New Bounds for Arithmetic Mean by the Seiffert-like Means

Abstract

By using the power series of the functions $1/{sin}^{n}t$ and $\left(cost\right)/\left({sin}^{n}t\right)$ ($n=1,2,3,4,5$), and the estimation of the ratio of two adjacent Bernoulli numbers, we obtained new bounds for arithmetic mean $\mathcal{A}$ by the weighted arithmetic means of ${\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}$ and $\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin},$${\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}$ and $\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh},$ where ${\mathcal{M}}_{tan}(x,y)$ and ${\mathcal{M}}_{sin}(x,y)$, ${\mathcal{M}}_{tanh}(x,y)$ and ${\mathcal{M}}_{sinh}(x,y)$ are the tangent mean, sine mean, hyperbolic tangent mean and hyperbolic sine mean, respectively. The upper and lower bounds obtained in this paper are compared in detail with the conclusions of the previous literature.

1. Introduction

Mean value inequality is an established topic in the field of inequality research. It is rooted in mathematics, and is applied to the fields of mathematics, statistics and engineering technology. Throughout the full text, we assume x and y to be two different positive numbers, and $\mathcal{H}(x,y),\mathcal{G}(x,y),\mathcal{L}(x,y),{\mathcal{M}}_{1/2}(x,y),{\mathcal{H}}_e(x,y),\mathcal{A}(x,y)$ and ${\mathcal{M}}_2(x,y)$ are the harmonic, geometric, logarithmic, $\frac{1}{2}-$ power mean, Heron, arithmetic and quadratic means, respectively, where

$$\begin{array}{cccc}\mathcal{H}(x,y)=\frac{2}{1/x+1/y},& \mathcal{G}(x,y)=\sqrt{xy},& \mathcal{L}(x,y)=\frac{y-x}{logy-logx},& {\mathcal{M}}_{1/2}(x,y)={\left(\frac{x^{1/2}+y^{1/2}}{2}\right)}^2,\\ {\mathcal{H}}_e(x,y)=\frac{x+\sqrt{xy}+y}{3},& \mathcal{A}(x,y)=\frac{x+y}{2},& {\mathcal{M}}_2(x,y)={\left(\frac{x^2+y^2}{2}\right)}^{1/2}.\end{array}$$

For convenience, we note that $\mathcal{M}<\mathcal{N}$ means $\mathcal{M}(x,y)<\mathcal{N}(x,y)$ holds for two means $\mathcal{M}$ and $\mathcal{N}$ with two different positive numbers x and y. In order to explore the various relationships between the above means, we can take the geometric mean $\mathcal{G}(x,y)$ as a reference and make two transformations $\sqrt{y/x}=\tau =e^w$ to obtain the following results:

$$\begin{array}{ccc}\hfill \frac{\mathcal{H}(x,y)}{\mathcal{G}(x,y)}& =& \frac{1}{coshw},\frac{\mathcal{L}(x,y)}{\mathcal{G}(x,y)}=\frac{sinhw}{w},\frac{{\mathcal{M}}_{1/2}(x,y)}{\mathcal{G}(x,y)}=\frac{1}{2}coshw+\frac{1}{2},\hfill \\ \hfill \frac{{\mathcal{H}}_e(x,y)}{\mathcal{G}(x,y)}& =& \frac{2}{3}coshw+\frac{1}{3},\frac{\mathcal{A}(x,y)}{\mathcal{G}(x,y)}=coshw,\frac{{\mathcal{M}}_2(x,y)}{\mathcal{G}(x,y)}=\sqrt{cosh2w},w>0.\hfill \end{array}$$

Then, via the relationships regarding the hyperbolic sine function and hyperbolic cosine function,

$$\frac{1}{coshw}<1<\frac{sinhw}{w}<\frac{1}{2}coshw+\frac{1}{2}<\frac{1}{3}+\frac{2}{3}coshw<coshw<\sqrt{cosh2w}$$

(1)

we can obtain the inequality chain

$$\mathcal{H}<\mathcal{G}<\mathcal{L}<{\mathcal{M}}_{1/2}<{\mathcal{H}}_e<\mathcal{A}<{\mathcal{M}}_2.$$

(2)

For a study of classical mean inequality, see [1,2,3,4,5,6,7,8].

The above is to determine the relationships of classical means through the relationships of hyperbolic functions. In recent years, scholars have been exploring other novel methods to study the relationships between classical means and newly introduced means. Among them, the so-called Seiffert function method is worth introducing.

In 1998, Kahlig and Matkowski [9] introduced a new concept in the field of means; the ratio of a homogeneous bivariable mean $\mathcal{M}$ in $(0,\infty )$ to a classical mean $\mathcal{N}$ can be expressed as a function of $(x-y)/(x+y)$, which is called the index function of $\mathcal{M}$ with respect to $\mathcal{N}$ or an $\mathcal{N}$-index of $\mathcal{M}$:

$$\frac{\mathcal{M}(x,y)}{\mathcal{N}(x,y)}=f_{\mathcal{M},\mathcal{N}}\left(\frac{x-y}{x+y}\right),$$

where $f_{\mathcal{M},\mathcal{N}}$: $\left(-1,1\right)⟶\left(0,2\right)$ is a unique single variable function (with the graph lying in a set of a butterfly shape).

Assuming $f:(0,1)\to {\mathbb{R}}^{+}$, Witkowski [10] constructed a new binary function:

$${\mathcal{M}}_f(x,y)=\left\{\begin{array}{cc}\frac{\left|x-y\right|}{2f\left(\frac{\left|x-y\right|}{x+y}\right)}& x\ne y\\ x& x=y\end{array}\right..$$

(3)

When the function $f\left(t\right)$ satisfies

$$\frac{t}{1+t}\le f\left(t\right)\le \frac{t}{1-t},$$

Witkowski [10] proved that ${\mathcal{M}}_f(x,y)$ is a symmetric and homogeneous mean. In this case, the function $f\left(t\right)$ with the property above produces a corresponding mean, and there is the following relationship between the two functions $f\left(t\right)$ and ${\mathcal{M}}_f(x,y)$:

$$f\left(t\right)=\frac{t}{{\mathcal{M}}_f(1-t,1+t)},$$

(4)

where

$$t=\frac{\left|x-y\right|}{x+y}.$$

Therefore, $f\left(t\right)$ and ${\mathcal{M}}_f(x,y)$ form a one-to-one correspondence via (3) and (4). For this reason, we can rewrite $f\left(t\right)=:f_{\mathcal{M}}\left(t\right)$. In general, $f\left(t\right)$ is called the Seiffert function.

In order to make the above new method of dealing with mean value inequality more useful, we make the following appropriate modifications to it. We change the parameters sign of a mean ${\mathcal{M}}_f$ about two parameters with u and v, and assume that $0<u<v$. Thus, there must be three positive numbers $x,y,$ and $\lambda$, so that

$$\left\{\begin{array}{c}u=\lambda \frac{2x}{x+y}\\ v=\lambda \frac{2y}{x+y}\end{array}\right.,0<x<y.$$

Then, we have that $0<t<1,$

$$t=\frac{y-x}{x+y},$$

and

$$\begin{array}{ccc}\hfill {\mathcal{M}}_f(u,v)& =& {\mathcal{M}}_f\left(\lambda \frac{2x}{x+y},\lambda \frac{2y}{x+y}\right)\hfill \\ & =& \lambda {\mathcal{M}}_f\left(\frac{2x}{x+y},\frac{2y}{x+y}\right)=\lambda {\mathcal{M}}_f\left(1-t,1+t\right)\hfill \\ & =& \lambda \frac{t}{f_{\mathcal{M}}\left(t\right)}.\hfill \end{array}$$

(5)

Via (4), we can obtain that

$$\begin{array}{ccc}\hfill f_{\mathcal{H}}\left(t\right)& =& \frac{t}{\frac{2}{\frac{1}{1-t}+\frac{1}{1+t}}}=\frac{t}{1-t^2}\hfill \\ \hfill f_{\mathcal{G}}\left(t\right)& =& \frac{t}{\sqrt{\left(1-t\right)\left(1+t\right)}}=\frac{t}{\sqrt{1-t^2}},\hfill \\ \hfill f_{\mathcal{L}}\left(t\right)& =& \frac{t}{\frac{\left(1+t\right)-\left(1-t\right)}{ln\left(1+t\right)-ln\left(1-t\right)}}=\frac{1}{2}ln\left(t+1\right)-\frac{1}{2}ln\left(1-t\right)\hfill \\ & =& \frac{1}{2}ln\frac{1+t}{1-t}={tanh}^{-1}t,\hfill \\ \hfill f_{{\mathcal{M}}_{1/2}}\left(t\right)& =& \frac{t}{\left(\frac{\left(1-t\right)+\left(1+t\right)}{2}+\sqrt{\left(1-t\right)\left(1+t\right)}\right)/2}\hfill \\ & =& \frac{2t}{1+\sqrt{1-t^2}},\hfill \\ \hfill f_{He}\left(t\right)& =& \frac{t}{\frac{2+\sqrt{1-t^2}}{3}}=\frac{3t}{2+\sqrt{1-t^2}},\hfill \\ \hfill f_{\mathcal{A}}\left(t\right)& =& \frac{t}{\frac{\left(1-t\right)+\left(1+t\right)}{2}}=t,\hfill \\ \hfill f_{{\mathcal{M}}_2}\left(t\right)& =& \frac{t}{\sqrt{\frac{{\left(1-t\right)}^2+{\left(1+t\right)}^2}{2}}}=\frac{t}{\sqrt{1+t^2}}.\hfill \end{array}$$

(6)

Then, by (5), we come to (2) when the following inequality chain holds:

$$\frac{t}{\sqrt{1+t^2}}<t<\frac{3t}{2+\sqrt{1-t^2}}<\frac{2t}{1+\sqrt{1-t^2}}<\frac{1}{2}ln\frac{1+t}{1-t}<\frac{t}{\sqrt{1-t^2}}<\frac{t}{1-t^2},$$

(7)

which is not difficult to prove. This is a surprising result of using the Seiffert function method.

In [10], Witkowski introduced the following two new means, one called the sine mean

$${\mathcal{M}}_{sin}(x,y)=\left\{\begin{array}{cc}\frac{\left|x-y\right|}{2sin\left(\frac{\left|x-y\right|}{x+y}\right)}& x\ne y\\ x& x=y\end{array}\right.,$$

and the other called the hyperbolic tangent mean

$${\mathcal{M}}_{tanh}(x,y)=\left\{\begin{array}{cc}\frac{\left|x-y\right|}{2tanh\left(\frac{\left|x-y\right|}{x+y}\right)}& x\ne y\\ x& x=y\end{array}\right..$$

In [11,12], Nowicka and Witkowski introduced the dual forms of the above two Seiffert-like means, which are called the tangent mean

$${\mathcal{M}}_{tan}(x,y)=\left\{\begin{array}{cc}\frac{\left|x-y\right|}{2tan\left(\frac{\left|x-y\right|}{x+y}\right)}& x\ne y\\ x& x=y\end{array}\right.,$$

and hyperbolic sine mean

$${\mathcal{M}}_{sinh}(x,y)=\left\{\begin{array}{cc}\frac{\left|x-y\right|}{2sinh\left(\frac{\left|x-y\right|}{x+y}\right)}& x\ne y\\ x& x=y\end{array}\right..$$

By (4), we have that

$$f_{{\mathcal{M}}_{tan}}\left(t\right)=tant,f_{{\mathcal{M}}_{sin}}\left(t\right)=sint,f_{{\mathcal{M}}_{tanh}}\left(t\right)=tanht,f_{{\mathcal{M}}_{sinh}}\left(t\right)=sinht.$$

(8)

For $t>0,$

$$tanht<sint<t<sinht<tant,$$

via (5), we can immediately obtain the following inequalities about these five means:

$${\mathcal{M}}_{tan}<{\mathcal{M}}_{sinh}<\mathcal{A}<{\mathcal{M}}_{sin}<{\mathcal{M}}_{tanh}.$$

(9)

Since the above new four Seiffert-like means have the Seiffert functions sin, tan, sinh and tanh, by (3), the inverse counterparts of which can produce the first Seiffert mean [13], second Seiffert mean [14], Neuman–Sandor mean [15] and logarithmic mean [16]. Inspired by this form of new mean, many scholars have presented achievements in the field of inequality in a short period of time. See the literature [17,18,19,20,21,22,23,24,25,26,27,28] for details.

Nowicka and Witkowski [17] determined various optimal bounds for $\mathcal{A}(x,y)$ by ${\mathcal{M}}_{tan}(x,y)$ and ${\mathcal{M}}_{sinh}(x,y)$, ${\mathcal{M}}_{tanh}(x,y)$ and ${\mathcal{M}}_{sin}(x,y)$, and obtained the following inequalities:

$$\begin{array}{ccc}\hfill {\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}& <& \mathcal{A}<\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin},\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill {\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}& <& \mathcal{A}<\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}.\hfill \end{array}$$

(11)

Recently, W.-M. Qian, T.-H. Zhao and Y.-P. Lv [25] considered the weighted geometric means of ${\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}$ and $\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}$, and the ones of ${\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}$ and $\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}$ as the bounds for $\mathcal{A}$, and obtained the following two inequality conclusions:

(i) the double inequality

$${\left({\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}\right)}^{1-{\lambda }_1}{\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right)}^{{\lambda }_1}<\mathcal{A}<{\left({\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}\right)}^{1-{\mu }_1}{\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right)}^{{\mu }_1}$$

(12)

holds if and only if ${\lambda }_1\le {\lambda }_1^{*}=4/5$ and

$${\mu }_1\ge {\mu }_1^{*}=\frac{3ln\left(sin1\right)-lncos1}{3ln\left[\left(2+cos1\right)/3\right]-ln\left(cos1\right)}\approx 0.8386;$$

(ii) the double inequality

$${\left({\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}\right)}^{1-{\lambda }_2}{\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)}^{{\lambda }_2}<\mathcal{A}<{\left({\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}\right)}^{1-{\mu }_2}{\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)}^{{\mu }_2}$$

(13)

holds if and only if

$${\lambda }_2\le {\lambda }_2^{*}=\frac{3ln(sinh1)-ln(cosh1)}{3ln\left[\left(2+cosh1\right)/3\right]-ln\left(cosh1\right)}\approx 0.7730$$

and ${\mu }_2$$\ge {\mu }_2^{*}=4/5.$

Since the arithmetic mean of two different positive numbers is greater than their geometric mean, in this paper, we consider the weighted arithmetic mean of ${\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}$ and $\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}$, and the ones of ${\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}$ and $\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}$ as the bounds for $\mathcal{A}$, and obtain the following two conclusions.

Theorem 1.

Let

$$\begin{array}{ccc}\hfill {\alpha }_1^{*}& =& \frac{4}{5},\hfill \\ \hfill {\beta }_1^{*}& =& \frac{1-{\left(\frac{1}{tan1}\right)}^{1/3}{\left(\frac{1}{sin1}\right)}^{2/3}}{\left(\frac{1}{3}\frac{1}{tan1}+\frac{2}{3}\frac{1}{sin1}\right)-{\left(\frac{1}{tan1}\right)}^{1/3}{\left(\frac{1}{sin1}\right)}^{2/3}}\approx 0.83597.\hfill \end{array}$$

Then, the double inequality

$$\left(1-{\alpha }_1\right){\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}+{\alpha }_1\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right)<\mathcal{A}<\left(1-{\beta }_1\right){\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}+{\beta }_1\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right)$$

(14)

holds if and only if ${\alpha }_1\le {\alpha }_1^{*}$ and ${\beta }_1\ge {\beta }_1^{*}.$

Theorem 2.

Let

$$\begin{array}{ccc}\hfill {\alpha }_2^{*}& =& \frac{1-{\left(\frac{1}{tanh1}\right)}^{1/3}{\left(\frac{1}{sinh1}\right)}^{2/3}}{\left(\frac{1}{3}\frac{1}{tanh1}+\frac{2}{3}\frac{1}{sinh1}\right)-{\left(\frac{1}{tanh1}\right)}^{1/3}{\left(\frac{1}{sinh1}\right)}^{2/3}}\approx 0.77116,\hfill \\ \hfill {\beta }_2^{*}& =& \frac{4}{5}.\hfill \end{array}$$

Then, the double inequality

$$\left(1-{\alpha }_2\right){\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}+{\alpha }_2\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)<\mathcal{A}<\left(1-{\beta }_2\right){\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}+{\beta }_2\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)$$

(15)

holds if and only if ${\alpha }_2\le {\alpha }_2^{*}$ and ${\beta }_2\ge {\beta }_2^{*}.$

In this paper, Theorem 2 is proven by using the power series of the functions $1/{sin}^{n}t$ and $\left(cost\right)/\left({sin}^{n}t\right)$ ($n=1,2,3,4,5$), and the estimation of the ratio of two adjacent Bernoulli numbers. At the same time, another main conclusion is proven by using the power series of some hyperbolic functions in parallel. Finally, the upper and lower bounds obtained in this paper are compared in detail with the conclusions of the previous literature.

2. Lemmas

In order to prove the main results of this paper, we need the following lemmas.

Lemma 1.

Let $B_{2n}$ be the even-indexed Bernoulli numbers. Then, we have the following power series expansions

$$\begin{array}{ccc}\hfill \frac{1}{sint}& =& \frac{1}{t}+\sum _{n=1}^{\infty }\frac{2(2^{2n-1}-1)\left|B_{2n}\right|}{\left(2n\right)!}{t}^{2n-1},\hfill \\ \hfill \frac{1}{{sin}^{2}t}& =& \frac{1}{t^2}+\sum _{n=1}^{\infty }\frac{2^{2n}\left(2n-1\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-2},\hfill \\ \hfill \frac{1}{{sin}^{3}t}& =& \frac{1}{t^3}+\frac{1}{2}\frac{1}{t}+\sum _{n=1}^{\infty }\frac{(2^{2n-1}-1)\left|B_{2n}\right|}{\left(2n\right)!}{t}^{2n-1}\hfill \\ & & +\sum _{n=2}^{\infty }\frac{(2^{2n-1}-1)\left(2n-1\right)\left(2n-2\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-3},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \frac{1}{{sin}^{4}t}& =& \frac{1}{t^4}+\frac{2}{3}\frac{1}{t^2}+\frac{2}{3}\sum _{n=1}^{\infty }\frac{2^{2n}\left(2n-1\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-2}\hfill \\ & & +\frac{1}{6}\sum _{n=2}^{\infty }\frac{2^{2n}\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-4},\hfill \\ \hfill \frac{1}{{sin}^{5}t}& =& \frac{1}{t^5}+\frac{5}{6t^3}+\frac{3}{8t}+\frac{3}{4}\sum _{n=1}^{\infty }\frac{(2^{2n-1}-1)\left|B_{2n}\right|}{\left(2n\right)!}{t}^{2n-1}\hfill \\ & & +\frac{5}{6}\sum _{n=2}^{\infty }\frac{(2^{2n-1}-1)\left(2n-1\right)\left(2n-2\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-3}\hfill \\ & & +\frac{1}{12}\sum _{n=3}^{\infty }\frac{(2^{2n-1}-1)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)\left(2n-4\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-5}\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \frac{cost}{sint}& =& cott=\frac{1}{t}-\sum _{n=1}^{\infty }\frac{2^{2n}}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-1},\hfill \\ \hfill \frac{cost}{{sin}^{2}t}& =& \frac{1}{t^2}-\sum _{n=1}^{\infty }\frac{2(2^{2n-1}-1)\left(2n-1\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-2},\hfill \\ \hfill \frac{cost}{{sin}^{3}t}& =& \frac{1}{t^3}-\frac{1}{2}\sum _{n=2}^{\infty }\frac{2^{2n}\left(2n-1\right)\left(2n-2\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-3},\hfill \\ \hfill \frac{cost}{{sin}^{4}t}& =& \frac{1}{t^4}+\frac{1}{6}\frac{1}{t^2}-\frac{1}{3}\sum _{n=1}^{\infty }\frac{(2^{2n-1}-1)\left(2n-1\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-2}\hfill \\ & & -\frac{1}{3}\sum _{n=2}^{\infty }\frac{(2^{2n-1}-1)\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-4},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \frac{cost}{{sin}^{5}t}& =& \frac{1}{t^5}+\frac{1}{3t^3}-\frac{1}{6}\sum _{n=2}^{\infty }\frac{2^{2n}\left(2n-1\right)\left(2n-2\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-3}\hfill \\ & & -\frac{1}{24}\sum _{n=3}^{\infty }\frac{2^{2n}\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)\left(2n-4\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n-5}\hfill \end{array}$$

hold for all $t\in \left(-\pi ,0)\cup (0,\pi \right)$.

Proof. 

The power series expansions of $1/sint$ and $cott$ can be found in [29,30]. Via the relations

$$\begin{array}{ccc}\hfill \frac{cost}{{sin}^{k+1}t}& =& -\frac{1}{k}{\left(\frac{1}{{sin}^{k}t}\right)}^{\prime },\hfill \\ \hfill \frac{1}{{sin}^{k}t}& =& \frac{k-2}{k-1}\frac{1}{{sin}^{k-2}t}-\frac{1}{k-1}{\left(\frac{cost}{{sin}^{k-1}t}\right)}^{\prime }\hfill \end{array}$$

we can obtain the desired results. □

Lemma 2

([31,32,33,34]). Let $B_{2n}$ be the even-indexed Bernoulli numbers. Then

$$\frac{2^{2n-1}-1}{2^{2n+1}-1}\frac{(2n+2)(2n+1)}{{\pi }^2}<\frac{|B_{2n+2}|}{|B_{2n}|}<\frac{2^{2n}-1}{2^{2n+2}-1}\frac{(2n+2)(2n+1)}{{\pi }^2}$$

holds.

Lemma 3.

Let $0<\left|t\right|<\pi /2$ and

$$q\left(t\right)=\frac{2t^{2}sint-6t{cos}^{2}t-3t+9costsint-2t^{2}costsint}{3\left(2sint-t-2tcost+costsint\right)}=:\frac{q_1\left(t\right)}{3q_2\left(t\right)}.$$

Then

$${\left[q\left(t\right)\right]}^3>{cos}^{2}t$$

(16)

holds.

Proof. 

Let

$$G\left(t\right)=\frac{3}{2}lnq\left(t\right)-lncost=\frac{3}{2}ln\frac{q_1\left(t\right)}{3q_2\left(t\right)}-lncost,0<t<\frac{\pi }{2}.$$

Then

$$G^{\prime }\left(t\right)=\frac{\left({sin}^{5}t\right)g\left(t\right)}{2q_1\left(t\right)q_2\left(t\right)},$$

where

$$\begin{array}{ccc}\hfill g\left(t\right)& =& 6t\frac{1}{{sin}^{3}t}-18t\frac{1}{sint}-36\frac{cost}{{sin}^{2}t}+34t^2\frac{1}{{sin}^{2}t}+4t^3\frac{1}{sint}-24t^2\frac{1}{{sin}^{4}t}\hfill \\ & & -14t^3\frac{1}{{sin}^{3}t}+6t^3\frac{1}{{sin}^{5}t}+4t^2+36\frac{1}{{sin}^{2}t}+24t\frac{cost}{sint}-6t\frac{cost}{{sin}^{3}t}\hfill \\ & & +14t^2\frac{cost}{{sin}^{2}t}+24t^2\frac{cost}{{sin}^{4}t}-16t^3\frac{cost}{{sin}^{3}t}-6t^3\frac{cost}{{sin}^{5}t}-36.\hfill \end{array}$$

(17)

All formulas in Lemma 1 are substituted into (17) and sorted out to draw the following conclusion

$$\begin{array}{ccc}\hfill g\left(t\right)& =& -\sum _{n=5}^{\infty }\frac{3}{4}\frac{2^{2n-2}-2}{\left(2n-2\right)!}\left|B_{2n-2}\right|t^{2n}\hfill \\ & & +\sum _{n=5}^{\infty }\frac{3\left(6\times 2^{2n}{n}^2-9\times 2^{2n}n-10\times 2^{2n}+12n^2+6n+4\right)}{\left(2n\right)!}\left|B_{2n}\right|t^{2n}\hfill \\ & & +\sum _{n=5}^{\infty }\frac{2\left(n+1\right)\left(2\left(n+1\right)-1\right)\left(-25\left(n+1\right)+2{\left(n+1\right)}^2+59\right)\left(2^{2n+2}-1\right)}{\left(2n+2\right)!}\left|B_{2n+2}\right|t^{2n}\hfill \\ \hfill =:& & \sum _{n=5}^{\infty }{a}_n{t}^{2n},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill a_n& =& -\frac{3}{4}\frac{2^{2n-2}-2}{\left(2n-2\right)!}\left|B_{2n-2}\right|\hfill \\ & & +\frac{3\left[\left(6n^2-9n-10\right)\times 2^{2n}+12n^2+6n+4\right]}{\left(2n\right)!}\left|B_{2n}\right|\hfill \\ & & +\frac{2\left(n+1\right)\left(2n+1\right)\left(2n^2-21n+36\right)\left(2^{2n+2}-1\right)}{\left(2n+2\right)!}\left|B_{2n+2}\right|.\hfill \end{array}$$

Especially,

$$\begin{array}{ccc}\hfill a_5& =& \frac{1}{1890},a_6=\frac{503}{1984500},a_7=\frac{3251}{43659000},a_8=\frac{41857}{2432430000},a_9=\frac{117756253}{34326452160000},\hfill \\ \hfill a_{10}& =& \frac{39666139957}{64190465539200000},a_{11}=\frac{21005560853}{203269807540800000},a_{12}=\frac{197638373689}{12107488900066560000}.\hfill \end{array}$$

Below, we will prove that $a_n>0$ for all $n\ge 13$. By Lemma 2, we have

$$\begin{array}{ccc}\hfill \frac{a_n}{|B_{2n}|}& =& -\frac{3}{4}\frac{2^{2n-2}-2}{\left(2n-2\right)!}\frac{|B_{2n-2}|}{|B_{2n}|}\hfill \\ & & +\frac{3\left[\left(6n^2-9n-10\right)\times 2^{2n}+12n^2+6n+4\right]}{\left(2n\right)!}\hfill \\ & & +\frac{2\left(n+1\right)\left(2\left(n+1\right)-1\right)\left(-25\left(n+1\right)+2{\left(n+1\right)}^2+59\right)\left(2^{2n+2}-1\right)}{\left(2n+2\right)!}\frac{|B_{2n+2}|}{B_{2n}|}\hfill \end{array}$$

$$\begin{array}{ccc}& >& -\frac{3}{4}\frac{2^{2n-2}-2}{\left(2n-2\right)!}\frac{2^{2n-1}-1}{2^{2n-3}-1}\frac{{\pi }^2}{\left(2n\right)(2n-1)}\hfill \\ & & +\frac{3\left[\left(6n^2-9n-10\right)\times 2^{2n}+12n^2+6n+4\right]}{\left(2n\right)!}\hfill \\ & & +\frac{2\left(n+1\right)\left(2\left(n+1\right)-1\right)\left(-25\left(n+1\right)+2{\left(n+1\right)}^2+59\right)\left(2^{2n+2}-1\right)}{\left(2n+2\right)!}\hfill \\ & & \times \frac{2^{2n-1}-1}{2^{2n+1}-1}\frac{(2n+2)(2n+1)}{{\pi }^2}\hfill \end{array}$$

$$\begin{array}{ccc}& =& -\frac{3}{4}\frac{2^{2n-2}-2}{\left(2n\right)!}\frac{2^{2n-1}-1}{2^{2n-3}-1}\frac{{\pi }^2}{1}\hfill \\ & & +\frac{3\left(6n+6\times 2^{2n}{n}^2-10\times 2^{2n}-9\times 2^{2n}n+12n^2+4\right)}{\left(2n\right)!}\hfill \\ & & +\frac{2\left(n+1\right)\left(2\left(n+1\right)-1\right)\left(-25\left(n+1\right)+2{\left(n+1\right)}^2+59\right)\left(2^{2n+2}-1\right)}{\left(2n\right)!}\hfill \\ & & \times \frac{2^{2n-1}-1}{2^{2n+1}-1}\frac{1}{{\pi }^2}\hfill \\ & =& :\frac{1}{4}\frac{H\left(n\right)}{{\pi }^2\left(2n\right)!\left(2\times 2^{2n}-1\right)},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill H\left(n\right)& =& 2\times 2^{4n}\left[32n^4-288n^3+\left(88+72{\pi }^2\right)n^2-\left(108{\pi }^2-696\right)n-120{\pi }^2-3{\pi }^4+288\right]\hfill \\ & & -3\times 2^{2n}\left[48n^4-432n^3-\left(72{\pi }^2-132\right)n^2+\left(1044-84{\pi }^2\right)n-72{\pi }^2-5{\pi }^4+432\right]\hfill \\ & & +32n^4-288n^3-\left(144{\pi }^2-88\right)n^2-\left(72{\pi }^2-696\right)n-48{\pi }^2-6{\pi }^4+288\hfill \\ & =& :2\times 2^{4n}u\left(n\right)-3\times 2^{2n}v\left(n\right)+w\left(n\right).\hfill \end{array}$$

Since $w\left(13\right)=305424-6{\pi }^4-25320{\pi }^2\approx 54941>0$ and

$$2^{2n}>\frac{3v\left(n\right)}{2u\left(n\right)}$$

(18)

for all $n\ge 13.$ The latter is not difficult to prove by mathematical induction. It is not difficult to verify that (18) is true for $n=13$ due to

$$\begin{array}{ccc}\hfill 2^{2\times 13}-\frac{3v\left(13\right)}{2u\left(13\right)}& =& \frac{1}{2}\frac{-476204512276{\pi }^2+134217723{\pi }^4-13664437994088}{-3548{\pi }^2+{\pi }^4-101808}\hfill \\ & \approx & 6.7109\times {10}^7>0.\hfill \end{array}$$

Assuming that (18) holds for $n\ge 14$, let us prove (18) holds for $n+1$. Since

$$2^{2n+2}=4\times 2^{2n}>4\times \frac{3v\left(n\right)}{2u\left(n\right)},$$

we complete when proving

$$\begin{array}{ccc}\hfill 4\times \frac{3v\left(n\right)}{2u\left(n\right)}& >& \frac{3v(n+1)}{2u(n+1)}\hfill \\ & ⟺& \\ \hfill \frac{4v\left(n\right)}{u\left(n\right)}& >& \frac{v(n+1)}{u(n+1)}.\hfill \end{array}$$

In fact, the last inequality holds for $n\ge 0$ due to

$$\begin{array}{ccc}& & 4u(n+1)v\left(n\right)-u\left(n\right)v(n+1)\hfill \\ & =& \left(1176{\pi }^4-292032{\pi }^2+2700{\pi }^6+45{\pi }^8+1057536\right)\hfill \\ & & -n\left(521568{\pi }^2+21092{\pi }^4+936{\pi }^6-2731968\right)\hfill \\ & & -n^2\left(432{\pi }^6-3156{\pi }^4-108144{\pi }^2+7776\right)-n^3\left(19696{\pi }^4-658800{\pi }^2+3040128\right)\hfill \\ & & -n^4\left(16464{\pi }^4-149472{\pi }^2+749664\right)-n^5\left(73152{\pi }^2-813312\right)\hfill \\ & & +n^6\left(3456{\pi }^2+135936\right)-64512n^7+4608n^8\hfill \end{array}$$

$$\begin{array}{ccc}& & =3\left(-1583905440{\pi }^2-171080024{\pi }^4-27492{\pi }^6+15{\pi }^8+213637979520\right)\hfill \\ & & +4\left(136075571136-38652635{\pi }^4-3042{\pi }^6-274369500{\pi }^2\right)\left(n-13\right)\hfill \\ & & +12\left(4234896{\pi }^2-1454957{\pi }^4-36{\pi }^6+16249252008\right){\left(n-13\right)}^2\hfill \\ & & +16\left(2291319{\pi }^2-54739{\pi }^4+2414279088\right){\left(n-13\right)}^3\hfill \\ & & +48\left(86574{\pi }^2-343{\pi }^4+96847902\right){\left(n-13\right)}^4+\left(196416{\pi }^2+349394688\right){\left(n-13\right)}^5\hfill \\ & & +\left(3456{\pi }^2+16070400\right){\left(n-13\right)}^6+414720{\left(n-13\right)}^7+4608{\left(n-13\right)}^8\hfill \\ & >& 0.\hfill \end{array}$$

Thus, $G^{\prime }\left(t\right)>0$ holds for all $t\in \left(0,\pi \right)$. This fact together with $G\left(0^{+}\right)=0$ directly gives (16).

This completes the proof of Lemma 3.

□

Lemma 4.

Let $t>0$ and

$$p\left(t\right)=\frac{9coshtsinht-3t-6t{cosh}^{2}t-2t^{2}sinht+2t^{2}coshtsinht}{3\left(2sinht-t-2tcosht+coshtsinht\right)}=:\frac{p_1\left(t\right)}{3p_2\left(t\right)}.$$

Then

$${\left[p\left(t\right)\right]}^3<{cosh}^{2}t$$

(19)

holds.

Proof. 

Let

$$H\left(t\right)=3lnp\left(t\right)-2lncosht=3ln\frac{p_1\left(t\right)}{3p_2\left(t\right)}-2lncosht,t>0.$$

Then

$$H^{\prime }\left(t\right)=-\frac{3}{4}\frac{h\left(t\right)}{\left(cosht\right)p_1\left(t\right)p_2\left(t\right)},$$

where

$$h\left(t\right)=\left(\begin{array}{c}2t^3+15t+36sinh2t-9sinh3t-18sinh4t\\ +9sinh5t-18sinht+20t^{3}cosh2t+16t^{3}cosh3t\\ +2t^{3}cosh4t+62t^{2}sinh2t-39t^{2}sinh3t-7t^{2}sinh4t\\ +t^{2}sinh5t-6tcosht-24tcosh2t+12tcosh3t\\ +9tcosh4t-6tcosh5t-40t^{3}cosht+16t^{2}sinht\end{array}\right).$$

(20)

The two formulas

$$sinhkt=\sum _{n=0}^{\infty }\frac{k^{2n+1}}{\left(2n+1\right)!}{t}^{2n+1},coshkt=\sum _{n=0}^{\infty }\frac{k^{2n}}{\left(2n\right)!}{t}^{2n}$$

are substituted into (20) and sorted out to draw the following conclusion

$$\begin{array}{ccc}\hfill h\left(t\right)& =& \sum _{n=6}^{\infty }\frac{\left[\begin{array}{c}2\left(2n+1\right)\left(10\times 2^{2n}+8\times 3^{2n}+4^{2n}-20\right)\\ +\left(124\times 2^{2n}-117\times 3^{2n}-28\times 4^{2n}+5\times 5^{2n}+16\right)\end{array}\right]}{\left(2n+1\right)!}{t}^{2n+3}\hfill \\ & & -\sum _{n=7}^{\infty }\frac{\left[\begin{array}{c}3\left(2n+1\right)\left(8\times 2^{2n}-4\times 3^{2n}-3\times 4^{2n}+2\times 5^{2n}+2\right)\\ -9\left(8\times 2^{2n}-3\times 3^{2n}-8\times 4^{2n}+5\times 5^{2n}-2\right)\end{array}\right]}{\left(2n+1\right)!}{t}^{2n+1}\hfill \\ & =& \sum _{n=6}^{\infty }\frac{\left[\begin{array}{c}2\left(2n+1\right)\left(10\times 2^{2n}+8\times 3^{2n}+4^{2n}-20\right)\\ +\left(124\times 2^{2n}-117\times 3^{2n}-28\times 4^{2n}+5\times 5^{2n}+16\right)\end{array}\right]}{\left(2n+1\right)!}{t}^{2n+3}\hfill \\ & & -\sum _{n=6}^{\infty }\frac{\left[\begin{array}{c}3\left(2n+3\right)\left(8\times 2^{2n+2}-4\times 3^{2n+2}-3\times 4^{2n+2}+2\times 5^{2n+2}+2\right)\\ -9\left(8\times 2^{2n+2}-3\times 3^{2n+2}-8\times 4^{2n+2}+5\times 5^{2n+2}-2\right)\end{array}\right]}{\left(2n+3\right)!}{t}^{2n+3}\hfill \\ & =& \sum _{n=6}^{\infty }\frac{\left[\begin{array}{c}2\left(2n+1\right)\left(10\times 2^{2n}+8\times 3^{2n}+4^{2n}-20\right)\\ +\left(124\times 2^{2n}-117\times 3^{2n}-28\times 4^{2n}+5\times 5^{2n}+16\right)\end{array}\right]}{\left(2n+1\right)!}{t}^{2n+3}\hfill \\ & & -\sum _{n=6}^{\infty }\frac{\left[\begin{array}{c}3\left(2n+3\right)\left(8\times 2^{2n+2}-4\times 3^{2n+2}-3\times 4^{2n+2}+2\times 5^{2n+2}+2\right)\\ -9\left(8\times 2^{2n+2}-3\times 3^{2n+2}-8\times 4^{2n+2}+5\times 5^{2n+2}-2\right)\end{array}\right]}{\left(2n+3\right)!}{t}^{2n+3}\hfill \\ \hfill =:& & \sum _{n=6}^{\infty }\frac{b_n}{\left(2n+3\right)!}{t}^{2n+3},\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill b_n& =& 5\times 5^{2n}\left(4n^2-50n+141\right)+4\times 4^{2n}\left(4n^3-16n^2+13n-219\right)\hfill \\ & & +3^{2n}\left(128n^3-84n^2-602n-525\right)+16\times 2^{2n}\left(10n^3+61n^2+93n+54\right)\hfill \\ & & -4\left(4n+3\right)\left(41n+20n^2+15\right)\hfill \\ & >& 0\hfill \end{array}$$

for all $n\ge 6$. So $H^{\prime }\left(t\right)<0$ holds for $t>0$. This fact together with $H\left(0^{+}\right)=0$ directly gives (19). □

3. Proofs of Main Results

Then, we can prove the main results of this paper as follows.

3.1. Proof of Theorem 1

By (5) and (6), we have

$$\begin{array}{ccc}\hfill {\alpha }_1& <& \frac{\mathcal{A}-{\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}}{\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right)-{\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}}<{\beta }_1\hfill \\ & ⟺& \\ \hfill {\alpha }_1& <& \frac{1-{\left(\frac{t}{tant}\right)}^{1/3}{\left(\frac{t}{sint}\right)}^{2/3}}{\left(\frac{1}{3}\frac{t}{tant}+\frac{2}{3}\frac{t}{sint}\right)-{\left(\frac{t}{tant}\right)}^{1/3}{\left(\frac{t}{sint}\right)}^{2/3}}<{\beta }_1.\hfill \end{array}$$

Let

$$F\left(t\right)=\frac{1-{\left(\frac{t}{tant}\right)}^{1/3}{\left(\frac{t}{sint}\right)}^{2/3}}{\left(\frac{1}{3}\frac{t}{tant}+\frac{2}{3}\frac{t}{sint}\right)-{\left(\frac{t}{tant}\right)}^{1/3}{\left(\frac{t}{sint}\right)}^{2/3}},0<t<1.$$

Then

$$F^{\prime }\left(t\right)=:\frac{\left(2sint-t-2tcost+costsint\right)\left[q\left(t\right)-{\left(cost\right)}^{2/3}\right]}{3\left({sin}^{2}t\right){\left(cost\right)}^{2/3}{\left[\left(\frac{1}{3}\frac{t}{tant}+\frac{2}{3}\frac{t}{sint}\right)-{\left(\frac{t}{tant}\right)}^{1/3}{\left(\frac{t}{sint}\right)}^{2/3}\right]}^2}.$$

From Lemma 3, we obtain that $F^{\prime }\left(t\right)>0$ for $0<t<1$, which means that the function $F\left(t\right)$ is increasing on $\left(0,1\right)$. Since

$$\begin{array}{ccc}\hfill {\alpha }_1^{*}=:& & F\left(0^{+}\right)=\frac{4}{5},\hfill \\ \hfill {\beta }_1^{*}=:& & F\left(1^{-}\right)=\frac{1-{\left(\frac{1}{tan1}\right)}^{1/3}{\left(\frac{1}{sin1}\right)}^{2/3}}{\left(\frac{1}{3}\frac{1}{tan1}+\frac{2}{3}\frac{1}{sin1}\right)-{\left(\frac{1}{tan1}\right)}^{1/3}{\left(\frac{1}{sin1}\right)}^{2/3}}\approx 0.83597,\hfill \end{array}$$

the proof of Theorem 1 is complete.

3.2. Proof of Theorem 2

By (5) and (6), we have

$$\begin{array}{ccc}\hfill {\alpha }_2& <& \frac{\mathcal{A}-{\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}}{\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)-{\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}}<{\beta }_2\hfill \\ & ⟺& \\ \hfill {\alpha }_2& <& \frac{1-{\left(\frac{t}{tanht}\right)}^{1/3}{\left(\frac{t}{sinht}\right)}^{2/3}}{\left(\frac{1}{3}\frac{t}{tanht}+\frac{2}{3}\frac{t}{sinht}\right)-{\left(\frac{t}{tanht}\right)}^{1/3}{\left(\frac{t}{sinht}\right)}^{2/3}}<{\beta }_2.\hfill \end{array}$$

Let

$$G\left(t\right)=\frac{1-{\left(\frac{t}{tanht}\right)}^{\frac{1}{3}}{\left(\frac{t}{sinht}\right)}^{\frac{2}{3}}}{\left(\frac{1}{3}\frac{t}{tanht}+\frac{2}{3}\frac{t}{sinht}\right)-{\left(\frac{t}{tanht}\right)}^{\frac{1}{3}}{\left(\frac{t}{sinht}\right)}^{\frac{2}{3}}},0<t<1.$$

Then

$$G^{\prime }\left(t\right)=\frac{\left(2sinht-t-2tcosht+coshtsinht\right)\left[p\left(t\right)-{\left(cosht\right)}^{2/3}\right]}{3\left({sinh}^{2}t\right){\left(cosht\right)}^{2/3}{\left[\left(\frac{1}{3}\frac{t}{tanht}+\frac{2}{3}\frac{t}{sinht}\right)-{\left(\frac{t}{tanht}\right)}^{1/3}{\left(\frac{t}{sinht}\right)}^{2/3}\right]}^2}.$$

By Lemma 4, we have $G^{\prime }\left(t\right)<0$ for $0<t<1$, which means that the function $G\left(t\right)$ is decreasing on $\left(0,1\right)$. Since

$$\begin{array}{ccc}\hfill {\beta }_2^{*}=:& & G\left(0^{+}\right)=\frac{4}{5},\hfill \\ \hfill {\alpha }_2^{*}=:& & G\left(1^{-}\right)=\frac{1-{\left(\frac{1}{tanh1}\right)}^{1/3}{\left(\frac{1}{sinh1}\right)}^{2/3}}{\left(\frac{1}{3}\frac{1}{tanh1}+\frac{2}{3}\frac{1}{sinh1}\right)-{\left(\frac{1}{tanh1}\right)}^{1/3}{\left(\frac{1}{sinh1}\right)}^{2/3}}\approx 0.77116,\hfill \end{array}$$

the proof of Theorem 2 is complete.

4. Remarks

In order to compare the conclusions of this paper with the similar ones in the previous literature, we briefly remember

$$\begin{array}{ccc}{\mathcal{M}}_1:=& {\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3},{\mathcal{M}}_2& =:\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin},\\ {\mathcal{M}}_3:=& {\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3},{\mathcal{M}}_4& =:\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}.\end{array}$$

(21)

Remark 1.

Letting ${\alpha }_i=0$ and ${\beta }_i=1$ for $i=1$ and 2 in (14) and (15), respectively, gives (10) and (11).

Since ${\mathcal{M}}_1<{\mathcal{M}}_2$ with $x\ne y,$ when letting ${\mathcal{M}}_2/{\mathcal{M}}_1=a>1,$ through numerical experiments, we obtain that

$$\left\{\begin{array}{ccc}{a}^{{\mu }_1^{*}}\ge \left(1-{\beta }_1^{*}\right)+{\beta }_1^{*}a& \mathrm{for}& 1<a\le 1.0396\\ a^{{\mu }_1^{*}}<\left(1-{\beta }_1^{*}\right)+{\beta }_1^{*}a& \mathrm{for}& a>1.0396\end{array}\right..$$

Then, we have

$${\mathcal{M}}_1^{1/5}{\mathcal{M}}_2^{4/5}<\frac{1}{5}{\mathcal{M}}_1+\frac{4}{5}{\mathcal{M}}_2<\mathcal{A}<\left\{\begin{array}{cc}\left(1-{\beta }_1^{*}\right){\mathcal{M}}_1+{\beta }_1^{*}{\mathcal{M}}_2\le {\mathcal{M}}_1^{\left(1-{\mu }_1^{*}\right)}{\mathcal{M}}_2^{{\mu }_1^{*}},& 1<{\mathcal{M}}_2/{\mathcal{M}}_1\le 1.0396\\ {\mathcal{M}}_1^{\left(1-{\mu }_1^{*}\right)}{\mathcal{M}}_2^{{\mu }_1^{*}}<\left(1-{\beta }_1^{*}\right){\mathcal{M}}_1+{\beta }_1^{*}{\mathcal{M}}_2,& {\mathcal{M}}_2/{\mathcal{M}}_1>1.0396\end{array}\right..$$

(22)

Similarly, because of ${\mathcal{M}}_3<{\mathcal{M}}_4$ with $x\ne y,$ when letting ${\mathcal{M}}_4/{\mathcal{M}}_3=b>1,$ through numerical experiments, we obtain that

$$\left\{\begin{array}{ccc}\left(1-{\alpha }_2^{*}\right)+{\alpha }_2^{*}b<b^{{\lambda }_2^{*}}& \mathrm{for}& 1<b<1.022\\ \left(1-{\alpha }_2^{*}\right)+{\alpha }_2^{*}b\ge b^{{\lambda }_2^{*}}& \mathrm{for}& a\ge 1.022\end{array}\right..$$

Thus, we have

$$\frac{1}{5}{\mathcal{M}}_3+\frac{4}{5}{\mathcal{M}}_4>{\mathcal{M}}_3^{1/5}{\mathcal{M}}_4^{4/5}>\mathcal{A}>\left\{\begin{array}{cc}{\mathcal{M}}_3^{\left(1-{\lambda }_2^{*}\right)}{\mathcal{M}}_4^{{\lambda }_2^{*}}>\left(1-{\alpha }_2^{*}\right){\mathcal{M}}_3+{\alpha }_2^{*}{\mathcal{M}}_4,& 1<{\mathcal{M}}_4/{\mathcal{M}}_3<1.022\\ \left(1-{\alpha }_2^{*}\right){\mathcal{M}}_3+{\alpha }_2^{*}{\mathcal{M}}_4>{\mathcal{M}}_3^{\left(1-{\lambda }_2^{*}\right)}{\mathcal{M}}_4^{{\lambda }_2^{*}},& {\mathcal{M}}_2/{\mathcal{M}}_1\ge 1.022\end{array}\right..$$

(23)

Remark 2.

Let ${\lambda }_1\le {\lambda }_1^{*}=4/5$ and ${\mu }_1={\mu }_1^{*},$${\lambda }_2\le {\lambda }_2^{*}$ and ${\mu }_2$$\ge {\mu }_2^{*}=4/5,$${\alpha }_1\le {\alpha }_1^{*}=4/5$ and ${\beta }_1\ge {\beta }_1^{*},$${\alpha }_2={\alpha }_2^{*}$ and ${\beta }_2={\beta }_2^{*}=4/5$ in (12)–(15), respectively. Then, we obtain that the left-hand side inequality of (14) is better than the one of (12), while the right-hand side inequality of (14) does not match the one of (12), and the right-hand side inequality of (13) is better than the one of (15) while the left-hand side inequality of (15) does not match the one of (13),

5. Conclusions

After reviewing the traditional study of mean inequality through hyperbolic functions, this paper introduced and modified a new method through the Seiffert function, and obtained two bilateral inequalities about the upper and lower bounds for arithmetic mean

$$\begin{array}{ccc}\hfill \left(1-{\alpha }_1\right){\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}+{\alpha }_1\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right)& <& \mathcal{A}<\left(1-{\beta }_1\right){\mathcal{M}}_{tan}^{1/3}{\mathcal{M}}_{sin}^{2/3}+{\beta }_1\left(\frac{1}{3}{\mathcal{M}}_{tan}+\frac{2}{3}{\mathcal{M}}_{sin}\right),\hfill \\ \hfill \left(1-{\alpha }_2\right){\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}+{\alpha }_2\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)& <& \mathcal{A}<\left(1-{\beta }_2\right){\mathcal{M}}_{tanh}^{1/3}{\mathcal{M}}_{sinh}^{2/3}+{\beta }_2\left(\frac{1}{3}{\mathcal{M}}_{tanh}+\frac{2}{3}{\mathcal{M}}_{sinh}\right)\hfill \end{array}$$

hold if and only if ${\alpha }_1\le {\alpha }_1^{*}=4/5$ and ${\beta }_1\ge {\beta }_1^{*}\approx 0.83597$, ${\alpha }_2\le {\alpha }_2^{*}\approx 0.77116$ and ${\beta }_2\ge {\beta }_2^{*}=4/5$, respectively. We have seen that the new method is particularly effective for the new means. In fact, merely studying the various relationships of the means involved in this paper is a good exploration theme, not to mention the classical means not involved in this paper. In addition, I hope that scholars can create better new methods for the study of mean value inequality.