In this section, we establish some new Lyapunov-type inequalities.

Denote

and for \lambda \in [0,1), denote

**Theorem 2.1** *Suppose that* (1.2) *and* (1.4) *hold*, *and let* n\in \mathbb{Z}[a,b] *with* a\le b-1. *Assume* (1.1) *has a real solution* (x(n),y(n)) *such that* (1.5) *holds*. *Then one has the following inequality*:

\sum _{n=a}^{b-1}\frac{\zeta (n)\eta (n)}{\zeta (n)+\eta (n)}{\gamma}^{+}(n)\ge 1.

(2.5)

*Proof* It follows from (1.5) that there exist {\xi}_{1},{\xi}_{2}\in [0,1) such that

(1-{\xi}_{1})x(a)+{\xi}_{1}x(a+1)=0

(2.6)

and

(1-{\xi}_{2})x(b)+{\xi}_{2}x(b+1)=0.

(2.7)

Multiplying the first equation of (1.1) by y(n) and the second one by x(n+1), and then adding, we get

\u25b3[x(n)y(n)]=\beta (n){|y(n)|}^{\mu}-\gamma (n){|x(n+1)|}^{\nu}.

(2.8)

Summing equation (2.8) from *a* to b-1, we can obtain

x(b)y(b)-x(a)y(a)=\sum _{n=a}^{b-1}\beta (n){|y(n)|}^{\mu}-\sum _{n=a}^{b-1}\gamma (n){|x(n+1)|}^{\nu}.

(2.9)

From the first equation of (1.1), we have

[1-\alpha (n)]x(n+1)=x(n)+\beta (n){|y(n)|}^{\mu -2}y(n).

(2.10)

Combining (2.10) with (2.6), we have

x(a)=-\frac{{\xi}_{1}\beta (a)}{1-(1-{\xi}_{1})\alpha (a)}{|y(a)|}^{\mu -2}y(a).

(2.11)

Similarly, it follows from (2.10) and (2.7) that

x(b)=-\frac{{\xi}_{2}\beta (b)}{1-(1-{\xi}_{2})\alpha (b)}{|y(b)|}^{\mu -2}y(b).

(2.12)

Substituting (2.11) and (2.12) into (2.9), we have

\begin{array}{c}\sum _{n=a}^{b-1}\beta (n){|y(n)|}^{\mu}-\sum _{n=a}^{b-1}\gamma (n){|x(n+1)|}^{\nu}\hfill \\ \phantom{\rule{1em}{0ex}}=-\frac{{\xi}_{2}\beta (b)}{1-(1-{\xi}_{2})\alpha (b)}{|y(b)|}^{\mu}+\frac{{\xi}_{1}\beta (a)}{1-(1-{\xi}_{1})\alpha (a)}{|y(a)|}^{\mu},\hfill \end{array}

which implies that

Denote that

\tilde{\beta}(a)=\frac{(1-{\xi}_{1})[1-\alpha (a)]}{1-(1-{\xi}_{1})\alpha (a)}\beta (a),\phantom{\rule{2em}{0ex}}\tilde{\beta}(b)=\frac{{\xi}_{2}}{1-(1-{\xi}_{2})\alpha (b)}\beta (b)

(2.14)

and

\tilde{\beta}(n)=\beta (n),\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a+1,b-1].

(2.15)

Then we can rewrite (2.13) as

\sum _{n=a}^{b}\tilde{\beta}(n){|y(n)|}^{\mu}=\sum _{n=a}^{b-1}\gamma (n){|x(n+1)|}^{\nu}.

(2.16)

From (2.10), (2.11), (2.14) and (2.15), we obtain

\begin{array}{rcl}x(n+1)& =& x(a)\prod _{s=a}^{n}{[1-\alpha (s)]}^{-1}+\sum _{\tau =a}^{n}\beta (\tau ){|y(\tau )|}^{\mu -2}y(\tau )\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-1}\\ =& -\frac{{\xi}_{1}\beta (a)}{1-(1-{\xi}_{1})\alpha (a)}{|y(a)|}^{\mu -2}y(a)\prod _{s=a}^{n}{[1-\alpha (s)]}^{-1}\\ +\sum _{\tau =a}^{n}\beta (\tau ){|y(\tau )|}^{\mu -2}y(\tau )\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-1}\\ =& \sum _{\tau =a}^{n}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu -2}y(\tau )\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-1},\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b-1].\end{array}

(2.17)

Similarly, from (2.10), (2.12), (2.14) and (2.15), we have

\begin{array}{rcl}x(n+1)& =& x(b)\prod _{s=n+1}^{b-1}[1-\alpha (s)]-\sum _{\tau =n+1}^{b-1}\beta (\tau ){|y(\tau )|}^{\mu -2}y(\tau )\prod _{s=n+1}^{\tau -1}[1-\alpha (s)]\\ =& -\frac{{\xi}_{2}\beta (b)}{1-(1-{\xi}_{2})\alpha (b)}{|y(b)|}^{\mu -2}y(b)\prod _{s=n+1}^{b-1}[1-\alpha (s)]\\ -\sum _{\tau =n+1}^{b-1}\beta (\tau ){|y(\tau )|}^{\mu -2}y(\tau )\prod _{s=n+1}^{\tau -1}[1-\alpha (s)]\\ =& -\sum _{\tau =n+1}^{b}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu -2}y(\tau )\prod _{s=n+1}^{\tau -1}[1-\alpha (s)],\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b-1].\end{array}

(2.18)

Since

0\le \tilde{\beta}(n)\le \beta (n),\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b],

(2.19)

it follows from (2.1) and (2.17) and the Hölder inequality that

\begin{array}{rcl}{|x(n+1)|}^{\nu}& \le & {\{\sum _{\tau =a}^{n}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu -1}\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-1}\}}^{\nu}\\ \le & {(\sum _{\tau =a}^{n}\tilde{\beta}(\tau )\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-\mu})}^{\frac{\nu}{\mu}}\sum _{\tau =a}^{n}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu}\\ \le & {(\sum _{\tau =a}^{n}\beta (\tau )\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-\mu})}^{\frac{\nu}{\mu}}\sum _{\tau =a}^{n}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu}\\ =& \zeta (n)\sum _{\tau =a}^{n}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu},\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b-1].\end{array}

(2.20)

Similarly, it follows from (2.2), (2.18), (2.19) and the Hölder inequality that

\begin{array}{rcl}{|x(n+1)|}^{\nu}& \le & {\{\sum _{\tau =n+1}^{b}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu -1}\prod _{s=n+1}^{\tau -1}[1-\alpha (s)]\}}^{\nu}\\ \le & {(\sum _{\tau =n+1}^{b}\tilde{\beta}(\tau )\prod _{s=n+1}^{\tau -1}{[1-\alpha (s)]}^{\mu})}^{\frac{\nu}{\mu}}\sum _{\tau =n+1}^{b}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu}\\ \le & {(\sum _{\tau =n+1}^{b}\beta (\tau )\prod _{s=n+1}^{\tau -1}{[1-\alpha (s)]}^{\mu})}^{\frac{\nu}{\mu}}\sum _{\tau =n+1}^{b}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu}\\ =& \eta (n)\sum _{\tau =n+1}^{b}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu},\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b-1].\end{array}

(2.21)

From (2.20) and (2.21), we obtain

{|x(n+1)|}^{\nu}\le \frac{\zeta (n)\eta (n)}{\zeta (n)+\eta (n)}\sum _{\tau =a}^{b}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu},\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b-1].

(2.22)

Combining (2.22) with (2.16), we have

\begin{array}{rcl}\sum _{n=a}^{b-1}{\gamma}^{+}(n){|x(n+1)|}^{\nu}& \le & \sum _{n=a}^{b-1}\frac{\zeta (n)\eta (n)}{\zeta (n)+\eta (n)}{\gamma}^{+}(n)\sum _{n=a}^{b}\tilde{\beta}(n){|y(\tau )|}^{\mu}\\ =& \sum _{n=a}^{b-1}\frac{\zeta (n)\eta (n)}{\zeta (n)+\eta (n)}{\gamma}^{+}(n)\sum _{n=a}^{b-1}\gamma (n){|x(n+1)|}^{\nu}\\ \le & \sum _{n=a}^{b-1}\frac{\zeta (n)\eta (n)}{\zeta (n)+\eta (n)}{\gamma}^{+}(n)\sum _{n=a}^{b-1}{\gamma}^{+}(n){|x(n+1)|}^{\nu}.\end{array}

(2.23)

We claim that

\sum _{n=a}^{b-1}{\gamma}^{+}(n){|x(n+1)|}^{\nu}>0.

(2.24)

If (2.24) is not true, then

\sum _{n=a}^{b-1}{\gamma}^{+}(n){|x(n+1)|}^{\nu}=0.

(2.25)

From (2.16) and (2.25), we have

0\le \sum _{n=a}^{b}\tilde{\beta}(n){|y(n)|}^{\mu}=\sum _{n=a}^{b-1}\gamma (n){|x(n+1)|}^{\nu}\le \sum _{n=a}^{b-1}{\gamma}^{+}(n){|x(n+1)|}^{\nu}=0.

(2.26)

It follows that

\tilde{\beta}(n){|y(n)|}^{\mu}=0,\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b].

(2.27)

Combining (2.20) with (2.27), we obtain that x(a+1)=x(a+2)=\cdots =x(b)=0, which together with (2.6) implies that x(a)=0. This contradicts (1.5). Therefore, (2.24) holds. Hence, it follows from (2.23) and (2.24) that (2.5) holds. □

In the case x(b)=0, *i.e.*, {\xi}_{2}=0, and so \tilde{\beta}(b)=0, we have the following equation:

\sum _{n=a}^{b-1}\tilde{\beta}(n){|y(n)|}^{\mu}=\sum _{n=a}^{b-2}\gamma (n){|x(n+1)|}^{\nu}

(2.28)

and inequality

\begin{array}{rcl}{|x(n+1)|}^{\nu}& \le & {\{\sum _{\tau =n+1}^{b-1}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu -1}\prod _{s=n+1}^{\tau -1}[1-\alpha (s)]\}}^{\nu}\\ \le & {(\sum _{\tau =n+1}^{b-1}\tilde{\beta}(\tau )\prod _{s=n+1}^{\tau -1}{[1-\alpha (s)]}^{\mu})}^{\frac{\nu}{\mu}}\sum _{\tau =n+1}^{b-1}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu}\\ \le & {(\sum _{\tau =n+1}^{b-1}\beta (\tau )\prod _{s=n+1}^{\tau -1}{[1-\alpha (s)]}^{\mu})}^{\frac{\nu}{\mu}}\sum _{\tau =n+1}^{b-1}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu}\\ =& {\eta}_{0}(n)\sum _{\tau =n+1}^{b-1}\tilde{\beta}(\tau ){|y(\tau )|}^{\mu},\phantom{\rule{1em}{0ex}}n\in \mathbb{Z}[a,b-1]\end{array}

(2.29)

instead of (2.16) and (2.21), respectively. Similar to the proof of Theorem 2.1, we have the following theorem.

**Theorem 2.2** *Suppose that* (1.2) *and* (1.4) *hold*, *and let* n\in \mathbb{Z}[a,b] *with* a\le b-2. *Assume* (1.1) *has a real solution* (x(n),y(n)) *such that* x(a)=0 *or* x(a)x(a+1)<0 *and* x(b)=0 *and* x(n) *is not identically zero on* \mathbb{Z}[a,b]. *Then one has the following inequality*:

\sum _{n=a}^{b-2}\frac{\zeta (n){\eta}_{0}(n)}{\zeta (n)+{\eta}_{0}(n)}{\gamma}^{+}(n)\ge 1,

(2.30)

*where* \zeta (n) *and* {\eta}_{0}(n) *are defined by* (2.1) *and* (2.4), *respectively*.

**Corollary 2.3** *Suppose that* (1.2) *and* (1.4) *hold*, *and let* n\in \mathbb{Z}[a,b] *with* a\le b-1. *Assume* (1.1) *has a real solution* (x(n),y(n)) *such that* (1.5) *holds*. *Then one has the following inequality*:

\sum _{n=a}^{b-1}{\gamma}^{+}(n){[\sum _{\tau =a}^{n}\beta (\tau )\sum _{\tau =n+1}^{b}\beta (\tau )]}^{\frac{\nu}{2\mu}}\ge 2{\left\{\prod _{n=a}^{b-1}\mathrm{\Theta}[\alpha (n)]\right\}}^{\frac{\nu}{2}},

(2.31)

*where*, *and in the sequel*,

\mathrm{\Theta}[\alpha (n)]=min\{1-{\alpha}^{+}(n),{[1+{\alpha}^{-}(n)]}^{-1}\}

*and*

{\alpha}^{+}(n)=max\{\alpha (n),0\},\phantom{\rule{2em}{0ex}}{\alpha}^{-}(n)=max\{-\alpha (n),0\}.

*Proof* Since

\zeta (n)+\eta (n)\ge 2{[\zeta (n)\eta (n)]}^{\frac{1}{2}},

it follows that

\begin{array}{rcl}1& \le & \sum _{n=a}^{b-1}\frac{\zeta (n)\eta (n)}{\zeta (n)+\eta (n)}{\gamma}^{+}(n)\\ \le & \frac{1}{2}\sum _{n=a}^{b-1}{[\zeta (n)\eta (n)]}^{\frac{1}{2}}{\gamma}^{+}(n)\\ =& \frac{1}{2}\sum _{n=a}^{b-1}{\gamma}^{+}(n){\{\sum _{\tau =a}^{n}\beta (\tau )\prod _{s=\tau}^{n}{[1-\alpha (s)]}^{-\mu}\sum _{\tau =n+1}^{b}\beta (\tau )\prod _{s=n+1}^{\tau -1}{[1-\alpha (s)]}^{\mu}\}}^{\frac{\nu}{2\mu}}\\ \le & \frac{1}{2}\sum _{n=a}^{b-1}{\gamma}^{+}(n){\{\sum _{\tau =a}^{n}\beta (\tau )\prod _{s=\tau}^{n}{[1-{\alpha}^{+}(s)]}^{-\mu}\sum _{\tau =n+1}^{b}\beta (\tau )\prod _{s=n+1}^{\tau -1}{[1+{\alpha}^{-}(s)]}^{\mu}\}}^{\frac{\nu}{2\mu}}\\ \le & \frac{1}{2}\sum _{n=a}^{b-1}{\gamma}^{+}(n){[\sum _{\tau =a}^{n}\beta (\tau )\sum _{\tau =n+1}^{b}\beta (\tau )]}^{\frac{\nu}{2\mu}}\prod _{s=a}^{n}{[1-{\alpha}^{+}(s)]}^{-\frac{\nu}{2}}\prod _{s=n+1}^{b-1}{[1+{\alpha}^{-}(s)]}^{\frac{\nu}{2}}\\ \le & \frac{1}{2}\sum _{n=a}^{b-1}{\gamma}^{+}(n){[\sum _{\tau =a}^{n}\beta (\tau )\sum _{\tau =n+1}^{b}\beta (\tau )]}^{\frac{\nu}{2\mu}}{\left\{\prod _{s=a}^{b-1}\mathrm{\Theta}[\alpha (s)]\right\}}^{-\frac{\nu}{2}},\end{array}

(2.32)

which implies (2.31) holds. □

Since

{[\sum _{\tau =a}^{n}\beta (\tau )\sum _{\tau =n+1}^{b}\beta (\tau )]}^{\frac{1}{2}}\le \frac{1}{2}\sum _{n=a}^{b}\beta (n),

we have the following result.

**Corollary 2.4** *Suppose that* (1.2) *and* (1.4) *hold*, *and let* n\in \mathbb{Z}[a,b] *with* a\le b-1. *Assume* (1.1) *has a real solution* (x(n),y(n)) *such that* (1.5) *holds*. *Then one has the following Lyapunov*-*type inequality*:

{(\sum _{n=a}^{b}\beta (n))}^{\frac{1}{\mu}}{(\sum _{n=a}^{b-1}{\gamma}^{+}(n))}^{\frac{1}{\nu}}\ge 2{\left\{\prod _{n=a}^{b-1}\mathrm{\Theta}[\alpha (n)]\right\}}^{\frac{1}{2}}.

(2.33)

In a fashion similar to the proofs of Corollaries 2.3 and 2.4, we can prove the following corollaries by using Theorem 2.2 instead of Theorem 2.1.

**Corollary 2.5** *Suppose that* (1.2) *and* (1.4) *hold*, *and let* n\in \mathbb{Z}[a,b] *with* a\le b-2. *Assume* (1.1) *has a real solution* (x(n),y(n)) *such that* x(a)=0 *or* x(a)x(a+1)<0 *and* x(b)=0 *and* x(n) *is not identically zero on* \mathbb{Z}[a,b]. *Then one has the following inequality*:

\sum _{n=a}^{b-2}{\gamma}^{+}(n){[\sum _{\tau =a}^{n}\beta (\tau )\sum _{\tau =n+1}^{b-1}\beta (\tau )]}^{\frac{\nu}{2\mu}}\ge 2{\left\{\prod _{n=a}^{b-2}\mathrm{\Theta}[\alpha (n)]\right\}}^{\frac{\nu}{2}}.

(2.34)

**Corollary 2.6** *Suppose that* (1.2) *and* (1.4) *hold*, *and let* n\in \mathbb{Z}[a,b] *with* a\le b-2. *Assume* (1.1) *has a real solution* (x(n),y(n)) *such that* x(a)=0 *or* x(a)x(a+1)<0 *and* x(b)=0 *and* x(n) *is not identically zero on* \mathbb{Z}[a,b]. *Then one has the following Lyapunov*-*type inequality*:

{(\sum _{n=a}^{b-1}\beta (n))}^{\frac{1}{\mu}}{(\sum _{n=a}^{b-2}{\gamma}^{+}(n))}^{\frac{1}{\nu}}\ge 2{\left\{\prod _{n=a}^{b-2}\mathrm{\Theta}[\alpha (n)]\right\}}^{\frac{1}{2}}.

(2.35)

**Remark 2.7** While the coefficient \alpha (n)\le 0 and \mu =\nu =2 in system (1.1), inequality (1.6) of Theorem 1.3 can be derived from inequality (2.33). Similarly, for inequality (1.7) of Theorem 1.4 and inequality (2.35), this result also holds.

On the one hand, when \alpha (n)\le 0, according to the definition of \mathrm{\Theta}[\alpha (n)], we have

\mathrm{\Theta}[\alpha (n)]=min\{1-{\alpha}^{+}(n),{[1+{\alpha}^{-}(n)]}^{-1}\}={[1+|\alpha (n)|]}^{-1}.

(2.36)

On the other hand, for any u\ge 0, we have

and

Using above inequalities (2.36), (2.37) and (2.38), we obtain

\begin{array}{rcl}2{\left\{\prod _{n=a}^{b-1}\mathrm{\Theta}[\alpha (n)]\right\}}^{\frac{1}{2}}& =& 2{\left\{\prod _{n=a}^{b-1}{[1+|\alpha (n)|]}^{-1}\right\}}^{\frac{1}{2}}\\ =& 2exp\{-ln\left[\prod _{n=a}^{b-1}{(1+|\alpha (n)|)}^{\frac{1}{2}}\right]\}\\ \ge & 2exp\{-\frac{1}{2}\sum _{n=a}^{b-1}|\alpha (n)|\}\\ \ge & 2[1-\frac{1}{2}\sum _{n=a}^{b-1}|\alpha (n)|]=2-\sum _{n=a}^{b-1}|\alpha (n)|.\end{array}

(2.39)

Then Remark 2.7 holds immediately.