Abstract
By means of the techniques of real analysis, applying some basic inequalities and formulas, a new reverse Mulholland’s inequality with one partial sum in the kernel is given. We obtain the equivalent conditions of the parameters related to the best value in the new inequality. As applications, we reduce to the equivalent forms and a few inequalities for particular parameters.
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1 Introduction
If \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(a_{m},b_{n} \ge 0\) are such that \(0 < \sum_{m = 1}^{\infty} a_{m}^{p} < \infty\) and \(0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty \), then we have the following Hardy–Hilbert’s inequality with the best value \(\frac{\pi}{\sin (\pi /p)}\) (cf. [1, Theorem 315]):
With regards to a similar assumption, the well-known Mulholland’s inequality was given as follows (cf. [1, Theorem 343]):
For \(\lambda _{i} \in (0,2]\) (\(i = 1,2\)), \(\lambda _{1} + \lambda _{2} = \lambda \in (0,4]\), a generalization of (1) was obtained (see [2]) in 2016 as follows:
where the constant \(B(\lambda _{1},\lambda _{2})\) is the best value and
is the Beta function related to the Gamma function. For \(\lambda = 1\), \(\lambda _{1} = \frac{1}{q}\), \(\lambda _{2} = \frac{1}{p}\), (3) reduces to (1); for \(p = q = 2\), \(\lambda _{1} = \lambda _{2} = \frac{\lambda}{2}\), (3) reduces to an inequality published in [3].
In 2019, by means of (3), Adiyasuren et al. [4] gave a generalization of (3) as follows: For \(\lambda _{i} \in (0,1] \cap (0,\lambda )\) (\(\lambda \in (0,2]\); \(i = 1,2\)), \(\lambda _{1} + \lambda _{2} = \lambda \), \(a_{m},b_{n} \ge 0\), we have
where \(\lambda _{1}\lambda _{2}B(\lambda _{1},\lambda _{2})\) is the best value, and two partial sums \(A_{m}: = \sum_{i = 1}^{m} a_{i}\) and \(B_{n}: = \sum_{k = 1}^{n} b_{k}\) (\(m,n \in \{ 1,2, \ldots \} \)) satisfy
Some generalizations of (1) and (2) were obtained in [5–15]. In 2021, Gu and Yang [16] gave an improvement of (4) with the kernel \(\frac{1}{(m^{\alpha} + n^{\beta} )^{\lambda}} \). But we find that the constant is not the best possible unless \(\alpha = \beta = 1\). In 2016, Hong et al. [17] gave a few equivalent conditions of the parameters related to the best value in the general form of (1). Some further works were provided in [18–29].
In this article, following the methods of [16, 17], by means of the techniques of analysis, several basic inequalities and formulas, a new reverse Mulholland’s inequality with one partial sum in the kernel is given. The equivalent conditions of the parameters related to the best value in the new inequality are obtained. We also deduce the equivalent forms and a few equivalent inequalities for particular parameters.
2 Some lemmas
In what follows, we assume that \(0 < p < 1\) (\(q < 0\)), \(\frac{1}{p} + \frac{1}{q} = 1\), \(\lambda > 0\), \(\lambda _{i} \in (0,2] \cap (0,\lambda )\) (\(i = 1,2\)), \(\hat{\lambda}_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} \), \(\hat{\lambda}_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\), \(\mathrm{N} = \{ 1,2, \ldots \}\), \(m,n \in \mathrm{N}\backslash \{ 1\}\), \(a_{m},b_{n} \ge 0\), \(A_{m}: = \sum_{k = 2}^{m} a_{k} = o(e^{t\ln m})\) (\(t > 0\); \(m \to \infty \)), and
Lemma 1
(cf. [5, (2.2.3)])
(i) If \(( - 1)^{i}\frac{d^{i}}{dt^{i}}h(t) > 0\), \(t \in [m,\infty )\) (\(m \in \mathrm{N}\)), \(h^{(i)}(\infty ) = 0\) (\(i = 0,1,2,3\)), \(P_{i}(t)\), \(B_{i}\) (\(i \in \mathrm{N}\)) are the Bernoulli functions and numbers, then
For \(q = 1\), \(B_{2} = \frac{1}{6}\), we have
If \(( - 1)^{i}\frac{d^{i}}{dt^{i}}h(t) > 0\), \(t \in [m,\infty )\), \(h^{(i)}(\infty ) = 0\) (\(i = 0,1\)), then we still have (cf. [5, (2.2.13)])
(ii) (cf. [5, (2.1.14)]) If \(n > m \in \mathrm{N}\), \(f(t)( > 0) \in C^{1}[m,\infty )\), \(f^{(i)}(\infty ) = 0\) (\(i = 0,1\)), then the following Euler–Maclaurin summation formulas are valid:
Lemma 2
If \(s > 0\), \(s_{2} \in (0,2] \cap (0,s)\), \(K_{s}(s_{2}): = B(s_{2},s - s_{2})\), and the weight coefficient is defined as follows:
then we have the following inequalities:
where \(O(\frac{1}{\ln ^{s_{2}}m}): = \frac{1}{k_{s}(s_{2})}\int _{0}^{\frac{\ln 2}{\ln m}} \frac{v^{s_{2} - 1}}{(1 + v)^{s}}\,dv > 0 \).
Proof
For a fixed \(m \in \mathrm{N}\backslash \{ 1\} \), we set \(g_{m}(t)\) as follows: \(g_{m}(t): = \frac{\ln ^{s_{2} - 1}t}{(\ln m + \ln t)^{s}t}\) (\(t > 1\)).
-
(i)
For \(s_{2} \in (0,1] \cap (0,s)\), in view of the decreasingness property of series, we have
$$ \int _{2}^{\infty} g_{m}(t)\,dt < \sum _{n = 2}^{\infty} g_{m}(n) < \int _{1}^{\infty} g_{m}(t)\,dt. $$(12) -
(ii)
For \(s_{2} \in (1,2] \cap (0,s)\), in view of (9), we have
$$\begin{aligned}& \begin{aligned} \sum_{n = 2}^{\infty} g_{m}(n)& = \int _{2}^{\infty} g_{m}(t)\,dt + \frac{1}{2} g_{m}(2) + \int _{2}^{\infty} P{}_{1}(t)g'_{m}(t)\,dt= \int _{1}^{\infty} g_{m}(t)\,dt - h(m), \end{aligned} \\& h(m): = \int _{1}^{2} g_{m}(t)\,dt - \frac{1}{2}g_{m}(2) - \int _{2}^{\infty} P_{1}(t)g'_{m}(t)\,dt. \end{aligned}$$We obtain \(- \frac{1}{2}g_{m}(2) = \frac{ - \ln ^{s_{2} - 1}2}{4(\ln m + \ln 2)^{s}}\). Setting \(v = \ln t\) and using integration by parts, we find
$$\begin{aligned} \int _{1}^{2} g_{m} (t)\,dt &= \int _{0}^{\ln 2} \frac{v^{s_{2} - 1}}{(\ln m + v)^{s}}\,dv = \int _{0}^{\ln 2} \frac{dv^{s_{2}}}{s_{2}(\ln m + v)^{s}} \\ &= \frac{v^{s_{2}}}{s_{2}(\ln m + v)^{s}}|_{0}^{\ln 2} - \int _{0}^{\ln 2} \frac{v^{s_{2}}}{s_{2}}\,d \frac{1}{(\ln m + v)^{s}} \\ &= \frac{\ln ^{s_{2}}2}{s_{2}(\ln m + \ln 2)^{s}} + \frac{s}{s_{2}(s_{2} + 1)} \int _{0}^{\ln 2} \frac{dv^{s_{2} + 1}}{(\ln m + v)^{s + 1}} \\ &> \frac{\ln ^{s_{2}}2}{s_{2}(\ln m + \ln 2)^{s}} + \frac{s}{s_{2}(s_{2} + 1)}\frac{\ln ^{s_{2} + 1}2}{(\ln m + \ln 2)^{s + 1}}. \end{aligned}$$Since \(\frac{\ln t}{t^{2}} > 0\), \((\frac{\ln t}{t^{2}})' = \frac{1 - 2\ln t}{t^{3}} < 0\) (\(t > 2\)), by (7), we have
$$\begin{aligned}& g'_{m}(t) = \frac{(s_{2} - 1)\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s}t^{2}} - \frac{s\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s + 1}}\frac{\ln t}{t^{2}} - \frac{\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s}}\frac{\ln t}{t^{2}}, \\& - \int _{2}^{\infty} P_{1}(t) \frac{(s_{2} - 1)\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s}t^{2}}\,dt \\& \quad = (1 - s_{2}) \int _{2}^{\infty} P_{1}(t) \frac{\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s}t^{2}}\,dt > 0 \quad (s_{2} \in (1,2]), \\& \int _{2}^{\infty} P_{1}(t)\biggl[ \frac{s\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s + 1}}\frac{\ln t}{t^{2}} + \frac{\ln ^{s_{2}2}t}{(\ln m + \ln t)^{s}} \frac{\ln t}{t^{2}}\biggr]\,dt\\& \quad > - \frac{1}{8}\biggl[ \frac{s\ln ^{s_{2} - 1}2}{4(\ln m + \ln 2)^{s + 1}} + \frac{\ln ^{s_{2} - 1}2}{4(\ln m + \ln 2)^{s}}\biggr], \\& - \int _{2}^{\infty} P_{1}(t)g'_{m}(t)\,dt > - \frac{s\ln ^{s_{2} - 1}2}{32(\ln m + \ln 2)^{s + 1}} - \frac{\ln ^{s_{2} - 1}2}{32(\ln m + \ln 2)^{s}}. \end{aligned}$$Hence, for \(s_{2} \in (1,2] \cap (0,s)\), we obtain
$$\begin{aligned} h(m) >{}& \frac{\ln ^{s_{2}}2}{s_{2}(\ln m + \ln 2)^{s}} + \frac{s}{s_{2}(s_{2} + 1)}\frac{\ln ^{s_{2} + 1}2}{(\ln m + \ln 2)^{s + 1}}\\ &{}- \frac{\ln ^{s_{2} - 1}2}{4(\ln m + \ln 2)^{s}} - \frac{s\ln ^{s_{2} - 1}2}{32(\ln m + \ln 2)^{s + 1}} - \frac{\ln ^{s_{2} - 1}2}{32(\ln m + \ln 2)^{s}}\\ ={}& \frac{\ln ^{s_{2} - 1}2}{(\ln m + \ln 2)^{s}}\biggl(\frac{\ln 2}{s_{2}} - \frac{1}{4} - \frac{1}{32}\biggr) + \frac{s\ln ^{s_{2} - 1}2}{(\ln m + \ln 2)^{s + 1}} \biggl[\frac{\ln ^{2}2}{s_{2}(s_{2} + 1)} - \frac{1}{32}\biggr] \\ \ge{}& \frac{\ln ^{s_{2} - 1}2}{(\ln m + \ln 2)^{s}}\biggl(\frac{\ln 2}{2} - \frac{9}{32} \biggr) + \frac{s\ln ^{s_{2} - 1}2}{(\ln m + \ln 2)^{s + 1}}\biggl(\frac{\ln ^{2}2}{6} - \frac{1}{32}\biggr)\\ >{}& 0\quad \bigl(\ln 2 = 0.6931^{ +} \bigr). \end{aligned}$$Therefore, we have \(h(m) > 0\). We still have
$$\begin{aligned}& \begin{aligned} \sum_{n = 2}^{\infty} g_{m}(n) &= \int _{2}^{\infty} g_{m}(t)\,dt + \frac{1}{2} g_{m}(2) + \int _{2}^{\infty} P_{1}(t)g'_{m}(t)\,dt \\ &= \int _{2}^{\infty} g_{m}(t)\,dt + h_{1}(m), \end{aligned} \\& h_{1}(m): = \frac{1}{2}g_{m}(2) + \int _{2}^{\infty} P_{1}(t)g'_{m}(t)\,dt. \end{aligned}$$For \(s_{2} \in (1,2] \cap (0,s)\), in view of (7), we find
$$\begin{aligned}& \int _{2}^{\infty} P_{1}(t) \frac{(s_{2} - 1)\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s}t^{2}}\,dt > - \frac{s_{2} - 1}{32}\frac{\ln ^{s_{2} - 2}2}{(\ln m + \ln 2)^{s}},\\& - \int _{2}^{\infty} P_{1}(t)\biggl[ \frac{s\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s + 1}}\frac{\ln t}{t^{2}} + \frac{s\ln ^{s_{2} - 2}t}{(\ln m + \ln t)^{s}} \frac{\ln t}{t^{2}}\biggr]\,dt > 0,\\& \begin{gathered} \int _{2}^{\infty} P_{1}(t)g'_{m}(t)\,dt > - \frac{s_{2} - 1}{32}\frac{\ln ^{s_{2} - 2}2}{(\ln m + \ln 2)^{s}}, \\ h_{1}(m) > \frac{\ln ^{s_{2} - 1}2}{4(\ln m + \ln 2)^{s}} - \frac{s_{2} - 1}{32} \frac{\ln ^{s_{2} - 2}2}{(\ln m + \ln 2)^{s}} \ge \frac{\ln ^{s_{2} - 2}2}{4(\ln m + \ln 2)^{s}}\biggl(\ln 2 - \frac{1}{8}\biggr) > 0. \end{gathered} \end{aligned}$$Hence, we have (12).
-
(iii)
For \(s_{2} \in (0,2] \cap (0,s)\), by (12), setting \(v = \frac{\ln t}{\ln m}\), it follows that
$$\begin{aligned}& \begin{aligned} \varpi _{s}(s_{2},m) &= \ln ^{s - s_{2}}m\sum_{n = 2}^{\infty} g_{m}(n) < \ln ^{s - s_{2}}m \int _{1}^{\infty} g_{m}(t)\,dt \\ &= \int _{0}^{\infty} \frac{v^{s_{2} - 1}\,dv}{(1 + v)^{s}} = B(s_{2},s - s_{2}) = k_{s}(s_{2}), \end{aligned}\\& \varpi _{s}(s_{2},m) > \ln ^{s - s_{2}}m \int _{2}^{\infty} g_{m}(t)\,dt = \int _{\frac{\ln 2}{\ln m}}^{\infty} \frac{v^{s_{2} - 1}\,dv}{(1 + v)^{s}} = k_{s}(s_{2}) \biggl(1 - O\biggl(\frac{1}{\ln ^{s_{2}}m} \biggr)\biggr) > 0, \end{aligned}$$where we indicate that \(O(\frac{1}{\ln ^{s_{2}}m}) = \frac{1}{k_{s}(s_{2})}\int _{0}^{\frac{\ln 2}{\ln m}} \frac{v^{s_{2} - 1}}{(1 + v)^{s}}\,dv\), satisfying
$$ 0 < \int _{0}^{\frac{\ln 2}{\ln m}} \frac{v^{s_{2} - 1}}{(1 + v)^{s}}\,dv \le \int _{0}^{\frac{\ln 2}{\ln m}} v^{s_{2} - 1}\,dv = \frac{1}{s_{2}} \biggl(\frac{\ln 2}{\ln m}\biggr)^{s_{2}}. $$Therefore, inequalities (11) follow.
This proves the lemma. □
Lemma 3
If \(a \in ( - 1,1)\), \(m \in \mathrm{N}\backslash \{ 1\}\), then there exists a constant C such that
Proof
We set \(f(t): = \frac{1}{t}\ln ^{a}t\) (\(t \ge 2\)). Then we find
where \(g_{1}(t) = \frac{1}{t^{2}}\ln ^{a - 1}t\), \(g_{2}(t) = \frac{1}{t^{2}}\ln ^{a}t\) (\(t \ge 2\)).
We obtain \(( - 1)^{i}g_{1}^{(i)}(t) > 0\) (\(t \ge 2\); \(i = 0,1\)). Since
it follows that \(( - 1)^{i}g_{2}^{(i)}(t) > 0\) (\(i = 0,1\)). In view of (2.2.12) in [5], we have
By (8), we have
By simplification, we obtain (13), where
This proves the lemma. □
Lemma 4
For \(t > 0\), the following inequality is valid:
Proof
Since \(A_{m}e^{ - t\ln m} = o(1)\) (\(m \to \infty \)), by Abel’s summation by parts formula, it follows that
For a fixed \(m \in \mathrm{N}\backslash \{ 1\}\), we set \(f(x): = e^{ - t\ln x}\), \(x \in [m,m + 1]\). Since \(f'(x) = - th(x)\), where \(h(x): = x^{ - 1}e^{ - t\ln x}\) is decreasing in \([m,m + 1]\), by the differentiation intermediate value theorem, there exists a constant \(\theta \in (0,1)\) such that
namely, inequality (14) follows.
This proves lemma. □
Lemma 5
For \(0 < p < 1\) (\(q < 0\)), the following reverse inequality is valid:
Proof
In view of the symmetry, for \(s_{1} \in (0,2] \cap (0,s)\), \(s > 0\), we set and obtain the next weight coefficient as follows:
where \(O(\frac{1}{\ln ^{s_{1}}n}): = \frac{1}{k_{s}(s_{1})}\int _{0}^{\frac{\ln 2}{\ln n}} \frac{v^{s_{1} - 1}}{(1 + v)^{s}}\,dv > 0\).
In view of the reverse Hölder’s inequality (cf. [30]), we find
By (11) and (16) (for \(s = \lambda \), \(s_{i} = \lambda _{i} \in (0,2] \cap (0,\lambda )\) (\(i = 1,2\))), we obtain (15).
This proves the lemma. □
3 Main results
Theorem 1
The following reverse Mulholland’s inequality with \(A_{m}\) in the kernel is valid:
In particular, for \(\lambda _{1} + \lambda _{2} = \lambda \), we have
and the following reverse inequality:
Proof
In view of the following expression related to the Gamma function:
by (14), it follows that
Then by (15), in view of \(\Gamma (\lambda + 1) = \lambda \Gamma (\lambda )\), we have (17). For \(\lambda _{1} + \lambda _{2} = \lambda \) in (17), we have (18).
This proves the theorem. □
Theorem 2
Assume that \(\lambda _{1} \in (0,2) \cap (0,\lambda )\), \(\lambda _{2} \in (0,2) \cap (0,\lambda )\). If \(\lambda _{1} + \lambda _{2} = \lambda \), then the constant \(\frac{1}{\lambda} (k_{\lambda} (\lambda _{2}))^{\frac{1}{p}}(k_{\lambda} (\lambda _{1}))^{\frac{1}{q}}\) in (17) is the best possible.
Proof
We now show that \(\frac{1}{\lambda} B(\lambda _{1},\lambda _{2})\) in (18) is the best value under the assumptions of this theorem.
For any \(0 < \varepsilon < \min \{ p\lambda _{1},|q|(2 - \lambda _{2})\}\), we set
For \(a = \lambda _{1} - \frac{\varepsilon}{p} - 1 \in ( - 1,1)\), by (13), we have
satisfying \(\tilde{A}_{m} = o(e^{t\ln m})\) (\(t > 0\); \(m \to \infty \)).
If there exists a constant \(M( \ge \frac{1}{\lambda} B(\lambda _{1},\lambda _{2}))\) such that (18) is valid when we replace \(\frac{1}{\lambda} B(\lambda _{1},\lambda _{2})\) by M, then for \(a_{m} = \tilde{a}_{m}\), \(b_{n} = \tilde{b}_{n}\), and \(A_{m} = \tilde{A}_{m}\), we have
We obtain
In view of (11) (for \(s = \lambda + 1 > 0\), \(s_{2} = \lambda _{2} - \frac{\varepsilon}{q} \in (0,2) \cap (0,\lambda )\)), we obtain
Then we have
Setting \(\varepsilon \to 0^{ +} \), in view of the continuity of the Beta function, we obtain
Therefore, \(M = \frac{1}{\lambda} B(\lambda _{1},\lambda _{2})\) is the best value in (18).
This proves the theorem. □
Theorem 3
Assume that \(\lambda > 0\), \(\lambda _{1} \in (0,2] \cap (0,\lambda )\), \(\lambda _{2} \in (0,2) \cap (0,\lambda )\). If the constant \(\frac{1}{\lambda} (k_{\lambda} (\lambda _{2}))^{\frac{1}{p}}(k_{\lambda} (\lambda _{1}))^{\frac{1}{q}}\) in (17) is the best possible, then for
we have \(\lambda _{1} + \lambda _{2} = \lambda \).
Proof
Since \(\hat{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} = \frac{\lambda - \lambda _{1} - \lambda _{2}}{p} + \lambda _{1}\), \(\hat{\lambda}_{2} = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} = \frac{\lambda - \lambda _{1} - \lambda _{2}}{q} + \lambda _{2}\), we find \(\hat{\lambda}_{1} + \hat{\lambda}_{2} = \lambda \). In view of (19), for \(\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))\), we have \(\hat{\lambda}_{1} \in (0,\lambda )\), \(\hat{\lambda}_{2} = \lambda - \hat{\lambda}_{1} \in (0,\lambda )\), and then \(B(\hat{\lambda}_{1},\hat{\lambda}_{2}) \in \mathrm{R}_{ +} \); for \(\lambda - \lambda _{1} - \lambda _{2} \le p(2 - \lambda _{1})\), we have \(\hat{\lambda}_{1} \le 2\); for \(\lambda - \lambda _{1} - \lambda _{2}\ \ge q(2 - \lambda _{2})\), we have \(\hat{\lambda}_{2} \le 2\). Then, for \(\lambda _{i} = \hat{\lambda}_{i}\) (\(i = 1,2\)) in (18), we still have
In view of the reverse Hölder’s inequality (cf. [30]), we find
If the constant \(\frac{1}{\lambda} (k_{\lambda} (\lambda _{2}))^{\frac{1}{p}}(k_{\lambda} (\lambda _{1}))^{\frac{1}{q}}\) in (17) is the best possible, then, comparing with the constants in (17) and (20), we have
namely, \(B(\hat{\lambda}_{1},\hat{\lambda}_{2}) \le (k_{\lambda} (\lambda _{2}))^{\frac{1}{p}}(k_{\lambda} (\lambda _{1}))^{\frac{1}{q}}\), and then (21) attains the form of an equality.
Inequality (21) becomes an equality if and only if there exist constants A and B such that they are not both zero and (cf. [30]) \(Au^{\lambda - \lambda _{2} - 1} = Bu^{\lambda _{1} - 1}\) a.e. in \(\mathrm{R}_{ +} \). Supposing that \(A \ne 0\), we have \(u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{B}{A}\) a.e. in \(\mathrm{R}_{ +} \). It follows that \(\lambda - \lambda _{2} - \lambda _{1} = 0\), namely, \(\lambda _{1} + \lambda _{2} = \lambda \).
This proves the theorem. □
4 Equivalent forms and some particular inequalities
Theorem 4
The following reverse inequality equivalent to (17) is valid:
Particularly, for \(\lambda _{1} + \lambda _{2} = \lambda \), the following reverse inequality equivalent to (18) is valid:
Proof
Assuming that (23) is valid, by the reverse Hölder’s inequality, we have
In view of (23), we have (17). Assuming that (17) is valid, we set
Then we find
If \(J = \infty \), then (23) is valid; if \(J = 0\), then it is impossible that makes (23) valid, namely, \(J > 0\). Assuming that \(0 < J < \infty \), by (17), it follows that
Hence, (23) is valid, which is equivalent to (17).
This proves the theorem. □
Theorem 5
Assume that \(\lambda _{1} \in (0,2) \cap (0,\lambda )\), \(\lambda _{2} \in (0,2] \cap (0,\lambda )\). If \(\lambda _{1} + \lambda _{2} = \lambda \), then the constant \(\frac{1}{\lambda} (k_{\lambda} (\lambda _{2}))^{\frac{1}{p}}(k_{\lambda} (\lambda _{1}))^{\frac{1}{q}}\) in (23) is the best possible. On the other hand, if the same constant in (23) is the best possible, then for \(\lambda - \lambda _{1} - \lambda _{2} \in [q(2 - \lambda _{2}),p(2 - \lambda _{1})]\), we have \(\lambda _{1} + \lambda _{2} = \lambda \).
Proof
We show that the constant \(\frac{1}{\lambda} B(\lambda _{1},\lambda _{2})\) in (24) is the best possible. Otherwise, by (25) (for \(\lambda _{1} + \lambda _{2} = \lambda \)), we would reach a contradiction that the same constant in (18) is not the best possible.
On the other hand, if the constant in (23) is the best possible, then the same constant in (17) is also the best possible. Otherwise, by (26) (for \(\lambda _{1} + \lambda _{2} = \lambda \)), we would reach a contradiction that the same constant in (24) is not the best possible.
This proves the theorem. □
Remark 1
For \(\lambda \in (0,4)\), \(\lambda _{1} = \lambda _{2} = \frac{\lambda}{2}( < 2)\) in (18) and (24), we have the following equivalent forms with the best value \(\frac{1}{\lambda} B(\frac{\lambda}{ 2},\frac{\lambda}{2})\):
Particularly, for \(\lambda = 1\), we have the following equivalent inequalities with the best value π:
5 Conclusions
In this article, by means of the techniques of analysis, applying the basic inequalities and formulas, a new reverse Mulholland’s inequality with one partial sum in the kernel is given in Theorem 1. The equivalent conditions of the best value related to parameters are obtained in Theorems 2 and 3. As applications, we deduce the equivalent forms in Theorems 4 and 5, and some new inequalities for particular parameters in Remark 1.
Data availability
The data used to support the findings of this study are included within the article.
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The authors thank the referees for useful comments which have improved this paper.
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This work was supported by the National Natural Science Foundation (No. 11561019), Guangxi Natural Science Foundation (No. 2020GXNSFAA159084), the National Natural Science Foundation of China (No. 12071491), and School-Level Quality Engineering Project (No. 2022ckjjd02).
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B.Y. and L.R. carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. X.Y.H. and X.S.H. participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Huang, X., Luo, R., Yang, B. et al. A new reverse Mulholland’s inequality with one partial sum in the kernel. J Inequal Appl 2024, 9 (2024). https://doi.org/10.1186/s13660-024-03080-x
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DOI: https://doi.org/10.1186/s13660-024-03080-x