Abstract
In the article, we prove that the double inequality
holds for \(a,b>0\) with \(a\ne b \) if and only if \(\alpha\ge1/4\) and \(\beta\le1-\pi/[4\log(1+\sqrt{2})]\), where \(\mathit{NS}(a,b)\), \(L(a,b)\) and \(T(a,b)\) denote the Neuman-Sándor, logarithmic and second Seiffert means of two positive numbers a and b, respectively.
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1 Introduction
For \(a,b>0\) with \(a\neq b\), the Neuman-Sándor mean \(\mathit {NS}(a,b)\) [1], the second Seiffert mean \(T(a,b)\) [2], and the logarithmic mean \(L(a,b)\) [1] are defined by
respectively. It can be observed that the logarithmic mean \(L(a,b)\) can be rewritten as (see as [1])
where \(\sinh^{-1}(x)=\log(x+\sqrt{1+x^{2}})\), \(\tanh^{-1}(x)= \log{\sqrt{(1+x)/(1-x)}}\) and \(\tan^{-1}(x)=\arctan(x)\), are the inverse hyperbolic sine, inverse hyperbolic tangent, and inverse tangent, respectively.
Recently, the means NS, T, L and other means have been the subject of extensive research. In particular, many remarkable inequalities for the Neuman-Sándor, second Seiffert and logarithmic means can be found in the literature [2–16].
Let \(P(a,b)=(a-b)/(2\sin^{-1}[(a-b)/(a+b)])\), \(S(a,b) = \sqrt{( {a^{2}} + {b^{2}})/2}\), \(A(a,b)=(a+b)/2\), \(I(a,b) = 1/e{({b^{b}}/ {a^{a}})^{1/(b - a)}}\), \(G(a,b)=\sqrt{ab}\), and \(H(a,b)=2ab/(a+b)\) denote the first Seiffert, root-square, arithmetic, identric, geometric, and the harmonic means of two positive numbers a and b with \(a\ne b\), respectively. Then it is well known that the inequality
holds for \(a,b>0\) with \(a\ne b\).
In [17] and [18], the authors proved that the double inequalities
hold for all \(a,b>0\) with \(a\neq b\) if and only if \(\alpha_{3}\leq1/3\), \(2(\log(2+\sqrt{2})-\log3)/\log2 \leq\beta_{3}\leq1\), \(\alpha_{4}\leq2/3\) and \(\beta_{4}\geq1/[\sqrt{2}\log(1+ \sqrt{2})]\).
In [19], it was showed that the inequality
holds for all \(a,b>0\) with \(a\ne b \) if and only if \(\alpha_{2}>1/3\) and
Let \(L_{p}(a,b)=(a^{p+1}+b^{p+1})/(a^{p}+b^{p})\) be the Lehmer mean of two positive numbers a and b with \(a\neq b\). In [10], the authors proved the double inequality
holds for all \(a,b>0\) with \(a\neq b\) if and only if \(\alpha_{1} =1.8435 \ldots\) is the unique solution of the equation \((p+1)^{1/p}=2\log(1+ \sqrt{2})\), and \(\beta_{1} =2\).
Let
be the pth power means of two positive numbers a and b with \(a\neq b\). In [20], the authors proved the sharp double inequality
holds.
Gao [21] proved the optimal double inequality
holds for all \(a,b>0\) with \(a\ne b\).
Yang [22] proved the inequality
holds for all \(a,b>0\) with \(a\ne b\) if and only if \(p\geq1/ \sqrt{5}\) and \(0< q\leq1/3\). And the inequality
was proved by Lin in [23].
In [24], the authors present bounds for L in terms of G and A
for all \(a,b>0\) with \(a\neq b\).
The purpose of this paper is to answer the following questions: What are the least value α and the greatest value β such that
holds for all \(a,b>0\) with \(a \ne b\) ?
2 Lemmas
It is well known that, for \(x \in(0,1) \),
To establish our main result, we need several lemmas as follows.
Lemma 2.1
[25]
Let
Then \(H(x)\) is strictly increasing on \((0,1)\). Moreover, the inequality
holds for any \(x \in(0,0.76) \) and the inequality
holds for any \(x \in(0,1) \).
Lemma 2.2
Let \(S(x) = 1/{\tanh^{ - 1}}x - x/[(1 - {x^{2}}){({\tanh^{ - 1}}x)^{2}}] \). Then
for any \(x \in(0,1) \) and
for any \(x \in(0,0.76)\).
Proof
Let
Then direct computation leads to
where \(g(x) = (2 - 7{x^{6}} - 3{x^{2}}){\tanh^{ - 1}}x + 2{x^{5}} + 2 {x^{3}} - 2x\). It follows that
where \({g_{1}}(x) = ( - 42{x^{5}} - 6x)(1 - {x^{2}}){\tanh^{ - 1}}x - 17{x^{6}} + 4{x^{4}} + 5{x^{2}}\). Considering (2.1), we have
for \(x \in(0,1) \). Thus, (2.9) and (2.10) as well as \(g(0)=0\) imply \(g(x)<0\) for \(x \in(0,1) \). Therefore, combining (2.7) and (2.8), we get \(G(x)<0\) for \(x \in(0,1) \). It means inequality (2.5) holds.
Let
Direct computation leads to
where
When \(x\in(0,0.7]\), considering (2.1) and the fact \(8+12x^{2}-45x ^{4}-21x^{6}=(3-21x^{6})+(5+12x^{2}-45x^{4})>0 \), we can get
When \(x\in(0.7,0.76)\), direct computation leads to
where \(q_{2}(x)=92x^{2}-87x^{4}-51x^{6}+(126x^{7}+54x^{5}-204x^{3}+24x) \tanh^{-1}x\). Considering (2.1) and the fact \(126x^{7}+54x^{5}-204x ^{3}+24x<12x(15x^{4}-17x^{2}+2)<0\), we can get
Thus, (2.13)-(2.15) imply that
holds for any \(x\in(0.7,0.76)\).
Therefore, \(Q(x)>0\) for \(x\in(0, 0.76)\) follows from (2.11), (2.12) and (2.16). That means inequality (2.6) holds. □
Lemma 2.3
Let \(T(x) = 1/{\tan^{ - 1}}x - x/[(1 + {x ^{2}}){({\tan^{ - 1}}x)^{2}}] \). Then
for any \(x\in(0,1)\) and
for any \(x\in(0,0.76)\).
Proof
Let
Differentiating \(M(x)\), we have \(M'(x) = [t(x){\tan^{ - 1}}x]/21 \), where
For \(x \in(0,1) \), we have \(- 42{x^{6}}+5{x^{4}} - 21{x^{2}} - 14 <-42x ^{6}-16x^{2}-14< 0 \). Thus from (2.2), we can get
Therefore \(M'(x)<0 \) for \(x\in(0,1) \). Considering the fact \(M(0)=0\), we get \(M(x)<0\) for \(x\in(0,1)\). So the inequality (2.17) holds.
Let
Differentiating \(N(x)\), we have \(N'(x) = n(x){\tan^{ - 1}}x \), where
Because of
for \(x\in(0,0.76)\), it follows that
Considering the fact \(N(0)=0\), the inequality (2.18) holds. □
Lemma 2.4
The function \(f(x) = \lambda S(x) + (1 - \lambda)T(x) - H(x) \) is strictly decreasing on \((0.76,1)\), where \(\lambda= 1 - \pi/[4\log(1 + \sqrt{2} )] = 0.1089 \ldots\) and \(H(x)\), \(S(x)\) and \(T(x)\) are defined as in Lemmas 2.1, 2.2 and 2.3, respectively.
Proof
Direct computation leads to
where \(\varphi(x) = 3(1 + {x^{2}}){\tanh^{ - 1}}x - 3x - 4x{({\tanh^{ - 1}}x)^{2}} \). It follows that
where \(R(x) = - 4(1 - {x^{2}}){({\tanh^{ - 1}}x)^{2}} - (6{x^{3}} + 2x) {\tanh^{ - 1}}x + 6{x^{2}} \). From (2.3), we can get
Thus \(\varphi(x)\) is strictly decreasing on \((0.76,1)\). Considering the fact \(\varphi(0.76) = - 0.5821\ldots<0 \), we have \(\varphi(x)<0 \) for any \(x \in(0.76,1) \). In other words, \(S'(x) \) is strictly decreasing on \((0.76,1) \).
Let \(\phi(x) = \lambda S(x) + (1 - \lambda)T(x) \). It was proved that \(T'(x) \) is strictly decreasing on \((0.7,1) \) in Lemma 5 of [26]. Thus, from the monotonicity of \(S'(x) \) and \(T'(x) \), we have
for any \(x \in(0.76,1) \). That is to say, \(\phi(x)\) is strictly decreasing on \((0.76,1)\). Considering the monotonicity of \(H(x)\) in Lemma 2.1, the proof is completed. □
Lemma 2.5
We have
for \(x \in(0,0.76)\), where \(\lambda= 1 - \pi/[4\log(1 + \sqrt{2} )]=0.1089\ldots\) .
Proof
Let
Then it is easy to verify that \(\eta(x)\) is decreasing on \((0,\mu)\), where
Considering \(\eta(0.76) = 0.01693 \ldots>0 \), we have \(\eta(x)>0\) for \(x\in(0,0.76)\). □
3 Main results
Theorem 3.1
The double inequality
holds for any \(a,b>0\) with \(a \ne b \) if and only if \(\alpha\ge1/4 \) and
Proof
Because \(\mathit {NS}(a,b)\), \(L(a,b)\), \(T(a,b)\) are symmetric and homogeneous of degree 1, without loss of generality, we can assume that \(a>b\) and \(x:=(a-b)/(a+b)\in(0,1)\). Let \(p\in(0,1)\) and \(\lambda = 1 - \pi/[4\log(1 + \sqrt{2} )] = 0.1089 \ldots\) . Then by (1.1)-(1.3), direct computation leads to
Let
Then it follows that
Differentiating \(F_{t}(x) \), we have
where \(H(x)\), \(S(x)\) and \(T(x)\) are defined as in Lemmas 2.1-2.3, respectively.
On one hand, from inequalities (2.4), (2.5) and (2.16), we clearly see that
for any \(x\in(0,1)\). It leads to
for any \(x\in(0,1)\). Thus, from (3.1) it follows that
for all \(a,b>0\) with \(a \ne b \). Considering \(L(a,b)<\mathit {NS}(a,b)<T(a,b)\), we can get
for all \(\alpha\geq1/4\) and \(a,b>0\) with \(a \ne b \).
On the other hand, from inequalities (2.3), (2.6) and (2.17), we have
for \(x\in(0,0.76)\). According to Lemma 2.5, we have
for \(x\in(0,0.76)\). Lemma 2.4 shows that \(F'_{\lambda}(x) \) is strictly decreasing on \((0.76,1)\). This fact and \(F'_{\lambda}(0.76)=0.0713 \ldots>0\) together with \(F'_{\lambda}(1^{-} ) = - \infty\) imply that there exists \({x_{0}}\in(0.76,1)\) such that \(F_{\lambda}(x) \) is strictly increasing on \((0,x_{0}]\) and strictly decreasing on \([x_{0},1)\). Equations (3.1) and (3.3) together with the piecewise monotonicity of \(F_{\lambda}(x) \) lead to the conclusion that
for all \(a,b>0\) with \(a\ne b\). Considering \(L(a,b)< M(a,b)< T(a,b)\), we can get
holds for \(\beta\leq\lambda\) and all \(a,b > 0\) with \(a \ne b\).
Finally, we prove that \(L(a,b)/4+3T(a,b)/4\) and \(\lambda L(a,b)+(1- \lambda)T(a,b)\) are the best possible lower and upper mean bound for the Neuman-Sándor mean \(M(a,b)\).
For any \(\epsilon_{1}, \epsilon_{2}>0\), let \(t_{1}=1/4-\epsilon_{1}\), \(t_{2}=\lambda+\epsilon_{2}\). Then one can get
Let \(x_{1}\rightarrow0^{+}\) and \(x_{2}\rightarrow1^{-}\), then the Taylor expansion leads to
Equations (3.8) and (3.10) imply that if \(\alpha<1/4\), then, for any \(\epsilon_{1}>0\), there exists \({\sigma_{1}} \in(0,1)\) such that \(\mathit {NS}(a,b) < (1/4-\epsilon_{1})L(a,b) + (3/4 - \epsilon_{1})T(a,b) \) for all a, b with \((a - b)/(a + b) \in(0,{\sigma_{1}}) \).
Equations (3.9) and (3.11) imply that if \(\beta>\lambda\), then, for any \(\epsilon_{2}>0\), there exists \({\sigma_{2}} \in(0,1)\) such that \(\mathit {NS}(a,b) > (\lambda+\epsilon_{2})L(a,b) + (1 - \lambda- \epsilon_{2})T(a,b) \) for all a, b with \((a - b)/(a + b) \in(1 - {\sigma_{2}},1) \). □
4 Conclusion
In the article, we give the sharp upper and lower bounds for Neuman-Sándor mean in terms of the linear convex combination of the logarithmic and second Seiffert means.
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Acknowledgements
This research was supported by Foundations of Anhui Educational Committee (KJ2017A029) and Anhui University (J10118520279, J01001901, Y01002428), China.
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Chen, JJ., Lei, JJ. & Long, BY. Optimal bounds for Neuman-Sándor mean in terms of the convex combination of the logarithmic and the second Seiffert means. J Inequal Appl 2017, 251 (2017). https://doi.org/10.1186/s13660-017-1516-7
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DOI: https://doi.org/10.1186/s13660-017-1516-7