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# Reverse Beckenbach-Dresher’s inequality

Journal of Inequalities and Applications20152015:153

https://doi.org/10.1186/s13660-015-0678-4

• Received: 11 September 2014
• Accepted: 27 April 2015
• Published:

## Abstract

In the paper, we establish an inverse of Beckenbach-Dresher’s integral inequality, which provides new estimates on inequality of this type.

## Keywords

• Beckenbach’s inequality
• Beckenbach-Dresher’s inequality

• 26D15

## 1 Introduction

The well-known inequality due to Beckenbach can be stated as follows (see , also see , p.27).

### Theorem A

If $$1\leq p\leq2$$, and $$x_{i},y_{i}>0$$ for $$i=1,2,\ldots,n$$, then
$$\frac{\sum_{i=1}^{n}(x_{i}+y_{i})^{p}}{\sum_{i=1}^{n}(x_{i}+y_{i})^{p-1}}\leq\frac{\sum_{i=1}^{n} x_{i}^{p}}{\sum_{i=1}^{n}x_{i}^{p-1}}+\frac{\sum_{i=1}^{n}y_{i}^{p}}{\sum_{i=1}^{n}y_{i}^{p-1}}.$$
(1.1)

An integral analogue of Beckenbach’s inequality easily follows.

### Theorem B

Let $$1\leq p\leq2$$. If f and g are positive and continuous functions on $$[a,b]$$, then
$$\frac{\int_{a}^{b}(f(x)+g(x))^{p}\,dx}{\int_{a}^{b}(f(x)+g(x))^{p-1}\,dx}\leq\frac{\int_{a}^{b}f(x)^{p}\,dx}{\int_{a}^{b}f(x)^{p-1}\,dx}+\frac{\int_{a}^{b}g(x)^{p}\,dx}{\int_{a}^{b}g(x)^{p-1}\,dx}.$$
(1.2)

An extension of Beckenbach’s inequality was obtained by Dresher  by an ingenious method using moment-space theory.

### Theorem C

Let f and g be positive and continuous functions on $$[a,b]$$. If $$p\geq1\geq r\geq0$$, then
$$\biggl(\frac{\int_{a}^{b}(f(x)+g(x))^{p}\,dx}{\int_{a}^{b}(f(x)+g(x))^{r}\,dx} \biggr)^{1/(p-r)}\leq \biggl(\frac{\int_{a}^{b} f^{p}(x)\,dx}{\int_{a}^{b} f^{r}(x)\,dx} \biggr) ^{1/(p-r)}+ \biggl(\frac{\int_{a}^{b}g^{p}(x)\,dx}{\int_{a}^{b} g^{r}(x)\,dx} \biggr)^{1/(p-r)}.$$
(1.3)

The inequality which we shall call Beckenbach-Dresher’s inequality. In fact, this result was also established by Danskin , who employed a combination of Hölder’s and Minkowski’s inequalities.

Beckenbach-Dresher’s inequality was studied extensively and numerous variants, generalizations, and extensions appeared in the literature (see  and the references cited therein). Research of reverse Beckenbach-Dresher’s integral inequality is rare (see  and ). The aim of this paper is to discuss reverse Beckenbach-Dresher’s integral inequality and establish the following reversed Beckenbach-Dresher integral inequality by deriving reverse Hölder’s, Minkowski’s and Radon’s integral inequalities.

### Theorem

Let f and g be continuous functions on $$[a,b]$$, $$0< m_{1}\leq f(x)\leq M_{1}$$ and $$0< m_{2}\leq g(x)\leq M_{2}$$. If $$p\geq1\geq r\geq0$$, then
$$\ell\cdot \biggl(\frac{\int_{a}^{b}(f(x)+g(x))^{p}\,dx}{\int_{a}^{b}(f(x)+g(x))^{r}\,dx} \biggr)^{1/(p-r)} \geq \biggl( \frac{\int_{a}^{b} f^{p}(x)\,dx}{\int_{a}^{b} f^{r}(x)\,dx} \biggr)^{1/(p-r)} + \biggl(\frac{\int_{a}^{b} g^{p}(x)\,dx}{\int_{a}^{b} g^{r}(x)\,dx} \biggr)^{1/(p-r)},$$
(1.4)
where
\begin{aligned}& \ell=\frac{L_{\alpha,\beta}(s,t,S,T)}{\Gamma_{\alpha,\beta }(m_{1},m_{2},M_{1},M_{2})}, \quad\frac{1}{\alpha}+\frac{1}{\beta}=1, \alpha>1, \end{aligned}
(1.5)
\begin{aligned}& L_{\alpha,\beta}(s,t,S,T)= \bigl(\Upsilon_{\alpha,\beta} \bigl(sT^{-\frac {m}{m+1}}, \bigl(stS^{-1}\bigr)^{\frac{m}{m+1}}, St^{-\frac{m}{m+1}}, \bigl(s^{-1}ST\bigr)^{\frac{m}{m+1}} \bigr) \bigr)^{m+1},\quad m>0, \end{aligned}
(1.6)
\begin{aligned}& s=\min\bigl\{ m_{1}(b-a)^{1/p},m_{2}(b-a)^{1/p} \bigr\} ,\qquad S=\max\bigl\{ M_{1}(b-a)^{1/p},M_{2}(b-a)^{1/p} \bigr\} , \\& t=\min\bigl\{ m_{1}(b-a)^{1/r},m_{2}(b-a)^{1/r} \bigr\} ,\qquad T=\max\bigl\{ M_{1}(b-a)^{1/r},M_{2}(b-a)^{1/r} \bigr\} , \\& \begin{aligned}[b] \Upsilon_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})={}& \max \biggl\{ C_{\alpha ,\beta} \biggl(\frac{M_{1}^{\alpha}}{m_{1}^{\alpha}(b-a)},\frac {m_{2}^{\beta}}{M_{2}^{\beta}(b-a)} \biggr),\\ &{} C_{\alpha,\beta} \biggl(\frac{m_{1}^{\alpha}}{M_{1}^{\alpha}(b-a)},\frac {M_{2}^{\beta}}{m_{2}^{\beta}(b-a)} \biggr) \biggr\} , \end{aligned} \end{aligned}
(1.7)
\begin{aligned}& C_{\alpha,\beta}(\xi,\eta)=\frac{\xi/\alpha+\eta/\beta}{\xi^{1/\alpha }\eta^{1/\beta}}, \end{aligned}
(1.8)
and
\begin{aligned} &\Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2}) \\ &\quad=\max \bigl\{ \Upsilon_{\alpha,\beta} \bigl(m_{1},(m_{1}+m_{2})^{\alpha -1},M_{1},(M_{1}+M_{2})^{\alpha-1} \bigr), \\ &\qquad{}\Upsilon_{\alpha,\beta} \bigl(m_{2},(m_{1}+m_{2})^{\alpha -1},M_{2},(M_{1}+M_{2})^{\alpha-1} \bigr) \bigr\} . \end{aligned}
(1.9)

## 2 Proof of theorem

### Lemma 2.1



If $$0< m_{1}\leq a\leq M_{1}$$, $$0< m_{2}\leq b\leq M_{2}$$, $$\frac{1}{\alpha}+\frac{1}{\beta}=1$$ and $$\alpha>1$$, then
$$\max\bigl\{ C_{\alpha,\beta}(M_{1},m_{2}),C_{\alpha,\beta}(m_{1},M_{2}) \bigr\} \cdot \alpha\beta a^{1/\alpha}b^{1/\beta}\geq a\beta+b \alpha,$$
(2.1)
with equality if and only if either $$(a,b)=(m_{1},M_{2})$$ or $$(a,b)=(M_{1},m_{2})$$, where $$C_{\alpha,\beta}(\xi,\eta)$$ is as in (1.8).

Obviously, by using a way similar to the proof of (2.1), we may find that inequality (2.1) is reversed if $$0<\alpha<1$$ or $$\alpha<0$$. Here, we omit the details.

### Lemma 2.2

Let f and g be positive continuous functions on $$[a,b]$$, $$\frac{1}{\alpha}+\frac{1}{\beta}=1$$, $$\alpha>1$$ and $$f^{\alpha}$$ and $$g^{\beta}$$ be integrable on $$[a,b]$$. If $$0< m_{1}\leq f(x)\leq M_{1}$$ and $$0< m_{2}\leq g(x)\leq M_{2}$$, then
$$\biggl(\int_{a}^{b}f^{\alpha}(x)\,dx \biggr)^{1/\alpha} \biggl(\int_{a}^{b}g^{\beta}(x)\,dx \biggr)^{1/\beta} \leq \Upsilon_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2}) \cdot\int_{a}^{b}f(x)g(x)\,dx,$$
(2.2)
with equality if and only if $$f^{\alpha}$$ and $$g^{\beta}$$ are proportional, where $$\Upsilon_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})$$ is as in (1.7).

The inequality is reversed if $$0<\alpha<1$$ or $$\alpha<0$$.

### Proof

If we set successively
\begin{aligned}& \bar{a}=\frac{f^{\alpha}(x)}{X},\qquad X=\int_{a}^{b}f^{\alpha}(x)\,dx, \\& \bar{b}=\frac{g^{\beta}(x)}{Y},\qquad Y=\int_{a}^{b}g^{\beta}(x)\,dx. \end{aligned}
Notice that
$$\frac{m_{1}^{\alpha}}{M_{1}^{\alpha}(b-a)}\leq\bar{a}\leq\frac {M_{1}^{\alpha}}{m_{1}^{\alpha}(b-a)},$$
and
$$\frac{m_{2}^{\beta}}{M_{2}^{\beta}(b-a)}\leq\bar{b}\leq\frac {M_{2}^{\beta}}{m_{2}^{\beta}(b-a)}.$$
By using Lemma 2.1, we have
\begin{aligned} &\max \biggl\{ C_{\alpha,\beta} \biggl(\frac{M_{1}^{\alpha}}{m_{1}^{\alpha }(b-a)},\frac{m_{2}^{\beta}}{M_{2}^{\beta}(b-a)} \biggr), C_{\alpha,\beta} \biggl(\frac{m_{1}^{\alpha}}{M_{1}^{\alpha}(b-a)},\frac {M_{2}^{\beta}}{m_{2}^{\beta}(b-a)} \biggr) \biggr\} \cdot\frac{f(x)g(x)}{X^{1/\alpha}Y^{1/\beta}} \\ &\quad\geq\frac{1}{\alpha}\frac{f^{\alpha}(x)}{X}+\frac{1}{\beta} \frac {g^{\beta}(x)}{Y}, \end{aligned}
with equality if and only if either
$$(\bar{a},\bar{b})= \biggl(\frac{m_{1}^{\alpha}}{M_{1}^{\alpha}(b-a)},\frac {M_{2}^{\beta}}{m_{2}^{\beta}(b-a)} \biggr)$$
or
$$(\bar{a},\bar{b})= \biggl(\frac{M_{1}^{\alpha}}{m_{1}^{\alpha}(b-a)},\frac {m_{2}^{\beta}}{M_{2}^{\beta}(b-a)} \biggr).$$
Therefore
$$\Upsilon_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2}) \cdot\frac{\int_{a}^{b}f(x)g(x)\,dx}{X^{1/\alpha}Y^{1/\beta}}\geq\frac{1}{\alpha} \frac{\int_{a}^{b}f^{\alpha}(x)\,dx}{X}+ \frac{1}{\beta} \frac{\int_{a}^{b}g^{\beta}(x)\,dx}{Y}=1.$$
(2.3)
From (2.3), inequality (2.2) easily follows.
In the following, we discuss the equality condition of (2.2). In view of the equality conditions of Lemma 2.1, the equality in (2.3) holds if and only if
$$\biggl(\frac{f^{\alpha}(x)}{\int_{a}^{b}f^{\alpha}(x)\,dx},\frac{g^{\beta }(x)}{\int_{a}^{b}g^{\beta}(x)\,dx} \biggr)= \biggl(\frac{m_{1}^{\alpha }}{M_{1}^{\alpha}(b-a)}, \frac{M_{2}^{\beta}}{m_{2}^{\beta}(b-a)} \biggr),$$
or
$$\biggl(\frac{f^{\alpha}(x)}{\int_{a}^{b}f^{\alpha}(x)\,dx},\frac{g^{\beta }(x)}{\int_{a}^{b}g^{\beta}(x)\,dx} \biggr)= \biggl(\frac{M_{1}^{\alpha }}{m_{1}^{\alpha}(b-a)}, \frac{m_{2}^{\beta}}{M_{2}^{\beta}(b-a)} \biggr).$$
Hence $$f^{\alpha}(x)=\mu g^{\beta}(x)$$, where
$$\mu=\frac{m_{1}^{\alpha}m_{2}^{\beta}}{M_{2}^{\beta}M_{1}^{\alpha}}\frac {\|f\|_{\alpha}^{\alpha}}{\|g\|_{\beta}^{\beta}},$$
or
$$\mu=\frac{M_{1}^{\alpha}M_{2}^{\beta}}{m_{2}^{\beta}m_{1}^{\alpha}}\frac {\|f\|_{\alpha}^{\alpha}}{\|g\|_{\beta}^{\beta}}$$
is a constant. It follows that the equality in (2.2) holds if and only if $$f^{\alpha}$$ and $$g^{\beta}$$ are proportional.

This proof is completed. □

### Lemma 2.3

Let f and g be non-negative continuous functions on $$[a,b]$$. If $$0< m_{1}\leq f(x)\leq M_{1}$$, $$0< m_{2}\leq g(x)\leq M_{2}$$ and $$\alpha>1$$, then
\begin{aligned} &\biggl(\int_{a}^{b} \bigl(f(x)+g(x) \bigr)^{\alpha}\,dx \biggr)^{1/\alpha} \\ &\quad\geq \Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2}) \biggl( \biggl(\int_{a}^{b}f^{\alpha}(x)\,dx \biggr)^{1/\alpha} + \biggl(\int_{a}^{b}g^{\alpha}(x)\,dx \biggr)^{1/\alpha} \biggr), \end{aligned}
(2.4)
with equality if and only if f and g are proportional, where $$\Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})$$ is as in (1.9).

The inequality is reversed if $$0<\alpha<1$$ or $$\alpha<0$$.

### Proof

From the hypotheses, we have
$$\bigl\| f(x)+g(x) \bigr\| ^{\alpha}_{\alpha}= \bigl\| f(x)\bigl[f(x)+g(x) \bigr]^{\alpha -1} \bigr\| _{1}+ \bigl\| g(x)\bigl[f(x)+g(x) \bigr]^{\alpha-1} \bigr\| _{1}.$$
(2.5)
By using Lemma 2.2, we obtain
\begin{aligned} \bigl\| f(x)\bigl[f(x)+g(x)\bigr]^{\alpha-1} \bigr\| _{1} \geq{}& \bigl[\Upsilon_{\alpha ,\beta} \bigl(m_{1},(m_{1}+m_{2})^{\alpha-1},M_{1},(M_{1}+M_{2})^{\alpha -1} \bigr) \bigr]^{-1} \\ &{}\times\bigl\| f(x)\bigr\| _{\alpha}\cdot\bigl\| f(x)+g(x)\bigr\| _{\alpha}^{\alpha/\beta}, \end{aligned}
(2.6)
with equality if and only if $$f^{\alpha}(x)$$ and $$(f(x)+g(x))^{\alpha}$$ are proportional. It follows that the equality holds if and only if $$f(x)$$ and $$g(x)$$ are proportional.
\begin{aligned}[b] \bigl\| g(x)\bigl[f(x)+g(x)\bigr]^{\alpha-1} \bigr\| _{1} \geq{}& \bigl[\Upsilon_{\alpha ,\beta} \bigl(m_{2},(m_{1}+m_{2})^{\alpha-1},M_{2},(M_{1}+M_{2})^{\alpha -1} \bigr) \bigr]^{-1} \\ &{}\times\bigl\| g(x)\bigr\| _{\alpha}\cdot \bigl\| f(x)+g(x)\bigr\| _{\alpha}^{\alpha/\beta}, \end{aligned}
(2.7)
with equality if and only if $$g^{\alpha}(x)$$ and $$(f(x)+g(x))^{\alpha}$$ are proportional. It follows that the equality holds if and only if $$f(x)$$ and $$g(x)$$ are proportional. Hence
$$\bigl\| f(x)+g(x)\bigr\| _{\alpha}^{\alpha}\geq\Gamma_{\alpha,\beta }(m_{1},m_{2},M_{1},M_{2}) \cdot\bigl\| f(x)+g(x)\bigr\| _{\alpha}^{\alpha/\beta} \bigl(\bigl\| f(x)\bigr\| _{\alpha}+ \bigl\| g(x)\bigr\| _{\alpha} \bigr),$$
(2.8)
where $$\Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})=\max\{M,N\}$$,
$$M=\Upsilon_{\alpha,\beta} \bigl(m_{1},(m_{1}+m_{2})^{\alpha -1},M_{1},(M_{1}+M_{2})^{\alpha-1} \bigr),$$
and
$$N=\Upsilon_{\alpha,\beta} \bigl(m_{2},(m_{1}+m_{2})^{\alpha -1},M_{2},(M_{1}+M_{2})^{\alpha-1} \bigr).$$
Dividing both sides of (2.8) by $$\|f(x)+g(x)\|_{\alpha}^{\alpha/\beta}$$, we have
$$\bigl\| f(x)+g(x)\bigr\| _{\alpha}\geq\Gamma_{\alpha,\beta }(m_{1},m_{2},M_{1},M_{2}) \cdot \bigl(\bigl\| f(x)\bigr\| _{\alpha}+\bigl\| g(x)\bigr\| _{\alpha } \bigr).$$
(2.9)
Moreover, in view of the equality conditions of (2.6) and (2.7), it follows that the equality in (2.4) holds if and only if $$f(x)$$ and $$g(x)$$ are proportional.

This proof is completed. □

### Lemma 2.4

Let f and g be continuous functions on $$[a,b]$$, $$0< m_{1}\leq f(x)\leq M_{1}$$ and $$0< m_{2}\leq g(x)\leq M_{2}$$. If $$m>0$$, then
$$\int_{a}^{b}\frac{f^{m+1}(x)}{g^{m}(x)}\,dx\leq L_{\alpha,\beta }(m_{1},m_{2},M_{1},M_{2}) \frac{ ( \int_{a}^{b}f(x)\,dx )^{m+1}}{ (\int_{a}^{b}g(x)\,dx )^{m}},$$
(2.10)
where $$L_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})$$ is as in (1.6).

### Proof

Let $$\alpha=m+1$$, $$\beta=(m+1)/m$$ and replacing $$f(x)$$ and $$g(x)$$ by $$u(x)$$ and $$v(x)$$ in (2.2), respectively, we have
\begin{aligned} &\biggl(\int_{a}^{b}u(x)^{m+1}\,dx \biggr)^{1/(m+1)} \biggl(\int_{a}^{b}v(x)^{(m+1)/m}\,dx \biggr)^{m/(m+1)} \\ &\quad\leq\Upsilon_{\alpha ,\beta}(m_{1},m_{2},M_{1},M_{2}) \cdot \int_{a}^{b}u(x)v(x)\,dx. \end{aligned}
(2.11)
Taking for
$$u(x)= \biggl(\frac{f(x)}{g(x)} \biggr)^{1/(m+1)},\qquad v(x)=f^{m/(m+1)}(x)g^{1/(m+1)}(x)$$
in (2.11), and in view of
$$\biggl(\frac{m_{1}}{M_{2}} \biggr)^{\frac{1}{m+1}}\leq u(x)\leq \biggl( \frac {M_{1}}{m_{2}} \biggr)^{\frac{1}{m+1}}$$
and
$$m_{1}^{\frac{m}{m+1}}m_{2}^{\frac{1}{m+1}}\leq v(x)\leq M_{1}^{\frac {m}{m+1}}M_{2}^{\frac{1}{m+1}},$$
we obtain
\begin{aligned} &\Upsilon_{\alpha,\beta} \bigl( \bigl(m_{1}M_{2}^{-1} \bigr)^{\frac {1}{m+1}},m_{1}^{\frac{m}{m+1}}m_{2}^{\frac{1}{m+1}}, \bigl(M_{1}m_{2}^{-1} \bigr)^{\frac{1}{m+1}}, M_{1}^{\frac{m}{m+1}}M_{2}^{\frac{1}{m+1}} \bigr)\int _{a}^{b} f(x)\,dx \\ &\quad\geq \biggl(\int_{a}^{b} \frac{f(x)}{g(x)}\,dx \biggr)^{1/(m+1)} \biggl(\int_{a}^{b}f(x)g^{1/m}(x)\,dx \biggr)^{m/(m+1)}. \end{aligned}
Hence
\begin{aligned} &\int_{a}^{b}\frac{f(x)}{g(x)}\,dx \\ &\quad\leq \frac{ [\Upsilon_{\alpha,\beta } ( (m_{1}M_{2}^{-1} )^{\frac{1}{m+1}},m_{1}^{\frac {m}{m+1}}m_{2}^{\frac{1}{m+1}}, (M_{1}m_{2}^{-1} )^{\frac{1}{m+1}}, M_{1}^{\frac{m}{m+1}}M_{2}^{\frac{1}{m+1}} )\int_{a}^{b}f(x)\,dx ]^{m+1}}{ (\int_{a}^{b}f(x)g^{1/m}(x)\,dx )^{m}}. \end{aligned}
(2.12)
On the other hand, in (2.12), replacing $$f(x)$$ and $$g(x)$$ by $$u(x)$$ and $$v(x)$$, respectively, and letting $$u(x)=f(x)$$ and $$v(x)= (\frac{g(x)}{f(x)} )^{m}$$, and in view of
$$m_{1}\leq u(x)\leq M_{1}$$
and
$$\biggl(\frac{m_{2}}{M_{1}} \biggr)^{m}\leq v(x)\leq \biggl( \frac {M_{2}}{m_{1}} \biggr)^{m},$$
we have
\begin{aligned} &\int_{a}^{b}\frac{f^{m+1}(x)}{g^{m}(x)}\,dx \\ &\quad\leq\frac{ [\Upsilon_{\alpha,\beta} (m_{1}M_{2}^{-\frac {m}{m+1}},(m_{1}m_{2}M_{1}^{-1})^{\frac{m}{m+1}}, M_{1}m_{2}^{-\frac{m}{m+1}}, (m_{1}^{-1}M_{1}M_{2})^{\frac{m}{m+1}} )\int_{a}^{b}f(x)\,dx ]^{m+1}}{ (\int_{a}^{b}g(x)\,dx )^{m}}\\ &\quad=\frac{ L_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2}) (\int_{a}^{b}f(x)\,dx )^{m+1}}{ (\int_{a}^{b}g(x)\,dx )^{m}}. \end{aligned}

This proof is completed. □

Let $$f(x)$$ and $$g(x)$$ reduce to positive real sequences $$a_{i}$$ and $$b_{i}$$ ($$i=1,\ldots,n$$), respectively, and with appropriate changes in the proof of (2.10), we have the following.

### Lemma 2.5

Let $$a_{i}$$ and $$b_{i}$$ be positive real sequences and $$0< m_{1}\leq a_{i}\leq M_{1}$$, $$0< m_{2}\leq b_{i}\leq M_{2}$$, $$i=1,\ldots,n$$. If $$m>0$$, then
$$\sum_{i=1}^{n}\frac{a_{i}^{m+1}}{b_{i}^{m}}\leq L_{\alpha,\beta }(m_{1},m_{2},M_{1},M_{2}) \frac{ ( \sum_{i=1}^{n}a_{i} )^{m+1}}{ (\sum_{i=1}^{n}b_{i} )^{m}},$$
(2.13)
where $$L_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})$$ is as in Lemma 2.4.
This is just an inverse of the following well-known Radon’s inequality , p.61
$$\sum_{i=1}^{n}\frac{a_{i}^{m+1}}{b_{i}^{m}}\geq \frac{ (\sum_{i=1}^{n} a_{i} )^{m+1}}{ (\sum_{i=1}^{n}b_{i} )^{m}},$$
where $$m>0$$, $$a_{i}\geq0$$ and $$b_{i}>0$$, $$i=1,2,\ldots,n$$.

### Proof of Theorem

Let
\begin{aligned}& \alpha_{1}= \biggl(\int_{a}^{b} f^{p}(x)\,dx \biggr)^{1/p}, \qquad \beta_{1}= \biggl( \int_{a}^{b} f^{r}(x)\,dx \biggr)^{1/r}, \\& \alpha_{2}= \biggl(\int _{a}^{b} g^{p}(x)\,dx \biggr)^{1/p}, \qquad \beta_{2}= \biggl(\int_{a}^{b} g^{r}(x)\,dx \biggr)^{1/r}, \end{aligned}
then
\begin{aligned}& 0< m_{1}(b-a)^{1/p}\leq\alpha_{1}\leq M_{1}(b-a)^{1/p}, \\& 0<m_{2}(b-a)^{1/p}\leq\alpha_{2}\leq M_{2}(b-a)^{1/p}, \\& 0<m_{1}(b-a)^{1/r}\leq\beta_{1}\leq M_{1}(b-a)^{1/r}, \end{aligned}
and
$$0< m_{2}(b-a)^{1/r}\leq\beta_{2}\leq M_{2}(b-a)^{1/r}.$$
Let
$$s=\min\bigl\{ m_{1}(b-a)^{1/p},m_{2}(b-a)^{1/p} \bigr\} ,\qquad S=\max\bigl\{ M_{1}(b-a)^{1/p},M_{2}(b-a)^{1/p} \bigr\}$$
and
$$t=\min\bigl\{ m_{1}(b-a)^{1/r},m_{2}(b-a)^{1/r} \bigr\} ,\qquad T=\max\bigl\{ M_{1}(b-a)^{1/r},M_{2}(b-a)^{1/r} \bigr\} .$$
From reverse Radon’s inequality (2.13) in Lemma 2.5, we have, for $$m>0$$,
$$\frac{\alpha_{1}^{m+1}}{\beta_{1}^{m}}+\frac{\alpha_{2}^{m+1}}{\beta_{2}^{m}} \leq L_{\alpha,\beta}(s,t,S,T) \frac{(\alpha_{1}+\alpha_{2})^{m+1}}{(\beta _{1}+\beta_{2})^{m}}.$$
(2.14)
If $$m=\frac{r}{p-r}$$, then
\begin{aligned}[b] & \biggl(\frac{\int f^{p}(x)\,dx}{\int f^{r}(x)\,dx} \biggr)^{1/(p-r)} + \biggl( \frac{\int g^{p}(x)\,dx}{\int g^{r}(x)\,dx} \biggr)^{1/(p-r)}\\ &\quad\leq L_{\alpha,\beta}(s,t,S,T)\frac{ [ (\int f^{p}(x)\,dx )^{1/p}+ (\int g^{p}(x)\,dx ) ^{1/p} ]^{p/(p-r)}}{ [ (\int f^{r}(x)\,dx )^{1/r}+ (\int g^{r}(x)\,dx )^{1/r} ]^{r/(p-r)}}. \end{aligned}
(2.15)

We have assumed $$p>r>0$$, since $$m=\frac{r}{p-r}>0$$.

On the other hand, by using the Minkowski inequality (2.4) and its reverse form, with $$p\geq1$$ and $$0< r\leq1$$, respectively,
\begin{aligned} &\Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})^{p} \biggl[ \biggl(\int f^{p}(x)\,dx \biggr)^{1/p}+ \biggl(\int g^{p}(x)\,dx \biggr)^{1/p} \biggr]^{p} \\ &\quad\leq\int \bigl(f(x)+g(x)\bigr)^{p}\,dx, \end{aligned}
(2.16)
with equality if and only if f and g are proportional, and
\begin{aligned} &\Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})^{r} \biggl[ \biggl(\int f^{r}(x)\,dx \biggr)^{1/r}+ \biggl(\int g^{r}(x)\,dx \biggr)^{1/r} \biggr]^{r} \\ &\quad\geq\int \bigl(f(x)+g(x)\bigr)^{r}(x)\,dx, \end{aligned}
(2.17)
with equality if and only if f and g are proportional.

From (2.15), (2.16) and (2.17), (1.4) follows. This proof is completed. □

## Declarations

### Acknowledgements

The first author’s research is supported by the Natural Science Foundation of China (11371334). The second author’s research is partially supported by a HKU Seed Grant for Basic Research. The authors express their grateful thanks to the two referees for their excellent suggestions and comments. 