1 Introduction

The calculus of time scales was accomplished by Stefan Hilger [7]. A time scale is an arbitrary nonempty closed subset of the real numbers. Let \(\mathbb{T}\) be a time scale, \(\xi ,\omega \in \mathbb{T}\) with \(\xi <\omega \), and an interval \([\xi ,\omega ]_{\mathbb{T}}\) means the intersection of the real interval with the given time scale. The major aim of the calculus of time scales is to establish results in general, comprehensive, unified, and extended forms. This hybrid theory is also widely applied in dynamic inequalities, see [2, 8–12]. The basic ideas about time scale calculus are given in the monographs [3, 4].

We state here the different versions of reverses of Callebaut, Rogers–Hölder, and Cauchy–Schwarz inequalities, see [5].

Let \(x_{k}>0\), \(y_{k}>0\), and \(w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\) with \(\sum^{\eta}_{k=1}w_{k}=1\). If there exist constants m, \(M>0\) such that \(0< m\leq \frac{x_{k}}{y_{k}}\leq M<\infty \) for any \(k\in \{1,2,\ldots ,\eta \}\), then

$$\begin{aligned} \sum^{\eta}_{k=1}w_{k}x^{2(1-v)}_{k}y^{2v}_{k} \sum^{ \eta}_{k=1}w_{k}x^{2v}_{k}y^{2(1-v)}_{k} &\leq \sum^{\eta}_{k=1}w_{k}x^{2}_{k} \sum^{\eta}_{k=1}w_{k}y^{2}_{k} \\ & \leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr)\sum ^{ \eta}_{k=1}w_{k}x^{2(1-v)}_{k}y^{2v}_{k} \sum^{\eta}_{k=1}w_{k}x^{2v}_{k}y^{2(1-v)}_{k}, \end{aligned}$$
(1.1)

for any \(v\in [0, 1]\) and, in particular,

$$ \Biggl(\sum^{\eta}_{k=1}w_{k}x_{k}y_{k} \Biggr)^{2} \leq \sum^{\eta}_{k=1}w_{k}x^{2}_{k} \sum^{\eta}_{k=1}w_{k}y^{2}_{k} \leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \Biggl(\sum ^{\eta}_{k=1}w_{k}x_{k}y_{k} \Biggr)^{2}. $$
(1.2)

Let \(\frac{1}{p}+\frac{1}{q}=1\) with \(p>1\). If there exist constants m, M, n, N such that \(0< m\leq x_{k}\leq M<\infty \) and \(0< n\leq y_{k}\leq N<\infty \) for any \(k\in \{1,2,\ldots ,\eta \}\), then we have the following reverse of Rogers–Hölder discrete inequality:

$$ \Biggl(\sum^{\eta}_{k=1}w_{k}x^{p} \Biggr)^{\frac{1}{p}} \Biggl(\sum^{\eta}_{k=1}w_{k}y^{q} \Biggr)^{\frac{1}{q}} \leq S \biggl( \biggl(\frac{M}{m} \biggr)^{p} \biggl(\frac{N}{n} \biggr)^{q} \biggr)\sum ^{\eta}_{k=1}w_{k}x_{k}y_{k}, $$
(1.3)

and, in particular, the reverse of Cauchy–Bunyakovsky–Schwarz inequality

$$ \Biggl(\sum^{\eta}_{k=1}w_{k}x^{2} \Biggr)^{\frac{1}{2}} \Biggl(\sum^{\eta}_{k=1}w_{k}y^{2} \Biggr)^{\frac{1}{2}} \leq S \biggl( \biggl(\frac{MN}{mn} \biggr)^{2} \biggr)\sum^{ \eta}_{k=1}w_{k}x_{k}y_{k}. $$
(1.4)

2 Preliminaries

First, we present a short introduction to the diamond-α derivative as given in [1, 13].

Let \(\mathbb{T}\) be a time scale and \(f(\tau )\) be differentiable on \(\mathbb{T}\) in the Δ and ∇ sense. For \(\tau \in \mathbb{T}\), the diamond-α dynamic derivative \(f^{\diamond _{\alpha}}(\tau )\) is defined by

$$ f^{\diamond _{\alpha}}(\tau )=\alpha f^{\Delta}(\tau )+(1-\alpha )f^{ \nabla}(\tau ),\quad 0 \leq \alpha \leq 1.$$

Thus f is diamond-α differentiable if and only if f is Δ and ∇ differentiable.

The diamond-α derivative reduces to the standard Δ-derivative for \(\alpha =1\), or the standard ∇-derivative for \(\alpha =0\). It represents a weighted dynamic derivative for \(\alpha \in (0,1)\).

The following definition is given in [13].

Let \(\xi ,\tau \in \mathbb{T}\) and \(h:\mathbb{T} \rightarrow \mathbb{R}\). Then the diamond-α integral from ξ to τ of h is defined by

$$ \int ^{\tau}_{\xi}h(\lambda )\diamond _{\alpha} \lambda =\alpha \int ^{ \tau}_{\xi}h(\lambda )\Delta \lambda +(1-\alpha ) \int ^{\tau}_{\xi}h( \lambda )\nabla \lambda ,\quad 0 \leq \alpha \leq 1,$$

provided that there exist delta and nabla integrals of h on \(\mathbb{T}\).

The following well-known Young inequality holds:

For \(\Phi , \Psi >0\) and \(v\in [0, 1]\), we have

$$ \Phi ^{1-v}\Psi ^{v}\leq (1-v)\Phi +v\Psi . $$
(2.1)

The following inequalities are given in [5].

For any \(\Phi , \Psi \in [m,M]\subset (0,\infty )\) and \(v\in [0,1]\), we have

$$ (1-v)\Phi +v\Psi \leq S \biggl(\frac{M}{m} \biggr)\Phi ^{1-v}\Psi ^{v}, $$
(2.2)

where Specht ratio [6, 14] is defined by

$$ S(h)=\frac{h^{\frac{1}{h-1}}}{e\log h^{\frac{1}{h-1}}}, $$

with \(h>0\), \(h\neq 1\).

Let \(v\in [0,1]\) and \(\Phi ,\Psi >0\). Then

$$ (1-v)\Phi +v\Psi \leq S(L)\Phi ^{1-v}\Psi ^{v}, $$
(2.3)

where \(0< L^{-1}\leq \frac{\Phi}{\Psi}\leq L<\infty \) and \(L>1\).

Let \(v\in [0,1]\) and \(\Phi ,\Psi >0\). Then

$$ (1-v)\Phi +v\Psi \leq \max \bigl\{ S(l),S(L) \bigr\} \Phi ^{1-v}\Psi ^{v}, $$
(2.4)

where \(0< l^{-1}\leq \frac{\Phi}{\Psi}\leq L<\infty \) and L, \(l>0\), with \(Ll>1\).

In this paper, it is assumed that all considered integrals exist and are finite.

3 Main results

In the following, we give an extension of reverse Callebaut inequality on time scales. Throughout this section, we assume that neither \(s\equiv 0\) nor \(t\equiv 0\).

Theorem 3.1

Let \(z,s,t\in C ([\xi , \omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq \frac{|s(\lambda )|}{|t(\lambda )|}\leq M<\infty \) on the set \([\xi , \omega ]_{\mathbb{T}}\). Let \(v\in [0, 1]\). Then the following inequalities hold true:

$$ \begin{aligned} & \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2(1-v)} \bigl\vert t( \lambda ) \bigr\vert ^{2v} \diamond _{\alpha}\lambda \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2v} \bigl\vert t( \lambda ) \bigr\vert ^{2(1-v)} \diamond _{\alpha} \lambda \\ &\quad\leq \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2} \diamond _{ \alpha} \lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2} \diamond _{\alpha} \lambda \\ &\quad\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2(1-v)} \bigl\vert t(\lambda ) \bigr\vert ^{2v} \diamond _{ \alpha}\lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2v} \bigl\vert t( \lambda ) \bigr\vert ^{2(1-v)} \diamond _{\alpha}\lambda . \end{aligned} $$
(3.1)

Proof

For \(\lambda ,\zeta \in [\xi , \omega ]_{\mathbb{T}}\), we observe that

$$ m^{2}\leq \frac{ \vert s(\lambda ) \vert ^{2}}{ \vert t(\lambda ) \vert ^{2}}, \frac{ \vert s(\zeta ) \vert ^{2}}{ \vert t(\zeta ) \vert ^{2}}\leq M^{2}. $$
(3.2)

Let \(\Phi (\lambda )=\frac{|s(\lambda )|^{2}}{|t(\lambda )|^{2}}\) and \(\Psi (\zeta )=\frac{|s(\zeta )|^{2}}{|t(\zeta )|^{2}}\), \(\lambda ,\zeta \in [\xi , \omega ]_{\mathbb{T}}\). Then using the inequalities (2.1) and (2.2), we have

$$ \begin{aligned} \biggl(\frac{ \vert s(\lambda ) \vert ^{2}}{ \vert t(\lambda ) \vert ^{2}} \biggr)^{1-v} \biggl(\frac{ \vert s(\zeta ) \vert ^{2}}{ \vert t(\zeta ) \vert ^{2}} \biggr)^{v} &\leq (1-v)\frac{ \vert s(\lambda ) \vert ^{2}}{ \vert t(\lambda ) \vert ^{2}}+v \frac{ \vert s(\zeta ) \vert ^{2}}{ \vert t(\zeta ) \vert ^{2}} \\ &\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \biggl( \frac{ \vert s(\lambda ) \vert ^{2}}{ \vert t(\lambda ) \vert ^{2}} \biggr)^{1-v} \biggl( \frac{ \vert s(\zeta ) \vert ^{2}}{ \vert t(\zeta ) \vert ^{2}} \biggr)^{v}. \end{aligned} $$
(3.3)

Multiplying by \(|t(\lambda )|^{2}|t(\zeta )|^{2}\), \(\lambda ,\zeta \in [\xi ,\omega ]_{\mathbb{T}}\), (3.3) takes the form

$$ \begin{aligned} &\bigl\vert s(\lambda ) \bigr\vert ^{2(1-v)} \bigl\vert t(\lambda ) \bigr\vert ^{2v} \bigl\vert s(\zeta ) \bigr\vert ^{2v} \bigl\vert t( \zeta ) \bigr\vert ^{2(1-v)} \\ &\quad \leq (1-v) \bigl\vert s(\lambda ) \bigr\vert ^{2} \bigl\vert t(\zeta ) \bigr\vert ^{2}+v \bigl\vert t(\lambda ) \bigr\vert ^{2} \bigl\vert s( \zeta ) \bigr\vert ^{2} \\ &\quad\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \bigl\vert s(\lambda ) \bigr\vert ^{2(1-v)} \bigl\vert t( \lambda ) \bigr\vert ^{2v} \bigl\vert s(\zeta ) \bigr\vert ^{2v} \bigl\vert t(\zeta ) \bigr\vert ^{2(1-v)}. \end{aligned} $$
(3.4)

Multiplying by \(|z(\lambda )|\) and integrating (3.4) with respect to λ from ξ to ω, we obtain

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2(1-v)} \bigl\vert t( \lambda ) \bigr\vert ^{2v} \diamond _{\alpha}\lambda \biggr) \bigl\vert s(\zeta ) \bigr\vert ^{2v} \bigl\vert t( \zeta ) \bigr\vert ^{2(1-v)} \\ &\quad\leq (1-v) \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2} \diamond _{\alpha} \lambda \biggr) \bigl\vert t(\zeta ) \bigr\vert ^{2} +v \biggl( \int ^{ \omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2}\diamond _{\alpha} \lambda \biggr) \bigl\vert s(\zeta ) \bigr\vert ^{2} \\ &\quad\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \biggl( \int ^{ \omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2(1-v)} \bigl\vert t(\lambda ) \bigr\vert ^{2v} \diamond _{\alpha}\lambda \biggr) \bigl\vert s(\zeta ) \bigr\vert ^{2v} \bigl\vert t(\zeta ) \bigr\vert ^{2(1-v)}. \end{aligned} $$
(3.5)

Again, multiplying by \(|z(\zeta )|\) and integrating (3.5) with respect to ζ from ξ to ω, we obtain the desired inequality (3.1). □

The following reverse of Callebaut inequality holds:

Corollary 3.1

Let \(z,s,t\in C ([\xi ,\omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq \frac{|s(\lambda )|}{|t(\lambda )|}\leq M<\infty \) on the set \([\xi ,\omega ]_{\mathbb{T}}\). Then the following inequalities hold true:

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert \bigl\vert t( \lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr)^{2} \\ &\quad\leq \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2} \diamond _{ \alpha} \lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2} \diamond _{\alpha} \lambda \\ &\quad\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \biggl( \int ^{ \omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert \diamond _{\alpha} \lambda \biggr)^{2}. \end{aligned} $$
(3.6)

Proof

Take \(v=\frac{1}{2}\) in Theorem 3.1, and the result follows. □

The following another reverse of Callebaut inequality holds:

Corollary 3.2

Let \(z,s,t\in C ([\xi ,\omega ]_{\mathbb{T}},\mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq \frac{|s(\lambda )|}{|t(\lambda )|}\leq M<\infty \) on the set \([\xi ,\omega ]_{\mathbb{T}}\). Let \(v\in [0, 1]\). Then the following inequalities hold true:

$$ \begin{aligned} & \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{1+v} \bigl\vert t( \lambda ) \bigr\vert ^{1-v} \diamond _{\alpha}\lambda \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{1-v} \bigl\vert t( \lambda ) \bigr\vert ^{1+v} \diamond _{\alpha} \lambda \\ &\quad\leq \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2} \diamond _{ \alpha} \lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2} \diamond _{\alpha} \lambda \\ &\quad\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{1+v} \bigl\vert t(\lambda ) \bigr\vert ^{1-v} \diamond _{ \alpha}\lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{1-v} \bigl\vert t( \lambda ) \bigr\vert ^{1+v} \diamond _{\alpha}\lambda . \end{aligned} $$
(3.7)

Proof

Replace v by \(\frac{1}{2}(1-v)\) in Theorem 3.1, and the result follows. □

The following another reverse of Callebaut inequality holds:

Corollary 3.3

Let \(z,s,t\in C ([\xi ,\omega ]_{\mathbb{T}},\mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq \frac{|s(\lambda )|}{|t(\lambda )|}\leq M<\infty \) on the set \([\xi ,\omega ]_{\mathbb{T}}\). Let \(\nu \in [0, 2]\). Then the following inequalities hold true:

$$ \begin{aligned} & \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2-\nu} \bigl\vert t( \lambda ) \bigr\vert ^{\nu} \diamond _{\alpha}\lambda \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{\nu} \bigl\vert t( \lambda ) \bigr\vert ^{2-\nu} \diamond _{\alpha} \lambda \\ &\quad\leq \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2} \diamond _{ \alpha} \lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2} \diamond _{\alpha} \lambda \\ &\quad\leq S \biggl( \biggl(\frac{M}{m} \biggr)^{2} \biggr) \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2-\nu} \bigl\vert t(\lambda ) \bigr\vert ^{\nu} \diamond _{ \alpha}\lambda \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{\nu} \bigl\vert t( \lambda ) \bigr\vert ^{2-\nu} \diamond _{\alpha}\lambda . \end{aligned} $$
(3.8)

Proof

Take \(v=\frac{1}{2}\nu \) in Theorem 3.1, and the result follows. □

In the following, we give an extension of reverse Rogers–Hölder inequality on time scales.

Theorem 3.2

Let \(z,s,t\in C ([\xi , \omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions satisfying \(\int ^{\omega}_{\xi}|z(\lambda )| \diamond _{\alpha}\lambda =1\). Assume further that \(0< m\leq |s(\lambda )|\leq M<\infty \) and \(0< n\leq |t(\lambda )|\leq N<\infty \) on the set \([\xi ,\omega ]_{\mathbb{T}}\). Let \(\frac{1}{p}+\frac{1}{q}=1\) with \(p>1\). Then the following inequality holds true:

$$\begin{aligned} &\biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{p}\diamond _{ \alpha} \lambda \biggr)^{\frac{1}{p}} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{q}\diamond _{\alpha}\lambda \biggr)^{ \frac{1}{q}} \\ &\quad \leq S \biggl( \biggl(\frac{M}{m} \biggr)^{p} \biggl( \frac{N}{n} \biggr)^{q} \biggr) \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda )t(\lambda ) \bigr\vert \diamond _{\alpha}\lambda . \end{aligned}$$
(3.9)

Proof

Using the given conditions, for \(\lambda \in [\xi ,\omega ]_{\mathbb{T}}\), we have

$$m^{p}\leq \bigl\vert s(\lambda ) \bigr\vert ^{p}\leq M^{p}\quad \text{and}\quad n^{q}\leq \bigl\vert t(\lambda ) \bigr\vert ^{q}\leq N^{q}, $$

which imply that

$$ \biggl(\frac{m}{M} \biggr)^{p}\leq \frac{ \vert s(\lambda ) \vert ^{p}}{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}\diamond _{\alpha}\lambda} \leq \biggl(\frac{M}{m} \biggr)^{p} $$
(3.10)

and

$$ \biggl(\frac{n}{N} \biggr)^{q}\leq \frac{ \vert t(\lambda ) \vert ^{q}}{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}\diamond _{\alpha}\lambda} \leq \biggl(\frac{N}{n} \biggr)^{q}. $$
(3.11)

Therefore,

$$ \begin{aligned} \biggl[ \biggl(\frac{M}{m} \biggr)^{p} \biggl(\frac{N}{n} \biggr)^{q} \biggr]^{-1} &\leq \biggl( \frac{ \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}}{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}\diamond _{\alpha}\lambda} \biggr) \biggl( \frac{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}\diamond _{\alpha}\lambda}{ \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}} \biggr) \\ &\leq \biggl(\frac{M}{m} \biggr)^{p} \biggl(\frac{N}{n} \biggr)^{q}. \end{aligned} $$
(3.12)

Using the inequality (2.3) with \(v=\frac{1}{q}\), \(L= (\frac{M}{m} )^{p} (\frac{N}{n} )^{q}\), \(\Phi (\lambda )= \frac{|z(\lambda )||s(\lambda )|^{p}}{\int ^{\omega}_{\xi}|z(\lambda )||s(\lambda )|^{p}\diamond _{\alpha}\lambda}\), and \(\Psi (\lambda )= \frac{|z(\lambda )||t(\lambda )|^{q}}{\int ^{\omega}_{\xi}|z(\lambda )||t(\lambda )|^{q}\diamond _{\alpha}\lambda}\), we get

$$\begin{aligned} &\frac{1}{p} \frac{ \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}}{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}\diamond _{\alpha}\lambda} +\frac{1}{q} \frac{ \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}}{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}\diamond _{\alpha}\lambda} \\ &\quad \leq S(L) \frac{ \vert z(\lambda ) \vert \vert s(\lambda )t(\lambda ) \vert }{ (\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}\diamond _{\alpha}\lambda )^{\frac{1}{p}} (\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}\diamond _{\alpha}\lambda )^{\frac{1}{q}}}. \end{aligned}$$
(3.13)

Integrating (3.13) with respect to λ from ξ to ω, we obtain

$$ 1\leq S(L) \frac{\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert s(\lambda )t(\lambda ) \vert \diamond _{\alpha}\lambda}{ (\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert s(\lambda ) \vert ^{p}\diamond _{\alpha}\lambda )^{\frac{1}{p}} (\int ^{\omega}_{\xi} \vert z(\lambda ) \vert \vert t(\lambda ) \vert ^{q}\diamond _{\alpha}\lambda )^{\frac{1}{q}}}. $$
(3.14)

This completes the proof of Theorem 3.2. □

Next, we give an extension of reverse Cauchy–Schwarz inequality on time scales.

Corollary 3.4

Let \(z,s,t\in C ([\xi ,\omega ]_{\mathbb{T}},\mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions satisfying \(\int ^{\omega}_{\xi}|z(\lambda )| \diamond _{\alpha}\lambda =1\). Assume further that \(0< m\leq |s(\lambda )|\leq M<\infty \) and \(0< n\leq |t(\lambda )|\leq N<\infty \) on the set \([\xi ,\omega ]_{\mathbb{T}}\). Then the following inequality holds true:

$$\begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2}\diamond _{ \alpha} \lambda \biggr)^{\frac{1}{2}} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2}\diamond _{\alpha}\lambda \biggr)^{ \frac{1}{2}} \\ &\quad \leq S \biggl( \biggl(\frac{MN}{mn} \biggr)^{2} \biggr) \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda )t(\lambda ) \bigr\vert \diamond _{\alpha}\lambda . \end{aligned}$$
(3.15)

Proof

Take \(p=q=2\) in Theorem 3.2, and the result follows. □

Remark 3.1

We have the following:

  1. (i)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(s(k)=x_{k}>0\), \(t(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\) with \(\sum^{\eta}_{k=1}w_{k}=1\). Then inequality (3.1) reduces to inequality (1.1).

  2. (ii)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(s(k)=x_{k}>0\), \(t(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\) with \(\sum^{\eta}_{k=1}w_{k}=1\). Then inequality (3.6) reduces to inequality (1.2).

  3. (iii)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(s(k)=x_{k}>0\), \(t(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\). Then inequality (3.9) reduces to inequality (1.3).

  4. (iv)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(s(k)=x_{k}>0\), \(t(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\). Then inequality (3.15) reduces to inequality (1.4).

Finally, we give another extension of reverse Rogers–Hölder dynamic inequality.

Theorem 3.3

Let \(z,u_{1},u_{2},s,t\in C ([\xi , \omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq |s(\lambda )|\leq M<\infty \) and \(0< n\leq |t(\lambda )|\leq N<\infty \) on the set \([\xi , \omega ]_{\mathbb{T}}\). Let \(\frac{1}{p}+\frac{1}{q}=1\) with \(p>1\). Then the following inequalities hold true:

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{1}( \lambda )s(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \biggl( \int ^{ \omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{2}(\lambda )t(\lambda ) \bigr\vert \diamond _{ \alpha}\lambda \biggr) \\ &\quad\leq \frac{1}{p} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{1}( \lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{p}\diamond _{\alpha}\lambda \biggr) \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{2}(\lambda ) \bigr\vert \diamond _{\alpha} \lambda \biggr) \\ &\quad\quad {}+\frac{1}{q} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{1}( \lambda ) \bigr\vert \diamond _{\alpha} \lambda \biggr) \biggl( \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{2}(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{q}\diamond _{\alpha} \lambda \biggr) \\ &\quad\leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert u_{1}(\lambda )s(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \\ &\quad\quad{}\times \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{2}(\lambda )t(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr). \end{aligned} $$
(3.16)

Proof

For \(\lambda , \zeta \in [\xi , \omega ]_{\mathbb{T}}\), it is clear that

$$ \frac{m^{p}}{N^{q}}\leq \frac{ \vert s(\lambda ) \vert ^{p}}{ \vert t(\zeta ) \vert ^{q}} \leq \frac{M^{p}}{n^{q}}. $$
(3.17)

Let \(l=\frac{N^{q}}{m^{p}}\), \(L=\frac{M^{p}}{n^{q}}\), \(\Phi (\lambda )=|s(\lambda )|^{p}\), \(\Psi (\zeta )=|t(\zeta )|^{q}\), and \(v=\frac{1}{q}\). Then using the inequalities (2.1) and (2.4), respectively, we have

$$ \bigl\vert s(\lambda ) \bigr\vert \bigl\vert t(\zeta ) \bigr\vert \leq \frac{1}{p} \bigl\vert s(\lambda ) \bigr\vert ^{p}+\frac{1}{q} \bigl\vert t( \zeta ) \bigr\vert ^{q} \leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \bigl\vert s(\lambda ) \bigr\vert \bigl\vert t(\zeta ) \bigr\vert . $$
(3.18)

Multiplying by \(|z(\lambda )||u_{1}(\lambda )|\) and integrating (3.18) with respect to λ from ξ to ω, we obtain

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{1}(\lambda )s( \lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \bigl\vert t(\zeta ) \bigr\vert \\ &\quad\leq \frac{1}{p} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{1}( \lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{p}\diamond _{\alpha}\lambda \biggr)+ \frac{1}{q} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert u_{1}(\lambda ) \bigr\vert \diamond _{\alpha} \lambda \biggr) \bigl\vert t(\zeta ) \bigr\vert ^{q} \\ &\quad\leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert u_{1}(\lambda )s(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \bigl\vert t(\zeta ) \bigr\vert . \end{aligned} $$
(3.19)

Multiplying by \(|z(\zeta )||u_{2}(\zeta )|\) and integrating (3.19) with respect to ζ from ξ to ω, we obtain the desired inequality (3.16). □

Next, we give an extension of reverse Rogers–Hölder inequality on time scales.

Corollary 3.5

Let \(z,s,t\in C ([\xi , \omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq |s(\lambda )|\leq M<\infty \) and \(0< n\leq |t(\lambda )|\leq N<\infty \) on the set \([\xi , \omega ]_{\mathbb{T}}\). Let \(\frac{1}{p}+\frac{1}{q}=1\) with \(p>1\). Then the following inequalities hold true:

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda )t( \lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr)^{2} \\ &\quad\leq \frac{1}{p} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert \bigl\vert s( \lambda ) \bigr\vert ^{p}\diamond _{\alpha}\lambda \biggr) \biggl( \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \\ &\quad\quad {}+\frac{1}{q} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t( \lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \biggl( \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{q}\diamond _{\alpha} \lambda \biggr) \\ &\quad\leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert s(\lambda )t(\lambda ) \bigr\vert \diamond _{\alpha} \lambda \biggr)^{2}. \end{aligned} $$
(3.20)

Proof

Put \(|u_{1}(\lambda )|=|t(\lambda )|\) and \(|u_{2}(\lambda )|=|s(\lambda )|\) on \([\xi , \omega ]_{\mathbb{T}}\) in Theorem 3.3, and then the result follows. □

Now, we give another extension of reverse Rogers–Hölder inequality on time scales.

Corollary 3.6

Let \(z,s,t\in C ([\xi , \omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions. Assume further that \(0< m\leq |s(\lambda )|\leq M<\infty \) and \(0< n\leq |t(\lambda )|\leq N<\infty \) on the set \([\xi , \omega ]_{\mathbb{T}}\). Let \(\frac{1}{p}+\frac{1}{q}=1\) with \(p>1\). Then the following inequalities hold true:

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2} \diamond _{\alpha} \lambda \biggr) \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2}\diamond _{\alpha}\lambda \biggr) \\ &\quad\leq \frac{1}{p} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{p+1} \diamond _{\alpha} \lambda \biggr) \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \\ &\quad\quad {}+\frac{1}{q} \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert s( \lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr) \biggl( \int ^{\omega}_{ \xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{q+1}\diamond _{\alpha} \lambda \biggr) \\ &\quad\leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert s(\lambda ) \bigr\vert ^{2}\diamond _{\alpha}\lambda \biggr) \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert t(\lambda ) \bigr\vert ^{2}\diamond _{\alpha} \lambda \biggr). \end{aligned} $$
(3.21)

Proof

Put \(|u_{1}(\lambda )|=|s(\lambda )|\) and \(|u_{2}(\lambda )|=|t(\lambda )|\) on \([\xi , \omega ]_{\mathbb{T}}\) in Theorem 3.3, and then the result follows. □

Next, we give another extension of reverse Rogers–Hölder inequality on time scales.

Corollary 3.7

Let \(z,f_{1},f_{2}\in C ([\xi , \omega ]_{\mathbb{T}}, \mathbb{R} )\) be \(\diamond _{\alpha}\)-integrable functions, with neither \(f_{1}\equiv 0\) nor \(f_{2}\equiv 0\). Assume further that \(0< m\leq \frac{|f_{1}(\lambda )|}{|f_{2}(\lambda )|}\leq M<\infty \) on the set \([\xi , \omega ]_{\mathbb{T}}\). Let \(\frac{1}{p}+\frac{1}{q}=1\) with \(p>1\). Then the following inequalities hold true:

$$ \begin{aligned} & \biggl( \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert f_{1}(\lambda )f_{2}( \lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr)^{2} \\ &\quad\leq \biggl[\frac{1}{p} \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert f_{1}( \lambda ) \bigr\vert ^{p} \bigl\vert f_{2}(\lambda ) \bigr\vert ^{2-p}\diamond _{\alpha} \lambda \\ &\quad\quad {}+\frac{1}{q} \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert f_{1}( \lambda ) \bigr\vert ^{q} \bigl\vert f_{2}(\lambda ) \bigr\vert ^{2-q}\diamond _{\alpha} \lambda \biggr] \int ^{\omega}_{\xi} \bigl\vert z(\lambda ) \bigr\vert \bigl\vert f_{2}(\lambda ) \bigr\vert ^{2} \diamond _{\alpha}\lambda \\ &\quad\leq \max \biggl\{ S \biggl(\frac{M^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{m^{q}} \biggr) \biggr\} \biggl( \int ^{\omega}_{\xi} \bigl\vert z( \lambda ) \bigr\vert \bigl\vert f_{1}(\lambda )f_{2}(\lambda ) \bigr\vert \diamond _{\alpha}\lambda \biggr)^{2}. \end{aligned} $$
(3.22)

Proof

Put \(|s(\lambda )|=|t(\lambda )|= \frac{|f_{1}(\lambda )|}{|f_{2}(\lambda )|}\), \(|u_{1}(\lambda )|=|u_{2}(\lambda )|=|f_{2}(\lambda )|^{2}\) on \([\xi , \omega ]_{\mathbb{T}}\), \(M=N\), and \(m=n\) in Theorem 3.3, and then the result follows. □

Remark 3.2

We have the following:

  1. (i)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(s(k)=x_{k}>0\), \(t(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\) with \(\sum^{\eta}_{k=1}w_{k}=1\). Then inequality (3.20) reduces to inequality [5]

    $$ \begin{aligned} \Biggl(\sum ^{\eta}_{k=1}w_{k}x_{k}y_{k} \Biggr)^{2} &\leq \frac{1}{p}\sum^{\eta}_{k=1}w_{k}y_{k}x^{p}_{k} \sum^{\eta}_{k=1}w_{k}x_{k}+ \frac{1}{q}\sum^{\eta}_{k=1}w_{k}y_{k} \sum^{\eta}_{k=1}w_{k}x_{k}y^{q}_{k} \\ &\leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \Biggl(\sum^{\eta}_{k=1}w_{k}x_{k}y_{k} \Biggr)^{2}. \end{aligned} $$
    (3.23)
  2. (ii)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(s(k)=x_{k}>0\), \(t(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\) with \(\sum^{\eta}_{k=1}w_{k}=1\). Then inequality (3.21) reduces to inequality [5]

    $$ \begin{aligned} \sum^{\eta}_{k=1}w_{k}x^{2}_{k} \sum^{ \eta}_{k=1}w_{k}y^{2}_{k} &\leq \frac{1}{p}\sum^{\eta}_{k=1}w_{k}x^{p+1}_{k} \sum^{\eta}_{k=1}w_{k}y_{k}+ \frac{1}{q}\sum^{\eta}_{k=1}w_{k}x_{k} \sum^{\eta}_{k=1}w_{k}y^{q+1}_{k} \\ &\leq \max \biggl\{ S \biggl(\frac{N^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{n^{q}} \biggr) \biggr\} \sum^{\eta}_{k=1}w_{k}x^{2}_{k} \sum^{\eta}_{k=1}w_{k}y^{2}_{k}. \end{aligned} $$
    (3.24)
  3. (iii)

    Let \(\alpha =1\), \(\mathbb{T}=\mathbb{Z}\), \(\xi =1\), \(\omega =\eta +1\), \(f_{1}(k)=x_{k}>0\), \(f_{2}(k)=y_{k}>0\), and \(z(k)=w_{k}\geq 0\) for any \(k\in \{1,2,\ldots ,\eta \}\) with \(\sum^{\eta}_{k=1}w_{k}=1\). Then inequality (3.22) reduces to inequality [5]

    $$ \begin{aligned} \Biggl(\sum ^{\eta}_{k=1}w_{k}x_{k}y_{k} \Biggr)^{2} &\leq \Biggl(\frac{1}{p}\sum ^{\eta}_{k=1}w_{k}x^{p}_{k}y^{2-p}_{k}+ \frac{1}{q}\sum^{\eta}_{k=1}w_{k}x^{q}_{k}y^{2-q}_{k} \Biggr) \sum^{\eta}_{k=1}w_{k}y^{2}_{k} \\ &\leq \max \biggl\{ S \biggl(\frac{M^{q}}{m^{p}} \biggr),S \biggl( \frac{M^{p}}{m^{q}} \biggr) \biggr\} \Biggl(\sum^{\eta}_{k=1}w_{k}x_{k}y_{k} \Biggr)^{2}. \end{aligned} $$
    (3.25)