$(29)

Let
$$V(t)={x}_{1}^{-\alpha}(t){x}_{2}^{\beta}(t).$$

From (27)-(29), for *t ≥* *Τ*_{1}, it follows that
$$\begin{array}{l}\begin{array}{l}\dot{V}(t)=V(t)\left[-\alpha ({r}_{1}(t)-{a}_{1}(t){x}_{1}(t)-\frac{{b}_{1}(t){x}_{2}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}\right.\\ \begin{array}{ll}& -{c}_{1}(t){x}_{1}(t){x}_{2}(t))+\beta ({r}_{2}(t)-{a}_{2}(t){x}_{2}(t)\\ & \left.-\frac{{b}_{2}(t){x}_{1}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}-{c}_{2}(t){x}_{1}(t){x}_{2}(t))\right]\end{array}\end{array}\\ \\ =V(t)[(-\alpha {r}_{1}(t)+\beta {r}_{2}(t))+(\alpha {c}_{1}(t)-\beta {c}_{2}(t)){x}_{1}(t){x}_{2}(t)\\ \begin{array}{ll}& +(\alpha {a}_{1}(t)-\frac{\beta {b}_{2}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}){x}_{1}(t)\\ & +(\alpha \frac{{b}_{1}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}-\beta {a}_{2}(t)){x}_{2}(t)]\end{array}\\ \le V(t)[(-\alpha {r}_{1L}+\beta {r}_{2M})+(\alpha {c}_{1M}-\beta {c}_{2L}){x}_{1}(t){x}_{2}(t)\\ \begin{array}{ll}& +(\alpha {a}_{1M}-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{1})+{f}_{2M}({M}_{2}+{\epsilon}_{1})}){x}_{1}(t)\\ & +(\alpha \frac{{b}_{1M}}{{d}_{1L}}-\beta {a}_{2L}){x}_{2}(t)]\end{array}\\ \le -{\delta}_{1}V(t),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{1}.\end{array}$$

Integrating this inequality from *Τ*_{1} to *t* (≥ *Τ*_{1} ), it follows
$$V(t)\le V({T}_{1})\mathrm{exp}(-{\delta}_{1}(t-{T}_{1})).$$(30)

By Lemma 2.2 we know that there exists *M > M*_{0}* >* 0 such that
$${x}_{i}(t)<M\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}i=1,2\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{1}.$$(31)

Therefore, (30) implies that
$${x}_{2}(t)<C\mathrm{exp}\left(-\frac{{\delta}_{1}}{\beta}(t-{T}_{1})\right),$$(32)

where
$$C={M}^{\alpha /\beta}({x}_{1}({T}_{1}){)}^{-\alpha /\beta}{x}_{2}({T}_{1})>0.$$(33)

Consequently, we have *x*_{2}(*t*) *→* 0 exponentially as *t* *→ +∞*. This ends the proof of Theorem 3.1.

*Proof of Theorem 3.3*. It follows from (16) and (17) that one could choose a positive constant *ε*_{2} > 0 small enough such that
$$\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{2})+{f}_{2M}({M}_{2}+{\epsilon}_{2})}}<\frac{{r}_{1L}-{c}_{1M}\left({M}_{1}+{\epsilon}_{2}\right)\left({M}_{2}+{\epsilon}_{2}\right)}{{r}_{2M}}$$(34)

and
$$\frac{{b}_{1M}}{{d}_{1L}{a}_{2L}}<\frac{{r}_{1L}-{c}_{1M}\left({M}_{1}+{\epsilon}_{2}\right)\left({M}_{2}+{\epsilon}_{2}\right)}{{r}_{2M}}$$(35)

hold. Therefore, there exist two constants *a, β* such that
$$\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{2})+{f}_{2M}({M}_{2}+{\epsilon}_{2})}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}-{c}_{1M}\left({M}_{1}+{\epsilon}_{2}\right)\left({M}_{2}+{\epsilon}_{2}\right)}{{r}_{2M}}$$(36)

and
$$\frac{{b}_{1M}}{{d}_{1L}{a}_{2L}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}-{c}_{1M}({M}_{1}+{\epsilon}_{2})({M}_{2}+{\epsilon}_{2})}{{r}_{2M}}$$(37)

hold. That is
$$\begin{array}{l}\alpha {a}_{1M}-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{2})+{f}_{2M}({M}_{2}+{\epsilon}_{2})}<0,\\ \alpha \frac{{b}_{1M}}{{d}_{1L}}-\beta {a}_{2L}<0,\\ -\alpha {r}_{1L}+\beta {r}_{2M}+\alpha {c}_{1M}({M}_{1}+{\epsilon}_{2})({M}_{2}+{\epsilon}_{2})\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}-{\delta}_{2}<0.\end{array}$$(38)

Let *x*(*t*)* =* (*x*_{1}(*t*)*, x*_{2}(*t*))^{T} be a solution of system (1) with *x*_{i} (0) > 0, *i* = 1, 2. For above *ε*_{2} > 0, from Lemma 2.2 there exists *T*_{2} large enough such that
$${x}_{1}(t)<{M}_{1}+{\epsilon}_{2},{x}_{2}(t)<{M}_{2}+{\epsilon}_{2}\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{2}.$$(39)

Let
$$V(t)={x}_{1}^{-\alpha}(t){x}_{2}^{\beta}(t).$$

From (29), (38) and (39), for *t ≥* *T*_{2}, it follows that
$$\begin{array}{lll}\dot{V}(t)& =V(t)& \left[-\alpha ({r}_{1}(t)-{a}_{1}(t){x}_{1}(t)-\frac{{b}_{1}(t){x}_{2}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}\right.\\ & & -{c}_{1}(t){x}_{1}(t){x}_{2}(t))+\beta ({r}_{2}(t)-{a}_{2}(t){x}_{2}(t)\\ & & \left.-\frac{{b}_{2}(t){x}_{1}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}-{c}_{2}(t){x}_{1}(t){x}_{2}(t))\right]\\ & \le V(t)& \left[(-\alpha {r}_{1}(t)+\beta {r}_{2}(t))+\alpha {c}_{1}(t){x}_{1}(t){x}_{2}(t)\right.\\ & & +(\alpha {a}_{1}(t)-\frac{\beta {b}_{2}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}){x}_{1}(t)\\ & & \left.+(\alpha \frac{{b}_{1}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}-\beta {a}_{2}(t)){x}_{2}(t)\right]\\ & \le V(t)& \left[-\alpha {r}_{1L}+\beta {r}_{2M}+\alpha {c}_{1M}({M}_{1}+{\epsilon}_{2})({M}_{2}+{\epsilon}_{2})\right.\\ & & +(\alpha {a}_{1M}-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{2})+{f}_{2M}({M}_{2}+{\epsilon}_{2})}){x}_{1}(t)\\ & & \left.+(\alpha \frac{{b}_{1M}}{{d}_{1L}}-\beta {a}_{2L}){x}_{2}(t)\right]\\ & & \le -{\delta}_{2}V(t),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{2}.\end{array}$$

Integrating this inequality from *T*_{2} to *t* (≥ *T*_{2}), it follows
$$V(t)\le V({T}_{2})\mathrm{exp}(-{\delta}_{2}(t-{T}_{2})).$$(40)

From (40), similarly to the analysis of (30)-(33), we have *x*_{2}(*t*) *→* 0 exponentially as *t →* +∞. This ends the proof of Theorem 3.3.

*Proof of Theorem 3.4*. It follows from (18) and (19) that one could choose a positive constant *ε*_{3} > 0 small enough such that
$$\frac{{a}_{1M}+{c}_{1M}\left({M}_{2}+{\epsilon}_{3}\right)}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{3})+{f}_{2M}({M}_{2}+{\epsilon}_{3})}}<\frac{{r}_{1L}}{{r}_{2M}}$$(41)

and
$$\frac{{b}_{1M}}{{d}_{1L}{a}_{2L}}<\frac{{r}_{1L}}{{r}_{2M}}.$$(42)

hold. Therefore, there exist two constants *α, β* such that
$$\frac{{a}_{1M}+{c}_{1M}\left({M}_{2}+{\epsilon}_{3}\right)}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{3})+{f}_{2M}({M}_{2}+{\epsilon}_{3})}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}}$$(43)

and
$$\frac{{b}_{1M}}{{d}_{1L}{a}_{2L}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}}$$(44)

hold. That is
$$\begin{array}{l}\alpha {a}_{1M}+\alpha {c}_{1M}({M}_{2}+{\epsilon}_{3})-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{3})+{f}_{2M}({M}_{2}+{\epsilon}_{3})}<0,\\ \alpha \frac{{b}_{1M}}{{d}_{1L}}-\beta {a}_{2L}<0,\phantom{\rule{thickmathspace}{0ex}}-\alpha {r}_{1L}+\beta {r}_{2M}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}-{\delta}_{3}<0.\end{array}$$(45)

Let *x*(*t*)* =* (*x*_{1}(*t*)*, x*_{2}(*t*))^{T} be a solution of system (1) with x_{i}(0) > 0, *i* = 1,2. For above *ε*_{3} > 0, from Lemma 2.2 there exists *T*_{3} large enough such that
$${x}_{1}(t)<{M}_{1}+{\epsilon}_{3},{x}_{2}(t)<{M}_{2}+{\epsilon}_{3}\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{3}.$$(46)

Let
$$V(t)={x}_{1}^{-\alpha}(t){x}_{2}^{\beta}(t).$$

From (29), (45) and (46), for *t* ≥ *T*_{3}, it follows that
$$\begin{array}{lll}\dot{V}(t)& =V(t)& \left[-\alpha ({r}_{1}(t)-{a}_{1}(t){x}_{1}(t)-\frac{{b}_{1}(t){x}_{2}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}\right.\\ & & -{c}_{1}(t){x}_{1}(t){x}_{2}(t))+\beta ({r}_{2}(t)-{a}_{2}(t){x}_{2}(t)\\ & & \left.-\frac{{b}_{2}(t){x}_{1}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}-{c}_{2}(t){x}_{1}(t){x}_{2}(t))\right]\\ & \le V(t)& \left[(-\alpha {r}_{1}(t)+\beta {r}_{2}(t))+(\alpha {a}_{1}(t)+\alpha {c}_{1}(t){x}_{2}(t)\right.\\ & & -\frac{\beta {b}_{2}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}){x}_{1}(t)\\ & & \left.+(\alpha \frac{{b}_{1}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}-\beta {a}_{2}(t)){x}_{2}(t)\right]\end{array}$$

$$\begin{array}{lll}& \le V(t)& \left[-\alpha {r}_{1L}+\beta {r}_{2M}+(\alpha {a}_{1M}+\alpha {c}_{1M}({M}_{2}+{\epsilon}_{3})\right.\\ & & -\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{3})+{f}_{2M}({M}_{2}+{\epsilon}_{3})}){x}_{1}(t)\\ & & \left.+(\alpha \frac{{b}_{1M}}{{d}_{1L}}-\beta {a}_{2L}){x}_{2}(t)\right]\\ & & \le -{\delta}_{3}V(t),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{3}.\end{array}$$

Integrating this inequality from *T*_{3} to *t* (≥ *T*_{3}), it follows
$$V(t)\le V({T}_{3})\mathrm{exp}(-{\delta}_{3}(t-{T}_{3})).$$(47)

From (47), similarly to the analysis of (30)-(33), we have *x*_{2}(*t*)* →* 0 exponentially as *t →* +∞. This ends the proof of Theorem 3.4.

*Proof of Theorem 3.5*. It follows from (20) and (21) that one could choose a positive constant *ε*_{4} > 0 small enough such that
$$\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{4})+{f}_{2M}({M}_{2}+{\epsilon}_{4})}}<\frac{{r}_{1L}}{{r}_{2M}}$$(48)

and
$$\frac{\frac{{b}_{1M}}{{d}_{1L}}+{c}_{1M}({M}_{1}+{\epsilon}_{4})}{{d}_{1L}{a}_{2L}}<\frac{{r}_{1L}}{{r}_{2M}}.$$(49)

hold. Therefore, there exist two constants *a, β* such that
$$\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{4})+{f}_{2M}({M}_{2}+{\epsilon}_{4})}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}}$$$$\frac{{a}_{1M}}{\frac{{b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{4})+{f}_{2M}({M}_{2}+{\epsilon}_{4})}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}}$$(50)

and
$$\frac{\frac{{b}_{1M}}{{d}_{1L}}+{c}_{1M}({M}_{1}+{\epsilon}_{4})}{{d}_{1L}{a}_{2L}}<\frac{\beta}{\alpha}<\frac{{r}_{1L}}{{r}_{2M}}$$(51)

hold. That is
$$\begin{array}{l}\alpha {a}_{1M}-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{4})+{f}_{2M}({M}_{2}+{\epsilon}_{4})}<0,\\ \alpha \frac{{b}_{1M}}{{d}_{1L}}+\alpha {c}_{1M}({M}_{2}+{\epsilon}_{4})-\beta {a}_{2L}<0,\\ -\alpha {r}_{1L}+\beta {r}_{2M}\stackrel{\mathrm{d}\mathrm{e}\mathrm{f}}{=}-{\delta}_{3}<0.\end{array}$$(52)

Let *x*(*t*)* =* (*x*_{1}(*t*)*, x*_{2}(*t*))^{T} be a solution of system (1) with *x*_{i} (0) > 0, *i* = 1,2. For above ε4 > 0, from Lemma 2.2 there exists *T*_{4}large enough such that
$${x}_{1}(t)<{M}_{1}+{\epsilon}_{4},\phantom{\rule{thinmathspace}{0ex}}{x}_{2}(t)<{M}_{2}+{\epsilon}_{4}\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{4}.$$(53)

Let
$$V(t)={x}_{1}^{-\alpha}(t){x}_{2}^{\beta}(t).$$

From (29), (52) and (53), for *t ≥* *T*_{4}, it follows that
$$\begin{array}{l}\begin{array}{lll}\dot{V}(t)& =V(t)& \left[-\alpha ({r}_{1}(t)-{a}_{1}(t){x}_{1}(t)-\frac{{b}_{1}(t){x}_{2}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}\right.\\ & & -{c}_{1}(t){x}_{1}(t){x}_{2}(t))+\beta ({r}_{2}(t)-{a}_{2}(t){x}_{2}(t)\\ & & \left.-\frac{{b}_{2}(t){x}_{1}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}-{c}_{2}(t){x}_{1}(t){x}_{2}(t))\right]\\ & \le V(t)& \left[(-\alpha {r}_{1}(t)+\beta {r}_{2}(t))+(\alpha {c}_{1}(t){x}_{1}(t)-\beta {a}_{2}(t)\right.\\ & & +\alpha \frac{{b}_{1}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}){x}_{2}(t)\\ & & \left.+(\alpha {a}_{1}(t)-\frac{\beta {b}_{2}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}){x}_{1}(t)\right]\end{array}\\ \le V(t)[(-\alpha {r}_{1L}+\beta {r}_{2M})\\ \begin{array}{ll}& +(\alpha {a}_{1M}-\frac{\beta {b}_{2L}}{{d}_{2M}+{e}_{2M}({M}_{1}+{\epsilon}_{4})+{f}_{2M}({M}_{2}+{\epsilon}_{4})}){x}_{1}(t)\\ & \left.+(\alpha \frac{{b}_{1M}}{{d}_{1L}}+\alpha {c}_{1M}({M}_{1}+{\epsilon}_{4})-\beta {a}_{2L}){x}_{2}(t)\right]\end{array}\\ \le -{\delta}_{4}V(t),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{4}.\end{array}$$

Integrating this inequality from *T*_{4}to *t* (≥ *T*_{4}), it follows
$$V(t)\le V({T}_{4})\mathrm{exp}(-{\delta}_{4}(t-{T}_{4})).$$(54)

From (54), similarly to the analysis of (30)-(33), we have *x*_{2}(*t*)* →* 0 exponentially as *t →* +∞. This ends the proof of Theorem 3.5.

*Proof of Theorem 3.8*. By applying Lemma 2.3 and 2.4, the proof of Theorem 3.8 is similar to that of the proof of Theorem in [4]. We omit the detail here.

*Proof of Theorem 3.9*. Conditions (22) and (23) imply that there exist two constants *α, β* and a positive constant *ε*_{5} small enough, such that
$$\begin{array}{l}\frac{{r}_{1M}}{{r}_{2L}}<\frac{\beta}{\alpha}<\frac{\frac{{b}_{1L}}{{d}_{1M}+{e}_{1M}({M}_{1}+{\epsilon}_{5})+{f}_{1M}({M}_{2}+{\epsilon}_{5})}}{{a}_{2M}},\\ \frac{{r}_{1M}}{{r}_{2L}}<\frac{\beta}{\alpha}<\frac{{a}_{1L}{d}_{2L}}{{b}_{2L}}.\end{array}$$(55)

That is
$$\begin{array}{l}-\alpha {a}_{1L}+\beta \frac{{b}_{2M}}{{d}_{2L}}<0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\alpha {r}_{1M}-\beta {r}_{2L}\stackrel{\text{def}}{=}-{\delta}_{4}<0,\\ -\frac{\alpha {b}_{1L}}{{d}_{1M}+{e}_{1M}({M}_{1}+{\epsilon}_{5})+{f}_{1M}({M}_{2}+{\epsilon}_{5})}+\beta {a}_{2M}<0.\end{array}$$(56)

For above *ε*_{5} > 0, from Lemma 2.2 there exists *T*_{5}large enough such that
$${x}_{2}(t)<{M}_{2}+{\epsilon}_{5}\phantom{\rule{thinmathspace}{0ex}}\text{for all}\phantom{\rule{thinmathspace}{0ex}}t\ge {T}_{5}.$$(57)

Let
$${V}_{1}(t)={x}_{1}^{\alpha}(t){x}_{2}^{-\beta}(t).$$

It follows from (56) and (57) that
$$\begin{array}{ll}{\dot{V}}_{1}(t)& ={V}_{1}(t)\left[\alpha ({r}_{1}(t)-{a}_{1}(t){x}_{1}(t)-\frac{{b}_{1}(t){x}_{2}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}\right.\\ & -{c}_{1}(t){x}_{1}(t){x}_{2}(t))\\ & \left.-\beta (r2(t)-\frac{{b}_{2}(t){x}_{1}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}-a2(t){x}_{2}(t))\right]\\ & ={V}_{1}(t)\left[(\alpha {r}_{1}(t)-\beta {r}_{2}(t)\right.\\ & +(-\alpha {a}_{1}(t)+\frac{\beta {b}_{2}(t)}{{d}_{2}(t)+{e}_{2}(t){x}_{1}(t)+{f}_{2}(t){x}_{2}(t)}){x}_{1}(t)\\ & \left.+(-\frac{\alpha {b}_{1}(t)}{{d}_{1}(t)+{e}_{1}(t){x}_{1}(t)+{f}_{1}(t){x}_{2}(t)}+\beta a2(t))x2(t)\right]\\ & \le {V}_{1}(t)\left[(\alpha {r}_{1M}-\beta {r}_{2L})+(-\alpha {a}_{1L}+\beta \frac{{b}_{2M}}{{d}_{2L}}){x}_{1}(t)\right.\\ & \left.+(-\frac{\alpha {b}_{1L}}{{d}_{1M}+{e}_{1M}({M}_{1}+{\epsilon}_{5})+{f}_{1M}({M}_{2}+{\epsilon}_{5})}+\beta {a}_{2M}){x}_{2}(t)\right]\\ & \le -{\delta}_{5}{V}_{1}(t)\phantom{\rule{thickmathspace}{0ex}}.\end{array}$$

Integrating this inequality from *T*_{5} to *t* (≥ *T*_{5}), it follows
$${V}_{1}\left(t\right)\le {V}_{1}\left({T}_{5}\right)\mathrm{exp}\left(-{\delta}_{5}\left(t-{T}_{5}\right)\right).$$(58)

From this, similarly to the analysis of (30)-(33), we have *x*_{1}(*t*)* →* 0 exponentially as *t →* +∞. The rest of the proof of Theorem 3.9 is similar to that of the proof of Theorem in [4]. We omit the detail here.

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