# Optimal bounds for Neuman-Sándor mean in terms of the geometric convex combination of two Seiffert means

## Abstract

In this paper, we find the least value α and the greatest value β such that the double inequality

$$P^{\alpha}(a,b)T^{1-\alpha}(a,b)< M(a,b)< P^{\beta}(a,b)T^{1-\beta}(a,b)$$

holds for all $$a,b>0$$ with $$a\neq b$$, where $$M(a,b)$$, $$P(a,b)$$, and $$T(a,b)$$ are the Neuman-Sándor, the first and second Seiffert means of two positive numbers a and b, respectively.

## 1 Introduction

For $$a,b>0$$ with $$a\neq b$$, the Neuman-Sándor mean $$M(a,b)$$ [1], the first Seiffert mean $$P(a,b)$$ [2], and the second Seiffert mean $$T(a,b)$$ [3] are defined by

\begin{aligned}& M(a,b)=\frac{a-b}{2\sinh^{-1}(\frac{a-b}{a+b})}, \end{aligned}
(1.1)
\begin{aligned}& P(a,b)=\frac{a-b}{4\tan^{-1}(\sqrt{a/b})-\pi}, \\& T(a,b)=\frac{a-b}{2\tan^{-1}(\frac{a-b}{a+b})}, \end{aligned}
(1.2)

respectively. It can be observed that the first Seiffert mean $$P(a,b)$$ can be rewritten as (see [1])

$$P(a,b)= \frac{a-b}{2\sin^{-1}(\frac{a-b}{a+b})},$$
(1.3)

where $$\sinh^{-1}(x)=\log(x+\sqrt{x^{2}+1})$$, $$\tan^{-1}(x)=\arctan (x)$$, and $$\sin^{-1}(x)=\arcsin(x)$$ are the inverse hyperbolic sine, inverse tangent, inverse sine functions, respectively.

Recently, the means M, P, and T and other means have been the subject of intensive research. Many remarkable inequalities for means can be found in the literature [410].

Let $$H(a,b)=2ab/(a+b)$$, $$G(a,b)=\sqrt{ab}$$, $$L(a, b)=(b-a)/(\log b-\log a)$$, $$I(a,b)=1/e(b^{b}/ a^{a})^{1/(b-a)}$$, $$A(a,b)=(a+b)/2$$, $$Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$$ and

$$M_{p}(a,b)=\left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{a^{p}+b^{p}}{2})^{1/p}, & p\neq0, \\ \sqrt{ab}, & p=0, \end{array}\displaystyle \right .$$

denote the harmonic, geometric, logarithmic, identric, arithmetic, root-square, and the pth power means of two positive numbers a and b with $$a\neq b$$, respectively. Then it is well known that the inequalities

$$H(a,b)< G(a,b)< L(a,b)< P(a,b)< I(a,b)< A(a,b)< M(a,b)< T(a,b)< Q(a,b)$$

hold for $$a,b>0$$ with $$a\neq b$$.

Neuman and Sándor [1] established

$$\frac{\pi}{2}P(a,b)>\sinh^{-1}(1)M(a,b)>\frac{\pi}{4}T(a,b)$$

for all $$a,b>0$$ with $$a\neq b$$.

Gao [11] proved that the optimal double inequalities

$$\frac{e}{\pi}I(a,b)< P(a,b)< I(a,b), \qquad I(a,b)< T(a,b)< \frac{2e}{\pi}I(a,b)$$

hold for all $$a,b>0$$ with $$a\neq b$$.

The following bounds for the Seiffert means $$P(a,b)$$ and $$T(a,b)$$ in terms of the power mean were presented by Jagers in [12]:

$$M_{\frac{1}{2}}< P(a,b)< M_{\frac{2}{3}}(a,b)$$

for all $$a,b>0$$ with $$a\neq b$$. Hästö [13] improved the results of [12] and found the sharp lower power mean bound for the Seiffert mean $$P(a,b)$$ as follows:

$$P(a,b)>M_{\frac{\log2}{\log\pi}}(a,b)$$

for all $$a,b>0$$ with $$a\neq b$$.

In [14], the authors proved that the sharp double inequality

$$M_{\frac{\log2}{\log\pi-\log2}}< T(a,b)< M_{\frac{5}{3}}(a,b)$$

holds for all $$a,b>0$$ with $$a\neq b$$.

Let $$\overline{L}_{p}(a,b)=(a^{p+1}+b^{p+1})/(a^{p}+b^{p})$$ be the Lehmer mean of two positive numbers a and b with $$a\neq b$$. In [15], the authors presented the following best possible Lehmer mean bounds for the Seiffert means $$P(a,b)$$ and $$T(a,b)$$:

$$\overline{L}_{-1/6}(a,b)< P(a,b)< \overline {L}_{0}(a,b) \quad \mbox{and}\quad \overline{L}_{0}(a,b)< T(a,b)< \overline{L}_{1/3}(a,b)$$

for all $$a,b>0$$ with $$a\neq b$$.

Let u, v, and w be bivariate means such that $$u(a,b)< v(a,b)< w(a,b)$$ for all $$a,b>0$$ with $$a\neq b$$. The problems of finding the best possible parameters α and β such that the inequalities $$\alpha u(a,b)+(1-\alpha)v(a,b)< w(a,b)<\beta u(a,b)+(1-\beta)v(a,b)$$ and $$u(a,b)^{\alpha}v^{1-\alpha}(a,b)< w(a,b)< u(a,b)^{\beta}v^{1-\beta }(a,b)$$ hold for all $$a,b>0$$ with $$a\neq b$$ have attracted the interest of many mathematicians.

In [16] and [17], the authors proved that the double inequalities

\begin{aligned}& \alpha_{1}Q(a,b)+(1-\alpha_{1})A(a,b)< T(a,b)< \beta _{1}Q(a,b)+(1-\beta_{1})A(a,b), \\& Q^{\alpha_{2}}(a,b)A^{1-\alpha_{2}}(a,b)< T(a,b)< Q^{\beta _{2}}(a,b)A^{1-\beta_{2}}(a,b) \end{aligned}

hold for all $$a,b>0$$ with $$a\neq b$$ if and only if $$\alpha_{1}\leq (4-\pi)/[(\sqrt{2}-1)\pi]$$, $$\beta_{1}\geq2/3$$, $$\alpha_{2}\leq2/3$$, $$\beta_{2}\geq4-2\log\pi/\log2$$.

In [1], Neuman and Sándor gave the inequality

$$Q(a,b)^{\frac{1}{3}}A(a,b)^{\frac{2}{3}}< M(a,b)< {\frac {1}{3}}Q(a,b)+{ \frac{2}{3}}A(a,b).$$

In [8], Sándor proved the inequality

$$G(a,b)^{\frac{1}{3}}A(a,b)^{\frac{2}{3}}< P(a,b)< {\frac {1}{3}}G(a,b)+{ \frac{2}{3}}A(a,b).$$

In [18] and [19], the authors proved that the double inequalities

\begin{aligned}& Q(a,b)^{\alpha_{3}}A^{1-\alpha_{3}}(a,b)< M(a,b)< Q(a,b)^{\beta _{3}}A^{1-\beta_{3}}(a,b), \\& \alpha_{4}Q(a,b)+(1-\alpha_{4})G(a,b)< M(a,b)< \beta _{4}Q(a,b)+(1-\beta_{4})G(a,b) \end{aligned}

hold for all $$a,b>0$$ with $$a\neq b$$ if and only if $$\alpha_{3}\leq 1/3$$, $$\beta_{3}\geq2(\log(2+\sqrt{2})-\log3)/\log2$$, $$\alpha _{4}\leq2/3$$, $$\beta_{4}\geq1/[\sqrt{2}\log(1+\sqrt{2})]$$.

In [20], the authors proved that the double inequality

$$\alpha_{5}A(a,b)+(1-\alpha_{5})G(a,b)< P(a,b)< \beta _{5}A(a,b)+(1-\beta_{5})G(a,b)$$

holds for all $$a,b>0$$ with $$a\neq b$$ if and only if $$\alpha_{5}\leq \pi/2$$, $$\beta_{5}\geq2/3$$.

The main purpose of this paper is to find the least value α and the greatest value β such that the double inequality

$$P^{\alpha}(a,b)T^{1-\alpha}(a,b)< M(a,b)< P^{\beta}(a,b)T^{1-\beta}(a,b)$$

holds for all $$a,b>0$$ with $$a\neq b$$.

Theorem 3.1 and Theorem 3.3 in [21] provide the inequality

$$A(a,b)T(a,b)\leq M^{2}(a,b), P(a,b)M(a,b)\leq A^{2}(a,b),$$

following which one can get $$P^{\frac{1}{3}}(a,b)T^{\frac{2}{3}}\leq M(a,b)$$. Then the lower bound of α in Theorem 3.1 of Section 3 is achieved.

## 2 Lemmas

To establish our main result, we need several lemmas, which we present in this section.

For $$x\in(0,1)$$, the power series expansions of the functions $$\tan ^{-1}(x)$$ and $$\sinh^{-1} (x)$$ are presented as follows:

\begin{aligned}& \sinh^{-1} (x) =x-\frac{x^{3}}{6}+\frac{3x^{5}}{40}- \frac {15x^{7}}{336}+\cdots=\sum^{\infty}_{n=0} \frac {(-1)^{n}(2n)!}{(2n+1)2^{2n}(n!)^{2}}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)

### Lemma 2.1

If $$x\in(0,1)$$, then one has

\begin{aligned}& 1+\frac{6x^{2}}{15}< \sqrt{1+x^{2}}< 1+\frac{x^{2}}{2}, \end{aligned}
(2.3)
\begin{aligned}& \sin^{-1}(x)\sqrt{1-x^{2}}< x-\frac{x^{3}}{3}- \frac{2x^{5}}{15}, \end{aligned}
(2.4)
\begin{aligned}& \sinh^{-1}(x)>x-\frac{x^{3}}{6}, \end{aligned}
(2.5)
\begin{aligned}& x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}< \tan^{-1}(x)< x- \frac {x^{3}}{3}+\frac{x^{5}}{5} , \end{aligned}
(2.6)

and

$$\tan^{-1}(x)>x-\frac{x^{3}}{3}+\frac{x^{5}}{9}.$$
(2.7)

### Proof

Square every terms of inequality (2.3) at the same time, then it is easy to prove it. Inequality (2.4) was proved in Lemma 2.3 of [22]. Inequalities (2.5) and (2.6) follow immediately from equations (2.1) and (2.2), respectively.

Let $$\Phi(x)=\tan^{-1}(x)-(x-\frac{x^{3}}{3}+\frac{x^{5}}{9})$$. Then $$\Phi'(x)=\frac{x^{4}(4-5x^{2})}{9(1+x^{2})}$$. Thus, $$\Phi(x)$$ is strictly increasing on $$(0,\frac{2\sqrt{5}}{5}]$$ and strictly decreasing on $$[\frac{2\sqrt{5}}{5},1)$$. Considering $$\Phi(0)=0$$ and $$\Phi(1)=0.0076\ldots>0$$, we can get $$\Phi(x)>0$$ for $$x \in(0,1)$$. Therefore, inequality (2.7) holds. □

### Lemma 2.2

If $$x\in(0,0.7)$$, then one has

\begin{aligned}& \sin^{-1}(x)\sqrt{1-x^{2}}>x-\frac{x^{3}}{3}- \frac{x^{5}}{5}, \end{aligned}
(2.8)
\begin{aligned}& \tan^{-1}(x)>x-\frac{x^{3}}{3}+\frac{x^{5}}{7}, \end{aligned}
(2.9)

and

$$\sinh^{-1}(x)< x-\frac{2x^{3}}{15}.$$
(2.10)

### Proof

Let

\begin{aligned}& \gamma_{1}(x)=\sin^{-1}(x)\sqrt{1-x^{2}}- \biggl(x-\frac{x^{3}}{3}-\frac{x^{5}}{5}\biggr), \end{aligned}
(2.11)
\begin{aligned}& \gamma_{2}(x)=\tan^{-1}(x)-\biggl(x-\frac{x^{3}}{3}+ \frac{x^{5}}{7}\biggr), \end{aligned}
(2.12)
\begin{aligned}& \gamma_{3}(x)=\biggl(x-\frac{2x^{3}}{15}\biggr)- \sinh^{-1}(x). \end{aligned}
(2.13)

Then

\begin{aligned}& \gamma_{1}'(x)=\frac{x\gamma_{1}^{*}(x)}{\sqrt{1-x^{2}}}, \end{aligned}
(2.14)
\begin{aligned}& \gamma_{2}'(x)=\frac{x^{4}(2-5x^{2})}{7+7x^{2}}, \end{aligned}
(2.15)
\begin{aligned}& \gamma_{3}'(x)=\frac{\gamma_{3}^{*}(x)}{5\sqrt{1+x^{2}}}, \end{aligned}
(2.16)

where

\begin{aligned}& \gamma_{1}^{*}(x)=\bigl(x+x^{3}\bigr) \sqrt{1-x^{2}}-\sin^{-1}(x), \end{aligned}
(2.17)
\begin{aligned}& \gamma_{3}^{*}(x)=\sqrt{1+x^{2}} \bigl(5-2x^{2}\bigr)-5. \end{aligned}
(2.18)

Differentiating $$\gamma_{1}^{*}(x)$$ and $$\gamma_{3}^{*}(x)$$, we have

\begin{aligned}& \gamma_{1}^{*\prime}(x)=\frac{x^{2}(1-4x^{2})}{\sqrt{1-x^{2}}}, \end{aligned}
(2.19)
\begin{aligned}& \gamma_{3}^{*\prime}(x)=\frac{x}{\sqrt{1+x^{2}}}\bigl(1-6x^{2} \bigr). \end{aligned}
(2.20)

Furthermore, direct or numerical computations lead to

\begin{aligned}& \gamma_{1}(0)=0, \qquad \gamma_{1}(0.7)=0.0017 \ldots>0, \end{aligned}
(2.21)
\begin{aligned}& \gamma_{1}^{*}(0)=0,\qquad \gamma_{1}^{*}(1)=-1.5708 \ldots< 0, \end{aligned}
(2.22)
\begin{aligned}& \gamma_{2}(0)=0, \qquad \gamma_{2}(0.7)=0.0010 \ldots>0, \end{aligned}
(2.23)
\begin{aligned}& \gamma_{3}(0)=0,\qquad \gamma_{3}(0.7)=0.0016 \ldots>0, \end{aligned}
(2.24)

and

$$\gamma_{3}^{*}(0)=0, \qquad \gamma_{3}^{*}(1)=-0.7574\ldots< 0.$$
(2.25)

From (2.19), we can easy to see that $$\gamma_{1}^{*}(x)$$ is strictly increasing on $$(0,\frac{1}{2}]$$ and strictly decreasing on $$[\frac {1}{2},1)$$. This fact and (2.22) together with (2.14) imply that there exists $$x_{0} \in(\frac{1}{2},1)$$, such that $$\gamma _{1}'(x)>0$$ on $$(0,x_{0})$$ and $$\gamma_{1}'(x)<0$$ on $$(x_{0},1)$$. The monotonicity of $$\gamma_{1}(x)$$ and (2.21) lead to

$$\gamma_{1}(x)>0$$

for x $$\in(0,0.7)$$. Therefore, inequality (2.8) holds.

Equation (2.15) shows that $$\gamma_{2}(x)>0$$ on $$(0,\frac{\sqrt {10}}{5})$$ and $$\gamma_{2}(x)<0$$ on $$(\frac{\sqrt{10}}{5},1)$$. This fact and (2.23) lead to

$$\gamma_{2}(x)>0$$

for x $$\in(0,0.7)$$. That is to say inequality (2.9) holds.

By (2.20), we know that $$\gamma_{3}^{*}(x)$$ is strictly increasing on $$(0,\frac{\sqrt{6}}{6}]$$ and strictly decreasing on $$[\frac{\sqrt {6}}{6},1)$$. This fact and (2.25) together with (2.16) imply that there must exist $$x_{1} \in(\frac{\sqrt{6}}{6},1)$$, such that $$\gamma_{3}'(x)>0$$ on $$(0,x_{1})$$ and $$\gamma_{3}'(x)<0$$ on $$(x_{1},1)$$. It follows from the monotonicity of $$\gamma_{3}(x)$$ and (2.24) that

$$\gamma_{3}(x)>0$$

for x $$\in(0,0.7)$$. This means the inequality (2.10) holds. □

### Lemma 2.3

If x $$\in(0.7,1)$$, the double inequality

$$x-\frac{x^{3}}{3}+\frac{2x^{5}}{17}< \tan^{-1}(x)< x- \frac{x^{3}}{3}+\frac {2x^{5}}{13}$$
(2.26)

holds.

### Proof

Let

\begin{aligned}& \xi_{1}(x)=\biggl(x-\frac{x^{3}}{3}+\frac{2x^{5}}{13}\biggr)- \tan^{-1}(x), \\& \xi_{2}(x)=\biggl(x-\frac{x^{3}}{3}+\frac{2x^{5}}{17}\biggr)- \tan^{-1}(x). \end{aligned}

Then

\begin{aligned}& \xi_{1}'(x)=\frac{x^{4}(10x^{2}-3)}{13(1+x^{2})}, \end{aligned}
(2.27)
\begin{aligned}& \xi_{2}'(x)=\frac{x^{4}(10x^{2}-7)}{17(1+x^{2})}. \end{aligned}
(2.28)

Equality (2.27) implies that $$\xi_{1}(x)$$ is strictly increasing on $$[\frac{\sqrt{30}}{10},1)$$. Additional numerical computations lead to $$\frac{\sqrt{30}}{10}<0.7$$ and $$\xi_{1}(0.7)=0.0007976\ldots>0$$. Therefore, we can get $$\xi_{1}(x)>0$$ for x $$\in(0.7,1)$$. This implies the right hand side of the double inequality (2.26) holds.

Equality (2.28) implies $$\xi_{2}(x)$$ is strictly decreasing on $$(0,\frac{\sqrt{70}}{10}]$$ and strictly increasing on $$[\frac{\sqrt {70}}{10},1)$$. Because of $$\xi_{2}(0)=0$$ and $$\xi _{2}(1)=-0.0011\ldots<0$$, it leads to $$\xi_{2}(x)<0$$ for $$x \in(0,1)$$. Specially, for $$x\in(0.7,1)$$. This means the left hand side of the double inequality (2.26) holds. □

### Lemma 2.4

Let

\begin{aligned}& \mu_{1}(x)=\frac{1+3x^{2}}{(x+x^{3})^{2}}-\frac {1}{(1+x^{2})[\sinh^{-1}(x)]^{2}}- \frac{x}{(1+x^{2})^{\frac{3}{2}}\sinh^{-1}(x)}, \end{aligned}
(2.29)
\begin{aligned}& \mu_{2}(x)=-\frac{1+3x^{2}}{(x+x^{3})^{2}}+\frac {1}{(1-x^{2})[\sin^{-1}(x)]^{2}}- \frac{x}{(1-x^{2})^{\frac{3}{2}}\sin^{-1}(x)}, \end{aligned}
(2.30)

and

$$\mu_{3}(x)=-\frac{2x}{(1+x^{2})^{2}\tan^{-1}(x)}-\frac{1}{[\tan ^{-1}(x)]^{2}(1+x^{2})^{2}}+ \frac{1+3x^{2}}{(x+x^{3})^{2}}.$$
(2.31)

Then, for any $$x\in(0.7,1)$$, we have

$$\mu_{1}(x)< 0.17, \qquad \mu_{2}(x)< -1.48,$$
(2.32)

and

$$\mu_{3}(x)>-0.05.$$
(2.33)

### Proof

From Lemmas 2.6 and 2.7 of [22], for any $$x\in [0.7,1)$$, we can get $$\mu_{1}'(x)\leq0.167\ldots<0.17$$ and $$\mu _{2}'(x)\leq-1.48798\ldots<-1.48$$, respectively.

Differentiating $$\mu_{3}(x)$$, we have

$$\mu_{3}'(x)=\frac{2\eta_{1}(x)+6x^{2}\tan^{-1}(x)\eta _{2}(x)}{[\tan^{-1}(x)(x+x^{3})]^{3}},$$
(2.34)

where

$$\eta_{1}(x)=x^{3}-x^{3}\bigl[ \tan^{-1}(x)\bigr]^{2}-\bigl[\tan^{-1}(x) \bigr]^{3}$$

and

$$\eta_{2}(x)=x^{2}+x^{3}\tan^{-1}(x)- \bigl(2x^{2}+1\bigr)\bigl[\tan^{-1}(x)\bigr]^{2}.$$

For any $$x \in[0.7,1)$$,

\begin{aligned} \eta_{1}(x) < &x^{3}-x^{3}\biggl(x- \frac{x^{3}}{3}+\frac{x^{5}}{9}\biggr)^{2}-\biggl(x- \frac{x^{3}}{3}+\frac {x^{5}}{9}\biggr)^{3} \\ =&-x^{9}\bigl(54+x^{6}\bigr)< 0 \end{aligned}
(2.35)

and

\begin{aligned} \eta_{2}(x) < &x^{2}+x^{3}\biggl(x- \frac{x^{3}}{3}+\frac{2x^{5}}{13}\biggr)-\bigl(2x^{2}+1\bigr) \biggl(x-\frac {x^{3}}{3}+\frac{2x^{5}}{17}\biggr)^{2} \\ =&\frac{x^{4}}{1\text{,}989}\bigl(-663+1\text{,}300x^{2}-637x^{4} \bigr)+\frac {x^{8}}{33\text{,}813}\bigl(-4\text{,}743+4\text{,}836x^{2}-936x^{4} \bigr) \end{aligned}

follow from inequalities (2.7) and (2.26), respectively. Because $$-663+1\text{,}300x^{2}-637x^{4}<0$$ and $$-4\text{,}743+4\text{,}836x^{2}-936x^{4}<0$$ for x $$\in (0.7,1)$$, we have

$$\eta_{2}(x)< 0$$
(2.36)

for $$x\in(0.7,1)$$. Thus $$\mu_{3}'(x)<0$$ for $$x\in(0.7,1)$$ follows from (2.34), (2.35), and (2.36). Therefore, we obtain $$\mu_{3}(x)>\mu_{3}(1)=-0.0419\ldots>-0.05$$ for $$x\in(0.7,1)$$. □

### Lemma 2.5

Let $$f(x)=\frac{1}{\sqrt{1+x^{2}}\sinh^{-1}(x)}-(1-\lambda_{0})\frac {1}{(1+x^{2})\tan^{-1}(x)}-\lambda_{0}\frac{1}{\sqrt {1-x^{2}}\sin^{-1}(x)}$$, where $$\lambda_{0}=\frac{\log(\frac{4\log (1+\sqrt{2})}{\pi})}{\log2}=0.1663\ldots$$ . Then the function $$f(x)$$ is strictly decreasing on $$(0.7,1)$$.

### Proof

It is obvious that

\begin{aligned} f(x) =&\biggl[\frac{1}{\sqrt{1+x^{2}}\sinh^{-1}(x)}-\frac{1}{x(1+x^{2})}\biggr] + \lambda_{0}\biggl[\frac{1}{x(1+x^{2})}-\frac{1}{\sqrt{1-x^{2}}\sin^{-1}(x)}\biggr] \\ &{}+(\lambda_{0}-1)\biggl[\frac{1}{(1+x^{2})\tan^{-1}(x)}-\frac{1}{x(1+x^{2})} \biggr] \\ :=&U_{1}(x)+\lambda_{0}U_{2}(x)+( \lambda_{0}-1)U_{3}(x). \end{aligned}

Differentiating $$f(x)$$, we have

\begin{aligned} f'(x) =&U'_{1}(x)+\lambda_{0}U'_{2}(x)+( \lambda_{0}-1)U'_{3}(x) \\ =&\mu_{1}(x)+\lambda_{0}\mu_{2}(x)+( \lambda_{0}-1)\mu_{3}(x), \end{aligned}
(2.37)

where $$\mu_{1}(x)$$, $$\mu_{2}(x)$$, and $$\mu_{3}(x)$$ are defined as in Lemma 2.4. Therefore, Lemma 2.4 and equation (2.37) yield

\begin{aligned} f'(x) < &0.17+\lambda_{0}(-1.48)+(\lambda_{0}-1) (-0.05) \\ =&-0.0345\ldots< 0 \end{aligned}

for $$x\in(0.7,1)$$. The proof is completed. □

## 3 Main results

### Theorem 3.1

The double inequality

$$P^{\alpha}(a,b)T^{1-\alpha}(a,b)< M(a,b)< P^{\beta}(a,b)T^{1-\beta}(a,b)$$

holds for all $$a,b > 0$$ with $$a\neq b$$ if and only if $$a\geq1/3$$ and $$\beta\leq\frac{\log(\frac{4\log(1+\sqrt{2})}{\pi})}{\log 2}=0.1663\ldots$$ .

### Proof

Because $$P(a,b)$$, $$M(a,b)$$, and $$T(a,b)$$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $$a>b$$. Let $$p\in(0,1)$$, $$\lambda_{0}=\frac{\log(\frac{4\log (1+\sqrt{2})}{\pi})}{\log2}$$, and $$x=(a-b)/(a+b)$$. Then $$x\in(0,1)$$ and

\begin{aligned}& p\log\bigl[P(a,b)\bigr]+(1-p)\log\bigl[T(a,b)\bigr]-\log \bigl[M(a,b)\bigr] \\& \quad =p\log\biggl[\frac{x}{\sin^{-1}(x)}\biggr]+(1-p)\log\biggl[\frac {x}{\tan^{-1}(x)} \biggr]-\log\biggl[\frac{x}{\sinh^{-1}(x)}\biggr] \\& \quad =\log\bigl[\sinh^{-1}(x)\bigr]-(1-p)\log\bigl[ \tan^{-1}(x)\bigr]-p\log\bigl[\sin^{-1}(x) \bigr]:=D_{p}(x). \end{aligned}
(3.1)

It follows that

$$D_{p}\bigl(0^{+}\bigr)=0 \quad \mbox{and} \quad D_{\lambda_{0}}\bigl(1^{-}\bigr)=0.$$
(3.2)

Differentiating $$D_{p}(x)$$, we have

\begin{aligned} D_{p}'(x) =&\frac{1}{\sinh^{-1}(x)\sqrt{1+x^{2}}}-(1-p) \frac {1}{\tan^{-1}(x)(1+x^{2})}-p\frac{1}{\sin^{-1}(x)\sqrt{1-x^{2}}} \\ =&\frac{g_{p}(x)}{\sinh^{-1}(x)(1+x^{2})\tan^{-1}(x)\sin^{-1}(x)\sqrt {1-x^{2}}}, \end{aligned}
(3.3)

where

\begin{aligned} g_{p}(x) =&\bigl[\sqrt{1+x^{2}} \tan^{-1}(x)-(1-p)\sinh^{-1}(x)\bigr]\sin ^{-1}(x) \sqrt{1-x^{2}} \\ &{}-p\sinh^{-1}(x) \bigl(1+x^{2}\bigr)\tan^{-1}(x). \end{aligned}
(3.4)

On one hand, when $$p=\frac{1}{3}$$, Lemma 2.1 and equation (3.4) lead to

\begin{aligned} g_{\frac{1}{3}}(x) =&\biggl[\sqrt{1+x^{2}} \tan^{-1}(x)-\frac{2}{3}\sinh ^{-1}(x)\biggr] \sin^{-1}(x)\sqrt{1-x^{2}} \\ &{}-\frac{1}{3}\sinh^{-1}(x) \bigl(1+x^{2}\bigr) \tan^{-1}(x) \\ < &\biggl[\biggl(1+\frac{x^{2}}{2}\biggr) \biggl(x-\frac{x^{3}}{3}+ \frac{x^{5}}{5}\biggr)-\frac {2}{3}\biggl(x-\frac{x^{3}}{6}\biggr) \biggr] \biggl(x-\frac{x^{3}}{3}-\frac{2x^{5}}{15}\biggr) \\ &{}-\frac{1}{3}\biggl(x-\frac{x^{3}}{6}\biggr) \bigl(1+x^{2} \bigr) \biggl(x-\frac{x^{3}}{3}+\frac {x^{5}}{5}-\frac{x^{7}}{7}\biggr) \\ =&\frac{x^{6}}{3\text{,}150}\bigl(-70+80x^{2}+41x^{4}-67x^{6} \bigr)< 0 \end{aligned}
(3.5)

for $$x\in(0,1)$$. According to (3.3) and (3.5), we can see that

$$D_{\frac{1}{3}}'(x)< 0$$
(3.6)

for $$x\in(0,1)$$.

On the other hand, when $$p=\lambda_{0}$$, the inequalities (2.3) and (2.6) and Lemma 2.2 together with equation (3.4) lead to

\begin{aligned} g_{\lambda_{0}}(x) =&\bigl[\sqrt{1+x^{2}} \tan^{-1}(x)-(1-\lambda_{0})\sinh ^{-1}(x)\bigr] \sin^{-1}(x)\sqrt{1-x^{2}} \\ &{}-\lambda_{0}\sinh^{-1}(x) \bigl(1+x^{2}\bigr) \tan^{-1}(x) \\ >&\biggl[\biggl(1+\frac{6x^{2}}{15}\biggr) \biggl(x-\frac{x^{3}}{3}+ \frac{x^{5}}{7}\biggr)-(1-\lambda _{0}) \biggl(x-\frac{2x^{3}}{15} \biggr)\biggr]\biggl(x-\frac{x^{3}}{3}-\frac{x^{5}}{5}\biggr) \\ &{}-\lambda_{0}\biggl(x-\frac{2x^{3}}{15}\biggr) \bigl(1+x^{2}\bigr) \biggl(x-\frac{x^{3}}{3}+\frac {x^{5}}{5} \biggr) \\ =&\frac{x^{4}}{1\text{,}575}F_{\lambda_{0}}(x) \end{aligned}
(3.7)

for $$x\in(0,0.7)$$, where

$$F_{\lambda_{0}}(x)=(315-1\text{,}575\lambda_{0})+(105\lambda _{0}-90)x^{2}+(22-301\lambda_{0})x^{4}+(42 \lambda_{0}-33)x^{6}-18x^{8}.$$

Because of $$42\lambda_{0}-33=-26.0142\ldots<0$$ and $$105\lambda _{0}-90=-72.5354\ldots<0$$, it follows that

\begin{aligned} F_{\lambda_{0}}(x) >&(315-1\text{,}575\lambda_{0})+(105 \lambda _{0}-90)x^{2}+(22-301\lambda_{0})x^{4}+(42 \lambda _{0}-33)x^{4}-18x^{4} \\ =&(315-1\text{,}575\lambda_{0})+(105\lambda_{0}-90)x^{2}+(-29-259 \lambda _{0})x^{4} \\ :=&F^{*}(x) \end{aligned}
(3.8)

and

$$F^{*\prime}(x)=2(105\lambda_{0}-90)x+4(-29-259 \lambda_{0})x^{3}< 0$$

for $$x\in(0,0.7)$$. Thus, we can get

$$F_{\lambda_{0}}(x)>F^{*}(x)>F^{*}(0.7)=0.1825 \ldots>0$$
(3.9)

for $$x\in(0,0.7)$$. Therefore, equation (3.3) and inequalities (3.7)-(3.9) imply

$$D_{\lambda_{0}}'(x)>0$$
(3.10)

for $$x\in(0,0.7)$$.

It follows from equation (3.3) and Lemma 2.5 that $$D_{\lambda _{0}}'(x)$$ is strictly decreasing on $$(0.7,1)$$. Then from equation (3.10) and $$D_{\lambda_{0}}'(1^{-})=-\infty$$, we know that there exists $$x_{*} \in(0.7,1)$$ such that $$D_{\lambda_{0}}(x)$$ is strictly increasing on $$(0,x_{*}]$$ and strictly decreasing on $$[x_{*},1)$$. This in conjunction with (3.2) means that

$$D_{\lambda_{0}}(x)>0$$
(3.11)

for x $$\in(0,1)$$.

Therefore, for all $$a,b>0$$ with $$a\neq b$$,

$$M(a,b)>P^{\frac{1}{3}}(a,b)T^{\frac{2}{3}}(a,b),$$
(3.12)

follows from equations (3.1), (3.2), and (3.6) as well as

$$M(a,b)< P^{\lambda_{0}}(a,b)T^{1-\lambda_{0}}(a,b)$$
(3.13)

follows from equations (3.1), (3.2), and (3.11).

Finally, by easy computations, equations (1.1), (1.2), and (1.3) lead to

\begin{aligned}& \frac{\log[T(a,b)]-\log[M(a,b)]}{\log[T(a,b)]-\log[P(a,b)]}=\frac {\log[\sinh^{-1}(x)]-\log[\tan^{-1}(x)]}{\log[\sin^{-1}(x)]-\log [\tan^{-1}(x)]}, \end{aligned}
(3.14)
\begin{aligned}& \lim_{x\rightarrow0^{+}}\frac{\log[\sinh^{-1}(x)]-\log [\tan^{-1}(x)]}{\log[\sin^{-1}(x)]-\log[\tan^{-1}(x)]}=\frac{1}{3} \end{aligned}
(3.15)

and

$$\lim_{x\rightarrow1^{-}}\frac{\log[\sinh^{-1}(x)]-\log [\tan^{-1}(x)]}{\log[\sin^{-1}(x)]-\log[\tan^{-1}(x)]}= \lambda_{0}.$$
(3.16)

Thus, we have the following two claims.

### Claim 1

If $$\alpha<\frac{1}{3}$$, then from (3.14) and (3.15), there must exist $$\delta_{1}\in(0,1)$$ such that $$M(a,b)< P^{\alpha}(a,b)T^{1-\alpha}(a,b)$$ for all $$a,b >0$$ with $$(a-b)/(a+b)\in(0,\delta_{1})$$.

### Claim 2

If $$\beta>\lambda_{0}$$, then from (3.14) and (3.16), there must exist $$\delta_{2} \in(0,1)$$ such that $$M(a,b)>P^{\beta}(a,b)T^{1-\beta}(a,b)$$ for all $$a,b >0$$ with $$(a-b)/(a+b)\in(1-\delta_{2},1)$$.

Inequalities (3.12) and (3.13) in conjunction with the above two claims mean the proof is completed.  □

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## Acknowledgements

This research was supported by the Program of the National Science Foundation of China (Grant No. 11501002), Doctoral Scientific Research Foundation of Anhui University and Scientific Research Training for Undergraduate of Anhui University (KYXL2014002), China.

## Author information

Authors

### Corresponding author

Correspondence to Bo-Yong Long.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

H-YH carried out the proof of Theorem 3.1. NW carried out the proof of Lemmas 2.1-2.3. B-YL provided the main idea and carried out the proof of Lemmas 2.4 and 2.5. All authors read and approved the final manuscript.

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Huang, HY., Wang, N. & Long, BY. Optimal bounds for Neuman-Sándor mean in terms of the geometric convex combination of two Seiffert means. J Inequal Appl 2016, 14 (2016). https://doi.org/10.1186/s13660-015-0955-2