and the following inequality holds: $$\label{estimate}
1\, >\, 1-\left(\f{1}{1-\mu(0)v(0)}+
\f{1}{2}\int_0^t\left(\g(s)-\f{\dot{\mu}(s)}{\mu(s)}\right)ds\right)^{-1}
\ge\mu(0)v(0).$$ From Lemma \[ineq\] and from formula (\[rc6\]) one gets: $$\label{rc7}
v(t)e^{\int_0^t \g(s)ds}:=
w(t)\le u(t)<\f{1}{\mu(t)}e^{\int_0^t \g(s)ds},$$ and thus estimate (\[ric5\]) is proved. $\Box$

[**Examples.**]{} To illustrate conditions (\[ric1\]) of Lemma \[ric\] consider the following examples of functions $\g$, $\s$, $\b$, satisfying (\[ric1\]).

[**1.**]{} Let $$\label{rcex} 
\g(t)=c_1(1+t)^{\nu_1},\q \s(t)=c_2(1+t)^{\nu_2},\q \b(t)=c_3(1+t)^{\nu_3},$$ where $c_2>0$, $c_3>0$. Choose $\mu(t):=c(1+t)^{\nu}$, $c>0$. From (\[ric1\]) one gets the following conditions $$c_2\le \f{cc_1}{2}(1+t)^{\nu+\nu_1-\nu_2}-\f{c\nu}{2}(1+t)^{\nu-1-\nu_2},\q$$ $$\label{rcex1}
c_3\le \f{c_1}{2c}(1+t)^{\nu_1-\nu-\nu_3}-\f{\nu}{2c}(1+t)^{-\nu-1-\nu_3},\q
cv(0)<1.$$ Thus one obtains the following conditions: $$\label{rcex2}
\nu_1\ge -1,\q \nu_2-\nu_1\le\nu\le\nu_1-\nu_3,$$ and $$\label{rcex3}
c_1>\nu,\q \f{2c_2}{c_1-\nu}\le c\le\f{c_1-\nu}{2c_3},\q
cv(0)<1.$$ Therefore for such $\g$, $\s$, $\b$ a function $\mu$ with the desired properties exists if $$\label{rcex4}
\nu_1\ge -1,\q \nu_2+\nu_3\le 2\nu_1,$$ and $$\label{rcex5}
c_1>\nu_2-\nu_1,\q 2\sqrt{c_2c_3}\le c_1+\nu_1-\nu_2,\q
2c_2v(0)<c_1+\nu_1-\nu_2.$$ In this case one can choose $\nu=\nu_2-\nu_1$, $c=\f{2c_2}{c_1+\nu_1-\nu_2}$. However in order to have $v(t)\to 0$ as $t\to+\infty$ (the case of interest in Theorem \[t1\]) one needs the following conditions: $$\label{rcex6}
\nu_1\ge -1,\q \nu_2+\nu_3\le 2\nu_1,\q \nu_1>\nu_3,$$ and $$\label{rcex7}
c_1>\nu_2-\nu_1,\q 2\sqrt{c_2c_3}\le c_1,\q 2c_2v(0)<c_1.$$

[**2.**]{} If $$\g(t)=\g_0,\q \s(t)=\s_0e^{\nu t},\q \b(t)=\b_0e^{-\nu t},\q \mu(t)=\mu_0e^{\nu t},$$ then conditions (\[ric1\]) are satisfied if $$0\le \s_0\le\f{\mu_0}{2}(\g_0-\nu),\q\b_0\le\f{1}{2\mu_0}(\g_0-\nu),\q\mu_0v(0)<1.$$

[**3.**]{} If $$\g(t)=\f{1}{\sqrt{\log(t+t_0)}},\q \mu(t)=c\log(t+t_0),$$ then conditions (\[ric1\]) are satisfied if $$0\le\s(t)\le\f{c}{2}\left(\sqrt{\log(t+t_0)}-\f{1}{t+t_0}\right),$$ $$\b(t)\le\f{1}{2c\log^2(t+t_0)}\left(\sqrt{\log(t+t_0)}-\f{1}{t+t_0}\right),\q
v(0)c\log t_0<1.$$

In all considered examples $\mu(t)$ can tend to infinity 