Abstract
In the paper, we establish an inverse of Beckenbach-Dresher’s integral inequality, which provides new estimates on inequality of this type.
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1 Introduction
The well-known inequality due to Beckenbach can be stated as follows (see [1], also see [2], p.27).
Theorem A
If \(1\leq p\leq2\), and \(x_{i},y_{i}>0\) for \(i=1,2,\ldots,n\), then
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
An extension of Beckenbach’s inequality was obtained by Dresher [3] 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
The inequality which we shall call Beckenbach-Dresher’s inequality. In fact, this result was also established by Danskin [4], 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 [3–12] and the references cited therein). Research of reverse Beckenbach-Dresher’s integral inequality is rare (see [13] and [14]). 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
where
and
2 Proof of theorem
Lemma 2.1
[15]
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
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
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
Notice that
and
By using Lemma 2.1, we have
with equality if and only if either
or
Therefore
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
or
Hence \(f^{\alpha}(x)=\mu g^{\beta}(x)\), where
or
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
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
By using Lemma 2.2, we obtain
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.
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
where \(\Gamma_{\alpha,\beta}(m_{1},m_{2},M_{1},M_{2})=\max\{M,N\}\),
and
Dividing both sides of (2.8) by \(\|f(x)+g(x)\|_{\alpha}^{\alpha/\beta}\), we have
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
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
Taking for
in (2.11), and in view of
and
we obtain
Hence
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
and
we have
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
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 [16], p.61
where \(m>0\), \(a_{i}\geq0\) and \(b_{i}>0\), \(i=1,2,\ldots,n\).
Proof of Theorem
Let
then
and
Let
and
From reverse Radon’s inequality (2.13) in Lemma 2.5, we have, for \(m>0\),
If \(m=\frac{r}{p-r}\), then
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,
with equality if and only if f and g are proportional, and
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. □
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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.
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C-JZ and W-SC jointly contributed to the main results. All authors read and approved the final manuscript.
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Zhao, CJ., Cheung, WS. Reverse Beckenbach-Dresher’s inequality. J Inequal Appl 2015, 153 (2015). https://doi.org/10.1186/s13660-015-0678-4
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DOI: https://doi.org/10.1186/s13660-015-0678-4