Open Access

Optimal bounds for Neuman-Sándor mean in terms of the convex combination of the logarithmic and the second Seiffert means

Journal of Inequalities and Applications20172017:251

https://doi.org/10.1186/s13660-017-1516-7

Received: 26 May 2017

Accepted: 13 September 2017

Published: 10 October 2017

Abstract

In the article, we prove that the double inequality
$$ \alpha L(a,b)+(1-\alpha)T(a,b)< \mathit{NS}(a,b)< \beta L(a,b)+(1-\beta)T(a,b) $$
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.

Keywords

Neuman-Sándor mean logarithmic mean the second Seiffert mean

MSC

26E60

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
$$\begin{aligned}& \mathit {NS}(a,b)=\frac{a-b}{2\sinh^{-1}[(a-b)/(a+b)]}, \end{aligned}$$
(1.1)
$$\begin{aligned}& T(a,b)=\frac{a-b}{2\tan^{-1}[(a-b)/(a+b)]}, \\& L(a,b)=\frac{a-b}{\log a-\log b}, \end{aligned}$$
(1.2)
respectively. It can be observed that the logarithmic mean \(L(a,b)\) can be rewritten as (see as [1])
$$\begin{aligned}& L(a,b)=\frac{a-b}{2\tanh^{-1}[(a-b)/(a+b)]}, \end{aligned}$$
(1.3)
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 [216].

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
$$ S(a,b)>T(a,b)>\mathit {NS}(a,b)>A(a,b)>I(a,b)>P(a,b)>L(a,b)>G(a,b)>H(a,b) $$
holds for \(a,b>0\) with \(a\ne b\).
In [17] and [18], the authors proved that the double inequalities
$$\begin{aligned}& S(a,b)^{\alpha_{3}}A^{1-\alpha_{3}}(a,b)< \mathit {NS}(a,b)< S(a,b)^{\beta_{3}}A ^{1-\beta_{3}}(a,b), \\& \alpha_{4}S(a,b)+(1-\alpha_{4})G(a,b)< \mathit {NS}(a,b)< \beta_{4}S(a,b)+(1-\beta _{4})G(a,b) \end{aligned}$$
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
$$ {P^{\alpha_{2}}}(a,b){T^{1 - \alpha_{2}}}(a,b) < \mathit {NS}(a,b) < {P^{\beta _{2}}}(a,b){T^{1 - \beta_{2}}}(a,b) $$
holds for all \(a,b>0\) with \(a\ne b \) if and only if \(\alpha_{2}>1/3\) and
$$ \beta_{2} \leq\log\biggl(\frac{4\log(1+\sqrt{2})}{\pi}\biggr)/\log2 =0.1663 \ldots. $$
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
$$ L_{\alpha_{1}}(a,b)< \mathit {NS}(a,b)< L_{\beta_{1}}(a,b) $$
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
$$ M_{p}(a,b)= \textstyle\begin{cases} (\frac{a^{p}+b^{p}}{2})^{1/p}, & p\neq0, \\ \sqrt{ab}, & p=0, \end{cases} $$
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
$$ {M_{\log2/(\log\pi- \log2)}}(a,b) < T(a,b) < {M_{5/3}}(a,b) $$
holds.
Gao [21] proved the optimal double inequality
$$ I(a,b) < T(a,b) < \frac{{2e}}{\pi}I(a,b) $$
holds for all \(a,b>0\) with \(a\ne b\).
Yang [22] proved the inequality
$$ A_{p}^{1/(3p)}(a,b){G^{1 - 1/(3p)}}(a,b) < L(a,b) < A_{q}^{1/(3q)}(a,b) {G^{1 - 1/(3q)}}(a,b) $$
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
$$ M_{0}(a,b)< L(a,b)< M_{1/3}(a,b) $$
was proved by Lin in [23].
In [24], the authors present bounds for L in terms of G and A
$$ G^{2/3}(a,b)A^{1/3}(a,b)< L(a,b)< \frac{2}{3}G(a,b)+ \frac{1}{3}A(a,b) $$
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
$$ \alpha L(a,b) + (1 - \alpha)T(a,b) < \mathit {NS}(a,b) < \beta L(a,b) + (1 - \beta)T(a,b) $$
holds for all \(a,b>0\) with \(a \ne b\) ?

2 Lemmas

It is well known that, for \(x \in(0,1) \),
$$\begin{aligned}& \tanh^{-1} (x) =x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots=\sum ^{ \infty}_{n=0}\frac{1}{2n+1}x^{2n+1}, \end{aligned}$$
(2.1)
$$\begin{aligned}& \tan^{-1}(x)=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}- \frac{x^{7}}{7}+\cdots =\sum^{\infty}_{n=0} \frac{(-1)^{n}}{2n+1}x^{2n+1}. \end{aligned}$$
(2.2)

To establish our main result, we need several lemmas as follows.

Lemma 2.1

[25]

Let
$$\begin{aligned}& H(x) = \frac{1}{\sinh^{-1}x}- \frac{x}{\sqrt{1+x^{2}}(\sinh^{-1}x)^{2}}. \end{aligned}$$
Then \(H(x)\) is strictly increasing on \((0,1)\). Moreover, the inequality
$$\begin{aligned}& H(x)< \frac{x}{3}-\frac{x^{3}}{9} \end{aligned}$$
(2.3)
holds for any \(x \in(0,0.76) \) and the inequality
$$\begin{aligned}& H(x)>\frac{x}{3}-\frac{17x^{3}}{90} \end{aligned}$$
(2.4)
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
$$\begin{aligned}& S(x)< -\frac{2}{3}x-\frac{1}{3}x^{3}- \frac{1}{3}x^{5} \end{aligned}$$
(2.5)
for any \(x \in(0,1) \) and
$$\begin{aligned}& S(x)>-\frac{2}{3}x-x^{3}-\frac{x^{5}}{4} \end{aligned}$$
(2.6)
for any \(x \in(0,0.76)\).

Proof

Let
$$\begin{aligned}& G(x)=\bigl(1-x^{2}\bigr) \bigl(\tanh^{-1}x \bigr)^{2}\biggl[S(x)+\frac{2}{3}x+\frac{1}{3}x^{3}+ \frac{1}{3}x^{5}\biggr]. \end{aligned}$$
Then direct computation leads to
$$\begin{aligned}& G(0)=0, \end{aligned}$$
(2.7)
$$\begin{aligned}& G'(x)=\frac{1}{3}g(x)\tanh^{-1}x, \end{aligned}$$
(2.8)
where \(g(x) = (2 - 7{x^{6}} - 3{x^{2}}){\tanh^{ - 1}}x + 2{x^{5}} + 2 {x^{3}} - 2x\). It follows that
$$\begin{aligned}& g'(x)=\frac{1}{1-x^{2}}g_{1}(x), \end{aligned}$$
(2.9)
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
$$\begin{aligned} {g_{1}}(x)& < \bigl( - 42{x^{5}} - 6x\bigr) \bigl(1 - {x^{2}}\bigr) \biggl( x + \frac{x^{3}}{3} + \frac{x^{5}}{5} \biggr) - 17{x^{6}} + 4{x^{4}} + 5{x^{2}} \\ & = \frac{1}{5}\bigl(42{x^{12}} + 28{x^{10}} + 146{x^{8}} - 291{x^{6}} + 40 {x^{4}} - 5{x^{2}}\bigr) \\ & < x^{2}\bigl(216x^{6}-291x^{4}+40x^{2}-5 \bigr)< 0, \end{aligned}$$
(2.10)
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
$$ Q(x)=\bigl(1-x^{2}\bigr) \bigl(\tanh^{-1}x \bigr)^{2}\biggl[S(x)+\frac{2}{3}x+x^{3}+ \frac{x^{5}}{4}\biggr]. $$
Direct computation leads to
$$\begin{aligned}& Q(0)=0, \end{aligned}$$
(2.11)
$$\begin{aligned}& Q'(x)=\frac{1}{12}q_{1}(x)\tanh^{-1}x, \end{aligned}$$
(2.12)
where
$$ q_{1}(x)=6x^{5}+24x^{3}-8x+ \bigl(8+12x^{2}-45x^{4}-21x^{6}\bigr) \tanh^{-1}x. $$
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
$$\begin{aligned} q_{1}(x)&>6x^{5}+24x^{3}-8x+ \bigl(8+12x^{2}-45x^{4}-21x^{6}\bigr) \biggl(x+ \frac{x ^{3}}{3}+\frac{x^{5}}{5}\biggr) \\ & =-\frac{21}{5}x^{11}-16x^{9}- \frac{168}{5}x^{7}-\frac{167}{5}x^{5}+ \frac{116}{3}x^{3} \\ & >x^{3}\biggl(-\frac{269}{5}x^{4}- \frac{167}{5}x^{2}+\frac{116}{3}\biggr)>0. \end{aligned}$$
When \(x\in(0.7,0.76)\), direct computation leads to
$$\begin{aligned}& q_{1}(0.76)=1.8639\ldots>0, \end{aligned}$$
(2.13)
$$\begin{aligned}& q_{1}'(x)=q_{2}(x)/\bigl(1-x^{2} \bigr), \end{aligned}$$
(2.14)
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
$$\begin{aligned} q_{2}(x)&< 92x^{2}-87x^{4}-51x^{6}+ \bigl(126x^{7}+54x^{5}-204x^{3}+24x\bigr) \biggl(x+\frac{x ^{3}}{3}+\frac{x^{5}}{5}\biggr) \\ & =\frac{126}{5}x^{12}+\frac{264}{5}x^{10}+ \frac{516}{5}x^{8}- \frac{301}{5}x^{6}-283x^{4}+116x^{2} \\ & < 2x^{4}\bigl(91x^{4}-30x^{2}-20 \bigr)+x^{2}\bigl(116-243x^{2}\bigr)< 0. \end{aligned}$$
(2.15)
Thus, (2.13)-(2.15) imply that
$$\begin{aligned}& q_{1}(x)>0 \end{aligned}$$
(2.16)
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
$$\begin{aligned}& T(x)< \frac{2}{3}x-\frac{1}{3}x^{3}+ \frac{2}{7}x^{5} \end{aligned}$$
(2.17)
for any \(x\in(0,1)\) and
$$\begin{aligned}& T(x)>\frac{2}{3}x-\frac{2}{5}x^{3}+\frac{x^{5}}{7} \end{aligned}$$
(2.18)
for any \(x\in(0,0.76)\).

Proof

Let
$$\begin{aligned}& M(x)=\biggl[T(x)-\frac{2}{3}x+\frac{x^{3}}{3}-\frac{2}{7}x^{5} \biggr]\bigl(1+x^{2}\bigr) \bigl( \tan^{-1}x \bigr)^{2}. \end{aligned}$$
Differentiating \(M(x)\), we have \(M'(x) = [t(x){\tan^{ - 1}}x]/21 \), where
$$ t(x) = 14x + 14{x^{3}} - 12{x^{5}} + \bigl( - 42{x^{6}}+5{x^{4}} - 21{x ^{2}} - 14\bigr){ \tan^{ - 1}}x. $$
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
$$\begin{aligned} t(x) &< 14x+14x^{3}-12x^{5}+\bigl(- 42{x^{6}}+5{x^{4}} - 21{x^{2}} - 14\bigr) \biggl(x-\frac{x ^{3}}{3}\biggr) \\ & = 14{x^{9}} - \frac{131}{3}x^{7} - \frac{7}{3}x^{3} \\ & < - \frac{89}{3}x^{7} -\frac{7}{3}x^{3} < 0. \end{aligned}$$

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
$$\begin{aligned}& N(x)=\biggl[T(x)-\frac{2}{3}x+\frac{2}{5}x^{3}- \frac{x^{5}}{7}\biggr]\bigl(1+x^{2}\bigr) \bigl( \tan^{-1}x \bigr)^{2}. \end{aligned}$$
Differentiating \(N(x)\), we have \(N'(x) = n(x){\tan^{ - 1}}x \), where
$$ n(x) = \frac{2}{3}x + \frac{4}{5}x^{3}- \frac{2}{7}x^{5}-\biggl(x^{6}- \frac{9}{7}x^{4}+ \frac{4}{5}x^{2}+\frac{2}{3}\biggr)\tan^{-1}x. $$
Because of
$$ \biggl(\frac{4}{5}x^{2}-\frac{9}{7}x^{4} \biggr)+x^{6}+\frac{2}{3}>0 $$
for \(x\in(0,0.76)\), it follows that
$$\begin{aligned} n(x)&>\frac{2}{3}x + \frac{4}{5}x^{3}- \frac{2}{7}x^{5}-\biggl(x^{6}- \frac{9}{7}x^{4}+ \frac{4}{5}x^{2}+\frac{2}{3}\biggr)x \\ & =x^{5}-x^{7}>0. \end{aligned}$$
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
$$\begin{aligned}& S'(x) = 2\frac{x-\tanh^{-1}x}{(1-x^{2})^{2}(\tanh^{-1}x)^{3}}, \\& S''(x) =2\frac{\varphi(x)}{(1-x^{2})^{3}(\tanh^{-1}x)^{4}}, \end{aligned}$$
where \(\varphi(x) = 3(1 + {x^{2}}){\tanh^{ - 1}}x - 3x - 4x{({\tanh^{ - 1}}x)^{2}} \). It follows that
$$ \varphi'(x) = \frac{R(x)}{1-x^{2}}, $$
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
$$\begin{aligned} R(x)&< -4\bigl(1-x^{2}\bigr) \biggl(x+\frac{x^{3}}{3} \biggr)^{2}-\bigl(6x^{3}+2x\bigr) \biggl(x+ \frac{x^{3}}{3}\biggr)+6x^{2} \\ &=\frac{4}{9}x^{8}+\frac{2}{9}x^{6}- \frac{16}{3}x^{4}< 0. \end{aligned}$$
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
$$ \phi'(x) < \lambda S'(0.76) + (1 - \lambda)T'(0.76) = - 0.0043 \ldots< 0 $$
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
$$ \frac{4-11\lambda}{28}x^{4} - \frac{27\lambda+13}{45}x^{2}+ \frac{1-4 \lambda}{3}>0 $$
for \(x \in(0,0.76)\), where \(\lambda= 1 - \pi/[4\log(1 + \sqrt{2} )]=0.1089\ldots\) .

Proof

Let
$$ \eta(x) = \frac{4-11\lambda}{28}x^{4} - \frac{27\lambda+13}{45}x ^{2}+\frac{1-4\lambda}{3}. $$
Then it is easy to verify that \(\eta(x)\) is decreasing on \((0,\mu)\), where
$$ \mu=\sqrt{\frac{14}{15}} \times\sqrt{\frac{160\log(1+ \sqrt{2})-27\pi}{11\pi-28\log(1+\sqrt{2})}}= 1.3303 \ldots. $$
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
$$ \alpha L(a,b) + (1 - \alpha)T(a,b) < \mathit {NS}(a,b) < \beta L(a,b) + (1 - \beta)T(a,b) $$
holds for any \(a,b>0\) with \(a \ne b \) if and only if \(\alpha\ge1/4 \) and
$$ \beta\le1 - \frac{\pi}{4\log(1+\sqrt{2})}= 0.1089 \ldots. $$

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
$$\begin{aligned}& \frac{\mathit {NS}(a,b)}{A(a,b)}=\frac{x}{\sinh^{-1}x}, \\& \frac{L(a,b)}{A(a,b)}=\frac{x}{\tanh^{-1}x}, \\& \frac{T(a,b)}{A(a,b)}=\frac{x}{\tan^{-1}x}. \end{aligned}$$
Let
$$\begin{aligned} F_{t}(x) &= \frac{tL(a,b) + (1 - t)T(a,b) - M(a,b)}{A(a,b)} \\ & =t\frac{x}{\tanh^{-1}x}+(1-t)\frac{x}{\tan^{-1}x}-\frac{x}{ \sinh^{-1}x}. \end{aligned}$$
(3.1)
Then it follows that
$$\begin{aligned}& F_{\frac{1}{4}}\bigl(0^{+}\bigr)=0, \end{aligned}$$
(3.2)
$$\begin{aligned}& F_{\lambda}\bigl(0^{+}\bigr)=F_{\lambda} \bigl(1^{-}\bigr)=0. \end{aligned}$$
(3.3)
Differentiating \(F_{t}(x) \), we have
$$\begin{aligned} F'_{t}(x)&=t\biggl[\frac{1}{\tanh^{-1}x}- \frac{x}{1-x^{2}}\frac{1}{(\tanh ^{-1}x)^{2}}\biggr] \\ &\quad{} +(1-t)\biggl[\frac{1}{\tan^{-1}x}-\frac{x}{1+x^{2}} \frac{1}{(\tan^{-1}x)^{2}}\biggr] \\ &\quad{} -\biggl[\frac{1}{\sinh^{-1}x}-\frac{x}{\sqrt{1+x^{2}}} \frac{1}{(\sinh^{-1}x)^{2}}\biggr] \\ &:=tS(x)+(1-t)T(x)-H(x), \end{aligned} $$
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
$$\begin{aligned} F'_{\frac{1}{4}}(x)&=\frac{1}{4}S(x)+\frac{3}{4}T(x)-H(x) \\ &< \frac{1}{4}\biggl(-\frac{2}{3}x- \frac{1}{3}x^{3}-\frac{1}{3}x^{5}\biggr)+ \frac{3}{4}\biggl(\frac{2}{3}x-\frac{1}{3}x^{3}+ \frac{2}{7}x^{5}\biggr) -\biggl( \frac{x}{3}- \frac{17}{90}x^{3}\biggr) \\ & =-\frac{13}{90}x^{3}+\frac{11}{84}x^{5}< 0 \end{aligned} $$
for any \(x\in(0,1)\). It leads to
$$\begin{aligned}& F_{\frac{1}{4}}(x)< F_{\frac{1}{4}}(0)=0 \end{aligned}$$
(3.4)
for any \(x\in(0,1)\). Thus, from (3.1) it follows that
$$\begin{aligned}& \mathit {NS}(a,b)>\frac{1}{4}L(a,b)+\frac{3}{4}T(a,b) \end{aligned}$$
for all \(a,b>0\) with \(a \ne b \). Considering \(L(a,b)<\mathit {NS}(a,b)<T(a,b)\), we can get
$$\begin{aligned}& \mathit {NS}(a,b)>\alpha L(a,b)+(1-\alpha)T(a,b) \end{aligned}$$
(3.5)
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
$$\begin{aligned} F'_{\lambda}(x)&>-\lambda\biggl(\frac{2}{3}x+x^{3}+ \frac{x^{5}}{4}\biggr)+(1- \lambda) \biggl(\frac{2}{3}x- \frac{2}{5}x^{3}+\frac{x^{5}}{7}\biggr)-\biggl( \frac{x}{3}-\frac{x^{3}}{9}\biggr) \\ &=x\biggl[\frac{4-11\lambda}{28}x^{4}-\frac{27\lambda +13}{45}x^{2}+ \frac{1-4 \lambda}{3}\biggr] \end{aligned}$$
for \(x\in(0,0.76)\). According to Lemma 2.5, we have
$$\begin{aligned}& F'_{\lambda}(x)>0 \end{aligned}$$
(3.6)
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
$$\begin{aligned} &\mathit {NS}(a,b) < \lambda L(a,b) + (1 - \lambda)T(a,b) \end{aligned}$$
for all \(a,b>0\) with \(a\ne b\). Considering \(L(a,b)< M(a,b)< T(a,b)\), we can get
$$\begin{aligned} &\mathit {NS}(a,b) < \beta L(a,b) + (1 - \beta)T(a,b) \end{aligned}$$
(3.7)
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
$$\begin{aligned}& F_{t_{1}}(x)=\biggl(\frac{1}{4}-\epsilon_{1}\biggr) \frac{x}{\tanh^{-1}x}+\biggl( \frac{3}{4}+\epsilon_{1}\biggr) \frac{x}{\tan^{-1}x}-\frac{x}{\sinh^{-1}x}, \end{aligned}$$
(3.8)
$$\begin{aligned}& F_{t_{2}}(x)=(\lambda+\epsilon_{2})\frac{x}{\tanh^{-1}x}+(1- \lambda -\epsilon_{2})\frac{x}{\tan^{-1}x}-\frac{x}{\sinh^{-1}x}. \end{aligned}$$
(3.9)
Let \(x_{1}\rightarrow0^{+}\) and \(x_{2}\rightarrow1^{-}\), then the Taylor expansion leads to
$$\begin{aligned}& F_{t_{1}}(x_{1})=\frac{2}{3}\epsilon_{1} x_{1}^{2}+O\bigl(x_{1}^{4}\bigr), \end{aligned}$$
(3.10)
$$\begin{aligned}& F_{t_{2}}(x_{2})=-4\epsilon_{2}/ \pi+O(x_{2}-1). \end{aligned}$$
(3.11)

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.

Declarations

Acknowledgements

This research was supported by Foundations of Anhui Educational Committee (KJ2017A029) and Anhui University (J10118520279, J01001901, Y01002428), China.

Authors’ contributions

All the authors worked jointly. All the authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematical Science, Anhui University

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