In this section, we formulate and prove our results. For reader’s convenience, we slightly modify Eq. (5) as follows. We consider the equation
$$\begin{array}{}{\left[{t}^{\alpha}{r}^{-\frac{p}{q}}(t)\mathit{\Phi}({x}^{\prime})\right]}^{\prime}+{t}^{\alpha -p}s(t)\mathit{\Phi}(x)=0,\phantom{\rule{1em}{0ex}}\mathit{\Phi}(x)=|x{|}^{p-1}\phantom{\rule{thinmathspace}{0ex}}\text{sgn\hspace{0.17em}}x,\phantom{\rule{1em}{0ex}}p>1,\end{array}$$(17)

#### Proof

In the both parts of the proof, we will consider such a number *a* > 2 for which (see Definition 2.1 together with (18) and (19))
$$\begin{array}{}{\displaystyle {\left(\frac{p}{p-\alpha -1}\right)}^{p}\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau \right){\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\text{d}\tau \right)}^{p-1}>1+\epsilon}\end{array}$$(20)

or
$$\begin{array}{}{\displaystyle {\left(\frac{p}{p-\alpha -1}\right)}^{p}\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau \right){\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\text{d}\tau \right)}^{p-1}<1-\epsilon}\end{array}$$(21)

for all considered *t* and for some *ε* ∈ (0, 1). We can rewrite (20) and (21) into the following forms
$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau >\left(1+\epsilon \right){\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\text{d}\tau \right)}^{1-p}}\end{array}$$(22)

and
$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau <\left(1-\epsilon \right){\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\text{d}\tau \right)}^{1-p}.}\end{array}$$(23)

Using (6), from (22) and (23), we obtain the existence of *L* > 0, for which
$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau -{\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\text{d}\tau \right)}^{1-p}>L}\end{array}$$(24)

and
$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau -{\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\text{d}\tau \right)}^{1-p}<-L}\end{array}$$(25)

for all considered *t*. Indeed, one can put
$$\begin{array}{}{\displaystyle L:=\epsilon {\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left[{r}^{+}\right]}^{1-p}}\end{array}$$

in the both cases.

We will consider the associated adapted generalized Riccati equation in the form of Eq. (10) which corresponds Eq. (17), i.e., the equation
$$\begin{array}{}{\displaystyle {\zeta}^{\prime}(t)=\frac{1}{t}\left[(p-\alpha -1)\zeta (t)+s(t)+(p-1)r(t)|\zeta (t){|}^{q}\right].}\end{array}$$(26)

At first, we show that any solution *ζ* of Eq. (26) defined for *t* ≥ *T* is bounded from below, i.e., we show that there exists *K* > 1 satisfying
$$\begin{array}{}\zeta (t)>-K,\phantom{\rule{1em}{0ex}}t\ge T.\end{array}$$(27)

On the contrary, let us assume that
$$\begin{array}{}\underset{t\to \mathrm{\infty}}{lim\u2006inf}\zeta (t)=-\mathrm{\infty}\end{array}$$(28)

or
$$\begin{array}{}\underset{t\to {T}_{0}^{-}}{lim\u2006inf}\zeta (t)=-\mathrm{\infty}\end{array}$$(29)

for some *T*_{0} ∈ (*T*, ∞). For given *a* and function *s*, let us consider *M*(*s*) from Lemma 2.2. Let *P* > 0 be an arbitrary number such that
$$\begin{array}{}(p-1){r}^{-}{y}^{q}-(p-\alpha -1)y>M(s),\phantom{\rule{1em}{0ex}}y\ge P.\end{array}$$(30)

In particular, considering inf_{t ≥ T} *ζ* (*t*) = −∞, the continuity of *ζ* implies the existence of an interval [*t*_{1}, *t*_{2}] such that *ζ* (*t*_{1}) ≤ − *P*, *ζ*(*t*) < −*P* for all *t* ∈ (*t*_{1}, *t*_{2}], and *t*_{2} − *t*_{1} ∈ (0, *a*]. Without loss of generality, we consider *T* > 2. Using (12) in Lemma 2.2, the form of Eq. (26), and (30), we have
$$\begin{array}{}\zeta ({t}_{2})-\zeta ({t}_{1})\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}& {\displaystyle =\underset{{t}_{1}}{\overset{{t}_{2}}{\int}}{\zeta}^{\mathrm{\prime}}(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau}\\ & {\displaystyle =\underset{{t}_{1}}{\overset{{t}_{2}}{\int}}\frac{(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau}\\ & {\displaystyle >\underset{{t}_{1}}{\overset{{t}_{2}}{\int}}\frac{M(s)+s(\tau )}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \ge \underset{{t}_{1}}{\overset{{t}_{2}}{\int}}\frac{M(s)}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\left|\underset{{t}_{1}}{\overset{{t}_{2}}{\int}}\frac{s(\tau )}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|}\\ & {\displaystyle >0-\frac{M(s)}{{t}_{1}}\ge -\frac{M(s)}{T}>-M(s).}\end{array}$$(31)

This inequality proves that (29) cannot be valid for any *T*_{0} ∈ ℝ (consider that *a* > 2 is given). Therefore, there exist arbitrarily long intervals, where *ζ*(*t*) ≤ −*P*. Let *I* = [*t*_{3}, *t*_{4}] be such an interval whose length is at least 2 (i.e., *t*_{4} − *t*_{3} ≥ 2) and *t*_{4} − *t*_{3} ≤ *a*. As in (31) (consider that *t*_{3} > 2), we obtain
$$\begin{array}{}\zeta ({t}_{4})-\zeta ({t}_{3})\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}& {\displaystyle =\underset{{t}_{3}}{\overset{{t}_{4}}{\int}}{\zeta}^{\mathrm{\prime}}(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau >\underset{{t}_{3}}{\overset{{t}_{4}}{\int}}\frac{M(s)}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\left|\underset{{t}_{3}}{\overset{{t}_{4}}{\int}}\frac{s(\tau )}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|}\\ & {\displaystyle \ge M(s)\mathrm{log}\frac{{t}_{4}}{{t}_{3}}-\frac{M(s)}{{t}_{3}}\ge M(s)\left(\mathrm{log}\frac{{t}_{3}+2}{{t}_{3}}-\frac{1}{{t}_{3}}\right)>0.}\end{array}$$(32)

Of course, (32) means that *ζ*(*t*_{4}) > *ζ*(*t*_{3}). In fact, (31) and (32) guarantee that
$$\begin{array}{}\underset{t\to \mathrm{\infty}}{lim\u2006inf}\zeta (t)>-\mathrm{\infty},\end{array}$$

which contradicts (28). Hence, (27) is valid.

In the both parts of the proof, we will also apply the estimation
$$\begin{array}{}{\displaystyle \left|\underset{t}{\overset{t+a}{\int}}\frac{s(\tau )}{t}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\underset{t}{\overset{t+a}{\int}}\frac{s(\tau )}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|\le \frac{a}{{t}^{2}}M(s)}\end{array}$$(33)

for all large *t* and *M*(*s*) from Lemma 2.2. We use the mean value theorem of the integral calculus to get this estimation. More precisely, considering *t* ∈ [*t*_{1}, *t*_{2}], where *t*_{1} is sufficiently large, since *s* is integrable and *x*(*t*) = *t*^{− 1} is monotone for *t* ∈ [*t*_{1}, *t*_{2}], there exists *t*_{3} ∈ [*t*_{1}, *t*_{2}] such that
$$\begin{array}{}{\displaystyle \underset{{t}_{1}}{\overset{{t}_{2}}{\int}}\frac{s(\tau )}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau =\frac{1}{{t}_{1}}\underset{{t}_{1}}{\overset{{t}_{3}}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau +\frac{1}{{t}_{2}}\underset{{t}_{3}}{\overset{{t}_{2}}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau .}\end{array}$$(34)

Immediately, from (34), we obtain (see (11) in Lemma 2.2)
$$\begin{array}{}{\displaystyle \left|\underset{t}{\overset{t+a}{\int}}\left(\frac{1}{t}-\frac{1}{\tau}\right)s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|=\left|\underset{t}{\overset{t+a}{\int}}\frac{s(\tau )}{t}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\frac{1}{t}\underset{t}{\overset{t+b}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\frac{1}{t+a}\underset{t+b}{\overset{t+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|}\\ {\displaystyle \phantom{\rule{2em}{0ex}}=\left|\frac{1}{t}\underset{t+b}{\overset{t+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\frac{1}{t+a}\underset{t+b}{\overset{t+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|=\frac{a}{t(t+a)}\left|\phantom{\rule{thinmathspace}{0ex}}\underset{t+b}{\overset{t+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|\le \frac{a}{{t}^{2}}M(s),}\end{array}$$(35)

where *b* ∈ [0, *a*]. It is seen that (35) gives (33).

Part (I). Evidently (see (18)), *s* > 0. By contradiction, let us suppose that Eq. (17) is non-oscillatory. From Lemma 2.5, we know that there exists a non-positive solution *ζ* of Eq. (26) on some interval [*T*, ∞). For this solution *ζ*, we introduce the averaging function *ζ*_{ave} by
$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}(t):=\frac{1}{a}\underset{t}{\overset{t+a}{\int}}\zeta (\tau )\text{d}\tau ,\phantom{\rule{1em}{0ex}}t\ge T.}\end{array}$$(36)

We know that (see (27))
$$\begin{array}{}{\zeta}_{\text{ave}}(t)\in (-K,0],\phantom{\rule{1em}{0ex}}t\ge T.\end{array}$$(37)

For *t* > *T* (see Eq. (26)), we have
$$\begin{array}{}{\displaystyle {\mathrm{\zeta !}}_{\text{ave}}^{\prime}(t)=\frac{1}{a}\underset{t}{\overset{t+a}{\int}}{\zeta}^{\prime}(\tau )\text{d}\tau =\frac{1}{a}\underset{t}{\overset{t+a}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\text{d}\tau .}\end{array}$$(38)

For *t* > *T*, we also have (see (27) and (33))
$$\begin{array}{}{\displaystyle \left|\underset{t}{\overset{t+a}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\text{d}\tau \right.}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left.-\frac{1}{t}\underset{t}{\overset{t+a}{\int}}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\text{d}\tau \right|}\\ {\displaystyle \le \underset{t}{\overset{t+a}{\int}}\left(\frac{1}{t}-\frac{1}{\tau}\right)\left[(p-\alpha -1)K+(p-1){r}^{+}{K}^{q}\right]\text{d}\tau +\left|\underset{t}{\overset{t+a}{\int}}\frac{s(\tau )}{t}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\underset{t}{\overset{t+a}{\int}}\frac{s(\tau )}{\tau}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right|}\\ {\displaystyle \le \frac{{a}^{2}}{t(t+a)}\left[(p-\alpha -1)K+(p-1){r}^{+}{K}^{q}\right]+\frac{a}{{t}^{2}}M(s)<\frac{Na}{{t}^{2}},}\end{array}$$(39)

where
$$\begin{array}{}N:=a(p-\alpha -1)K+a(p-1){r}^{+}{K}^{q}+M(s).\end{array}$$(40)

For *t* > *T*, we obtain (see (38), (39), and (40))
$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}(t)\ge \frac{1}{at}\underset{t}{\overset{t+a}{\int}}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}-\frac{N}{t}\right]\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau .}\end{array}$$(41)

If we put
$$\begin{array}{}{\displaystyle X(t):=\frac{(p-\alpha -1{)}^{p}}{pt}{\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{-\frac{p}{q}},\phantom{\rule{1em}{0ex}}Y(t):=\frac{|{\zeta}_{\text{ave}}(t){|}^{q}}{qt}\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}\end{array}$$(42)

for *t* > *T*, then we have (see (41))
$$\begin{array}{}{\zeta}_{\text{ave}}^{\prime}(t)\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}& {\displaystyle \ge \frac{1}{at}\underset{t}{\overset{t+a}{\int}}(p-\alpha -1)\zeta (\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau +X(t)+Y(t)-\frac{N}{{t}^{2}}}\\ & {\displaystyle +\frac{1}{at}\underset{t}{\overset{t+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -X(t)+\frac{1}{at}\underset{t}{\overset{t+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -Y(t),\phantom{\rule{1em}{0ex}}t>T.}\end{array}$$(43)

Taking into account (43), for large *t*, we will show the inequalities
$$\begin{array}{}{\displaystyle \frac{N}{{t}^{2}}\le \frac{L}{3t},}\end{array}$$(44)
$$\begin{array}{}{\displaystyle \frac{1}{at}\underset{t}{\overset{t+a}{\int}}(p-\alpha -1)\zeta (\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau +X(t)+Y(t)\ge 0,}\end{array}$$(45)
$$\begin{array}{}{\displaystyle \frac{1}{at}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau -X(t)\ge \frac{L}{t},}\end{array}$$(46)
$$\begin{array}{}{\displaystyle \left|\frac{1}{at}\underset{t}{\overset{t+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -Y(t)\right|\le \frac{L}{3t}.}\end{array}$$(47)

The first inequality (44) is valid for all *t* ≥ 3*N*/*L*. Hence, we can approach to (45). It holds (see (36) and (42))
$$\begin{array}{}{\displaystyle \frac{1}{at}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}}& {\displaystyle \underset{t}{\overset{t+a}{\int}}(p-\alpha -1)\zeta (\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau +X(t)+Y(t)}\\ & {\displaystyle =\frac{1}{t}\left[(p-\alpha -1){\zeta}_{\text{ave}}(t)+\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{-\frac{p}{q}}\right.}\\ & \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}{\displaystyle \left.+\frac{|{\zeta}_{\text{ave}}(t){|}^{q}}{q}\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)\right].}\end{array}$$(48)

We recall the well-known Young inequality which says that
$$\begin{array}{}{\displaystyle \frac{{A}^{p}}{p}+\frac{{B}^{q}}{q}-AB\ge 0}\end{array}$$(49)

holds for all non-negative numbers *A*, *B*. We take *A* = (*p t X*(*t*))^{1/p} and *B* = (*qtY*(*t*))^{1/q}. Hence,
$$\begin{array}{c}{\displaystyle \frac{{A}^{p}}{p}=tX(t)=\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{-\frac{p}{q}},}\\ \\ {\displaystyle \frac{{B}^{q}}{q}=tY(t)=\frac{|{\zeta}_{\text{ave}}(t){|}^{q}}{q}\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right),}\end{array}$$

and (see (37))
$$\begin{array}{}{\displaystyle AB\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}}& ={\left(ptX(t)\right)}^{\frac{1}{p}}{\left(qtY(t)\right)}^{\frac{1}{q}}\\ & {\displaystyle =(p-\alpha -1){\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{-\frac{1}{q}}|{\zeta}_{\text{ave}}(t)|{\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{\frac{1}{q}}=-(p-\alpha -1){\zeta}_{\text{ave}}(t).}\end{array}$$

Finally, considering (48) and (49), we have
$$\begin{array}{}{\displaystyle \frac{1}{at}\underset{t}{\overset{t+a}{\int}}(p-\alpha -1)\zeta (\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau +X(t)+Y(t)=\frac{1}{t}\left[\frac{{A}^{p}}{p}+\frac{{B}^{q}}{q}-AB\right]\ge 0,}\end{array}$$

which proves (45).

Next, (46) is valid. Indeed, we have (see (24) and (42))
$$\begin{array}{}{\displaystyle \frac{1}{at}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau -X(t)\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}}& ={\displaystyle \frac{1}{t}\left[\frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau -\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{-\frac{p}{q}}\right]}\\ & {\displaystyle =\frac{1}{t}\left[\frac{1}{a}\underset{t}{\overset{t+a}{\int}}s(\tau )\text{d}\tau -{\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left(\frac{1}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)}^{1-p}\phantom{\rule{thinmathspace}{0ex}}\right]>\frac{L}{t}}\end{array}$$

for all considered *t*.

To prove (47), we use the form of Eq. (26) together with (6), (12) from Lemma 2.2, and with (27) which immediately give
$$\begin{array}{}{\displaystyle \left|\phantom{\rule{thinmathspace}{0ex}}\underset{t+i}{\overset{t+j}{\int}}{\zeta}^{\prime}(\tau )\text{d}\tau \right|\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}}& {\displaystyle \le \left|\phantom{\rule{thinmathspace}{0ex}}\underset{t+i}{\overset{t+j}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\text{d}\tau \right|}\\ & {\displaystyle \le \underset{t+i}{\overset{t+j}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)K+(p-1){r}^{+}{K}^{q}\right]\text{d}\tau +\left|\phantom{\rule{thinmathspace}{0ex}}\underset{t+i}{\overset{t+j}{\int}}\frac{s(\tau )}{\tau}\text{d}\tau \right|}\\ & {\displaystyle \le \frac{a}{t}\left[(p-\alpha -1)K+(p-1){r}^{+}{K}^{q}\right]+\frac{M(s)}{t+i}\le \frac{Q}{t}}\end{array}$$(50)

for *t* > *T* and *i*, *j* ∈ [0, *a*], *i* ≤ *j*, where
$$\begin{array}{}Q:=a\left[(p-\alpha -1)K+(p-1){r}^{+}{K}^{q}\right]+M(s).\end{array}$$

Hence, we have
$$\begin{array}{}{\displaystyle \left|\zeta (t+j)-\zeta (t+i)\right|\le \frac{Q}{t},\phantom{\rule{1em}{0ex}}t>T,\phantom{\rule{thinmathspace}{0ex}}i,j\in [0,a],}\end{array}$$

which implies (see (36))
$$\begin{array}{}{\displaystyle \left|\zeta (\tau )-{\zeta}_{\text{ave}}(t)\right|\le \frac{Q}{t}}\end{array}$$(51)

for all *t* > *T* and *τ* ∈ [*t*, *t*+*a*].

Further, since the function *x*(*t*) = |*t*|^{q} is continuously differentiable on [−*K*, 0], there exists *C* > 0 for which
$$\begin{array}{}\left||y{|}^{q}-|z{|}^{q}\right|\le C|y-z|,\phantom{\rule{1em}{0ex}}y,z\in [-K,0].\end{array}$$(52)

Thus, we have (see (6), (7), (27), (37), (42), (51), and (52))
$$\begin{array}{}{\displaystyle \left|\frac{1}{at}\underset{t}{\overset{t+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -Y(t)\right|}\\ {\displaystyle =\frac{1}{t}\left|\frac{1}{a}\underset{t}{\overset{t+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau -\frac{|{\zeta}_{\text{ave}}(t){|}^{q}}{q}\left(\frac{p}{a}\underset{t}{\overset{t+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \right)\right|}\\ {\displaystyle \le \frac{p-1}{at}\underset{t}{\overset{t+a}{\int}}||\zeta (\tau ){|}^{q}-|{\zeta}_{\text{ave}}(t){|}^{q}|\phantom{\rule{thinmathspace}{0ex}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \le \frac{(p-1){r}^{+}}{at}\underset{t}{\overset{t+a}{\int}}||\zeta (\tau ){|}^{q}-|{\zeta}_{\text{ave}}(t){|}^{q}|\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau}\\ {\displaystyle \le \frac{(p-1){r}^{+}}{at}\underset{t}{\overset{t+a}{\int}}C|\zeta (\tau )-{\zeta}_{\text{ave}}(t)|\phantom{\rule{thinmathspace}{0ex}}\text{d}\tau \le \frac{(p-1){r}^{+}CQ}{{t}^{2}}}\end{array}$$

for *t* > *T* which gives (47) for all sufficiently large *t*.

Altogether, (43) together with (44), (45), (46), and (47) guarantee
$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}(t)\ge 0-\frac{L}{3t}+\frac{L}{t}-\frac{L}{3t}=\frac{L}{3t}}\end{array}$$(53)

for all large *t*. From (53) it follows that lim_{t → ∞} *ζ*_{ave} (*t*) = ∞. In particular (see (36)), *ζ* is positive at least in one point which is a contradiction. The proof of part (I) is complete.

Part (II). Without loss of generality, we can assume that *s* > 0. Indeed, for *s* ≤ 0, it suffices to replace function *s* by function *s* + *k* for a constant *k* > 0 such that *s* + *k* > 0 and
$$\begin{array}{}{\displaystyle {\left(\frac{p}{p-\alpha -1}\right)}^{p}\left(\overline{s}+k\right){\overline{r}}^{p-1}<1}\end{array}$$

and to use Theorem 2.3.

Let *t*_{0} be a sufficiently large number. We denote (see (6))
$$\begin{array}{}{\displaystyle Z:={\left(\frac{p\phantom{\rule{thinmathspace}{0ex}}{r}^{-}}{p-\alpha -1}\right)}^{1-p}.}\end{array}$$

Let *ζ* be the solution of the adapted generalized Riccati equation (26) satisfying
$$\begin{array}{}{\displaystyle \zeta ({t}_{0})=-{\left(\frac{p}{(p-\alpha -1)a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\text{d}\tau \right)}^{1-p}\in \left(-2Z,0\right).}\end{array}$$(54)

Based on Lemma 2.6, it suffices to prove that this solution exists for all *t* ∈ [*t*_{0}, ∞). Let [*t*_{0}, *T*) be the maximal interval, where the solution *ζ* exists. Note that *T* ∈ (*t*_{0}, ∞) ∪ {∞}. In fact, considering the continuity and the boundedness from below of *ζ* (see (27)), it suffices to show that *ζ*(*t*) < 0 for all *t* ∈ [*t*_{0}, *T*).

Let us consider an interval *J* := [*t*_{0}, *t*_{1}) such that *ζ*(*t*) ∈ (−2 *Z*, 0) for *t* ∈ *J*. For any *t*_{2}, *t*_{3} ∈ *J* ∩ [*t*_{0}, *t*_{0} + *a*], *t*_{2} ≤ *t*_{3}, we have (see (6) and (12) in Lemma 2.2, cf. (50))
$$\begin{array}{}\left|\phantom{\rule{thinmathspace}{0ex}}\zeta ({t}_{3})-\zeta ({t}_{2})\right|\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}& {\displaystyle =\left|\phantom{\rule{thinmathspace}{0ex}}\underset{{t}_{2}}{\overset{{t}_{3}}{\int}}{\zeta}^{\prime}(\tau )\text{d}\tau \right|}\\ & {\displaystyle =\left|\phantom{\rule{thinmathspace}{0ex}}\underset{{t}_{2}}{\overset{{t}_{3}}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\text{d}\tau \right|}\\ & {\displaystyle \le \underset{{t}_{2}}{\overset{{t}_{3}}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)2Z+(p-1){r}^{+}{\left(2Z\right)}^{q}\right]\text{d}\tau +\left|\phantom{\rule{thinmathspace}{0ex}}\underset{{t}_{2}}{\overset{{t}_{3}}{\int}}\frac{s(\tau )}{\tau}\text{d}\tau \right|}\\ & {\displaystyle \le \frac{1}{{t}_{0}}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}\left[(p-\alpha -1)2Z+(p-1){r}^{+}{\left(2Z\right)}^{q}\right]\text{d}\tau +\frac{M(s)}{{t}_{2}}\le \frac{\stackrel{~}{Q}}{{t}_{0}},}\end{array}$$(55)

where
$$\begin{array}{}\stackrel{~}{Q}:=a\left[(p-\alpha -1)2Z+(p-1){r}^{+}{\left(2Z\right)}^{q}\right]+M(s).\end{array}$$

Since *t*_{0} is given as a sufficiently large number, from (55), we see that
$$\begin{array}{}\zeta (t)\in (-2Z,0),\phantom{\rule{1em}{0ex}}t\in [{t}_{0},{t}_{0}+a].\end{array}$$(56)

In addition, (55) means that
$$\begin{array}{}{\displaystyle |\zeta ({t}_{0})-\zeta (\tau )|<\frac{\stackrel{~}{Q}}{{t}_{0}},\phantom{\rule{1em}{0ex}}\tau \in [{t}_{0},{t}_{0}+a].}\end{array}$$(57)

Similarly as in the first part of the proof, we introduce
$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}(t):=\frac{1}{a}\underset{t}{\overset{t+a}{\int}}\zeta (\tau )\text{d}\tau}\end{array}$$(58)

for *t* from a neighbourhood of *t*_{0}. It holds (see (56), (57), and (58))
$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}({t}_{0})\in (-2Z,0),\phantom{\rule{2em}{0ex}}|{\zeta}_{\text{ave}}({t}_{0})-\zeta ({t}_{0})|<\frac{\stackrel{~}{Q}}{{t}_{0}}.}\end{array}$$(59)

We have (see Eq. (26), cf. (38))

$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}({t}_{0})=\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}{\zeta}^{\prime}(\tau )\mathrm{d}\tau =\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\mathrm{d}\tau .}\end{array}$$(60)

In addition, as in the first part of the proof (see (39)), we have

$$\begin{array}{}{\displaystyle |\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}\frac{1}{\tau}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\mathrm{d}\tau}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}-\frac{1}{{t}_{0}}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}\right]\mathrm{d}\tau |<\frac{Na}{{t}_{0}^{2}},}\end{array}$$(61)

where *N* is defined in (40) (we can also put *K* = 2*Z*).

Hence, from (60) and (61), we obtain (cf. (41))

$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}({t}_{0})\le \frac{1}{a{t}_{0}}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}\left[(p-\alpha -1)\zeta (\tau )+s(\tau )+(p-1)r(\tau )|\zeta (\tau ){|}^{q}+\frac{N}{{t}_{0}}\right]\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau .}\end{array}$$(62)

If we put (cf. (42))

$$\begin{array}{}{\displaystyle X({t}_{0}):=\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p},}\end{array}$$(63)

$$\begin{array}{}{\displaystyle Y({t}_{0}):=\frac{(p-\alpha -1{)}^{p}}{q}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p},}\end{array}$$(64)

then we have (see (62))

$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}({t}_{0})\le \frac{1}{a{t}_{0}}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}s(\tau )-X({t}_{0})\mathrm{d}\tau +\frac{1}{a{t}_{0}}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}-Y({t}_{0})\mathrm{d}\tau}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}+\frac{1}{a{t}_{0}}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-\alpha -1)\zeta (\tau )\mathrm{d}\tau +\frac{X({t}_{0})}{{t}_{0}}+\frac{Y({t}_{0})}{{t}_{0}}+\frac{N}{{t}_{0}^{2}}.}\end{array}$$(65)

The aim of our process is to prove that $\begin{array}{}{\zeta}_{\text{ave}}^{\prime}({t}_{0})<0.\end{array}$ To obtain this inequality, it suffices to show (see (65))

$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}s(\tau )-X({t}_{0})\mathrm{d}\tau \le -L,}\end{array}$$(66)

$$\begin{array}{}{\displaystyle \left|\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}-Y({t}_{0})\mathrm{d}\tau \right|\le \frac{L}{4},}\end{array}$$(67)

$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-\alpha -1)\zeta (\tau )\mathrm{d}\tau +X({t}_{0})+Y({t}_{0})\le \frac{L}{4},}\end{array}$$(68)

$$\begin{array}{}{\displaystyle \frac{N}{{t}_{0}}\le \frac{L}{4}.}\end{array}$$(69)

Evidently, (69) is valid, because *t*_{0} is sufficiently large. We show that (66) is valid. We have (see (25), (63))

$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}s(\tau )-X({t}_{0})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau =\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau -\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}=\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}s(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau -{\left(\frac{p-\alpha -1}{p}\right)}^{p}{\left(\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}<-L,}\end{array}$$

i.e., we obtain (66).

Now we prove (67). Let *D* > 0 be such that (cf. (52))

$$\begin{array}{}{\displaystyle \left||y{|}^{q}-|z{|}^{q}\right|\le D|y-z|,\phantom{\rule{2em}{0ex}}y,z\in [-2Z,0].}\end{array}$$(70)

Using (6), (7), (54), (56), (57), (64), and (70), we have

$$\begin{array}{}{\displaystyle \left|\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}-Y({t}_{0})\mathrm{d}\tau \right|}\\ {\displaystyle =\left|\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}\mathrm{d}\tau -\frac{(p-\alpha -1{)}^{p}}{q}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}\right|}\\ {\displaystyle =\left|\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-1)r(\tau )|\zeta (\tau ){|}^{q}\mathrm{d}\tau -\frac{p}{q}{\left|\zeta ({t}_{0})\right|}^{q}\left(\frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)\right|}\\ {\displaystyle \le \frac{p-1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )||\zeta (\tau ){|}^{q}-|{\zeta ({t}_{0})}^{q}|\mathrm{d}\tau \le \frac{(p-1){r}^{+}}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}||\zeta (\tau ){|}^{q}-|{\zeta ({t}_{0})}^{q}|\mathrm{d}\tau}\\ {\displaystyle \le \frac{(p-1){r}^{+}}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}D\left|\zeta (\tau )-\zeta ({t}_{0})\right|\mathrm{d}\tau \le \frac{(p-1){r}^{+}}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}D\phantom{\rule{thinmathspace}{0ex}}\frac{\stackrel{~}{Q}}{{t}_{0}}\mathrm{d}\tau =\frac{D\stackrel{~}{Q}(p-1){r}^{+}}{{t}_{0}}.}\end{array}$$(71)

For a sufficiently large number *t*_{0}, inequality (67) follows from (71).

It remains to prove (68). We have (see (58), (63), and (64))

$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-\alpha -1)\zeta (\tau )\mathrm{d}\tau +X({t}_{0})+Y({t}_{0})=(p-\alpha -1){\zeta}_{\text{ave}}({t}_{0})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}+\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}+\frac{(p-\alpha -1{)}^{p}}{q}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}.}\end{array}$$(72)

Let us assume that (see (54))

$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}({t}_{0})=\zeta ({t}_{0})=-{\left(\frac{p}{(p-\alpha -1)a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\mathrm{d}\tau \right)}^{1-p}.}\end{array}$$(73)

Then, (72) gives

$$\begin{array}{}{\displaystyle \frac{1}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}(p-\alpha -1)\zeta (\tau )\mathrm{d}\tau +X({t}_{0})+Y({t}_{0})}\\ {\displaystyle =-(p-\alpha -1){\left(\frac{p}{(p-\alpha -1)a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\mathrm{d}\tau \right)}^{1-p}}\\ {\displaystyle +\frac{(p-\alpha -1{)}^{p}}{p}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}+\frac{(p-\alpha -1{)}^{p}}{q}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}\tau \right)}^{1-p}}\\ {\displaystyle =(p-\alpha -1{)}^{p}{\left(\frac{p}{a}\underset{{t}_{0}}{\overset{{t}_{0}+a}{\int}}r(\tau )\mathrm{d}\tau \right)}^{1-p}\left(-1+\frac{1}{p}+\frac{1}{q}\right)=0.}\end{array}$$(74)

Using (59) (consider (73)), one can see that (74) gives (68) for large *t*_{0}.

Finally, applying (66), (67), (68), and (69) in (65), we have

$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}({t}_{0})\le \frac{1}{{t}_{0}}\left(-L+\frac{L}{4}+\frac{L}{4}+\frac{L}{4}\right)=-\frac{L}{4{t}_{0}}<0,}\end{array}$$

i.e., we have (see (58))

$$\begin{array}{}{\displaystyle {\zeta}_{\text{ave}}^{\prime}({t}_{0})=\frac{\zeta ({t}_{0}+a)-\zeta ({t}_{0})}{a}<0,}\end{array}$$

i.e., *ζ* (*t*_{0} + *a*) < *ζ* (*t*_{0}). Since we can replace *t*_{0} by an arbitrary number *t* > *t*_{0} in the process above, we obtain *ζ* (*t* + *a*) < *ζ*(*t*) and *ζ*(*τ*) < 0 for all *τ* ∈ [*t*, *t* + *a*] if

$$\begin{array}{}{\displaystyle \zeta (t)=-{\left(\frac{p}{(p-\alpha -1)a}\underset{t}{\overset{t+a}{\int}}r(\tau )\mathrm{d}\tau \right)}^{1-p}}\end{array}$$

for some *t* ∈ (*t*_{0}, ∞). Considering the fact that *a* can be arbitrarily large (see Definition 2.1 and (6)) together with (55) and $\begin{array}{}{\zeta}_{\text{ave}}^{\prime}(t)<0,\end{array}$ we obtain that *ζ* (*t*) < 0 for all *t* > *t*_{0}. The proof is complete.∈□

We repeat that we obtain new results even for linear equations. With regard to the importance of this fact, we formulate the corresponding consequence of Theorem 3.1 as the corollary below.

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