$\psi(u):=u-u^2$. This function goes to $-\infty$ as $u$ goes to $+\infty$, so it does not satisfy the positivity condition imposed in [@al1].

Unlike in the case of Bihari integral inequality ([@bb]) one cannot separate variables in the right hand side of inequality (\[ric4\]) and estimate $v(t)$ by a solution of the Cauchy problem for a differential equation with separating variables. The proof below is based on a special choice of the solution to the Riccati equation majorizing a solution of integral inequality (\[ric4\]). \[bihari\]

[**Proof**]{}[**of Lemma \[ric\]**]{} Denote: $$\label{rc1}
w(t):=v(t)e^{\int_0^t \g(s)ds},$$ then (\[ric4\]) implies: $$\label{rc2}
\dot{w}(t)\le a(t)w^2(t)+b(t),\q w(0)=v(0),$$ where $$a(t)=\s(t)e^{-\int_0^t \g(s)ds},\q b(t)=\b(t)e^{\int_0^t \g(s)ds}.$$ Consider Riccati’s equation: $$\label{rc3}
\dot{u}(t)=\f{\dot{f}(t)}{g(t)}u^2(t)-\f{\dot{g}(t)}{f(t)}.$$ One can check by a direct calculation that the the solution to problem (\[rc3\]) is given by the following formula [@kam eq. 1.33]: $$\label{rc5}
u(t)=-\f{g(t)}{f(t)}+\left[f^2(t)\left(C-
\int_0^t\f{\dot{f}(s)}{g(s)f^2(s)}ds\right)\right]^{-1}.$$ Define $f$ and $g$ as follows: $$\label{rc4}
f(t):=\mu^{\f{1}{2}}(t)e^{-\f{1}{2}\int_0^t \g(s)ds},\q
g(t):=-\mu^{-\f{1}{2}}(t)e^{\f{1}{2}\int_0^t \g(s)ds},$$ and consider the Cauchy problem for equation (\[rc3\]) with the initial condition $u(0)=v(0)$. Then $C$ in (\[rc5\]) takes the form: $$C=\f{1}{\mu(0)v(0)-1}.$$ From (\[ric1\]) one gets $$a(t)\le \f{\dot{f}(t)}{g(t)},\q b(t)\le -\f{\dot{g}(t)}{f(t)}.$$ Since $fg=-1$ one has: $$\int_0^t\f{\dot{f}(s)}{g(s)f^2(s)}ds=-\int_0^t\f{\dot{f}(s)}{f(s)}ds=
\f{1}{2}\int_0^t\left(\g(s)-\f{\dot{\mu}(s)}{\mu(s)}\right)ds.$$ Thus $$\label{rc6}
u(t)=\f{e^{\int_0^t \g(s)ds}}{\mu(t)}\left[1-\left(\f{1}{1-\mu(0)v(0)}+
\f{1}{2}\int_0^t\left(\g(s)-\f{\dot{\mu}(s)}{\mu(s)}\right)ds\right)^{-1}\right].$$ It follows from condition (\[ric1\]) that the solution to problem (\[rc3\]) exists for all $t\in[0,\infty)$ 