the initial Cauchy problem that has a unique solution. The inequality $u_{n_k}(t)\geq w(t),$ $k=0,1,...$ implies $u(t)\geq w(t).$ If the solution to the Cauchy problem (\[Cauchy\]) is not unique, the inequality $w(t)\leq u(t)$ holds for the maximal solution to (\[Cauchy\]). $\Box$

Our second lemma is a key to the basic result of this section, namely to Theorem \[t1\].

Let $\g(t), \s(t), \b(t)\in C[0,\infty)$. If there exists a positive function $\mu(t)\in C^1[0,\infty)$ such that $$\label{ric1}
0\le\s(t)\le\f{\mu(t)}{2}\left(\g(t)-\f{\dot{\mu}(t)}{\mu(t)}\right),\q
\b(t)\le\f{1}{2\mu(t)}\left(\g(t)-\f{\dot{\mu}(t)}{\mu(t)}\right),\q
\mu(0)v(0)<1,$$ then a nonnegative solution to the following inequality $$\label{ric4}
\dot{v}(t)\le -\g(t) v(t)+\s(t)v^2(t)+\b(t).$$ satisfies the estimate: $$\label{ric5}
v(t)\,<\, \f{1}{\mu(t)}.$$ \[ric\]

Without loss of generality one can assume $\b(t)\geq 0$.

In [@al1] (see also Appendix) a differential inequality $\dot{u}\le -a(t)\psi(u(t))+b(t)$ was studied under assumptions 1) - 3) of Lemma \[apineq\] of the Appendix. These assumptions alone, as we show in the Appendix, do not imply the desired conclusion (\[descon\]). We have added assumption 4) in order to prove conclusion (\[descon\]). In Lemma \[ric\] the term $-\g(t)v(t)+\s(t)v^2(t)$ (which is analogous to some extent to the term $-a(t)\psi(u(t))$) can change sign. Our Lemma \[ric\] is not covered by the result in [@al1]. In particular, in Lemma \[ric\] an analog of $\psi(u)$, for the case $\g(t)=\s(t)=a(t)$, is the function 