Qi Type Diamond-Alpha Integral Inequalities

Abstract

In this paper, we establish sufficient conditions for Qi type diamond-alpha integral inequalities and its generalized form on time scales.

1. Introduction

In 2000, Qi provided an open problem following [1], “Qi type integral inequality” for short in this paper.

Theorem 1

(Open problem).Under what conditions does the following inequality hold,

$${\int }_{\alpha }^{\beta }{q}^p\left(t\right)\mathrm{d}t\ge {\left({\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t\right)}^{p-1},$$

for $p>1$.

Yu and Qi [2] deduced the following theorem via Jensen’s inequality.

Theorem 2.

If $q\in C\left([\alpha ,\beta ]\right)$, ${\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t\ge {(\beta -\alpha )}^{p-1}$ for given $p>1$, then

$${\int }_{\alpha }^{\beta }{q}^p\left(t\right)\mathrm{d}t\ge {\left({\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t\right)}^{p-1}.$$

(1)

The open problem has attracted the interest of many authors [3,4,5,6,7]. The analytic method and employing the Jensen’s inequality are two powerful methods for the study of Qi type integral inequality. Meanwhile, studies in the past two decades have provided some promotions of the inequality.

Pogány [3] posed the following inequality

$${\int }_{\alpha }^{\beta }{q}^{p_1}\left(t\right)\mathrm{d}t\ge {\left({\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t\right)}^{p_2},p_2>0,p_1>max\{p_2,1\},$$

(2)

and gave a sufficient condition for (2) by using the Hölder inequality.

In [4], the authors proved the following results which strengthen the Qi type integral inequality.

Theorem 3.

If $q:[\alpha ,\beta ]\to \mathbb{R}$ is non-negative and increasing, $q^{\prime }\left(t\right)>(p-2){(t-\alpha )}^{p-3}$ for all $p>3$, then

$${\int }_{\alpha }^{\beta }{q}^p\left(t\right)\mathrm{d}t-{\left({\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t\right)}^{p-1}\ge q^{p-1}\left(\alpha \right){\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t.$$

Theorem 4.

If $q:[\alpha ,\beta ]\to \mathbb{R}$ is non-negative and increasing, $q^{\prime }\left(t\right)>p{((t-\alpha )/(\beta -\alpha ))}^{p-1}$ for given $p\ge 1$, then

$${\int }_{\alpha }^{\beta }{q}^{p+2}\left(t\right)\mathrm{d}t-\frac{1}{{(b-a)}^{p-1}}{\left({\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t\right)}^{p+1}\ge q^{p+1}\left(\alpha \right){\int }_{\alpha }^{\beta }q\left(t\right)\mathrm{d}t.$$

(3)

Since the theory of time scales was established by Hilger [8] in 1988, it has been used widely by many branches of sciences such as finance, statistics, physics. Moreover, many researches about the theory which unifies and gives a generalization of the discrete theory and the continuous theory have been published, such as [9,10,11,12,13,14,15,16,17,18].

As a generalization of the differential in calculus, $\Delta \left(\mathrm{delta}\right)$ and $\nabla \left(\mathrm{nabla}\right)$ dynamic derivatives play a foundational role in the time scales. Recently, researchers also have provided ${\diamond }_{\alpha }$ as a weighting between $\Delta$ and ∇ dynamic derivatives. It was defined as a linear combination of $\Delta$ and ∇ dynamic derivatives. Readers can consult [19] to find out more basic rules of ${\diamond }_{\alpha }$ dynamic derivatives.

Some works in recent years established the Qi type integral inequality on time scales [5,20].

Theorem 5

(Qi type$\mathbf{\Delta }$-integral inequality).([20]) If $p\ge 3$ and ϕ is a monotonic non-negative function defined on ${[\alpha ,\beta ]}_{\mathbb{T}}$ which satisfies

$${\varphi }^{p-2}\left(s\right){\varphi }^{\Delta }\left(s\right)\ge {\sigma }^{\Delta }\left(s\right)(p-2){({\sigma }^2\left(s\right)-\alpha )}^{p-3}{\varphi }^{p-2}\left({\sigma }^2\left(s\right)\right),$$

for all $s\in {[\alpha ,\beta ]}_{\mathbb{T}}$. Then

$${\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\Delta s\ge {\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\right)}^{p-1}.$$

Theorem 6

(Qi type ∇-integral inequality).([20]) If $p\ge 3$ and ϕ is a monotonic non-negative function defined on ${[\alpha ,\beta ]}_{\mathbb{T}}$ which satisfies

$${\varphi }^{\nabla }\left(s\right)\ge (p-2){(s-\alpha )}^{p-3},$$

for all $s\in {[\alpha ,\beta ]}_{\mathbb{T}}$. Then

$${\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\nabla s\ge {\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s\right)}^{p-1}.$$

Theorem 7

([5]) If $p\ge 3$, $\varphi :{[\alpha ,\beta ]}_{\mathbb{T}}\to [0,+\infty )$ is Δ-differential and increasing function satisfies

$${\varphi }^{p-2}\left(s\right){\varphi }^{\Delta }\left(s\right)\ge (p-2){\left(\varphi \left({\sigma }^2\left(s\right)\right)\right)}^{p-2}{({\sigma }^2\left(s\right)-\alpha )}^{p-3}{\sigma }^{\Delta }\left(s\right).$$

Then

$$\begin{array}{c}\hfill {\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\Delta s-{\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\right)}^{p-1}\ge {\varphi }^{p-2}\left(\alpha \right)\left(\varphi \left(\alpha \right)-(p-1){\mu }^{p-2}\left(\alpha \right)\right){\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s.\end{array}$$

Theorem 8

([5]) If $p\ge 3$, $\varphi :{[\alpha ,\beta ]}_{\mathbb{T}}\to [0,+\infty )$ is ∇-differential and increasing function satisfies

$${\left(\varphi \left(\rho \right(s\left)\right)\right)}^{p-2}{\varphi }^{\nabla }\left(s\right)\ge (p-2){\varphi }^{p-2}\left(s\right){(s-\alpha )}^{p-3}.$$

Then

$${\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\nabla s-{\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s\right)}^{p-1}\ge {\varphi }^{p-1}\left(\alpha \right){\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s.$$

However, generalizing the Qi type integral inequality to the diamond-alpha integral had been a largely under explored domain which none of works has been devoted to it.

The first aim of this paper is to determine a sufficient condition for inequality (5) via analytic method in Theorem 9.

Then we will consider the inequalities (2) and (3) generalized to diamond-alpha integral cases, that is, we will determine the sufficient conditions for inequalities (7) and (8) in Theorems 10 and 11.

Meanwhile we also consider a sufficient condition for the reverse of inequality (7) in Theorem 12.

Last but not least, we will give concise solutions of the open Problem 1 generalized on time scales via Jensen’s inequalities. Meantime, we will consider the cases including n variables, more precisely, special cases $\alpha =0,1,\frac{1}{2},\frac{1}{3}$ will be considered.

In the following part of this paper, some important and fundamental properties of time scales will be given in the Section 2. In Section 3 we will deduce Theorems 9–12 via analysis method. A concise method will be used to prove the Qi type high dimensional integral inequalities on time scales in Section 4.

2. Preliminaries

We introduce some definitions and algorithms of time scales in this section. Time scales is an arbitrary nonempty closed subset of the real number and we regard ${[\alpha ,\beta ]}_{\mathbb{T}}$ as $[\alpha ,\beta ]\cap \mathbb{T}$. In what follows, we always suppose $\alpha ,\beta \in \mathbb{T}$. We refer the readers to [9] for more details.

Definition 1.

For any $s\in \mathbb{T}$, the forward jump operator σ: $\mathbb{T}\to \mathbb{T}$ is defined by

$$\sigma \left(s\right)=inf\{t\in \mathbb{T}:t>s\},$$

and the backward jump operator ρ: $\mathbb{T}\to \mathbb{T}$ is defined by

$$\rho \left(s\right)=sup\{t\in \mathbb{T}:t<s\}.$$

As a complement, set

$$inf\varnothing =sup\mathbb{T},sup\varnothing =inf\mathbb{T}.$$

It is obvious that $\sigma \left(s\right)\ge s\ge \rho \left(s\right)$.

Definition 2.

The graininess function μ: $\mathbb{T}\to [0,\infty )$ is defined by

$$\mu \left(s\right)=\sigma \left(s\right)-s.$$

Accordingly, υ: $\mathbb{T}\to [0,\infty )$ is defined by

$$\upsilon \left(s\right)=s-\rho \left(s\right).$$

Definition 3.

${\mathbb{T}}_k$, ${\mathbb{T}}^k$ is defined as follows:

$${\mathbb{T}}_k=\left\{\begin{array}{cc}\mathbb{T}\backslash [inf\mathbb{T},\sigma (inf\mathbb{T}\left)\right),\hfill & \mathrm{if}inf\mathbb{T}>-\infty ,\hfill \\ \mathbb{T},\hfill & \mathrm{if}inf\mathbb{T}=-\infty .\hfill \end{array}\right.$$

$${\mathbb{T}}_k=\left\{\begin{array}{cc}\mathbb{T}\backslash \left(\rho \right(sup\mathbb{T}),sup\mathbb{T}),\hfill & \mathrm{if}sup\mathbb{T}<\infty ,\hfill \\ \mathbb{T},\hfill & \mathrm{if}sup\mathbb{T}=\infty .\hfill \end{array}\right.$$

Property 1.

If q: $\mathbb{T}\to \mathbb{R}$ is a continuous function.

• If $\sigma \left(s\right)>s$, then q is $\Delta$-differentiable at $s\in {\mathbb{T}}^k$ and

  $$q^{\Delta }\left(s\right)=\frac{q\left(\sigma \right(s\left)\right)-q\left(s\right)}{\mu \left(s\right)}.$$
• If $\rho \left(s\right)<s$, then q is ∇-differentiable at $s\in {\mathbb{T}}_k$ and

  $$q^{\nabla }\left(s\right)=\frac{q\left(s\right)-q\left(\rho \right(s\left)\right)}{\upsilon \left(s\right)}.$$
• If $\rho \left(s\right)=s=\sigma \left(s\right)$, then

  $$q^{\nabla }\left(s\right)=q^{\Delta }\left(s\right)=q^{\prime }\left(s\right).$$

Property 2.

Suppose $q_1$, $q_2$ are differentiable at $s\in {\mathbb{T}}_k^k$. Then the following holds:

• The sum $q_1$+$q_2$ is differentiable at s and

  $${(q_1+q_2)}^{\Delta }\left(s\right)=q_1^{\Delta }\left(s\right)+q_2^{\Delta }\left(s\right).$$

  $${(q_1+q_2)}^{\nabla }\left(s\right)=q_1^{\nabla }\left(s\right)+q_2^{\nabla }\left(s\right).$$
• For $\alpha \in \mathbb{R}$. $\alpha q$ is differentiable at s and

  $${\left(\alpha q\right)}^{\Delta }\left(s\right)=\alpha q^{\Delta }\left(s\right).$$

  $${\left(\alpha q\right)}^{\nabla }\left(s\right)=\alpha q^{\nabla }\left(s\right).$$

Definition 4.

Let $Q,q$: $\mathbb{T}\to \mathbb{R}$. If

$$Q^{\Delta }\left(s\right)=q\left(s\right),$$

holds for any $s\in {\mathbb{T}}^k$. Then Q is called a $delta$ $antiderivative$ of q. Moreover

$${\int }_a^{s}q\left(t\right)\Delta t=Q\left(s\right)-Q\left(a\right).$$

If for any $s\in {\mathbb{T}}_k$ satisfies

$$Q^{\nabla }\left(s\right)=q\left(s\right).$$

Then Q is called a $nabla$ $antiderivative$ of q. Moreover

$${\int }_a^{s}q\left(t\right)\nabla t=Q\left(s\right)-Q\left(a\right).$$

Property 3.

If $q_1$, $q_2$: $\mathbb{T}\to \mathbb{R}$ are integrable on$[\alpha ,\beta ]$. Then

• The sum $q_1$+$q_2$ is integrable on $(\alpha ,\beta )$ and

  $${\int }_{\alpha }^{\beta }(q_1+q_2)\left(s\right)\Delta s={\int }_{\alpha }^{\beta }{q}_1\left(s\right)\Delta s+{\int }_{\alpha }^{\beta }{q}_2\left(s\right)\Delta s.$$

  $${\int }_{\alpha }^{\beta }(q_1+q_2)\left(s\right)\nabla s={\int }_{\alpha }^{\beta }{q}_1\left(s\right)\nabla s+{\int }_{\alpha }^{\beta }{q}_2\left(s\right)\nabla s.$$
• For any const k, $kq$ is integrable on $(\alpha ,\beta )$ and

  $${\int }_{\alpha }^{\beta }kq\left(s\right)\Delta s=k{\int }_{\alpha }^{\beta }q\left(s\right)\Delta s.$$

  $${\int }_{\alpha }^{\beta }kq\left(s\right)\nabla s=k{\int }_{\alpha }^{\beta }q\left(s\right)\nabla s.$$

Property 4.

If $q_1,q_2,q$: $\mathbb{T}\to \mathbb{R}$ are integrable on$[{\alpha }_1,{\beta }_1]$, then

$${\int }_{{\alpha }_1}^{{\beta }_1}(q_1+q_2)\left(s\right){\diamond }_{\alpha }s={\int }_{{\alpha }_1}^{{\beta }_1}{q}_1\left(s\right){\diamond }_{\alpha }s+{\int }_{{\alpha }_1}^{{\beta }_1}{q}_2\left(s\right){\diamond }_{\alpha }s,$$

and

$${\int }_{{\alpha }_1}^{{\beta }_1}kq\left(s\right){\diamond }_{\alpha }s=k{\int }_{{\alpha }_1}^{{\beta }_1}q\left(s\right){\diamond }_{\alpha }s.$$

Next are two particularly useful formulas.

Property 5.

Let $s\in {\mathbb{T}}^k$.

• If $q\in C_{rd}$, then

  $${\int }_s^{\sigma \left(s\right)}q\left(t\right)\Delta t=\mu \left(s\right)q\left(s\right),$$

  where $q\in C_{rd}$ mean q is continuous at right-dense points, and its left-sided limits exist at left-dense points.
• If $q\in C_{ld}$, then

  $${\int }_{\rho \left(s\right)}^{s}q\left(s\right)\nabla s=\upsilon \left(s\right)q\left(s\right),$$

  where $q\in C_{ld}$ mean f is continuous at left-dense points, and its right-sided limits exist at right-dense points.

Corollary 2.47 in [9] shows that there exists the relationship between monotonicity and the $\Delta$-differential or the ∇-differential as follows.

Property 6.

If $q\in C\left([\alpha ,\beta )\right)$ are delta derivative at $(\alpha ,\beta )$, then q is increasing (decreasing) if and only if $q^{\Delta }\left(s\right)\ge 0(\le 0)$ for all $s\in [\alpha ,\beta )$.

Property 7.

If $q\in C\left([\alpha ,\beta )\right)$ are nabla derivative at $(\alpha ,\beta )$, then q is increasing (decreasing) if and only if $q^{\nabla }\left(s\right)\ge 0(\le 0)$ for all $s\in (\alpha ,\beta ]$.

The following two propositions can be found in [19].

Property 8.

If $q:\mathbb{T}\to \mathbb{R}$ is continuous at t, $t\in {[\alpha ,\beta ]}_{{\mathbb{T}}_k}$, then

$${\left({\int }_{\alpha }^{s}q\left(t\right)\Delta t\right)}^{\nabla }=q\left(\rho \left(s\right)\right).$$

Property 9.

If $q:\mathbb{T}\to \mathbb{R}$ is continuous at t, $t\in {[\alpha ,\beta ]}_{{\mathbb{T}}^k}$, then

$${\left({\int }_{\alpha }^{s}q\left(t\right)\nabla t\right)}^{\Delta }=q\left(\sigma \left(s\right)\right).$$

Finally, we list some useful properties which can be found in [9].

Definition 5.

If g is Δ-integrable on

$$R=[{\alpha }_1,{\beta }_1]\times [{\alpha }_2,{\beta }_2]\times \cdots \times [{\alpha }_n,{\beta }_n],$$

then set

$${\int }_{R}q(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1\dots {\Delta }_n{v}_n={\int }_{{\alpha }_1}^{{\beta }_1}\cdots {\int }_{{\alpha }_n}^{{\beta }_n}q(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1\dots {\Delta }_n{v}_n.$$

Property 10.

If $q_1,q_2$ are bounded Δ-integral over

$$R=[{\alpha }_1,{\beta }_1)\times [{\alpha }_2,{\beta }_2)\times \cdots \times [{\alpha }_n,{\beta }_n)$$

and $k_1,k_2$$\in \mathbb{R}$, then

$$\begin{array}{ccc}& & {\int }_R(k_1{q}_1+k_2{q}_2)(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1{\Delta }_2{v}_2\dots {\Delta }_n{v}_n\hfill \\ & =& k_1{\int }_R{q}_1(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1{\Delta }_2{v}_2\dots {\Delta }_n{v}_n+k_2{\int }_R{q}_2(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1{\Delta }_2{v}_2\dots {\Delta }_n{v}_n.\hfill \end{array}$$

Property 11.

If $q_1$ and $q_2$ are bounded functions that are Δ-integral over R with

$$q_1(v_1,v_2,\dots ,v_n)\le q_2(v_1,v_2,\dots ,v_n),\forall (v_1,v_2,\dots ,v_n)\in R,$$

then

$${\int }_R{q}_1(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1{\Delta }_2{v}_2\dots {\Delta }_n{v}_n\le {\int }_R{q}_2(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1{\Delta }_2{v}_2\dots {\Delta }_n{v}_n.$$

Remark 1.

In particular, if $q_1(v_1,v_2,\dots ,v_n)=0$, then

$${\int }_R{q}_2(v_1,v_2,\dots ,v_n){\Delta }_1{v}_1{\Delta }_2{v}_2\dots {\Delta }_n{v}_n\ge 0.$$

3. Qi Type Diamond-Alpha Integral Integral and Its Generalized Form

In this section, analysis method will be used to deduce sufficient conditions for Qi type diamond-alpha integral inequalities and its generalized forms.

We need the following lemmas which give an estimation to the differential of the power of f.

Lemma 1.

([20]). Suppose g: ${[\alpha ,\beta ]}_{\mathbb{T}}\to [0,\infty )$ is a increasing function, and if $p\ge 1$, then

$$p{g}^{p-1}\left(t\right)g^{\Delta }\left(t\right)\le {\left(g^p\left(t\right)\right)}^{\Delta }\le p{g}^{p-1}\left(\sigma \left(t\right)\right)g^{\Delta }\left(t\right),$$

where σ is forward jump operate.

Lemma 2.

([20]). Suppose g: ${[\alpha ,\beta ]}_{\mathbb{T}}\to [0,\infty )$ is a increasing function, and if $p\ge 1$, then

$$p{g}^{p-1}\left(\rho \left(t\right)\right)g^{\nabla }\left(t\right)\le {\left(g^p\left(t\right)\right)}^{\nabla }\le p{g}^{p-1}\left(t\right)g^{\nabla }\left(t\right),$$

where ρ is backward jump operate.

Following we consider $G\left(h\right)$ as the difference between the left hand and their right hand side, and take its nabla differential. According to the Proposition 7, we can complete the proof with analysis.

Theorem 9.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$${\varphi }^{p-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge (p-2){(t-\beta )}^{p-3}{\varphi }^{p-3}\left(t\right)\left(\alpha \varphi \left(\rho \right(t\left)\right)+(1-\alpha )\varphi \left(t\right)\right),$$

(4)

for all $t\in [\beta ,\gamma ]$, where ρ is backward jump operator. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p\ge 3$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^p\left(s\right){\diamond }_{\alpha }s\ge {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p-1}.$$

(5)

Proof. 

Set the difference

$$G\left(h\right)={\int }_{\beta }^h{\varphi }^p\left(s\right){\diamond }_{\alpha }s-{\left({\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p-1},$$

and let

$$g\left(h\right)={\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s.$$

It follows from Proposition 8 and Lemma 2 that

$$\begin{array}{ccc}\hfill G^{\nabla }\left(h\right)& =& {\left(\alpha {\int }_{\beta }^h{\varphi }^p\left(s\right)\Delta s+(1-\alpha ){\int }_{\beta }^h{\varphi }^p\left(s\right)\nabla s\right)}^{\nabla }-{\left(g^{p-1}\left(h\right)\right)}^{\nabla }\hfill \\ & \ge & \alpha {\varphi }^p\left(\rho \left(h\right)\right)+(1-\alpha ){\varphi }^p\left(h\right)-(p-1)g^{p-2}\left(h\right)g^{\nabla }\left(h\right)\hfill \\ & =& \alpha {\varphi }^p\left(\rho \left(h\right)\right)+(1-\alpha ){\varphi }^p\left(h\right)-(p-1)g^{p-2}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & =& \alpha \varphi \left(\rho \left(h\right)\right)\left({\varphi }^{p-1}\left(\rho \left(h\right)\right)-(p-1)g^{p-2}\left(h\right)\right)\hfill \\ & +& (1-\alpha )\varphi \left(h\right)\left({\varphi }^{p-1}\left(h\right)-(p-1)g^{p-2}\left(h\right)\right).\hfill \end{array}$$

Since $\alpha ,1-\alpha ,\varphi \left(h\right)$ are non-negative, and ${\varphi }^{p-1}\left(h\right)\ge {\varphi }^{p-1}\left(\rho \left(h\right)\right)$, it is sufficient to prove that ${\varphi }^{p-1}\left(\rho \left(h\right)\right)-(p-1)g^{p-2}\left(h\right)\ge 0$ (define ${\varphi }^{p-1}\left(\rho \left(h\right)\right)-(p-1)g^{p-2}\left(h\right)$ as $G_1\left(h\right)$). By Lemmas 1 and 2 again, we can get

$$\begin{array}{ccc}\hfill G_1^{\nabla }\left(h\right)& =& {\left({\varphi }^{p-1}\left(\rho \left(h\right)\right)-(p-1)g^{p-2}\left(h\right)\right)}^{\nabla }\hfill \\ & \ge & (p-1){\varphi }^{p-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)-(p-1)(p-2)g^{p-3}\left(h\right)g^{\nabla }\left(h\right)\hfill \\ & =& (p-1){\varphi }^{p-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)-(p-1)(p-2)g^{p-3}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right).\hfill \end{array}$$

Due to $\varphi$ is increasing,

$$\begin{array}{ccc}\hfill g\left(h\right)& =& {\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s\hfill \\ & =& \alpha {\int }_{\beta }^h\varphi \left(s\right)\Delta s+(1-\alpha ){\int }_{\beta }^h\varphi \left(s\right)\nabla s\hfill \\ & \le & \alpha (h-\beta )\varphi \left(h\right)+(1-\alpha )(h-\beta )\varphi \left(h\right)\hfill \\ & =& (h-\beta )\varphi \left(h\right).\hfill \end{array}$$

We immediately get

$$\begin{array}{ccc}\hfill G_1^{\nabla }\left(h\right)& =& (p-1){\varphi }^{p-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)-(p-1)(p-2)g^{p-3}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & \ge & (p-1){\varphi }^{p-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)\hfill \\ & & -(p-1)(p-2){(h-\beta )}^{p-3}{\varphi }^{p-3}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & \ge & 0.\hfill \end{array}$$

Clearly, $G_1\left(\beta \right)={\varphi }^{p-1}\left(\rho \left(\beta \right)\right)-(p-1)g^{p-2}\left(\beta \right)={\varphi }^{p-1}\left(\rho \left(\beta \right)\right)={\varphi }^{p-1}\left(\beta \right)\ge 0$, so $G_1\left(h\right)\ge 0$. Because $G\left(\beta \right)=0$, we deduce $G\left(h\right)\ge 0$, thereby completes the proof. □

Theorem 10.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$(p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(t\right)\left(\alpha \varphi \left(\rho \right(t\left)\right)+(1-\alpha )\varphi \left(t\right)\right),$$

(6)

for all $t\in [\beta ,\gamma ]$, where ρ is backward jump operator. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right){\diamond }_{\alpha }s\ge {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p_2}.$$

(7)

Proof. 

Set the difference

$$G\left(h\right)={\int }_{\beta }^h{\varphi }^{p_1}\left(s\right){\diamond }_{\alpha }s-{\left({\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p_2},$$

and let

$$g\left(h\right)={\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s.$$

It follows from Proposition 8 and Lemma 2 that

$$\begin{array}{ccc}\hfill G^{\nabla }\left(h\right)& =& {\left(\alpha {\int }_{\beta }^h{\varphi }^{p_1}\left(s\right)\Delta s+(1-\alpha ){\int }_{\beta }^h{\varphi }^{p_1}\left(s\right)\nabla s\right)}^{\nabla }-{\left(g^{p_2}\left(h\right)\right)}^{\nabla }\hfill \\ & \ge & \alpha {\varphi }^{p_1}\left(\rho \left(h\right)\right)+(1-\alpha ){\varphi }^{p_1}\left(h\right)-p_2{g}^{p_2-1}\left(h\right)g^{\nabla }\left(h\right)\hfill \\ & =& \alpha {\varphi }^{p_1}\left(\rho \left(h\right)\right)+(1-\alpha ){\varphi }^{p_1}\left(h\right)-p_2{g}^{p_2-1}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & =& \alpha \varphi \left(\rho \left(h\right)\right)\left({\varphi }^{p_1-1}\left(\rho \left(h\right)\right)-p_2{g}^{p_2-1}\left(h\right)\right)\hfill \\ & +& (1-\alpha )\varphi \left(h\right)\left({\varphi }^{p_1-1}\left(h\right)-p_2{g}^{p_2-1}\left(h\right)\right).\hfill \end{array}$$

Since $\alpha ,1-\alpha ,\varphi \left(h\right)$ are non-negative, and ${\varphi }^{p_1-1}\left(h\right)\ge {\varphi }^{p_1-1}\left(\rho \left(h\right)\right)$, it is suffices to prove that ${\varphi }^{p_1-1}\left(\rho \left(h\right)\right)-p_2{g}^{p_2-1}\left(h\right)\ge 0$ (define ${\varphi }^{p_1-1}\left(\rho \left(h\right)\right)-p_2{g}^{p_2-1}\left(h\right)$ as $G_1\left(h\right)$). By Lemmas 1 and 2 again, we can get

$$\begin{array}{ccc}\hfill G_1^{\nabla }\left(h\right)& =& {\left({\varphi }^{p_1-1}\left(\rho \left(h\right)\right)-p_2{g}^{p_2-1}\left(h\right)\right)}^{\nabla }\hfill \\ & \ge & (p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)-p_2(p_2-1)g^{p_2-2}\left(h\right)g^{\nabla }\left(h\right)\hfill \\ & =& (p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)\hfill \\ & & -p_2(p_2-1)g^{p_2-2}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right).\hfill \end{array}$$

Due to $\varphi$ is increasing,

$$\begin{array}{ccc}\hfill g\left(h\right)& =& {\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s\hfill \\ & =& \alpha {\int }_{\beta }^h\varphi \left(s\right)\Delta s+(1-\alpha ){\int }_{\beta }^h\varphi \left(s\right)\nabla s\hfill \\ & \le & \alpha (h-\beta )\varphi \left(h\right)+(1-\alpha )(h-\beta )\varphi \left(h\right)\hfill \\ & =& (h-\beta )\varphi \left(h\right).\hfill \end{array}$$

We immediately get

$$\begin{array}{ccc}\hfill G_1^{\nabla }\left(h\right)& =& (p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)\hfill \\ & & -p_2(p_2-1)g^{p_2-2}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & \ge & (p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(h\right)\right){\varphi }^{\nabla }\left(h\right)\hfill \\ & & -p_2(p_2-1){(h-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(h\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & \ge & 0.\hfill \end{array}$$

Clearly, $G_1\left(\beta \right)={\varphi }^{p_1-1}\left(\rho \left(\beta \right)\right)-p_2{g}^{p_2-1}\left(\beta \right)={\varphi }^{p_1-1}\left(\rho \left(\beta \right)\right)={\varphi }^{p_1-1}\left(\beta \right)\ge 0$, so $G_1\left(h\right)\ge 0$. According to that $G\left(\beta \right)=0$, we deduce $G\left(h\right)\ge 0$, thereby completes the proof. □

If take $p_2=p_1-1$ in Theorem 10, we can deduce Theorem 9 immediately.

By virtue Theorem 10, we obtain the following corollaries by setting $\alpha =0$ and 1.

Corollary 1.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$(p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-1}\left(t\right),$$

for all $t\in [\beta ,\gamma ]$, where ρ is the backward jump operator. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right)\Delta s\ge {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right)\Delta s\right)}^{p_2}.$$

Corollary 2.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$(p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(t\right)\varphi \left(\rho \left(t\right)\right),$$

for all $t\in [\beta ,\gamma ]$, where ρ is backward jump operator. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right)\nabla s\ge {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right)\nabla s\right)}^{p_2}.$$

Theorem 11.

Suppose ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$(p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(t\right)\left(\alpha \varphi \left(\rho \right(t\left)\right)+(1-\alpha )\varphi \left(t\right)\right),$$

for all $t\in [\beta ,\gamma ]$, where ρ is backward jump operator. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right){\diamond }_{\alpha }s-{\left({\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p_2}\ge {\varphi }^{p_1-1}\left(\beta \right){\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s.$$

(8)

Proof. 

Set the difference

$$G\left(h\right)={\int }_{\beta }^h{\varphi }^{p_1}\left(s\right){\diamond }_{\alpha }s-{\left({\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p_2}-C{\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s,$$

where $C={\varphi }^{p_1-1}\left(\beta \right)$, and let

$$g\left(h\right)={\int }_{\beta }^h\varphi \left(s\right){\diamond }_{\alpha }s.$$

It follows from Proposition 8 and Lemma 2 that

$$\begin{array}{ccc}\hfill G^{\nabla }\left(h\right)& =& {\left(\alpha {\int }_{\beta }^h{\varphi }^{p_1}\left(s\right)\Delta s+(1-\alpha ){\int }_{\beta }^h{\varphi }^{p_1}\left(s\right)\nabla s\right)}^{\nabla }-{\left(g^{p_2}\left(h\right)\right)}^{\nabla }-C{g}^{\nabla }\left(h\right)\hfill \\ & \ge & \alpha {\varphi }^{p_1}\left(\rho \left(h\right)\right)+(1-\alpha ){\varphi }^{p_1}\left(h\right)-p_2{g}^{p_2-1}\left(h\right)g^{\nabla }\left(h\right)-C{g}^{\nabla }\left(h\right)\hfill \\ & =& \alpha {\varphi }^{p_1}\left(\rho \left(h\right)\right)+(1-\alpha ){\varphi }^{p_1}\left(h\right)-\left(p_2{g}^{p_2-1}\left(h\right)+C\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & =& \alpha \varphi \left(\rho \left(h\right)\right)\left({\varphi }^{p_1-1}\left(\rho \left(h\right)\right)-p_2{g}^{p_2-1}\left(h\right)-C\right)\hfill \\ & +& (1-\alpha )\varphi \left(h\right)\left({\varphi }^{p_1-1}\left(h\right)-p_2{g}^{p_2-1}\left(h\right)-C\right).\hfill \end{array}$$

Since $\alpha ,1-\alpha ,\varphi \left(h\right)$ are non-negative, and ${\varphi }^{p_1-1}\left(h\right)\ge {\varphi }^{p_1-1}\left(\rho \left(h\right)\right)$, it is suffices to prove that

$$G_2\left(h\right):={\varphi }^{p_1-1}\left(\rho \left(h\right)\right)-p_2{g}^{p_2-1}\left(h\right)-C\ge 0.$$

Note that $G_2\left(h\right)+C$ equal to $G_1\left(h\right)$, thus

$$G_2^{\nabla }\left(h\right)=G_1^{\nabla }\left(h\right)>0.$$

In this sense,

$$\begin{array}{ccc}\hfill G_1\left(\beta \right)& =& {\varphi }^{p_1-1}\left(\rho \left(\beta \right)\right)-p_2{g}^{p_2-1}\left(\beta \right)-C\hfill \\ & =& {\varphi }^{p_1-1}\left(\rho \left(\beta \right)\right)-C\hfill \\ & =& {\varphi }^{p_1-1}\left(\beta \right)-C=0,\hfill \end{array}$$

so $G_1\left(h\right)\ge 0$. Since $G\left(\beta \right)=0$, we deduce $G\left(h\right)\ge 0$, thereby completes the proof. □

Some special cases, if take $\alpha =0$ and 1, we obtain the following corollaries.

Corollary 3.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$(p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-1}\left(t\right),$$

for all $t\in [\beta ,\gamma ]$. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right)\Delta s-{\left({\int }_{\beta }^{\gamma }\varphi \left(s\right)\Delta s\right)}^{p_2}\ge {\varphi }^{p_1-1}\left(\beta \right){\int }_{\beta }^{\gamma }\varphi \left(s\right)\Delta s.$$

Corollary 4.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$(p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(h-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(t\right)\varphi \left(\rho \left(t\right)\right),$$

for all $t\in [\beta ,\gamma ]$. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right)\nabla s-{\left({\int }_{\beta }^{\gamma }\varphi \left(s\right)\nabla s\right)}^{p_2}\ge {\varphi }^{p_1-1}\left(\beta \right){\int }_{\beta }^{\gamma }\varphi \left(s\right)\nabla s.$$

Theorem 12.

Suppose ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(\beta \right)\left(\alpha \varphi \left(\rho \right(t\left)\right)+(1-\alpha )\varphi \left(t\right)\right)\ge (p_1-1){\varphi }^{p_1-2}\left(\rho \left(t\right)\right){\varphi }^{\nabla }\left(t\right)$$

(9)

for all $t\in [\beta ,\gamma ]$, where ρ is backward jump operator. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 3$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right){\diamond }_{\alpha }s\le {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p_2}.$$

(10)

Proof. 

Set the difference

$$R\left(h\right)={\left({\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p_2}-{\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right){\diamond }_{\alpha }s,$$

and

$$r\left(h\right)={\int }_{\beta }^{\gamma }\varphi \left(s\right){\diamond }_{\alpha }s.$$

It follows from Proposition 8 and Lemma 2 that

$$\begin{array}{ccc}\hfill R^{\nabla }\left(h\right)& =& {\left(r^{p_2}\left(h\right)\right)}^{\nabla }-{\left(\alpha {\int }_{\beta }^h{\varphi }^{p_1}\left(s\right)\Delta s+(1-\alpha ){\int }_{\beta }^h{\varphi }^{p_1}\left(s\right)\nabla s\right)}^{\nabla }\hfill \\ & \ge & p_2{r}^{p_2-1}\left(\rho \left(h\right)\right)r^{\nabla }\left(h\right)-\alpha {\varphi }^{p_1}\left(\rho \left(h\right)\right)-(1-\alpha ){\varphi }^{p_1}\left(h\right)\hfill \\ & =& p_2{r}^{p_2-1}\left(\rho \left(h\right)\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & & -\alpha {\varphi }^{p_1}\left(\rho \left(h\right)\right)-(1-\alpha ){\varphi }^{p_1}\left(h\right)\hfill \\ & =& \alpha \varphi \left(\rho \left(h\right)\right)\left(p_2{r}^{p_2-1}\left(\rho \left(h\right)\right)-{\varphi }^{p_1-1}\left(\rho \left(h\right)\right)\right)\hfill \\ & & +(1-\alpha )\varphi \left(h\right)\left(p_2{r}^{p_2-1}\left(\rho \left(h\right)\right)-{\varphi }^{p_1-1}\left(h\right)\right).\hfill \end{array}$$

Since $\alpha ,1-\alpha ,\varphi \left(h\right)$ are non-negative, and ${\varphi }^{p_1-1}\left(h\right)\ge {\varphi }^{p_1-1}\left(\rho \left(h\right)\right)$, it is suffices to prove that $p_2{r}^{p_2-1}\left(h\right)-{\varphi }^{p_1-1}\left(\rho \left(h\right)\right)\ge 0$ (define $p_2{r}^{p_2-1}\left(h\right)-{\varphi }^{p_1-1}\left(\rho \left(h\right)\right)$ as $R_1\left(h\right)$). By Lemmas 1 and 2 again, we can yield

$$\begin{array}{ccc}\hfill R_1^{\nabla }\left(h\right)& =& {\left(p_2{r}^{p_2-1}\left(h\right)-{\varphi }^{p_1-1}\left(\rho \left(h\right)\right)\right)}^{\nabla }\hfill \\ & \ge & p_2(p_2-1)r^{p_2-2}\left(\rho \left(h\right)\right)r^{\nabla }\left(h\right)-(p_1-1){\varphi }^{p_1-2}\left(\rho \left(h\right)\right){\varphi }^{\nabla }\left(h\right)\hfill \\ & =& p_2(p_2-1)r^{p_2-2}\left(\rho \left(h\right)\right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & & -(p_1-1){\varphi }^{p_1-2}\left(\rho \left(h\right)\right){\varphi }^{\nabla }\left(h\right).\hfill \end{array}$$

According with monotony of $\varphi$,

$$r\left(h\right)\ge (h-\beta )\varphi \left(\beta \right).$$

We immediately get

$$\begin{array}{ccc}\hfill R_1^{\nabla }\left(h\right)& \ge & p_2(p_2-1){(h-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(\beta \right)\left(\alpha \varphi \left(\rho \right(h\left)\right)+(1-\alpha )\varphi \left(h\right)\right)\hfill \\ & & -(p_1-1){\varphi }^{p_1-2}\left(\rho \left(h\right)\right){\varphi }^{\nabla }\left(h\right)\hfill \\ & \ge & 0.\hfill \end{array}$$

Taking $t=\beta$ in (9) we get

$$\begin{array}{ccc}\hfill 0& =& p_2(p_2-1){(\beta -\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(\beta \right)\left(\alpha \varphi \left(\rho \right(\beta \left)\right)+(1-\alpha )\varphi \left(\beta \right)\right)\hfill \\ & \le & (p_1-1){\varphi }^{p_1-2}\left(\rho \left(\beta \right)\right){\varphi }^{\nabla }\left(\beta \right),\hfill \end{array}$$

hence we obtain

$$\varphi \left(\rho \right(\beta \left)\right)=\varphi \left(\beta \right)\le 0.$$

Clearly,

$$\begin{array}{ccc}\hfill R_1\left(\beta \right)& =& p_2{r}^{p_2-1}\left(\beta \right)-{\varphi }^{p_1-1}\left(\rho \left(\beta \right)\right)\hfill \\ & =& -{\varphi }^{p_1-1}\left(\rho \left(\beta \right)\right)\ge 0,\hfill \end{array}$$

so $G_1\left(h\right)\ge 0$. According to that $G\left(\beta \right)=0$, we deduce $G\left(h\right)\ge 0$, thereby completes the proof. □

If we choose $\alpha =0$ or 1, we obtain the following results.

Corollary 5.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(\beta \right)\varphi \left(t\right)\ge (p_1-1){\varphi }^{p_1-2}\left(\rho \left(t\right)\right){\varphi }^{\nabla }\left(t\right)$$

for all $t\in [\beta ,\gamma ]$. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right)\Delta s\le {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right)\Delta s\right)}^{p_2}.$$

Corollary 6.

If ϕ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, satisfies

$$p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(\beta \right)\ge (p_1-1){\varphi }^{p_1-3}\left(\rho \left(t\right)\right){\varphi }^{\nabla }\left(t\right)$$

for all $t\in [\beta ,\gamma ]$. Then for all $s\in {[\beta ,\gamma ]}_{{\mathbb{T}}_k^k}$ and $p_1,p_2\ge 2$, the following inequality holds,

$${\int }_{\beta }^{\gamma }{\varphi }^{p_1}\left(s\right)\nabla s\le {\left({\int }_{\beta }^{\gamma }\varphi \left(s\right)\nabla s\right)}^{p_2}.$$

Under the basic assumptions that $p_1,p_2\ge 3$, $\varphi$ is a non-negative and continuous function defined on ${[\beta ,\gamma ]}_{\mathbb{T}}$, based on Theorems 10 and 12, we obtain the following results for all $t\in [\beta ,\gamma ]$.

Remark 2.

$$\begin{array}{ccc}& & (p_1-1){\varphi }^{p_1-2}\left({\rho }^2\left(t\right)\right){\varphi }^{\nabla }\left(t\right)\ge p_2(p_2-1){(h-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(t\right)\left(\alpha \varphi \left(\rho \right(t\left)\right)+(1-\alpha )\varphi \left(t\right)\right),\hfill \\ & & p_2(p_2-1){(t-\beta )}^{p_2-2}{\varphi }^{p_2-2}\left(\beta \right)\left(\alpha \varphi \left(\rho \right(t\left)\right)+(1-\alpha )\varphi \left(t\right)\right)\ge (p_1-1){\varphi }^{p_1-2}\left(\rho \left(t\right)\right){\varphi }^{\nabla }\left(t\right).\hfill \end{array}$$

If condition (11) holds, then (7) established; if condition (11) holds, then the reverse of (7) established.

If $\varphi$ is decreasing, we can obtain the similar results using the same method. However, if $\varphi$ satisfying neither (11) nor (11), whether inequality (7) or the reverse of inequality (7) holds needs further research.

4. Qi Type Integral Inequalities of N Variables

In Section 3, we use differential to observe the monotonicity of G. It is surely a useful method that can be used in Qi type integral of n variables. However, it will produce complex conditions. In fact, once we take the differential of one variable among n variables, in order to ensure it is greater than or equal to 0, we need a condition. Regardless of the initial conditions, it also need n conditions.

In this section, we use Jensen’s inequalities on time scales to deduce concise condition for Qi type inequality. It’s worth to point out that Jensen’s inequalities and other related topics are still a research hot spot recent years [21,22,23,24,25,26,27,28,29,30,31,32,33].

4.1. Qi Type Integral Inequalities of One Variable on Time Scales

Firstly, we list three Jensen’s inequalities of one variable on time scales, all of them have been given in [34,35,36].

Lemma 3.

([34]). If $\varphi \in C_{rd}([\alpha ,\beta ],(c,d))$ and q is a continuous and convex function defined on $(c,d)$, then

$$q\left(\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s}{\beta -\alpha }\right)\le \frac{{\int }_{\alpha }^{\beta }q\left(\varphi \left(s\right)\right)\Delta s}{\beta -\alpha }.$$

Lemma 4.

([35]). If $\varphi \in C_{ld}([\alpha ,\beta ],(c,d))$ and q is a continuous and convex function defined on $(c,d)$, then

$$q\left(\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s}{\beta -\alpha }\right)\le \frac{{\int }_{\alpha }^{\beta }q\left(\varphi \left(s\right)\right)\nabla s}{\beta -\alpha }.$$

Lemma 5.

([36]). Let ${\alpha }_1,{\beta }_1\in \mathbb{T}$ and $c,d\in \mathbb{R}$. If $\varphi \in C({[{\alpha }_1,{\beta }_1]}_{\mathbb{T}},(c,d))$ and q is a continuous and convex function defined on $(c,d)$, then

$$q\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}\varphi \left(s\right){\diamond }_{\alpha }s}{{\beta }_1-{\alpha }_1}\right)\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}q\left(\varphi \left(s\right)\right){\diamond }_{\alpha }s}{{\beta }_1-{\alpha }_1}.$$

Using lemmas, we can get following three theorems.

Theorem 13

(Qi type$\mathbf{\Delta }$integral inequality).If $\varphi \in C_{rd}\left({[\alpha ,\beta ]}_{\mathbb{T}}\right)$ is a non-negative function, and for given $p>1$ satisfies

$${\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\ge {(\beta -\alpha )}^{p-1}.$$

Then,

$${\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\Delta s\ge {\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\right)}^{p-1}.$$

Proof. 

Consider the function $q\left(x\right)=x^p$ defined on $[c,d]\subset \mathbb{R}$. If $x\ge 0$, we know that q is convex. Since $\varphi$ is non-negative,

$$\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s}{\beta -\alpha }\ge 0,$$

using Lemma 3 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s}{\beta -\alpha }\right)}^p\le \frac{{\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\Delta s}{\beta -\alpha }.$$

According to the condition,

$${\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\ge {(\beta -\alpha )}^{p-1}.$$

Thus,

$$\begin{array}{ccc}\hfill {\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\Delta s& \ge & (\beta -\alpha ){\left(\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s}{\beta -\alpha }\right)}^p\hfill \\ & =& \frac{{\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\right)}^p}{{(\beta -\alpha )}^{p-1}}\hfill \\ & =& {\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\right)}^{p-1}\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s}{{(\beta -\alpha )}^{p-1}}\hfill \\ & \ge & {\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\Delta s\right)}^{p-1}.\hfill \end{array}$$

Thereby we complete the proof. □

In the same way, we can deduce other two inequalities.

Theorem 14

(Qi type ∇ integral inequality).If $\varphi \in C_{ld}\left({[\alpha ,\beta ]}_{\mathbb{T}}\right)$ is a non-negative function, and for given $p>1$ satisfies

$${\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s\ge {(\beta -\alpha )}^{p-1}.$$

Then

$${\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\nabla s\ge {\left({\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s\right)}^{p-1}.$$

Proof. 

We know that $q\left(x\right)=x^p$ where $p>1$ is convex on $x\ge 0$. It is obvious that

$$\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s}{\beta -\alpha }\ge 0,$$

using Lemma 4 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s}{\beta -\alpha }\right)}^p\le \frac{{\int }_{\alpha }^{\beta }{\varphi }^p\left(s\right)\nabla s}{\beta -\alpha }.$$

(11)

According to the condition,

$${\int }_{\alpha }^{\beta }\varphi \left(s\right)\nabla s\ge {(\beta -\alpha )}^{p-1}.$$

(12)

Combining the inequalities (11 and (12), we complete the proof. □

Theorem 15

(Qi type diamond-$\mathbf{\alpha }$integral inequality).If $\varphi \in C\left({[{\alpha }_1,{\beta }_1]}_{\mathbb{T}}\right)$ is a non-negative function, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}\varphi \left(s\right){\diamond }_{\alpha }s\ge {({\beta }_1-{\alpha }_1)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\varphi }^p\left(s\right){\diamond }_{\alpha }s\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}\varphi \left(s\right){\diamond }_{\alpha }s\right)}^{p-1}.$$

Proof. 

Based on that

$$\frac{{\int }_{{\alpha }_1}^{{\beta }_1}\varphi \left(s\right){\diamond }_{\alpha }s}{{\beta }_1-{\alpha }_1}\ge 0,$$

and use Lemma 5 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}\varphi \left(s\right){\diamond }_{\alpha }s}{{\beta }_1-{\alpha }_1}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\varphi }^p\left(s\right){\diamond }_{\alpha }s}{{\beta }_1-{\alpha }_1}.$$

(13)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}\varphi \left(s\right){\diamond }_{\alpha }s\ge {({\beta }_1-{\alpha }_1)}^{p-1}.$$

(14)

Combining the inequalities (13) and (14), we complete the proof. □

In particularly, if we let $\mathbb{T}=\mathbb{R}$ in arbitrary one among the above three theorems, then $\Delta =\nabla ={\diamond }_{\alpha }=\mathrm{d}$, it will deduce Theorem 2.

Comparing the proofs in Section 3 and this subsection, we find that Jensen’s inequalities not only can simplify the condition, but also it can simplify the proof. Most importantly, it keeps the condition similar. This means that we can generalize it to the case of n dimensions.

4.2. Qi Type Integral Inequalities of Several Variables on Time Scales

In the same way, we generalize Qi type integral inequalities to higher dimensions. Firstly, we write down Jensen’s inequalities of n variables as lemmas, them can be found in [37,38].

Lemma 6.

([37]). If ϕ: $R\to (m_1,m_2)$ is a non-negative function where n have been given, $n\ge 3$, and $q\in C((m_1,m_2),\mathbb{R})$ is convex, then

$$q\left(\frac{{\int }_R\varphi (v_1,v2,\cdots ,v_n){\Delta }_1{v}_1\cdots {\Delta }_n{v}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}\right)\le \frac{{\int }_{R}q\left(\varphi (v_1,v2,\cdots ,v_n)\right){\Delta }_1{v}_1\cdots {\Delta }_n{v}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)},$$

where R is $[{\alpha }_1,{\beta }_1]\times [{\alpha }_2,{\beta }_2]\times \cdots \times [{\alpha }_n,{\beta }_n]$.

Lemma 7.

If ϕ: $R\to (m_1,m_2)$ is a non-negative function where n have been given, $n\ge 3$, and $q\in C((m_1,m_2),\mathbb{R})$ is convex, then

$$q\left(\frac{{\int }_R\varphi (v_1,v_2,\cdots ,v_n){\nabla }_1{v}_1\cdots {\nabla }_n{v}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}\right)\le \frac{{\int }_{R}q\left(\varphi (v_1,v_2,\cdots ,v_n)\right){\nabla }_1{v}_1\cdots {\nabla }_n{v}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}.$$

Lemma 8.

([38]). If ϕ: $R\to (m_1,m_2)$ is a non-negative function where n have been given, $n\ge 3$, and $q\in C((m_1,m_2),\mathbb{R})$ is convex, then

$$q\left(\frac{{\int }_R\varphi (v_1,v_2,\cdots ,v_n){\diamond }_{\alpha }{v}_1\cdots {\diamond }_{\alpha }{v}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}\right)\le \frac{{\int }_{R}q\left(\varphi (v_1,v_2,\cdots ,v_n)\right){\diamond }_{\alpha }{v}_1\cdots {\diamond }_{\alpha }{v}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}.$$

Next, we deduce Qi type $\Delta$-integral inequalities of two, three, n variables.

Theorem 16

(Qi type$\mathbf{\Delta }$-integral inequalities of two variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\to (m_1,m_2)$ is continuous with $m_1\ge 0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2\right)}^{p-1}.$$

Proof. 

Since $q\left(x\right)=x^p$ is convex and

$$\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}\ge 0,$$

using Lemma 6 with $q\left(x\right)=x^p$ and $n=2$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}.$$

(15)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\Delta }_1{s}_1{\Delta }_2{s}_2\right)}^{p-1}.$$

(16)

Combining the inequalities (15) and (16), we complete the proof. □

Theorem 17

(Qi type$\mathbf{\Delta }$-integral inequalities of three variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times {[{\alpha }_3,{\beta }_3]}_{{\mathbb{T}}_3}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3{s}_3\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3{s}_3\right)}^{p-1}.$$

Proof. 

Since $q\left(x\right)=x^p$ is convex and

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3\ge 0,$$

using Lemma 6 with $q\left(x\right)=x^p$ and $n=3$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)({\beta }_3-{\alpha }_3)}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)({\beta }_3-{\alpha }_3)}.$$

(17)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\Delta }_1{s}_1{\Delta }_2{s}_2{\Delta }_3{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

(18)

Combining the inequalities (17) and (18), we complete the proof. □

In the same way, we can generalize it to n dimensions.

Theorem 18

(Qi type$\mathbf{\Delta }$-integral inequalities of$\mathit{n}$variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times \cdots \times {[{\alpha }_n,{\beta }_n]}_{{\mathbb{T}}_n}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n)\Delta s_1\Delta s_2\cdots \Delta s_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

Then

$${\int }_R{\varphi }^p(s_1,s_2,\cdots ,s_n)\Delta s_1\Delta s_2\cdots \Delta s_n\ge {\left({\int }_R\varphi (s_1,s_2,\cdots ,s_n)\Delta s_1\Delta s_2\cdots \Delta s_n\right)}^{p-1}.$$

Proof. 

According to Remark 1,

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\Delta }_1{s}_1\cdots {\Delta }_n{s}_n\ge 0,$$

using Lemma 4 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_R\varphi (s_1,s_2,\cdots ,s_n){\Delta }_1{s}_1\cdots {\Delta }_n{s}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}\right)}^p\le \frac{{\int }_R{\left(\varphi (s_1,s_2,\cdots ,s_n)\right)}^p{\Delta }_1{s}_1\cdots {\Delta }_n{s}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}.$$

Together with the condition

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\Delta }_1{s}_1{\Delta }_2{s}_2\cdots {\Delta }_n{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

we can complete the proof. □

Next three inequalities is about Qi type ∇-integral inequalities of two, three, and n variables.

Theorem 19

(Qi type ∇-integral inequalities of two variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2\right)}^{p-1}.$$

Proof. 

Since $\varphi$ is non-negative, then

$$\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}\ge 0,$$

using Lemma 7 with $q\left(x\right)=x^p$ and $n=2$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}.$$

(19)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\nabla }_1{s}_1{\nabla }_2{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

(20)

Combining the inequalities (19) and (20), we complete the proof. □

Theorem 20

(Qi type ∇-integral inequalities of three variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times {[{\alpha }_3,{\beta }_3]}_{{\mathbb{T}}_3}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3\right)}^{p-1}.$$

Proof. 

Since $\varphi$ is non-negative,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3\ge 0,$$

using Lemma 7 where $q\left(x\right)=x^p$ and $n=3$, for $q\left(x\right)=x^p$ is convex when $x\ge 0$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)({\beta }_3-{\alpha }_3)}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)({\beta }_3-{\alpha }_3)}.$$

(21)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\nabla }_1{s}_1{\nabla }_2{s}_2{\nabla }_3{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

(22)

Combining the inequalities (21) and (22), we complete the proof. □

We can get Qi type ∇-integral inequalities of n variables in the same way.

Theorem 21

(Qi type ∇-integral inequalities of$\mathit{n}$variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times \cdots \times {[{\alpha }_n,{\beta }_n]}_{{\mathbb{T}}_n}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

Then

$${\int }_R{\varphi }^p(s_1,s_2,\cdots ,s_n){\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n\ge {\left({\int }_R\varphi (s_1,s_2,\cdots ,s_n){\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n\right)}^{p-1}.$$

Proof. 

In the same way, we find

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n\ge 0,$$

using Lemma 7 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_R\varphi (s_1,s_2,\cdots ,s_n){\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}\right)}^p\le \frac{{\int }_R{\left(\varphi (s_1,s_2,\cdots ,s_n)\right)}^p{\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}.$$

Using now

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\nabla }_1{s}_1{\nabla }_2{s}_2\cdots {\nabla }_n{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1},$$

(23)

the relation (23) is satisfied. □

Next three inequalities is about Qi type diamond-$\alpha$ integral inequalities of two, three, and n variables.

Theorem 22

(Qi type diamond-$\mathbf{\alpha }$integral inequalities of two variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\right)}^{p-1}.$$

Proof. 

We can find that

$$\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}\ge 0.$$

In Lemma 8, take $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)}.$$

(24)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

(25)

Combining the inequalities (24) and (25), we complete the proof. □

If set $\alpha$ equal to $\frac{1}{2}$ or $\frac{1}{3}$, then we have following corollaries.

Corollary 7.

If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

Then,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2\right)}^{p-1}.$$

Corollary 8.

If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\varphi }^p(s_1,s_2){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}\varphi (s_1,s_2){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\right)}^{p-1}.$$

Theorem 23

(Qi type diamond-$\mathbf{\alpha }$ integral inequalities of three variables). If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times {[{\alpha }_3,{\beta }_3]}_{{\mathbb{T}}_3}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3\right)}^{p-1}.$$

Proof. 

In the same way, the following inequality holds.

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3\ge 0,$$

using Lemma 8 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)({\beta }_3-{\alpha }_3)}\right)}^p\le \frac{{\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3}{({\beta }_1-{\alpha }_1)({\beta }_2-{\alpha }_2)({\beta }_3-{\alpha }_3)}.$$

(26)

According to the condition,

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2{\diamond }_{\alpha }{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

(27)

Combining the inequalities (26) and (27), we complete the proof. □

In the same way, if set $\alpha$ equal to $\frac{1}{2}$ or $\frac{1}{3}$ in the theorem above, then following corollaries hold.

Corollary 9.

If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times {[{\alpha }_3,{\beta }_3]}_{{\mathbb{T}}_3}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2{\diamond }_{\frac{1}{2}}{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2{\diamond }_{\frac{1}{2}}{s}_3\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2{\diamond }_{\frac{1}{2}}{s}_3\right)}^{p-1}.$$

Corollary 10.

If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times {[{\alpha }_3,{\beta }_3]}_{{\mathbb{T}}_3}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2{\diamond }_{\frac{1}{3}}{s}_3\ge {({\beta }_1-{\alpha }_1)}^{p-1}{({\beta }_2-{\alpha }_2)}^{p-1}{({\beta }_3-{\alpha }_3)}^{p-1}.$$

Then

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{\varphi }^p(s_1,s_2,s_3){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2{\diamond }_{\frac{1}{3}}{s}_3\ge {\left({\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}\varphi (s_1,s_2,s_3){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2{\diamond }_{\frac{1}{3}}{s}_3\right)}^{p-1}.$$

Employing Lemma 8, we can deduce

Theorem 24

(Qi type diamond-$\mathbf{\alpha }$integral inequalities of$\mathit{n}$variables).If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times \cdots \times {[{\alpha }_n,{\beta }_n]}_{{\mathbb{T}}_n}\to (m_1,m_2)$ is a continuous function with $m_1>0$, and for given $p>1$ satisfies

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

(28)

Then

$${\int }_R{\varphi }^p(s_1,s_2,\cdots ,s_n){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n\ge {\left({\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n\right)}^{p-1}.$$

Proof. 

We can find that

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n\ge 0,$$

using Lemma 8 with $q\left(x\right)=x^p$, we have

$${\left(\frac{{\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}\right)}^p\le \frac{{\int }_R{\left(\varphi (s_1,s_2,\cdots ,s_n)\right)}^p{\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n}{{\prod }_{i=1}^n({\beta }_i-{\alpha }_i)}.$$

Using now

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\alpha }{s}_1{\diamond }_{\alpha }{s}_2\cdots {\diamond }_{\alpha }{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

We can complete the proof. □

If we consider $\alpha =\frac{1}{2}$, we obtain the following corollary.

Corollary 11.

If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times \cdots \times {[{\alpha }_n,{\beta }_n]}_{{\mathbb{T}}_n}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2\cdots {\diamond }_{\frac{1}{2}}{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

Then

$${\int }_R{\varphi }^p(s_1,s_2,\cdots ,s_n){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2\cdots {\diamond }_{\frac{1}{2}}{s}_n\ge {\left({\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\frac{1}{2}}{s}_1{\diamond }_{\frac{1}{2}}{s}_2\cdots {\diamond }_{\frac{1}{2}}{s}_n\right)}^{p-1}.$$

If we consider $\alpha =\frac{1}{3}$, we obtain the following corollary.

Corollary 12.

If ϕ: ${[{\alpha }_1,{\beta }_1]}_{{\mathbb{T}}_1}\times {[{\alpha }_2,{\beta }_2]}_{{\mathbb{T}}_2}\times \cdots \times {[{\alpha }_n,{\beta }_n]}_{{\mathbb{T}}_n}\to (m_1,m_2)$ is continuous with $m_1>0$, and for given $p>1$ satisfies

$${\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\cdots {\diamond }_{\frac{1}{3}}{s}_n\ge \prod _{i=1}^n{({\beta }_i-{\alpha }_i)}^{p-1}.$$

(29)

Then

$${\int }_R{\varphi }^p(s_1,s_2,\cdots ,s_n){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\cdots {\diamond }_{\frac{1}{3}}{s}_n\ge {\left({\int }_R\varphi (s_1,s_2,\cdots ,s_n){\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\cdots {\diamond }_{\frac{1}{3}}{s}_n\right)}^{p-1}.$$

5. Examples

In this section, we give some examples which applied the conclusions in Section 3 and Section 4.

Example 1.

Consider the inequality:

$$\sum _{k=1}^{N-1}{2}^{3k}+\sum _{k=2}^N{2}^{3k}\ge \frac{1}{2}{\left(\sum _{k=1}^{N-1}{2}^k+\sum _{k=2}^N{2}^k\right)}^2,$$

(30)

where $N\in \mathbb{N}\ge 2$.

Proof. 

We take $\mathbb{T}=\mathbb{N},\varphi \left(t\right)=2^t,p=3$ and $\alpha =\frac{1}{2}$ in (4), then it transforms into

$${\left(2^{t-2}\right)}^{p-2}{2}^{t-1}\ge \frac{3}{2}{2}^{t(p-3)}{2}^t,$$

it is always true when $t\ge 1$. Based on Theorem 9, we have

$${\int }_1^N{2}^{3t}{\diamond }_{\frac{1}{2}}t\ge ({\int }_1^N{2}^t{\diamond }_{\frac{1}{2}}{t)}^2,N\in \mathbb{N}\ge 2.$$

Noting that

$${\int }_1^{N}f\left(t\right){\diamond }_{\frac{1}{2}}t=\frac{1}{2}{\int }_1^{N}f\left(t\right)\Delta t+\frac{1}{2}{\int }_1^{N}f\left(t\right)\nabla t=\frac{1}{2}\sum _{k=1}^{N-1}f\left(k\right)+\frac{1}{2}\sum _{k=2}^{N}f\left(k\right).$$

Thereby we can arrive to inequality (30). □

Example 2.

Consider the inequality:

$$\alpha \sum _{k=2}^{N-1}{a}^{p_{1}k}+(1-\alpha )\sum _{k=3}^N{a}^{p_{1}k}\ge {\left(\alpha \sum _{k=2}^{N-1}{a}^k+(1-\alpha )\sum _{k=3}^N{a}^k\right)}^2,$$

(31)

where $1\ge \alpha \ge 0,a\ge 2,p_1\ge 2$ and $N\ge 3$.

Proof. 

We take $\mathbb{T}=\mathbb{N},\varphi \left(t\right)=a^t,p_2=2$ in (6), then it transforms into

$$a^{(p_1-1)t-2p_1+4}\ge \frac{2(\alpha +(1-\alpha \left)a\right)}{(p_1-1)(a-1)},$$

(32)

noting that $a^{(p_1-1)t-2p_1+4}\ge a^2\ge \frac{2a}{a-1}\ge \frac{2(\alpha +(1-\alpha \left)a\right)}{(p_1-1)(a-1)}$ for $t\ge 2$, so (32) holds for $t\ge 2$. Based on Theorem 10, we have

$${\int }_2^N{a}^{p_{1}t}{\diamond }_{\alpha }t\ge ({\int }_2^N{a}^t{\diamond }_{\alpha }{t)}^2,N\in \mathbb{N}\ge 3.$$

Noting that

$${\int }_2^{N}f\left(t\right){\diamond }_{\alpha }t=\alpha {\int }_2^{N}f\left(t\right)\Delta t+(1-\alpha ){\int }_2^{N}f\left(t\right)\nabla t=\alpha \sum _{k=2}^{N-1}f\left(k\right)+(1-\alpha )\sum _{k=3}^{N}f\left(k\right).$$

Thereby we can arrive at inequality (31). □

Example 3.

Consider the inequality:

$$\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}\cdots \sum _{k_n=1}^{N-1}{16}^{(k_1+k_2+\cdots +k_n)}\ge {\left(\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}\cdots \sum _{k_n=1}^{N-1}{2}^{(k_1+k_2+\cdots +k_n)}\right)}^3,$$

where $N\ge 10$.

Proof. 

Consider the condition (28) with $R={\mathbb{N}}^n,\alpha =1,\varphi (t_1,t_2,\cdots ,t_n)=2^{(t_1+t_2+\cdots +t_n)},p=4$ and ${\alpha }_i=1,{\beta }_i=N(i=1,2,\cdots ,n)$, then it change into

$$\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}\cdots \sum _{k_n=1}^{N-1}{2}^{(t_1+t_2+\cdots +t_n)}\ge {(N-1)}^{3n}.$$

(33)

Noting that ${\sum }_{k_1=1}^{N-1}{\sum }_{k_2=1}^{N-1}\cdots {\sum }_{k_n=1}^{N-1}{2}^{(t_1+t_2+\cdots +t_n)}={(2^N-2)}^n$, so (33) is hold for $N\ge 10$. Based on Theorem 24, we get the desired inequality. □

Example 4.

Consider the inequalities:

(i) 

$$\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}\cdots \sum _{k_n=1}^{N-1}{t}_1^{p^2}{t}_2^{p^2}\cdots t_n^{p^2}\ge {\left(\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}\cdots \sum _{k_n=1}^{N-1}{t}_1^p{t}_2^p\cdots t_n^p\right)}^{p-1},$$

where $N\ge 2$.

(ii) 

$$\begin{array}{ccc}\hfill & & {\int }_1^N{\int }_1^N\cdots {\int }_1^N{t}_1^{p^2}{t}_2^{p^2}\cdots t_n^{p^2}{\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\cdots {\diamond }_{\frac{1}{3}}{s}_n\hfill \\ & \ge & {\left({\int }_1^N{\int }_1^N\cdots {\int }_1^N{t}_1^p{t}_2^p\cdots t_n^p{\diamond }_{\frac{1}{3}}{s}_1{\diamond }_{\frac{1}{3}}{s}_2\cdots {\diamond }_{\frac{1}{3}}{s}_n\right)}^{p-1},\hfill \end{array}$$

(34)

where $N\ge 2$ and ${\mathbb{T}}_i=\mathbb{N}$.

Proof. 

Consider the condition (28) with $R={\mathbb{N}}^n,\varphi (t_1,t_2,\cdots ,t_n)=t_1^p{t}_2^p\cdots t_n^p$ and ${\alpha }_i=1,{\beta }_i=N(i=1,2,\cdots ,n)$, then it change into

$${\int }_{{\alpha }_1}^{{\beta }_1}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_3}^{{\beta }_3}{t}_1^p{t}_2^p\cdots t_n^p{\diamond }_{\alpha }{t}_1{\diamond }_{\alpha }{t}_2\cdots {\diamond }_{\alpha }{t}_n\ge {(N-1)}^{n(p-1)}.$$

(35)

1. If $\alpha =1$, then the left hand of (35) become

$$\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}\cdots \sum _{k_n=1}^{N-1}{k}_1^p{k}_2^p\cdots k_n^p={\left(\frac{N(N-1)}{2}\right)}^{np},$$

(36)

thus (35) hold for $N\ge 2$. Based on Theorem 24, we have the first inequality.

2. If $\alpha =\frac{1}{3}$, then the left hand of (35) is bigger than (36), thus (35) hold for $N\ge 2$. Based on Theorem 24 or Corollary 12, we have the second inequality. □

Remark 3.

We take $n=2$ in (34) for example, (34) will transforms into

$$\begin{array}{ccc}& & \frac{1}{9}\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}{k}_1^{p^2}{k}_2^{p^2}+\frac{2}{9}\sum _{k_1=1}^{N-1}\sum _{k_2=2}^N{k}_1^{p^2}{k}_2^{p^2}+\frac{2}{9}\sum _{k_1=2}^N\sum _{k_2=1}^{N-1}{k}_1^{p^2}{k}_2^{p^2}+\frac{4}{9}\sum _{k_1=2}^N\sum _{k_2=2}^N{k}_1^{p^2}{k}_2^{p^2}\hfill \\ & \ge & {\left(\frac{1}{9}\sum _{k_1=1}^{N-1}\sum _{k_2=1}^{N-1}{k}_1^p{k}_2^p+\frac{2}{9}\sum _{k_1=1}^{N-1}\sum _{k_2=2}^N{k}_1^p{k}_2^p+\frac{2}{9}\sum _{k_1=2}^N\sum _{k_2=1}^{N-1}{k}_1^p{k}_2^p+\frac{4}{9}\sum _{k_1=2}^N\sum _{k_2=2}^N{k}_1^p{k}_2^p\right)}^{p-1}.\hfill \end{array}$$

6. Conclusions

In this paper, we greatly promoted the research of Qi type inequality on time scales theory. More completely, we generalize Qi type inequality and its two generalized forms through the Diamond-Alpha integral. The sufficient condition for the reverse of Qi type inequality is also considered. Then we generalize Qi type inequality to higher dimension via Jessen’s inequality. In this method, concise conditions are deduced. Qi type high dimension inequality has been studied in great detail and some special cases are given as corollaries. Moreover, some examples are given to show our conclusions are useful in the end.