Now let us state several lemmas which will be useful in proving the main results.
Lemma 3.1
Assume that
\(x_{2}(\theta)\geq0\), \(y_{2}(\theta)\geq0\)
are continuous on
\(\theta\in[\tau,0]\), and
\(x_{2}(0)>0\), \(y_{2}(0)>0\). Let
\((x_{2}(t),y_{2}(t))^{T}\)
be a any solution of system (1.6), then
\(x_{2}(t)>0\), \(y_{2}(t)>0\)
for all
\(t>0\).
The proof of Lemma 3.1 is similar to the proof of Theorem 1 in [1], so we omit its proof.
Lemma 3.2
[2]
Consider the following equation:
$$\begin{aligned}& x'(t)=bx(t\delta)a_{1}x(t)a_{2}x^{2}(t), \\& x(t)=\phi(t)>0, \quad \delta\leq t\leq0, \end{aligned}$$
and assume that
\(b, a_{2}>0\), \(a_{1}\geq0\), and
\(\delta\geq0\)
is a constant. Then

(i)
if
\(b\geq a_{1}\), then
\(\lim_{t\rightarrow+\infty }x(t)= \frac{ba_{1}}{a_{2}}\);

(ii)
if
\(b\leq a_{1}\), then
\(\lim_{t\rightarrow+\infty}x(t)=0\).
Lemma 3.3
(Fluctuation lemma [23])
Let
\(x(t)\)
be a bounded differentiable function on
\((\alpha,\infty)\), Then there exist sequences
\(\gamma_{n}\rightarrow\infty\), \(\sigma _{n}\rightarrow\infty\)
such that

(i)
\(x'(\gamma_{n})\rightarrow0 \)
and
\(x(\gamma_{n})\rightarrow\limsup_{t\rightarrow+\infty }x(t)=\overline{x}\)
as
\(n\rightarrow\infty\),

(ii)
\(x'(\sigma_{n})\rightarrow0 \)
and
\(x(\sigma_{n})\rightarrow\liminf_{t\rightarrow+\infty }x(t)=\underline{x}\)
as
\(n\rightarrow\infty\).
Lemma 3.4
Assume that
\(x_{2}(\theta), y_{2}(\theta )\geq0\)
are continuous on
\(\theta\in[\tau,0]\), and
\(x_{2}(0)>0\), \(y_{2}(0)>0\). Let
\((x_{2}(t),y_{2}(t))^{T}\)
be a any solution of system (1.6). If
\(\lambda_{2}>0\), then
$$\liminf_{t\rightarrow+\infty}y_{2}(t)\geq\frac{k_{2}\lambda_{2}}{a_{2}}. $$
Proof
From the second equation of system (1.6), we have
$$y'_{2}(t)\geq r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau _{2})d_{21}y_{2}(t)\frac{a_{2}y^{2}_{2}(t)}{k_{2}}. $$
Since \(\lambda_{2}>0\), and by applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\liminf_{t\rightarrow+\infty}y_{2}(t)\geq\frac {k_{2}(r_{2}e^{d_{22}\tau_{2}}d_{21})}{a_{2}}= \frac{k_{2}\lambda_{2}}{a_{2}}>0. $$
This completes the proof of Lemma 3.4. □
Now we start to prove the above results.
Proof of Theorem 2.2
From the first equation of system (1.6), we have
$$x'_{2}(t)< r_{1}e^{d_{11}\tau_{1}}x_{2}(t \tau _{1})d_{12}x_{2}(t)bx_{2}^{2}(t). $$
According to condition (\(\mathrm{H}_{1}\)), we know \(r_{1}e^{d_{11}\tau_{1}}d_{12}>0\). By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\limsup_{t\rightarrow+\infty}x_{2}(t)\leq\frac {\lambda_{1}}{b}. $$
So, for any small constant \(\varepsilon>0\), there exists a \(T_{1}>0\) such that
$$ x_{2}(t)\leq\frac{\lambda_{1}}{b}+\varepsilon \stackrel{\mathrm{def}}{=}M^{(1)}_{1},\quad t> T_{1}. $$
(3.1)
For \(t>T_{1}+\tau_{2}\), substituting (3.1) into the second equation of system (1.6), we have
$$y'_{2}(t)< r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau_{2})d_{21}y_{2}(t)\frac {a_{2}y^{2}_{2}(t)}{M^{(1)}_{1}+k_{2}}. $$
According to condition (\(\mathrm{H}_{1}\)), we have \(r_{2}e^{d_{22}\tau_{2}}d_{21}>0\). By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\limsup_{t\rightarrow+\infty}y_{2}(t)\leq\frac {\lambda_{2} (M^{(1)}_{1}+k_{2} )}{a_{2}}. $$
Then, for the above ε, there exists a \(T_{2}>T_{1}+\tau_{2}\), such that
$$ y_{2}(t)< \frac{\lambda_{2} (M^{(1)}_{1}+k_{2} )}{a_{2}} +\varepsilon\stackrel{\mathrm{def}}{=}M^{(1)}_{2},\quad t> T_{2}. $$
(3.2)
For \(t>T_{2}+\tau_{1}\), substituting (3.2) into the first equation of system (1.6), we have
$$\begin{aligned} x'_{2}(t) >& r_{1}e^{d_{11}\tau_{1}}x_{2}(t \tau _{1})d_{12}x_{2}(t)bx_{2}^{2}(t) \frac{a_{1}M^{(1)}_{2} x_{2}(t)}{k_{1}} \\ =& r_{1}e^{d_{11}\tau_{1}}x_{2}(t\tau_{1}) \biggl(d_{12}+\frac {a_{1}M^{(1)}_{2}}{k_{1}} \biggr)x_{2}(t)bx_{2}^{2}(t). \end{aligned}$$
Let
$$r_{1}e^{d_{11}\tau_{1}} \biggl(d_{12}+\frac {a_{1}M^{(1)}_{2}}{k_{1}} \biggr)\stackrel{\mathrm{def}}{=}\Delta. $$
Then substituting (3.1) and (3.2) into Δ, we have
$$\Delta= \frac{b(a_{2}k_{1}\lambda_{1}a_{1}k_{2}\lambda_{2})a_{1}\lambda_{1}\lambda_{2}}{a_{2}k_{1}b} \frac{a_{1}}{k_{1}} \biggl(\frac{\lambda_{2}}{a_{2}}+1 \biggr)\varepsilon. $$
Then, for small enough \(\varepsilon>0\) and condition (\(\mathrm{H}_{2}\)), we have
$$ \Delta=r_{1}e^{d_{11}\tau_{1}} \biggl(d_{12}+ \frac {a_{1}M^{(1)}_{2}}{k_{1}} \biggr)>0. $$
(3.3)
By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\liminf_{t\rightarrow+\infty}x_{2}(t)\geq\frac {\lambda_{1}\frac{a_{1}M^{(1)}_{2}}{k_{1}}}{b}. $$
Then, for the above \(\varepsilon>0\), there exists a \(T_{3}>T_{2}+\tau_{1}\), such that
$$ x_{2}(t)> \frac{\lambda_{1}\frac{a_{1}M^{(1)}_{2}}{k_{1}}}{b} \varepsilon\stackrel{\mathrm{def}}{=}m^{(1)}_{1}, \quad t>T_{3}. $$
(3.4)
For \(t>T_{3}+\tau_{2}\), substituting (3.4) into the second equation of system (1.6), we have
$$y'_{2}(t) > r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau _{2})d_{21}y_{2}(t)\frac{a_{2}y^{2}_{2}(t)}{m^{(1)}_{1}+k_{2}}. $$
By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\liminf_{t\rightarrow+\infty}y_{2}(t)\geq\frac {\lambda_{2} (m^{(1)}_{1}+k_{2} )}{a_{2}}. $$
Then for the above \(\varepsilon>0\), there exists a \(T_{4}>T_{3}+\tau_{2}\), such that
$$ y_{2}(t)> \frac{\lambda_{2} (m^{(1)}_{1}+k_{2} )}{a_{2}}\varepsilon \stackrel{\mathrm{def}}{=}m^{(1)}_{2}, \quad t>T_{4}. $$
(3.5)
According to (3.1), (3.2), (3.4), and (3.5), we obtain
$$ 0< m^{(1)}_{1}< x_{2}(t)< M^{(1)}_{1}, \qquad 0< m^{(1)}_{2}< y_{2}(t)< M^{(1)}_{2}, \quad t>T_{4}. $$
(3.6)
Then for \(t>T_{4}+\tau_{1}\), substituting (3.1) and (3.5) into the first equation of system (1.6), we have
$$\begin{aligned} x'_{2}(t) < & r_{1}e^{d_{11}\tau_{1}}x_{2}(t \tau _{1})d_{12}x_{2}(t)bx_{2}^{2}(t) \frac{a_{1}m^{(1)}_{2} x_{2}(t)}{k_{1}+ M^{(1)}_{1}} \\ =& r_{1}e^{d_{11}\tau_{1}}x_{2}(t\tau_{1}) \biggl(d_{12}+\frac {a_{1}m^{(1)}_{2}}{k_{1}+M^{(1)}_{1}} \biggr)x_{2}(t)bx_{2}^{2}(t). \end{aligned}$$
According to the inequalities (3.3) and (3.6), we have \(r_{1}e^{d_{11}\tau_{1}} (d_{12}+\frac {a_{1}m^{(1)}_{2}}{k_{1}+M^{(1)}_{1}} )>0\). By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\limsup_{t\rightarrow+\infty}x_{2}(t)\leq \frac{\lambda_{1}\frac{a_{1}m^{(1)}_{2}}{k_{1}+M^{(1)}_{1}}}{b}. $$
Then, for the above \(\varepsilon>0\), there exists a \(T_{5}>T_{4}+\tau_{1}\), such that
$$ x_{2}(t)< \frac{\lambda_{1}\frac {a_{1}m^{(1)}_{2}}{k_{1}+M^{(1)}_{1}}}{b}+\frac{\varepsilon}{2} \stackrel{ \mathrm{def}}{=}M^{(2)}_{1}, \quad t>T_{5}. $$
(3.7)
From inequalities (3.1) and (3.7), we obtain
$$ x_{2}(t)< M^{(2)}_{1}< M^{(1)}_{1}, \quad t>T_{5}. $$
(3.8)
For \(t>T_{5}+\tau_{2}\), substituting (3.7) into the second equation of system (1.6), we have
$$y'_{2}(t)< r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau_{2})d_{21}y_{2}(t)\frac {a_{2}y^{2}_{2}(t)}{M^{(2)}_{1}+k_{2}} . $$
By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\limsup_{t\rightarrow+\infty}y_{2}(t)\leq \frac{\lambda_{2} (M^{(2)}_{1}+k_{2} )}{a_{2}}. $$
Then for above \(\varepsilon>0\), there exists a \(T_{6}>T_{5}+\tau_{2}\), such that
$$ y_{2}(t)< \frac{\lambda_{2} (M^{(2)}_{1}+k_{2} )}{a_{2}}+\frac {\varepsilon}{2} \stackrel{\mathrm{def}}{=}M^{(2)}_{2},\quad t>T_{6}. $$
(3.9)
From inequalities (3.2), (3.8), and (3.9), we have
$$ y_{2}(t)< M^{(2)}_{2}< M^{(1)}_{2}, \quad t>T_{6}. $$
(3.10)
For \(t>T_{6}+\tau_{1} \), substituting inequalities (3.4) and (3.9) into the first equation of system (1.6), we have
$$\begin{aligned} x'_{2}(t) >& r_{1}e^{d_{11}\tau_{1}}x_{2}(t \tau _{1})d_{12}x_{2}(t)bx_{2}^{2}(t) \frac{a_{1}M^{(2)}_{2} x_{2}(t)}{k_{1}+ m^{(1)}_{1}} \\ =& r_{1}e^{d_{11}\tau_{1}}x_{2}(t\tau_{1}) \biggl(d_{12}+\frac {a_{1}M^{(2)}_{2}}{k_{1}+m^{(1)}_{1}} \biggr)x_{2}(t)bx_{2}^{2}(t). \end{aligned}$$
According to inequalities (3.3) and (3.10), we can obtain \(r_{1}e^{d_{11}\tau_{1}}d_{12}\frac{a_{1}M^{(2)}_{2}}{k_{1}+m^{(1)}_{1}}>0\). By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\liminf_{t\rightarrow+\infty}x_{2}(t)\geq \frac{\lambda_{1}\frac{a_{1}M^{(2)}_{2}}{k_{1}+m^{(1)}_{1}}}{b}. $$
Then, for the above \(\varepsilon>0\), there exists a \(T_{7}>T_{6}+\tau_{1}\), such that
$$ x_{2}(t)>\frac{\lambda_{1}\frac{a_{1}M^{(2)}_{2}}{k_{1}+m^{(1)}_{1}}}{b} \frac{\varepsilon}{2}\stackrel{\mathrm{def}}{=}m^{(2)}_{1}, \quad t>T_{7}. $$
(3.11)
According to the inequalities (3.4), (3.10), and (3.11), we can obtain
$$ x_{2}(t)>m^{(2)}_{1}>m^{(1)}_{1}, \quad t>T_{7}. $$
(3.12)
Substituting inequality (3.11) into the second equation of system (1.6), we have
$$y'_{2}(t) > r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau_{2})d_{21}y_{2}(t) \frac{a_{2}y^{2}_{2}(t)}{m^{(2)}_{1}+k_{2}},\quad t>T_{7}+\tau_{2} . $$
By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\liminf_{t\rightarrow+\infty}y_{2}(t)\geq\frac {\lambda_{2} (m^{(2)}_{1}+k_{2} )}{a_{2}}. $$
Then, for the above \(\varepsilon>0\), there exists a \(T_{8}>T_{7}+\tau_{2}\), such that
$$ y_{2}(t)>\frac{\lambda_{2} (m^{(2)}_{1}+k_{2} )}{a_{2}}\frac {\varepsilon}{2} \stackrel{\mathrm{def}}{=}m^{(2)}_{2},\quad t>T_{8}. $$
(3.13)
According to the inequalities (3.5), (3.12), and (3.13), we can obtain
$$ y_{2}(t)>m^{(2)}_{2}>m^{(1)}_{2}, \quad t>T_{8}. $$
(3.14)
For \(t>T_{8}\), according to (3.8), (3.10), (3.12), and (3.14), we have
$$ \begin{aligned} &m^{(1)}_{1}< m^{(2)}_{1}< x_{2}(t)< M^{(2)}_{1}< M^{(1)}_{1}, \\ &m^{(1)}_{2}< m^{(2)}_{2}< y_{2}(t)< M^{(2)}_{2}< M^{(1)}_{2}. \end{aligned} $$
(3.15)
Repeating the above process, we get four sequences
$$ \begin{aligned} &M_{1}^{(n)}= \frac{\lambda_{1}\frac {a_{1}m^{(n1)}_{2}}{k_{1}+M^{(n1)}_{1}}}{b}+ \frac{\varepsilon}{n}, \qquad M_{2}^{(n)}= \frac{\lambda_{2} (M^{(n)}_{1}+k_{2} )}{a_{2}}+ \frac{\varepsilon}{n}, \\ &m_{1}^{(n)}= \frac{\lambda_{1}\frac {a_{1}M^{(n)}_{2}}{k_{1}+m^{(n1)}_{1}}}{b}\frac{\varepsilon}{n}, \qquad m_{2}^{(n)}= \frac{\lambda_{2} (m^{(n)}_{1}+k_{2} )}{a_{2}}\frac{\varepsilon}{n}. \end{aligned} $$
(3.16)
For \(i=1,2\), we claim that \(M^{(n)}_{i}\) are monotonic decreasing sequences, and \(m^{(n)}_{i}\) are monotone increasing sequences. In the following we will prove this claim by induction. First of all, according to inequalities (3.15), we have
$$m^{(1)}_{i}< m^{(2)}_{i}, \qquad M^{(2)}_{i}< M^{(1)}_{i}, \quad i=1,2. $$
Second, we suppose that our claim is true for n, that is,
$$ m^{(n1)}_{i}< m^{(n)}_{i}, \qquad M^{(n)}_{i}< M^{(n1)}_{i},\quad i=1,2. $$
(3.17)
Noting that
$$ \begin{aligned} &M_{1}^{(n+1)}= \frac{\lambda_{1}\frac {a_{1}m^{(n)}_{2}}{k_{1}+M^{(n)}_{1}}}{b}+ \frac{\varepsilon}{n+1}, \qquad M_{2}^{(n+1)}= \frac{\lambda_{2} (M^{(n+1)}_{1}+k_{2} )}{a_{2}}+ \frac{\varepsilon}{n+1}, \\ &m_{1}^{(n+1)}= \frac{\lambda_{1}\frac {a_{1}M^{(n+1)}_{2}}{k_{1}+m^{(n)}_{1}}}{b}\frac{\varepsilon}{n+1},\qquad m_{2}^{(n+1)}= \frac{\lambda_{2} (m^{(n+1)}_{1}+k_{2} )}{a_{2}}\frac{\varepsilon}{n+1}. \end{aligned} $$
(3.18)
According to inequalities (3.16), (3.17), and (3.18), one could easily see that
$$M^{(n+1)}_{i}< M^{(n)}_{i}, \qquad m^{(n)}_{i}< m^{(n+1)}_{i}, \quad i=1,2. $$
Then for \(t>T_{4n}\), we have
$$\begin{aligned}& 0< m^{(1)}_{1}< m^{(2)}_{1}< \cdots< x_{2}(t)< M^{(n)}_{1}< \cdots< M^{(2)}_{1}< M^{(1)}_{1}, \\& 0< m^{(1)}_{2}< m^{(2)}_{2}< \cdots< y_{2}(t)< M^{(n)}_{2}< \cdots< M^{(2)}_{2}< M^{(1)}_{2}. \end{aligned}$$
Therefore the limits of \(M^{(n)}_{i}\), \(m^{(n)}_{i}\) (\(i=1,2\), \(n=1,2,\ldots\)) exist. Denote that
$$\begin{aligned}& \lim_{t\rightarrow+\infty}M^{(n)}_{1}=\overline {x}_{2},\qquad \lim_{t\rightarrow+\infty}m^{(n)}_{1}= \underline {x}_{2}, \\& \lim_{t\rightarrow+\infty}M^{(n)}_{2}= \overline {y}_{2},\qquad \lim_{t\rightarrow+\infty}m^{(n)}_{2}= \underline{y}_{2}. \end{aligned}$$
Consequently, \(\overline{x}_{2} \geq\underline{x}_{2}\), \(\overline{y}_{2} \geq\underline{y}_{2}\). In order to complete the proof, we just need to show that \(\overline {x}_{2}= \underline{x}_{2}\), \(\overline{y}_{2}=\underline{y}_{2}\). Letting \(n\rightarrow+\infty\) in (3.16), we have
$$\begin{aligned}& b\overline{x}_{2}=\lambda_{1}\frac{a_{1}\underline {y}_{2}}{\overline{x}_{2}+k_{1}},\qquad a_{2}\overline{y}_{2}=\lambda_{2}( \overline{x}_{2}+k_{2}), \\& b\underline{x}_{2}=\lambda_{1}\frac{a_{1}\overline {y}_{2}}{\underline{x}_{2}+k_{1}},\qquad a_{2}\underline{y}_{2}=\lambda_{2}( \underline{x}_{2}+k_{2}). \end{aligned}$$
It follows from the above four equations that
$$ \begin{aligned} &a_{1}\lambda_{2} ( \underline{x}_{2}+k_{2} )=a_{2} ( \lambda_{1}b\overline{x}_{2} ) (\overline{x}_{2}+k_{1} ), \\ &a_{1}\lambda_{2} (\overline{x}_{2}+k_{2} )=a_{2}(\lambda _{1}b\underline{x}_{2} ) ( \underline{x}_{2}+k_{1} ). \end{aligned} $$
(3.19)
Subtracting the first equation of (3.19) from the second equation, we get
$$\bigl(a_{2}\lambda_{1}+a_{1} \lambda_{2}a_{2}b (\overline{x}_{2}+\underline {x}_{2} )k_{1}a_{2}b \bigr) ( \overline{x}_{2}\underline{x}_{2} )=0. $$
Suppose that \(\overline{x}_{2}\neq\underline{x}_{2}\), it follows from the above equation that
$$ a_{2}\lambda_{1}+a_{1}\lambda_{2}k_{1}a_{2}b=a_{2}b (\overline {x}_{2}+\underline{x}_{2} ). $$
(3.20)
Substituting (3.20) into (3.19), we find \(\overline {x}_{2} \) and \(\underline{x}_{2}\) both satisfy the following equation:
$$ a^{2}_{2}b (\lambda_{1}bx ) (x+k_{1} )=a_{1}\lambda_{2} (a_{1}\lambda_{2} +a_{2}\lambda_{1}a_{2}k_{1}ba_{2}bx+a_{2}k_{2}b ). $$
(3.21)
Simplifying equality (3.21), we get
$$ a^{2}_{2}b^{2}x^{2}+a_{2}b (a_{2}k_{1}ba_{2}\lambda_{1}a_{1} \lambda_{2} )x+D=0, $$
(3.22)
where
$$D=a_{2} (a_{1}\lambda_{1}\lambda_{2}+ba_{1}k_{2} \lambda _{2}ba_{2}k_{1}\lambda_{1} )+a_{1}\lambda_{2} (a_{1}\lambda_{2}a_{2}k_{1}b ). $$
According to condition (\(\mathrm{H}_{2}\)), we can immediately obtain
$$\lambda_{3}=a_{1}\lambda_{1} \lambda_{2}+ba_{1}k_{2}\lambda_{2}ba_{2}k_{1} \lambda _{1}< 0. $$
It implies that
$$a_{1}\lambda_{2}ba_{2}k_{1}< 0. $$
Therefore, we have \(D<0\), that is, equation (3.22) has only one positive root. Then \(\overline{x}_{2}=\underline{x}_{2}\), and consequently, \(\overline{y}_{2}=\underline{y}_{2}\). Obviously, conditions (\(\mathrm{H}_{1}\)), (\(\mathrm{H}_{2}\)) imply inequality (2.1), so system (1.6) has a unique positive equilibrium \(E(x^{*}_{2},y^{*}_{2})\). That is,
$$\lim_{t\rightarrow +\infty}x_{2}(t)=x^{*}_{2}, \qquad \lim _{t\rightarrow +\infty}y_{2}(t)=y^{*}_{2} . $$
This completes the proof of Theorem 2.2. □
Proof of Theorem 2.3
It follows from the first equation of system (1.6) that
$$x'_{2}(t)< r_{1}e^{d_{11}\tau_{1}}x_{2}(t \tau _{1})d_{12}x_{2}(t)bx_{2}^{2}(t). $$
According to first inequality of condition (\(\mathrm{H}_{3}\)), we have \(r_{1}e^{d_{11}\tau_{1}}d_{12}<0\). By applying Lemma 3.2(ii) and the standard comparison theorem, we have \(\limsup_{t\rightarrow+\infty}x_{2}(t)\leq0\). That is,
$$\lim_{t\rightarrow+\infty}x_{2}(t)= 0. $$
Then, for any \(\varepsilon>0\), there exists a \(T>0\) such that
$$0< x_{2}(t)< \varepsilon. $$
Therefore, it follows from the second equation of system (1.6) that
$$y'_{2}(t)< r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau_{2})d_{21}y_{2}(t)\frac {a_{2}y^{2}_{2}(t)}{\varepsilon+k_{2}}, \quad t>T+\tau_{2}. $$
Similar to the above analysis, we also have
$$\lim_{t\rightarrow+\infty}y_{2}(t)= 0. $$
Therefore, \(E_{0}=(0,0)\) is globally attractive. This completes the proof of Theorem 2.3. □
Proof of Theorem 2.4
According to the first inequality of condition (\(\mathrm{H}_{4}\)), we have \(r_{1}e^{d_{11}\tau_{1}}d_{12}>0\). Therefore from the proof of Theorem 2.2, we know that
$$ \limsup_{t\rightarrow+\infty}x_{2}(t)\leq\frac{\lambda_{1}}{b}. $$
(3.23)
And for any small positive constant \(\varepsilon>0\), there exists a \(T_{1}>0\) such that
$$y'_{2}(t)< r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau_{2})d_{21}y_{2}(t)\frac {a_{2}y^{2}_{2}(t)}{M^{(1)}_{1}+k_{2}}, \quad t>T_{1}+\tau_{2}. $$
According the second inequality of condition (\(\mathrm{H}_{4}\)), we have \(r_{2}e^{d_{22}\tau_{2}}d_{21}<0\). By applying Lemma 3.2(ii) and the standard comparison theorem, we have \(\limsup_{t\rightarrow+\infty}y_{2}(t)\leq0\). That is,
$$\lim_{t\rightarrow+\infty}y_{2}(t)= 0. $$
Then, for any small \(\varepsilon>0\), there exists \(T_{2}>T_{1}+\tau_{2}\), such that
$$ 0< y_{2}(t)< \varepsilon,\quad t>T_{2}. $$
(3.24)
Substituting inequality (3.24) into the first equation of system (1.6), we have
$$\begin{aligned} x'_{2}(t) >& r_{1}e^{d_{11}\tau_{1}}x_{2}(t \tau _{1})d_{12}x_{2}(t)bx_{2}^{2}(t) \frac{a_{1}\varepsilon x_{2}(t)}{k_{1}} \\ =& r_{1}e^{d_{11}\tau_{1}}x_{2}(t\tau_{1}) \biggl(d_{12}+\frac{a_{1}\varepsilon }{k_{1}} \biggr)x_{2}(t)bx_{2}^{2}(t), \quad t>T_{2}+\tau_{1}. \end{aligned}$$
Since \(r_{1}e^{d_{11}\tau_{1}}d_{12}>0\), we can choose sufficiently small \(\varepsilon>0\) such that \(r_{1}e^{d_{11}\tau _{1}}d_{12} \frac{a_{1}\varepsilon}{k_{1}}>0\). By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\liminf_{t\rightarrow+\infty}x_{2}(t)\geq \frac{r_{1}e^{d_{11}\tau_{1}}d_{12}\frac{a_{1}\varepsilon}{k_{1}}}{b}= \frac{\lambda_{1}\frac{a_{1}\varepsilon}{k_{1}}}{b}. $$
For the above formula, letting \(\varepsilon\rightarrow0\), we have
$$ \liminf_{t\rightarrow+\infty}x_{2}(t)\geq\frac{\lambda_{1}}{b}. $$
(3.25)
From inequalities (3.23) and (3.25), we get
$$\frac{\lambda_{1}}{b}\leq\liminf_{t\rightarrow+\infty }x_{2}(t)\leq \limsup_{t\rightarrow+\infty}x_{2}(t)\leq\frac{\lambda_{1}}{b}. $$
Then we have
$$\lim_{t\rightarrow+\infty}x_{2}(t)=\frac{\lambda_{1}}{b}=x_{2*}. $$
Therefore, \(E_{1}(x_{2*},0)\) is globally attractive. This completes the proof of Theorem 2.4. □
Proof of Theorem 2.5
According to the fluctuation lemma, there exist two sequences \(\gamma_{n}\rightarrow\infty\), \(\sigma_{n}\rightarrow\infty\) such that \(x'_{2}(\gamma_{n})\rightarrow0\), \(x_{2}(\gamma_{n})\rightarrow\limsup_{t\rightarrow+\infty}x(t)=\overline{x}_{2}\), and \(y'_{2}(\sigma_{n})\rightarrow0 \), \(y_{2}(\sigma_{n})\rightarrow\liminf_{t\rightarrow+\infty}y_{2}(t)=\underline{y}_{2}\) as \(n\rightarrow\infty\). Since from Lemma 3.1, we know \(\overline{x}_{2}\geq0\). In order to prove \(\lim_{t\rightarrow\infty}x_{2}(t)=0\), we only need to prove \(\overline{x}_{2}=0\), so for getting a contradiction, we suppose that \(\overline{x}_{2}>0\). Since for \(\lambda_{2}>0\), and according to Lemma 3.4, we know \(\underline{y}_{2}>0\), it follows from the first equation of the system (1.6) that
$$\begin{aligned} x'_{2}(\gamma_{n}) =& r_{1}e^{d_{11}\tau_{1}}x_{2}( \gamma_{n}\tau_{1})d_{12}x_{2}(\gamma _{n})bx_{2}^{2}(\gamma_{n})  \frac{a_{1}y_{2}(\gamma_{n})x_{2}(\gamma_{n})}{x_{2}(\gamma_{n})+k_{1}} \\ \leq& r_{1}e^{d_{11}\tau_{1}}\sup_{t\geq\gamma_{n}\tau_{1}} x_{2}(t)d_{12}x_{2}(\gamma_{n})bx_{2}^{2}( \gamma_{n}) \frac{a_{1}x_{2}(\gamma_{n})}{x_{2}(\gamma_{n})+k_{1}}\inf_{t\geq \gamma_{n}}y_{2}( \gamma_{n}). \end{aligned}$$
Letting \(n\rightarrow\infty\) in the above inequality, we obtain
$$0\leq\lambda_{1}b\overline{x}_{2}\frac{a_{1}\underline{y}_{2}}{\overline {x}_{2}+k_{1}}, $$
that is,
$$ 0\leq\lambda_{1} (\overline{x}_{2}+k_{1} )b \overline{x}_{2} (\overline{x}_{2}+k_{1} )a_{1}\underline{y}_{2}. $$
(3.26)
From the second equation of system (1.6), by a similar argument, we have
$$ 0\geq k_{2}\lambda_{2}a_{2}\underline{y}_{2}. $$
(3.27)
It follows from inequalities (3.26) and (3.27) that
$$0\leq a_{2}\lambda_{1} (k_{1}+ \overline{x}_{2} ) a_{2}b\overline {x}_{2} (k_{1}+\overline{x}_{2} )a_{1}k_{2} \lambda_{2}. $$
Simplifying the above inequality, we have
$$ a_{2}b\overline{x}^{2}_{2}+ (a_{2}k_{1}ba_{2} \lambda_{1} )\overline {x}_{2}+a_{1}k_{2} \lambda_{2}a_{2}k_{1}\lambda_{1}\leq0. $$
(3.28)
According to the second inequality of condition (\(\mathrm{H}_{5}\)), we know \(a_{1}k_{2}\lambda_{2}a_{2}k_{1}\lambda_{1}>0\), \(a_{2}k_{1}ba_{2}\lambda_{1}>0\), then only \(\overline{x}_{2}<0 \) can ensure (3.28) holds. And \(\overline{x}_{2}<0\) contradicts the hypothesis \(\overline{x}_{2}>0\), then we get
$$\lim_{t\rightarrow\infty}x_{2}(t)=0. $$
Then, for any small enough \(\varepsilon>0\), there exists a \(T>0\), such that
$$ 0< x_{2}(t)< \varepsilon, \quad t>T. $$
(3.29)
Substituting inequality (3.29) into the second equation of system (1.6), we have
$$y'_{2}(t) < r_{2}e^{d_{22}\tau_{2}}y_{2}(t \tau_{2})d_{21}y_{2}(t) \frac{a_{2}y^{2}_{2}(t)}{\varepsilon+k_{2}}, \quad t>T+\tau_{2} . $$
According to second inequality of condition (\(\mathrm{H}_{5}\)), we know \(r_{2}e^{d_{22}\tau_{2}}d_{21}>0\). By applying Lemma 3.2(i), and the standard comparison theorem, we have
$$\limsup_{t\rightarrow+\infty}y_{2}(t)\leq\frac{\lambda _{2}(\varepsilon+k_{2})}{a_{2}}. $$
Letting \(\varepsilon\rightarrow0\) in the above inequality, we have
$$ \limsup_{t\rightarrow+\infty}y_{2}(t)\leq\frac{\lambda_{2}k_{2}}{a_{2}}. $$
(3.30)
According to Lemma 3.4 and (3.30), we obtain
$$\frac{\lambda_{2}k_{2}}{a_{2}}\leq\liminf_{t\rightarrow+\infty}y_{2}(t) \leq \limsup_{t\rightarrow+\infty}y_{2}(t)\leq\frac{\lambda _{2}k_{2}}{a_{2}}. $$
Then we have
$$\lim_{t\rightarrow+\infty}y_{2}(t)= \frac{\lambda _{2}k_{2}}{a_{2}}=y_{2*}. $$
This completes the proof of Theorem 2.5. □