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# 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:

## 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

• 26E60

## 1 Introduction

For $$a,b>0$$ with $$a\neq b$$, the Neuman-Sándor mean $$\mathit {NS}(a,b)$$ , the second Seiffert mean $$T(a,b)$$ , and the logarithmic mean $$L(a,b)$$  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 )
\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 .

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  and , 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 , 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 , 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 , the authors proved the sharp double inequality
$${M_{\log2/(\log\pi- \log2)}}(a,b) < T(a,b) < {M_{5/3}}(a,b)$$
holds.
Gao  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  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 .
In , 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



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 . 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. 