1 Introduction
To unify and generalize discrete and continuous analysis, in 1998, Hilger [1] introduced the theory of time scales. Since then, many researchers have studied various aspects of the theory and obtained a lot of interesting results on time scales [1‐10]. The first purpose of this paper is to give the definition of the triple Diamond-Alpha integral (triple diamond-α integral or triple \(\diamond_{\alpha}\)-integral) for functions of three variables on time scales.
Let \(u(x)\) and \(v(x)\) be continuous real-valued functions on \([a, b]\), and let \(\frac{1}{p}+ \frac{1}{q}=1\).
(I)
if \(p>1\) and if \(u(x)\geq0\), \(v(x)\geq0\), then the classical Hölder inequality holds (see [11]):
$$ \int_{a}^{b}u(x)v(x)\,dx\leq \biggl( \int_{a}^{b}u^{p}(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}v^{q}(x)\,dx \biggr)^{\frac{1}{q}}. $$
(1)
(II)
if \(0< p<1\) and if \(u(x)>0\), \(v(x)>0\), then the following reverse Hölder inequality (e.g., see [12]) holds:
$$ \int_{a}^{b}u(x)v(x)\,dx \geq \biggl( \int_{a}^{b}u^{p}(x)\,dx \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b}v^{q}(x)\,dx \biggr)^{\frac{1}{q}}. $$
(2)
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The classical Hölder and reverse Hölder inequalities play a very important role and have wide applications in different branches of modern mathematics. A large number of papers dealing with refinements, generalizations, and applications of the Hölder and reverse Hölder inequalities and their series analogues in different ares of mathematics have appeared. For example, Agahi et al. [13] gave generalizations of the Hölder and reverse Hölder inequalities for the pseudo-integral. Zhao et al. [14] found that the Hölder inequality for the pan-integral holds if the monotone measurer is subadditive. Tian [15‐18] gave some new properties and refinements of the Hölder and reverse Hölder inequalities. For more detail, the reader may consult [19‐25].
Among various extensions of (1) and (2), Agarwal, Bohner, and Peterson first presented the time scale versions of (1) via the Delta-integral (Δ-integral).
Theorem A
Assume that
\(\mathbb{T}\)
is a time scale, \(a,b\in \mathbb{T}\), and
\(a< b\). If
\(p>1\)
with
\(\frac{1}{p}+\frac{1}{q}=1\)
and if
\(u, v\in C_{rd}([a, b], \mathbb{R})\), then
$$ \int_{a}^{b} \bigl\vert u(x)v(x) \bigr\vert \Delta x \leq \biggl( \int_{a}^{b} \bigl\vert u(x) \bigr\vert ^{p}\Delta x \biggr)^{\frac {1}{p}} \biggl( \int_{a}^{b} \bigl\vert v(x) \bigr\vert ^{q}\Delta x \biggr)^{\frac{1}{q}}. $$
(3)
Later, in 2005, Wong et al. [26] gave the following Hölder-type inequalities via the Delta-integral.
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Theorem B
Assume that
\(\mathbb{T}\)
is a time scale, \(a,b\in \mathbb{T}\), and
\(a< b\). Let
\(\frac{1}{p}+\frac{1}{q}=1\)
and
\(u, v, \omega\in C_{rd}([a, b], \mathbb{R})\). If
\(p>1\), then
If
\(p<0\)
or
\(q<0\), then inequality (4) is reversed.
$$ \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert u(x)v(x) \bigr\vert \Delta x \leq \biggl( \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert u(x) \bigr\vert ^{p}\Delta x \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert v(x) \bigr\vert ^{q}\Delta x \biggr)^{\frac{1}{q}} . $$
(4)
In 2008, Özkan et al. [27] presented the following time scale versions of inequalities (1) and (2) via the Nabla-integral (∇-integral) and Diamond-Alpha integral (\(\diamond_{\alpha}\)-integral).
Theorem C
Assume that
\(\mathbb{T}\)
is a time scale, \(a,b\in\mathbb{T}\), and
\(a< b\). Let
\(\frac{1}{p}+\frac{1}{q}=1\)
and
\(u, v, \omega\in C_{ld}([a, b], \mathbb{R})\). If
\(p>1\), then
If
\(p<0\)
or
\(q<0\), then inequality (5) is reversed.
$$ \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert u(x)v(x) \bigr\vert \nabla x\leq \biggl( \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert u(x) \bigr\vert ^{p}\nabla x \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert v(x) \bigr\vert ^{q}\nabla x \biggr)^{\frac{1}{q}}; $$
(5)
Theorem D
Assume that
\(\mathbb{T}\)
is a time scale, \(a,b\in\mathbb{T}\), and
\(a< b\). Let
\(\frac{1}{p}+\frac{1}{q}=1\), and let
\(u, v, \omega: [a, b]\rightarrow\mathbb{R}\)
be
\(\diamond_{\alpha}\)-integrable functions. If
\(p>1\), then
If
\(p<0\)
or
\(q<0\), then inequality (6) is reversed.
$$ \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert u(x)v(x) \bigr\vert \diamond_{\alpha} x\leq \biggl( \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert u(x) \bigr\vert ^{p}\diamond_{\alpha} x \biggr)^{\frac{1}{p}} \biggl( \int_{a}^{b} \bigl\vert \omega(x) \bigr\vert \bigl\vert v(x) \bigr\vert ^{q} \diamond _{\alpha}x \biggr)^{\frac{1}{q}} . $$
(6)
Remark 1.1
The second purpose of this paper is to give the time scale versions of the Hölder and reverse Hölder inequalities for the triple Diamond-Alpha integral. Then we obtain some new generalizations of the Hölder and reverse Hölder inequalities for the triple Diamond-Alpha integral on time scales. Moreover, using the obtained results, we present a new generalization of the Minkowski inequality for the triple Diamond-Alpha integral on time scales.
2 Main results
For details on time scales theory, the readers may consult [1, 3‐9] and the references therein. Now we give the definition of triple Diamond-Alpha integral for functions of three variables.
The triple Diamond-Alpha integral is defined as an iterated integral. Suppose that \(\mathbb{T}\) is a time scale and \(a_{i}, b_{i}\in\mathbb {T}\) with \(a_{i}< b_{i}\)
\((i=1,2,3)\). Let \(f(x_{1},x_{2},x_{3})\) be a real-valued function on \(\mathbb{T}\times\mathbb{T}\times\mathbb{T}\). Because we need the notation of partial derivatives with respect to variables \(x_{i}\), we denote the time scale partial derivatives of \(f(x_{1},x_{2},x_{3})\) with respect to \(x_{i}\) by \(f^{\diamond_{\alpha}^{i}} (x_{1},x_{2},x_{3})\), \(i=1,2,3\). We now give the definition of these partial derivatives. Fixing \(x_{2},x_{3}\in\mathbb{T}\), the diamond-α derivative of a function
is denoted by \(f^{{\diamond}_{\alpha}^{1}}\). Next, fixing \(x_{1},x_{3}\in\mathbb{T}\), the diamond-α derivative of a function
is denoted by \(f^{{\diamond}_{\alpha}^{2}}\). Finally, fixing \(x_{1}, x_{2}\in\mathbb{T}\), the diamond-α derivative of a function
is denoted by \(f^{{\diamond}_{\alpha}^{3}}\).
$$\begin{aligned} &\mathbb{T}\rightarrow\mathbb{R}, \\ &x_{1}\rightarrow f(x_{1},x_{2},x_{3}) \end{aligned}$$
$$\begin{aligned} &\mathbb{T}\rightarrow\mathbb{R}, \\ &x_{2}\rightarrow f(x_{1},x_{2},x_{3}) \end{aligned}$$
$$\begin{aligned} &\mathbb{T}\rightarrow\mathbb{R}, \\ &x_{3}\rightarrow f(x_{1},x_{2},x_{3}) \end{aligned}$$
If a function f has a \(\diamond_{\alpha}^{1}\) antiderivative \(F_{1}\), \(F_{1}\) has a \(\diamond_{\alpha}^{2}\) antiderivative \(F_{2}\), and \(F_{2}\) has a \(\diamond_{\alpha}^{3}\) antiderivative \(F_{3}\), that is, \(F^{{\diamond}_{\alpha}^{1}}_{1}=f\), \(F^{{\diamond}_{\alpha }^{2}}_{2}=F_{1}\), and \(F^{{\diamond}_{\alpha}^{3}}_{3}=F_{2}\), then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}f(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad:= \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl(F_{1}(b_{1},x_{2},x_{3})-F_{1}(a_{1},x_{2},x_{3}) \bigr)\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad= \int_{a_{3}}^{b_{3}} \bigl[F_{2}(b_{1},b_{2},x_{3})-F_{2}(b_{1},a_{2},x_{3}) - \bigl(F_{2}(a_{1},b_{2},x_{3})-F_{2}(a_{1},a_{2},x_{3}) \bigr) \bigr]\diamond_{\alpha} x_{3} \\ &\quad=F_{3}(b_{1},b_{2},b_{3})-F_{3}(b_{1},b_{2},a_{3}) -F_{3}(b_{1},a_{2},b_{3})+F_{3}(b_{1},a_{2},a_{3}) \\ &\qquad{} -F_{3}(a_{1},b_{2},b_{3})+F_{3}(a_{1},b_{2},a_{3}) +F_{3}(a_{1},a_{2},b_{3})-F_{3}(a_{1},a_{2},a_{3}) . \end{aligned}$$
(7)
By this definition it is easy to obtain the following property for the triple Diamond-Alpha integral.
Proposition 2.1
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i}, b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and
\(f(x_{1},x_{2},x_{3})\)
and
\(g(x_{1},x_{2},x_{3})\)
are
\(\diamond_{\alpha}\)-integrable functions on
\([a_{i}, b_{i}]^{3}_{\mathbb{T}}\)
\((i=1,2,3)\).
(i)
If
\(f(x_{1},x_{2},x_{3})\geq0\)
for all
\(x_{i}\in[a_{i}, b_{i}]_{\mathbb{T}}\)
\((i=1,2,3)\), then
$$\int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}f(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2} \diamond_{\alpha} x_{3}\geq0; $$
(ii)
If
\(f(x_{1},x_{2},x_{3})\leq g(x_{1},x_{2},x_{3})\)
for all
\(x_{i}\in[a_{i}, b_{i}]_{\mathbb{T}}\)
\((i=1,2,3)\), then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}f(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad\leq \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}g(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2} \diamond_{\alpha} x_{3}; \end{aligned}$$
(iii)
If
\(f(x_{1},x_{2},x_{3})\geq0\)
for all
\(x_{i}\in[a_{i}, b_{i}]_{\mathbb{T}}\)
\((i=1,2,3)\), then
\(f(x_{1},x_{2},x_{3})=0\)
if and only if
$$\int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}f(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2} \diamond_{\alpha} x_{3}=0. $$
To prove the main results, we need the following lemmas.
Lemma 2.2
(Bernoulli’s inequality; see [28])
If
\(x>0\)
and
\(p>1\), then
$$ x^{p}\geq px+1-p. $$
(8)
Lemma 2.3
(Young inequality; see [11])
Let
\(a,b>0\).
(i)
If
\(p,q>0\)
with
\(\frac{1}{p}+\frac{1}{q}=1\), then
$$ ab \leq\frac{1}{p}a^{p}+\frac{1}{q}b^{q}. $$
(9)
(ii)
If
\(p>0\), \(q<0\)
with
\(\frac{1}{p}+\frac{1}{q}=1\), then
$$ ab \geq\frac{1}{p}a^{p}+\frac{1}{q}b^{q}. $$
(10)
Lemma 2.4
(AG inequality; see [11])
Let
\(\alpha_{i}>0\)
\((i=1,2,3)\).
(i)
If
\(1<\lambda_{1}\), \(\lambda_{2}, \ldots, \lambda_{n}<\infty\)
with
\(\sum_{i=1}^{n}\frac{1}{\lambda_{i}}=1\), then
$$ \prod_{i=1}^{n} \alpha_{i} \leq\sum_{i=1}^{n} \frac{\alpha_{i}^{\lambda_{i}}}{\lambda_{i}}. $$
(11)
(ii)
If
\(\lambda_{1}>0\), \(\lambda_{2}, \dots,\lambda_{m}<0\)
with
\(\sum_{i=1}^{m}\frac{1}{\lambda_{i}}=1\), then
$$ \prod_{i=1}^{m} \alpha_{i} \geq\sum_{i=1}^{m} \frac{\alpha_{i}^{\lambda_{i}}}{\lambda_{i}}. $$
(12)
Lemma 2.5
(Schlömilch’s inequality for triple Diamond-Alpha integral)
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i}, b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and
\(f(x_{1},x_{2},x_{3}), \omega (x_{1},x_{2},x_{3}):[a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow[0,+\infty )\)
are
\(\diamond_{\alpha}\)-integrable functions with
\(\int _{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}>0\). Then, for
\(s>r>0\), we have
$$\begin{aligned} & \biggl(\frac{\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3} )f^{s}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}}{ \int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}} \biggr)^{\frac{1}{s}} \\ &\quad\geq \biggl(\frac{\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}}{ \int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}} \biggr)^{\frac{1}{r}}. \end{aligned}$$
Proof
Without loss of generality, we may suppose that \(\int _{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int _{a_{3}}^{b_{3}}h(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3}=1\). If \(s>r>0\), then \(\frac{s}{r}>1\). Therefore, by inequality (8) we have
which implies
Replacing f by \(f^{r}\) in (13), we find
Then
Thus, the proof of Lemma 2.5 is completed. □
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \biggl[\biggl( \frac{f(x_{1},x_{2},x_{3})}{ \int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3})f(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}} \biggr)^{\frac{s}{r}} \\ &\qquad{} \times \omega(x_{1},x_{2},x_{3})\biggr] \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad\geq \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \biggl[\biggl( \frac{\frac{s}{r}f(x_{1},x_{2},x_{3})}{\int_{a}^{b}\int_{a}^{b} \cdots\int_{a}^{b}\omega(x_{1},x_{2},x_{3})f(x_{1},x_{2},x_{3})\diamond _{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}}+1- \frac {s}{r} \biggr) \\ &\qquad{} \times \omega(x_{1},x_{2},x_{3})\biggr] \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad= 1, \end{aligned}$$
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3} )f^{\frac{s}{r}}(x_{1},x_{2},x_{3})\diamond _{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad \geq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3})f(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\frac{s}{r}}. \end{aligned}$$
(13)
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3} )f^{s}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad \geq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3} )f^{r}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\frac{s}{r}}. \end{aligned}$$
$$\begin{aligned} & \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3} )f^{s}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\frac{1}{s}} \\ &\quad \geq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3} )f^{r}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\frac{1}{r}}. \end{aligned}$$
Now, we give the following Hölder and reverse Hölder inequalities for triple Diamond-Alpha integral.
Theorem 2.6
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and
\(f(x_{1},x_{2},x_{3})\), \(g(x_{1},x_{2},x_{3})\), \(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow \mathbb{R}\)
are
\(\diamond_{\alpha}\)-integrable functions.
(i)
If
\(p, q>0\)
are such that
\(\frac{1}{p}+\frac{1}{q}=1\), then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f(x_{1},x_{2},x_{3})g(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ & \qquad{}\times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}} . \end{aligned}$$
(14)
(ii)
If
\(q<0\)
is such that
\(\frac{1}{p}+\frac{1}{q}=1\), then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f(x_{1},x_{2},x_{3})g(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\geq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ & \qquad{}\times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}} . \end{aligned}$$
(15)
Proof
Case (i): Let \(p,q>0\) with \(\frac{1}{p}+\frac{1}{q}=1\). Without loss of generality, we may suppose that
and
Let
and
From the Young inequality (9) we get
Therefore, we get the desired inequality (14).
$$\int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \neq0 $$
$$\int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \neq0. $$
$$\mu =\frac{ \vert \omega(x_{1},x_{2},x_{3}) \vert ^{\frac {1}{p}} \vert f(x_{1},x_{2},x_{3}) \vert }{ (\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \vert \omega(x_{1},x_{2},x_{3} ) \vert \vert f(x_{1},x_{2},x_{3}) \vert ^{p}\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} )^{\frac{1}{p}}} $$
$$\nu =\frac{ \vert \omega(x_{1},x_{2},x_{3}) \vert ^{\frac {1}{q}} \vert g(x_{1},x_{2},x_{3}) \vert }{ (\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \vert \omega(x_{1},x_{2},x_{3} ) \vert \vert g(x_{1},x_{2},x_{3}) \vert ^{q}\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} )^{\frac{1}{q}}}. $$
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\mu\nu \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \biggl(\frac{\mu^{p}}{p}+ \frac{\nu^{q}}{q} \biggr) \diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad=\frac{1}{p} \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \biggl(\frac{ \vert \omega(x_{1},x_{2},x_{3}) \vert }{ \int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \vert \omega(x_{1},x_{2},x_{3} ) \vert \vert f(x_{1},x_{2},x_{3}) \vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3}} \\ &\qquad{}\times \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\biggr)\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\qquad{} +\frac{1}{q} \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \biggl(\frac{ \vert \omega(x_{1},x_{2},x_{3}) \vert }{ \int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \vert \omega(x_{1},x_{2},x_{3} ) \vert \vert f(x_{1},x_{2},x_{3}) \vert ^{q}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3}} \\ &\qquad{}\times \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{q}\biggr)\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad=\frac{1}{p}+\frac{1}{q}=1. \end{aligned}$$
Case (ii). Let \(q<0\) with \(\frac{1}{p}+\frac{1}{q}=1\). Write \(\alpha =-\frac{p}{q}, \beta=\frac{1}{q}\). Then \(\frac{1}{\alpha}+\frac{1}{\beta }=1\) with \(\alpha, \beta>0\). Setting \(f(x_{1},x_{2},x_{3})=F(x_{1},x_{2},x_{3})\) and \(g(x_{1},x_{2},x_{3})=G(x_{1},x_{2},x_{3})\) in (14), we get
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) F(x_{1},x_{2},x_{3})G(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3} ) \bigr\vert \bigl\vert F(x_{1},x_{2},x_{3}) \bigr\vert ^{\alpha}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{\alpha}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert G(x_{1},x_{2},x_{3}) \bigr\vert ^{\beta} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{\beta}} . \end{aligned}$$
(16)
Theorem 2.7
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and
\(f(x_{1},x_{2},x_{3})\), \(g(x_{1},x_{2},x_{3})\), \(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow \mathbb{R}\)
are
\(\diamond_{\alpha}\)-integrable functions.
(i)
Let
\(\frac{1}{p}+\frac{1}{q}=\frac{1}{r}\)
with
\(p,q,r\in\mathbb {R}-\{0\}\), \(p>0\)
and
\(q>0\)
or
\(p>0\), \(q<0\), and
\(r<0\). Then
$$\begin{aligned} & \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{r}} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}} . \end{aligned}$$
(17)
(ii)
Let
\(\frac{1}{p}+\frac{1}{q}=\frac{1}{r}\)
with
\(p,q,r\in\mathbb {R}-\{0\}\), \(p>0\), \(q<0\), and
\(r>0\)
or
\(p<0\)
and
\(q<0\). Then
$$\begin{aligned} & \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{r}} \\ &\quad\geq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}} . \end{aligned}$$
(18)
Proof
(i) Case 1. When \(p>0\), \(q>0\), by \(\frac{1}{p}+\frac{1}{q}=\frac {1}{r}\) we get
Then, by (14) we find that
Therefore
Case 2. When \(p>0\), \(q<0\), and \(r<0\), by the same method as in Case 1, we can obtain inequality (17).
$$\frac{p}{r}>1, \qquad\frac{1}{p/r}+\frac{1}{q/r}=1. $$
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3}) \bigr\vert ^{\frac{p}{r}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{r}{p}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert ^{\frac{q}{r}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{r}{q}}. \end{aligned}$$
$$\begin{aligned} &\biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{r}} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3}) \bigr\vert ^{\frac{p}{r}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ & \qquad{}\times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert ^{\frac{q}{r}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}}. \end{aligned}$$
(ii) Case I. When \(p>0\), \(q<0\), and \(r>0\), we find from \(\frac {1}{p}+\frac{1}{q}=\frac{1}{r}\) that
Then, by inequality (14) we get
which implies
Thus from inequality (19) we obtain
Case II. When \(p<0\) and \(q<0\), by the same method as in Case I, we can obtain the desired inequality (18). The proof of Theorem 2.7 is completed. □
$$\frac{r}{p}>1, \qquad \frac{1}{r/p}+\frac{1}{-q/p} = \frac{p}{r}- \frac{p}{q}=p \biggl(\frac{1}{r}- \frac{1}{q} \biggr)=1. $$
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})f^{p}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad= \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \biggl( \bigl\vert \omega(x_{1},x_{2},x_{3})f^{p}(x_{1},x_{2},x_{3})\bigr\vert \\ &\qquad{}\times\bigl\vert g^{p}(x_{1},x_{2},x_{3})g^{-p}(x_{1},x_{2},x_{3}) \bigr\vert \biggr) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f^{p}(x_{1},x_{2},x_{3})g^{p}(x_{1},x_{2},x_{3}) \bigr\vert ^{\frac{r}{p}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{p}{r}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g^{-p}(x_{1},x_{2},x_{3}) \bigr\vert ^{-\frac{q}{p}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{-\frac{p}{q}} \\ &\quad= \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{p}{r}} \\ & \qquad{}\times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g^{q}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{-\frac{p}{q}}, \end{aligned}$$
$$\begin{aligned} & \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})f^{p}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{r}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g^{q}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{-\frac{1}{q}}. \end{aligned}$$
(19)
$$\begin{aligned} & \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f^{r}(x_{1},x_{2},x_{3})g^{r}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{r}} \\ &\quad\geq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}} . \end{aligned}$$
We present another generalization of inequality (14).
Theorem 2.8
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and that
\(f(x_{1},x_{2},x_{3})\), \(g(x_{1},x_{2},x_{3})\), \(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow \mathbb{R}\)
are
\(\diamond_{\alpha}\)-integrable functions. If
\(p>0\), \(q>0\)
are such that
\(0<\frac{1}{p}+\frac{1}{q}<1\), then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) f(x_{1},x_{2},x_{3})g(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\qquad{}\times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha}x_{3} \biggr)^{\frac{1}{q}} . \end{aligned}$$
(20)
Proof
Denote \(\gamma:=\frac{1}{p}+\frac{1}{q}\), \(\zeta=\gamma p\), \(\delta =q\gamma\). Then \(\zeta>1\), \(\delta>1\), and \(\frac{1}{\zeta}+\frac {1}{\delta}=1\). Hence from Lemma 2.5 with \(\zeta< p\), \(\delta< q\) and inequality (14) we find that
The proof of Theorem 2.8 is completed. □
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega (x_{1},x_{2},x_{3})f(x_{1},x_{2},x_{3})g(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{\zeta}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac {1}{\zeta}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{\delta} \diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac {1}{\delta}} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert g(x_{1},x_{2},x_{3}) \bigr\vert ^{q} \diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\cdots\diamond_{\alpha}x_{n} \biggr)^{\frac{1}{q}}. \end{aligned}$$
Theorem 2.9
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and that
\(f_{i}(x_{1},x_{2},x_{3})\ (i=1,2,\ldots,m)\), \(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow \mathbb{R}\)
are
\(\diamond_{\alpha}\)-integrable functions.
(I)
Let
\(1<\lambda_{1}\), \(\lambda_{2}, \ldots, \lambda_{m}<\infty\)
such that
\(\sum_{i=1}^{m}\frac{1}{\lambda_{i}}=1\). Then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \Biggl\vert \omega(x_{1},x_{2},x_{3}) \Biggl(\prod_{i=1}^{m}f_{i}(x_{1},x_{2},x_{3}) \Biggr) \Biggr\vert \diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\quad\leq\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{\lambda_{i}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{\lambda_{i}}}. \end{aligned}$$
(21)
(II)
Let
\(\lambda_{1}>0\), \(\lambda_{2}, \ldots, \lambda_{m}<0\)
be such that
\(\sum_{i=1}^{m}\frac{1}{\lambda_{i}}=1\). Then inequality (21) is reversed.
Proof
Case (I). Without loss of generality, we may suppose that
Write
for \(i=1,2,\ldots,m\).
$$\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{\lambda_{i}}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{\lambda_{i}}}\neq0. $$
$$\alpha_{i} =\frac{ \vert \omega(x_{1},x_{2},x_{3}) \vert ^{\frac {1}{\lambda_{i}}} \vert f_{i}(x_{1},x_{2},x_{3}) \vert }{ (\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \vert \omega(x_{1},x_{2},x_{3}) \vert \vert f_{i}(x_{1},x_{2},x_{3}) \vert ^{\lambda_{i}}\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} )^{\frac{1}{\lambda_{i}}}} $$
From AG inequality (11) we have
Therefore, we get the desired inequality (21).
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\prod_{i=1}^{m} \alpha_{i} \diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \Biggl(\sum _{i=1}^{m}\frac{\alpha_{i}^{\lambda_{i}}}{\lambda_{i}} \Biggr) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad=\sum_{i=1}^{m}\frac{1}{\lambda_{i}} \int_{a_{1}}^{b_{1}} \int _{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\biggl( \frac{ \vert \omega(x_{1},x_{2},x_{3}) \vert }{ \int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int_{a_{3}}^{b_{3}} \vert \omega(x_{1},x_{2},x_{3}) \vert \vert f_{i}(x_{1},x_{2},x_{3}) \vert ^{\lambda_{i}}\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}} \\ &\qquad{}\times \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{\lambda_{i}}\biggr)\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad=\sum_{i=1}^{m}\frac{1}{\lambda_{i}}=1. \end{aligned}$$
Case (II). By the same method as in Case (I) and using the reversed inequality (11), we can obtain the desired result. □
Theorem 2.10
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and that
\(f_{i}(x_{1},x_{2},x_{3})\), \(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow [0, +\infty)\)
\((i=1,2,\ldots,m)\)
are
\(\diamond_{\alpha}\)-integrable functions with
\(\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int _{a_{3}}^{b_{3}}\omega(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3}=1\). Let
\(0<\lambda_{1}\), \(\lambda _{2}, \ldots, \lambda_{m}<1\)
be such that
\(\lambda_{1}+\lambda _{2}+\cdots+\lambda_{m}=k<1\). Then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\omega (x_{1},x_{2},x_{3}) \Biggl(\prod_{i=1}^{m}f_{i}^{\lambda_{i}}(x_{1},x_{2},x_{3}) \Biggr)\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad \leq\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}\omega(x_{1},x_{2},x_{3})f_{i}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\lambda_{i}}. \end{aligned}$$
(22)
Proof
Denote \(\xi_{i}=\frac{\lambda_{i}}{k}\)
\((i=1,2,\ldots,m)\). Then
Write
From Theorem 2.9 and Lemma 2.5 we have
for \(k<1\). Therefore the proof of Theorem 2.10 is completed. □
$$\xi_{1}+\xi_{2}+\cdots+\xi_{m}=1. $$
$$\psi_{i}(x_{1},x_{2},x_{3})=f_{i}^{k}(x_{1},x_{2},x_{3}) \quad (i=1,2,\ldots,m). $$
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\omega (x_{1},x_{2},x_{3}) \prod_{i=1}^{m}f_{i}^{\lambda_{i}}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad= \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\omega (x_{1},x_{2},x_{3}) \prod_{i=1}^{m}\psi_{i}^{\xi_{i}}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \\ &\quad\leq\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \omega(x_{1},x_{2},x_{3}) \psi_{i}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\xi_{i}} \\ &\quad=\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}\omega(x_{1},x_{2},x_{3})f_{i}^{k}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\frac{\lambda_{i}}{k}} \\ &\quad\leq\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}}\omega(x_{1},x_{2},x_{3})f_{i}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\lambda_{i}} \end{aligned}$$
(23)
Let \(f_{i}^{\lambda_{i}}(x_{1},x_{2},x_{3})\)=\(\mu _{i}(x_{1},x_{2},x_{3})\), that is, \(f_{i}(x_{1},x_{2},x_{3})\)=\(\mu _{i}^{\frac{1}{\lambda_{i}}}(x_{1},x_{2},x_{3})\) for \(i=1,2,\ldots,m\). Then from Theorem 2.10 we get the following Hölder-type inequality.
Corollary 2.11
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and that
\(g_{i}(x_{1},x_{2},x_{3})\), \(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow [0, +\infty)\)
\((i=1,2,\ldots,m)\)
are
\(\diamond_{\alpha}\)-integrable functions with
\(\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int _{a_{3}}^{b_{3}}\omega(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond _{\alpha} x_{2}\diamond_{\alpha} x_{3}=1\). Let
\(0<\lambda_{1}\), \(\lambda _{2}, \ldots, \lambda_{m}<1\)
be such that
\(\lambda_{1}+\lambda _{2}+\cdots+\lambda_{m}=k<1\). Then
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\omega (x_{1},x_{2},x_{3}) \Biggl(\prod_{i=1}^{m}g_{i}(x_{1},x_{2},x_{3}) \Biggr)\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \\ &\quad\leq\prod_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}}\omega (x_{1},x_{2},x_{3})g_{i}^{\frac{1}{\lambda_{i}}}(x_{1},x_{2},x_{3}) \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2} \diamond_{\alpha} x_{3} \biggr)^{\lambda_{i}}. \end{aligned}$$
(24)
3 Application
In this section, using the obtained results, we give the following generalization of the Minkowski inequality for the triple Diamond-Alpha integral on time scales.
Theorem 3.1
Suppose that
\(\mathbb{T}\)
is a time scale, \(a_{i},b_{i}\in\mathbb{T}\)
with
\(a_{i}< b_{i}\)
\((i=1,2,3)\), and that
\(f_{i}(x_{1},x_{2},x_{3})\ (i=1,2,\ldots,m)\)
and
\(\omega(x_{1},x_{2},x_{3}): [a_{i},b_{i}]_{\mathbb{T}}^{3}\rightarrow \mathbb{R}\)
are
\(\diamond_{\alpha}\)-integrable functions.
(I)
If
\(p>1\), then
$$\begin{aligned} & \Biggl[ \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \Biggl\vert \omega(x_{1},x_{2},x_{3}) \Biggl(\sum_{i=1}^{m}f_{i}(x_{1},x_{2},x_{3}) \Biggr)^{p} \Biggr\vert \diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \Biggr]^{\frac{1}{p}} \\ &\quad\leq\sum_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{p}}. \end{aligned}$$
(25)
Proof
We prove only case (I). Write \(\Phi(x_{1},x_{2},x_{3})=\sum_{i=1}^{m}f_{i}(x_{1},x_{2},x_{3})\). Without loss of generality, we may assume that \(\int_{a_{1}}^{b_{1}}\int_{a_{2}}^{b_{2}}\int _{a_{3}}^{b_{3}}\omega(x_{1},x_{2},x_{3})\Phi ^{p}(x_{1},x_{2},x_{3})\diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3}\neq0\). From inequality (14) for \(p,q>0\) and \(\frac{1}{p}+\frac {1}{q}=1\) we have
Therefore we get
Thus we obtain the desired inequality (25). □
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \omega(x_{1},x_{2},x_{3})\Phi^{p-1}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ &\qquad{} \times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert \Phi(x_{1},x_{2},x_{3}) \bigr\vert ^{(p-1)q}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond_{\alpha} x_{3} \biggr)^{\frac{1}{q}} \\ &\quad= \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{p}} \\ & \qquad{}\times \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert \Phi(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{q}}. \end{aligned}$$
$$\begin{aligned} & \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \Phi^{p}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\quad= \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \Phi(x_{1},x_{2},x_{3})\Phi ^{p-1}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\quad= \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})f_{1}(x_{1},x_{2},x_{3}) \Phi ^{p-1}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\qquad{} +\cdots+ \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3})f_{m}(x_{1},x_{2},x_{3}) \Phi^{p-1}(x_{1},x_{2},x_{3}) \bigr\vert \diamond_{\alpha} x_{1}\diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \\ &\quad\leq \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int_{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert \Phi(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{q}} \\ & \qquad{}\times \Biggl[\sum_{i=1}^{m} \biggl( \int_{a_{1}}^{b_{1}} \int_{a_{2}}^{b_{2}} \int _{a_{3}}^{b_{3}} \bigl\vert \omega(x_{1},x_{2},x_{3}) \bigr\vert \bigl\vert f_{i}(x_{1},x_{2},x_{3}) \bigr\vert ^{p}\diamond_{\alpha} x_{1} \diamond_{\alpha} x_{2}\diamond _{\alpha} x_{3} \biggr)^{\frac{1}{p}} \Biggr]. \end{aligned}$$
4 Conclusions
As is well known, the Hölder inequality and its various extensions play a very important role in mathematical analysis. In this paper, based on the definition of the triple Diamond-Alpha integral for functions of three variables, we have presented the Hölder and reverse Hölder inequalities for the triple Diamond-Alpha integral on time scales. Moreover, we gave some new generalizations of the Hölder and reverse Hölder inequalities for the triple Diamond-Alpha integral. Finally, using the obtained results, we have obtained a new generalization of the Minkowski inequality for the triple Diamond-Alpha integral on time scales. In the future research, we will continue to explore other inequalities for the triple Diamond-Alpha integral on time scales.
Acknowledgements
This work was supported by the Fundamental Research Funds for the Central Universities (Nos. 2015ZD29 and 13ZD19) and the Higher School Science Research Funds of Hebei Province of China (No. Z2015137).
Competing interests
The author declares that there is no conflict of interests regarding the publication of this paper.
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