In this section, we state and prove some lemmas that will be useful in establishing our main results. For completeness, we start by recalling the adaptation of the generalized Kiguradze lemma. Next, we will provide some important monotonic properties of a positive solution *y* of (1.1) in cases (2.1) and (2.2), respectively.

For the sake of brevity, we define

$$ L_{0}y(t) = y(t), \quad \quad L_{i}y(t) = r_{i}(t) \bigl( L_{i-1}y(t) \bigr) ', \quad i = 1,2, \quad\quad L_{3}y(t) = \bigl( L_{2}y(t) \bigr) ', $$

for \(t\in I\). Using this notation, equation (1.1) takes the form

$$ L_{3}y(t) + q(t)y \bigl(\tau(t) \bigr) = 0. $$

### Lemma 1

(See [28, Lemma 2])

*Assume that*
\(y(t)\)
*is an eventually positive solution of* (1.1). *Then*
\(y(t)\)
*satisfies one of the following two cases*:

$$\begin{aligned}& y(t)>0, \quad\quad L_{1}y(t)>0, \quad\quad L_{2}y(t)>0, \quad\quad L_{3}y(t)< 0, \end{aligned}$$

(2.1)

$$\begin{aligned}& y(t)>0, \quad\quad L_{1}y(t)< 0, \quad\quad L_{2}y(t)>0, \quad\quad L_{3}y(t)< 0, \end{aligned}$$

(2.2)

*eventually*.

### Lemma 2

*Assume that*
\(y(t)\)
*is a positive solution of* (1.1) *which satisfies* (2.1). *Denote*

$$ \begin{gathered} R_{1}(t,t_{*}) := \int_{t_{*}}^{t}\frac{1}{r_{1}(v)} \int_{t_{*}}^{v}\frac{1}{r _{2}(s)}\,\mathrm {d}s \,\mathrm {d}v, \\ R_{m+1}(t,t_{*}) := \int_{t_{*}}^{t}\frac{1}{r_{1}(v)} \int_{t_{*}} ^{v}\frac{1}{r_{2}(s)}\exp \biggl( \int_{s}^{t}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s \,\mathrm {d}v , \quad m \in \mathbb {N}, \end{gathered} $$

(2.3)

*for*
\(t\ge t_{*}\), *where*
\(t_{*}\in I\)
*is large enough*. *Then*

$$ y \bigl(\tau(t) \bigr)\ge L_{2}y \bigl(\tau(t) \bigr)R_{m} \bigl(\tau(t),t_{*} \bigr), \quad t\ge t _{*}. $$

(2.4)

### Proof

Let \(y(t)\) be a solution of (1.1) satisfying (2.1) on \([t_{1},\infty)\), \(t_{1}\in I\). Since \(L_{2}y(t)\) is nonincreasing, it is easy to see that

$$ L_{1}y(t) \ge \int_{t_{1}}^{t} \bigl( L_{1}y(s) \bigr) '\,\mathrm {d}s = \int_{t_{1}}^{t}\frac{1}{r_{2}(s)}L_{2}y(s) \,\mathrm {d}s\ge L_{2}y(t) \int_{t _{1}}^{t}\frac{1}{r_{2}(s)}\,\mathrm {d}s $$

or

$$ y'(t)\ge L_{2}y(t)\frac{1}{r_{1}(t)} \int_{t_{1}}^{t} \frac{1}{r_{2}(s)}\,\mathrm {d}s, \quad t\ge t_{1}. $$

Integrating the above inequality from \(t_{1}\) to *t* and using (2.3), we have

$$ y(t)\ge \int_{t_{1}}^{t}L_{2}y(v)\frac{1}{r_{1}(v)} \int_{t_{1}}^{v}\frac{1}{r _{2}(s)}\,\mathrm {d}s \,\mathrm {d}v \ge L_{2}y(t)R_{1}(t,t_{1}). $$

Obviously, there exists \(t_{2}\ge t_{1}\) such that for all \(t\ge t _{2}\),

$$ y \bigl(\tau(t) \bigr)\ge L_{2}y \bigl(\tau(t) \bigr)R_{1} \bigl(\tau(t),t_{2} \bigr). $$

That is, (2.4) is satisfied for \(m = 1\).

Now, proceeding to the next induction step, we assume that (2.4) holds for some \(m>1\), which means that

$$ y \bigl(\tau(t) \bigr)\ge L_{2}y \bigl(\tau(t) \bigr)R_{m} \bigl(\tau(t),t_{*} \bigr), \quad t\ge t _{*}\ge t_{2}. $$

(2.5)

Combining (2.5) and (1.1), we get

$$ L_{3}y(t)+q(t)R_{m} \bigl(\tau(t),t_{*} \bigr)L_{2}y \bigl(\tau(t) \bigr) \le0 $$

or

$$ x'(t)+q(t)R_{m} \bigl(\tau(t),t_{*} \bigr)x \bigl(\tau(t) \bigr)\le0, $$

(2.6)

where \(x(t): = L_{2}y(t)\). In view of \(x(\tau(t))\ge x(t)\), we can write the last inequality in the form

$$ x'(t)+q(t)R_{m} \bigl(\tau (t),t_{*} \bigr)x(t)\le0. $$

(2.7)

Applying the Grönwall inequality in (2.7), we obtain

$$ x(s)\ge x(t)\exp \biggl( \int_{s}^{t}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) , \quad t\ge s\ge t_{*}, $$

or,

$$ L_{2}y(s)\ge L_{2}y(t)\exp \biggl( \int_{s}^{t}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) , \quad t\ge s\ge t_{*}. $$

(2.8)

Therefore,

$$ L_{1}y(t)\ge \int_{t_{*}}^{t}\frac{1}{r_{2}(s)}L_{2}y(s) \,\mathrm {d}s \ge L _{2}y(t) \int_{t_{*}}^{t}\frac{1}{r_{2}(s)}\exp \biggl( \int_{s}^{t}q(u)R _{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s, $$

that is,

$$ y'(t)\ge L_{2}y(t)\frac{1}{r_{1}(t)} \int_{t_{*}}^{t} \frac{1}{r_{2}(s)}\exp \biggl( \int_{s}^{t}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s. $$

Integrating the above inequality from \(t_{*}\) to *t* and using (2.8) again, we obtain

$$ \begin{aligned} y(t) &\ge \int_{t_{*}}^{t}L_{2}y(v)\frac{1}{r_{1}(v)} \int_{t_{*}}^{v}\frac{1}{r _{2}(s)}\exp \biggl( \int_{s}^{v}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s \,\mathrm {d}v \\ &\ge L_{2}y(t) \int_{t_{*}}^{t}\exp \biggl( \int_{v}^{t}q(u)R_{m} \bigl( \tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \\ &\quad{}\times \frac{1}{r_{1}(v)} \int_{t_{*}}^{v}\frac{1}{r _{2}(s)}\exp \biggl( \int_{s}^{v}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s \,\mathrm {d}v \\ & = L_{2}y(t) \int_{t_{*}}^{t}\frac{1}{r_{1}(v)} \int_{t_{*}}^{v}\frac{1}{r _{2}(s)}\exp \biggl( \int_{s}^{t}q(u)R_{m} \bigl(\tau (u),t_{*} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s \,\mathrm {d}v. \end{aligned} $$

Hence, there exists \(t_{**}\ge t_{*}\) such that for all \(t\ge t_{**}\),

$$ y \bigl(\tau(t) \bigr)\ge L_{2}y \bigl(\tau(t) \bigr) \int_{t_{**}}^{\tau(t)} \frac{1}{r_{1}(v)} \int_{t_{**}}^{v}\frac{1}{r_{2}(s)}\exp \biggl( \int_{s}^{\tau(t)}q(u)R_{m} \bigl(\tau (u),t_{**} \bigr)\,\mathrm {d}u \biggr) \,\mathrm {d}s \,\mathrm {d}v, $$

(2.9)

which, in view of (2.3), becomes

$$ y \bigl(\tau(t) \bigr)\ge L_{2}y \bigl(\tau(t) \bigr)R_{m+1} \bigl(\tau(t),t_{**} \bigr), \quad t \ge t_{**}. $$

This completes the induction step and the proof of the lemma. □

### Remark 3

Note that, for \(m=1\), (2.4) reduces to (1.6).

### Lemma 3

*Assume that*
\(y(t)\)
*is an eventually positive solution of* (1.1) *which satisfies* (2.2). *Denote*

$$ \begin{gathered} \tilde{R}_{1}(v,u) := \int_{u}^{v}\frac{1}{r_{1}(x)} \int_{x}^{v}\frac{1}{r _{2}(s)}\,\mathrm {d}s \,\mathrm {d}x, \\ \tilde{R}_{n+1}(v,u) := \int_{u}^{v}\frac{1}{r_{1}(x)} \int_{x}^{v}\frac{1}{r _{2}(s)}\exp \biggl( \int_{s}^{v}q(z)R_{n} \bigl(z,\tau(z) \bigr)\,\mathrm {d}z \biggr) \,\mathrm {d}s \,\mathrm {d}x, \quad n\in \mathbb {N}, \end{gathered} $$

(2.10)

*for*
\(v\ge u\ge t_{*}\), *where*
\(t_{*}\in I\)
*is large enough*. *Then*

$$ y(u)\ge L_{2}y(v)\tilde{R}_{n}(v,u),\quad v \ge u\ge t_{*}. $$

(2.11)

### Proof

Let \(y(t)\) be a solution of (1.1) satisfying (2.2) on \([t_{1},\infty)\), \(t_{1}\in I\). Since \(L_{2}y(t)\) is nonincreasing, we may write

$$ \begin{aligned} -L_{1}y(u) &\ge L_{1}y(v) - L_{1}y(u) = \int_{u}^{v}\frac{1}{r_{2}(s)}L _{2}y(s) \,\mathrm {d}s \ge L_{2}y(v) \int_{u}^{v}\frac{1}{r_{2}(s)}\,\mathrm {d}s \end{aligned} $$

(2.12)

for some \(v\ge u \ge t_{1}\), and hence,

$$ - y'(u)\ge L_{2}y(v)\frac{1}{r_{1}(u)} \int_{u}^{v}\frac{1}{r_{2}(s)}\,\mathrm {d}s. $$

Integrating the above inequality from *u* to \(v\ge u\ge t_{1}\), we get

$$ y(u)\ge L_{2}y(v) \int_{u}^{v}\frac{1}{r_{1}(x)} \int_{x}^{v}\frac{1}{r _{2}(s)}\,\mathrm {d}s\,\mathrm {d}x = L_{2}y(v)\tilde{R}_{1}(v,u), $$

(2.13)

which means that (2.11) is satisfied for \(n = 1\).

Now, assume that (2.11) holds for some \(n>1\), i.e.,

$$ y(u)\ge L_{2}y(v)\tilde{R}_{n}(v,u), \quad v\ge u\ge t_{*}\ge t_{1}. $$

(2.14)

Clearly, using (2.14) with \(u = \tau(t)\) and \(v = t\) in (1.1) yields

$$ L_{3}y(t)+q(t)\tilde{R}_{n} \bigl(t,\tau(t) \bigr)L_{2}y(t)\le0. $$

Proceeding as in the proof of Lemma 2, we get

$$ L_{2}y(s)\ge L_{2}y(v)\exp \biggl( \int_{s}^{v}q(z)\tilde{R}_{n} \bigl(z, \tau(z) \bigr)\,\mathrm {d}z \biggr) , \quad v\ge s\ge t_{*}. $$

(2.15)

Using (2.15) in (2.12), we have

$$ -L_{1}y(u)\ge L_{2}y(v) \int_{u}^{v}\frac{1}{r_{2}(s)}\exp \biggl( \int_{s}^{v}q(z)\tilde{R}_{n} \bigl(z, \tau(z) \bigr)\,\mathrm {d}z \biggr) \,\mathrm {d}s, $$

and therefore,

$$ - y'(u)\ge L_{2}y(v)\frac{1}{r_{1}(u)} \int_{u}^{v}\frac{1}{r_{2}(s)} \exp \biggl( \int_{s}^{v}q(z)\tilde{R}_{n} \bigl(z, \tau(z) \bigr)\,\mathrm {d}z \biggr) \,\mathrm {d}s. $$

(2.16)

Finally, by integrating (2.16) from *u* to \(v\ge u\ge t_{*}\), we conclude that

$$ \begin{aligned} y(u) &\ge L_{2}y(v) \int_{u}^{v}\frac{1}{r_{1}(x)} \int_{x}^{v}\frac{1}{r _{2}(s)}\exp \biggl( \int_{s}^{v}q(z)\tilde{R}_{n} \bigl(z, \tau(z) \bigr)\,\mathrm {d}z \biggr) \,\mathrm {d}s \,\mathrm {d}x \\ & = L_{2}y(v)\tilde{R}_{n+1}(v,u). \end{aligned} $$

Thus, relation (2.11) holds for any \(n\in \mathbb {N}\). The proof is complete. □

### Lemma 4

*Assume that*
\(y(t)\)
*is an eventually positive solution of* (1.1) *which satisfies* (2.2). *If*

$$ \int_{t_{0}}^{\infty}\frac{1}{r_{1}(v)} \int_{v}^{\infty}\frac{1}{r _{2}(u)} \int_{u}^{\infty}q(s)\,\mathrm {d}s \,\mathrm {d}u \,\mathrm {d}v = \infty, $$

(2.17)

*then*
\(y(t)\)
*tends to zero as*
\(t\to\infty\).

### Proof

Since the proof is similar to that of [17, Lemma 2], we omit it. □