A new proof of a double inequality of Masjed-Jamei type

Fen Wang; Fen Wang

doi:10.3934/math.2024425

AIMS Mathematics

2024, Volume 9, Issue 4: 8768-8775. doi: 10.3934/math.2024425

Previous Article Next Article

Research article

A new proof of a double inequality of Masjed-Jamei type

Fen Wang ^,

School of Mathematics and Statistics, Hubei University of Education, Wuhan 430205, China

Received: 23 November 2023 Revised: 30 January 2024 Accepted: 26 February 2024 Published: 29 February 2024
MSC : 26D05, 26D15

In this paper, we provide a new simple proof of a double inequality of Masjed-Jamei type proved by L. Zhu ^[1].

Keywords:

Citation: Fen Wang. A new proof of a double inequality of Masjed-Jamei type[J]. AIMS Mathematics, 2024, 9(4): 8768-8775. doi: 10.3934/math.2024425

Related Papers:

[1]	Ling Zhu . New inequalities of Wilker’s type for circular functions. AIMS Mathematics, 2020, 5(5): 4874-4888. doi: 10.3934/math.2020311
[2]	Ling Zhu, Branko Malešević . Optimal bounds for the sine and hyperbolic tangent means by arithmetic and centroidal means in exponential type. AIMS Mathematics, 2021, 6(12): 13024-13040. doi: 10.3934/math.2021753
[3]	Wei-Mao Qian, Tie-Hong Zhao, Yu-Pei Lv . Refinements of bounds for the arithmetic mean by new Seiffert-like means. AIMS Mathematics, 2021, 6(8): 9036-9047. doi: 10.3934/math.2021524
[4]	Ling Zhu . New inequalities of Wilker's type for hyperbolic functions. AIMS Mathematics, 2020, 5(1): 376-384. doi: 10.3934/math.2020025
[5]	Sarah Elahi, Muhammad Aslam Noor . Integral inequalities for hyperbolic type preinvex functions. AIMS Mathematics, 2021, 6(9): 10313-10326. doi: 10.3934/math.2021597
[6]	François Dubeau . Alternating reflection method on conics leading to inverse trigonometric and hyperbolic functions. AIMS Mathematics, 2022, 7(7): 11708-11717. doi: 10.3934/math.2022652
[7]	Wenzheng Hu, Jian Deng . Hankel determinants, Fekete-Szegö inequality, and estimates of initial coefficients for certain subclasses of analytic functions. AIMS Mathematics, 2024, 9(3): 6445-6467. doi: 10.3934/math.2024314
[8]	Mohamed Jleli, Bessem Samet . On $ \theta $-hyperbolic sine distance functions and existence results in complete metric spaces. AIMS Mathematics, 2024, 9(10): 29001-29017. doi: 10.3934/math.20241407
[9]	Muhammad Ghaffar Khan, Sheza.M. El-Deeb, Daniel Breaz, Wali Khan Mashwani, Bakhtiar Ahmad . Sufficiency criteria for a class of convex functions connected with tangent function. AIMS Mathematics, 2024, 9(7): 18608-18624. doi: 10.3934/math.2024906
[10]	Huo Tang, Muhammad Abbas, Reem K. Alhefthi, Muhammad Arif . Problems involving combinations of coefficients for the inverse of some complex-valued analytical functions. AIMS Mathematics, 2024, 9(10): 28931-28954. doi: 10.3934/math.20241404

Abstract

In this paper, we provide a new simple proof of a double inequality of Masjed-Jamei type proved by L. Zhu ^[1].

1. Introduction

In 2010, Masjed-Jamei ^[2] obtained some interesting inequalities for several special functions, one of which is about the relation of the inverse tangent function $\arctan x$ and inverse hyperbolic sine function $\sinh^{-1}(x)$ as follows:

$\begin{equation} \left(\arctan x\right)^2\leqslant\frac{x\sinh^{-1}(x)}{\sqrt{1+x^2}}, \ \ x\in(-1, 1). \end{equation}$

(1.1)

The study related to (1.1) attracted much attention in last decade. At first, Zhu and Male $\breve{\rm{s}}$ ević ^[3] proved that (1.1) holds for any $x\in(-\infty, +\infty)$ . They also obtained some refinements of (1.1) as follows:

Proposition 1.1. [, Theorem 1.3] For any $x\in (-\infty, +\infty)$ , we have

$\begin{equation} -\frac{1}{45}x^6\leqslant \left(\arctan x\right)^2-\frac{x\sinh^{-1} x}{\sqrt{1+x^2}}\leqslant -\frac{1}{45}x^6+\frac{4}{105}x^8, \end{equation}$

(1.2)

$\begin{equation} \begin{aligned} -\frac{1}{45}x^6+\frac{4}{105}x^8-\frac{11}{225}x^{10}&\leqslant\left(\arctan x\right)^2-\frac{x\sinh^{-1} x}{\sqrt{1+x^2}}\\ &\leqslant -\frac{1}{45}x^6+\frac{4}{105}x^8-\frac{11}{225}x^{10}+\frac{586}{10395}x^{12}. \end{aligned} \end{equation}$

(1.3)

Define

$\begin{equation} v_n = \frac{1}{n}\left[\frac{n!2^{n-1}}{(2n-1)!!}-\left(1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\right)\right], \; n\geqslant3. \end{equation}$

(1.4)

By using flexible analysis tools, Zhu and Male $\breve{\rm{s}}$ ević ^[4] extended (1.2) and (1.3) to a general form as follows:

Proposition 1.2. [, Theorem 1.1] For any $x\in (-\infty, +\infty)$ , we have

$\begin{equation} \sum\limits_{n = 3}^{2m+1}(-1)^nv_nx^{2n}\leqslant \left(\arctan x\right)^2-\frac{x\sinh^{-1} x}{\sqrt{1+x^2}}\leqslant \sum\limits_{n = 3}^{2m+2}(-1)^nv_nx^{2n}. \end{equation}$

(1.5)

Proposition 1.3. [5, Theorem 2.1] The double inequality

$\begin{equation} \frac{x\sinh^{-1}(x)}{\sqrt{1+x^2+\frac{1}{45}x^2}} < \left(\arctan x\right)^2 < \frac{x\sinh^{-1}(x)}{\sqrt{1+x^2}} \end{equation}$

(1.6)

holds for any $x\in(0, +\infty)$ with best constants $0$ and $1/45$ .

Please see ^[6,7] for more generalizations.

Motivated by (1.1)–(1.6), Zhu and Male $\breve{\rm{s}}$ ević ^[3] also studied the relation of the inverse hyperbolic tangent function $\tanh^{-1}(x)$ and inverse sine function $\arcsin x$ as follows:

Proposition 1.4. [3, Theorem 1.4] The inequality

$\begin{equation} \left[\tanh^{-1}(x)\right]^2 < \frac{x\arcsin x}{\sqrt{1-x^2}} \end{equation}$

(1.7)

holds for any $x\in (0, 1)$ with the the best power number $2$ .

Proposition 1.5. [3, Theorem 1.6] The inequality

$\begin{equation} \frac{x\arcsin x}{\sqrt{1-x^2}}-\left[\tanh^{-1}(x)\right]^2 < \sum\limits_{n = 3}^{N}v_nx^{2n} \end{equation}$

(1.8)

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

Moreover, by investigating the power series of the following function:

$\begin{equation} \frac{\left[\tanh^{-1}(x)\right]^2}{\dfrac{\arcsin x}{\sqrt{1-x^2}}} = x-\frac{1}{45}x^5-\frac{22}{945}x^7-\frac{61}{2835}x^9+O(x^{10}), \end{equation}$

(1.9)

Zhu ^[1] obtained the following interesting double inequality of Masjed-Jamei type.

Theorem 1.1. [1, Theorem 1] The double inequality

$\begin{equation} \frac{\left(x-x^5\right)\arcsin x}{\sqrt{1-x^2}} < \left[\tanh^{-1}(x)\right]^2 < \frac{\left(x-\frac{1}{45}x^5\right)\arcsin x}{\sqrt{1-x^2}} \end{equation}$

(1.10)

holds for any $x\in (0, 1)$ with best constants $-1$ and $-\frac{1}{45}$ .

The goal of this paper is to give a new and elementary proof of Theorem 1.1, which is much simpler than the proof of Zhu ^[1]. Zhu's proof ^[1] used the power series of the functions $1/\cos^nx$ and $\sin x/\cos^nx$ and properties of the Bernoulli numbers and Euler numbers. Our proof only relies on the power series of hyperbolic sine and cosine functions and some elementary computations.

2. A new proof of Theorem 1.1

We first establish two lemmas about the monotonicity of two functions.

Lemma 2.1. Let

$f(x) = \frac{\left[\tanh^{-1}(x)\right]^2\sqrt{1-x^2}}{x-x^5}-\arcsin x,$

then $f(x)$ is strictly increasing on $(0, 1)$ .

Proof. Let $t = \tanh^{-1}(x)\in(0, +\infty)$ , then $x = \tanh(t)$ . Define

$\begin{align*} F(t): = f\left(\tanh(t)\right)& = \frac{\frac{t^2}{\cosh(t)}}{\tanh(t)-\tanh^5(t)}-\arcsin\left(\tanh(t)\right)\\ & = \frac{t^2\cosh^4t}{\left(\cosh^2t+\sinh^2t\right)\sinh t}-\arcsin\left(\tanh(t)\right). \end{align*}$

In order to prove that $f(x)$ is strictly increasing on $(0, 1)$ , we only need to prove $F(t)$ is strictly increasing on $(0, +\infty)$ . In fact,

$\begin{equation} \begin{split} \varphi(t):& = \left[\left(\cosh^2t+\sinh^2t\right)^2\cosh t\sinh^2 t\right]F^{\prime}(t)\\ & = t^2\left(2\cosh^4t-7\cosh^2t+4\right)\cosh^4t+2t\left(\cosh^2t+\sinh^2t\right)\sinh t\cosh^5t\\ & \quad -\left(\cosh^2t+\sinh^2t\right)^2\sinh^2t. \end{split} \end{equation}$

(2.1)

Since

$\begin{align*} A_1: = &\left(2\cosh^4t-7\cosh^2t+4\right)\cosh^4t\\ = &\frac{1}{2^6}\left[\cosh(8t)-6\cosh(6t)-24\cosh(4t)-26\cosh(2t)-9\right]\\ = &\frac{1}{2^6}\left[\sum\limits_{n = 0}^{\infty}\frac{(8t)^{2n}}{(2n)!}-6\cdot\sum\limits_{n = 0}^{\infty}\frac{(6t)^{2n}}{(2n)!} -24\cdot\sum\limits_{n = 0}^{\infty}\frac{(4t)^{2n}}{(2n)!}-26\cdot\sum\limits_{n = 0}^{\infty}\frac{(2t)^{2n}}{(2n)!}-9\right], \end{align*}$

$\begin{align*} B_1: = &2\left(\cosh^2t+\sinh^2t\right)\sinh t\cosh^5t\\ = &\frac{1}{2^5}\left[\sinh(8t)+4\sinh(6t)+6\sinh(4t)+4\sinh(2t)\right]t\\ = &\frac{1}{2^5}\left[\sum\limits_{n = 0}^{\infty}\frac{(8t)^{2n+1}}{(2n+1)!}+4\cdot\sum\limits_{n = 0}^{\infty}\frac{(6t)^{2n+1}}{(2n+1)!} +6\cdot\sum\limits_{n = 0}^{\infty}\frac{(4t)^{2n+1}}{(2n+1)!}+4\cdot\sum\limits_{n = 0}^{\infty}\frac{(2t)^{2n+1}}{(2n+1)!}\right] \end{align*}$

and

$\begin{align*} C_1: = &-\left(\cosh^2t+\sinh^2t\right)^2\sinh^2t\\ = &\frac{1}{2^3}\left[\cosh(6t)-2\cosh(4t)+3\cosh(2t)-2\right]\\ = &\frac{1}{2^3}\left[\sum\limits_{n = 0}^{\infty}\frac{(6t)^{2n}}{(2n)!} -2\cdot\sum\limits_{n = 0}^{\infty}\frac{(4t)^{2n}}{(2n)!}+3\cdot\sum\limits_{n = 0}^{\infty}\frac{(2t)^{2n}}{(2n)!}-2\right], \end{align*}$

we have

$\begin{equation} \varphi(t) = A_1t^2+B_1t+C_1 = \frac{1}{2^6}\sum\limits_{n = 1}^{\infty}a_{2n+2}t^{2n+2}, \end{equation}$

(2.2)

where

$\begin{align*} a_{2n+2} = &\frac{8^{2n}}{(2n)!}-6\cdot\frac{6^{2n}}{(2n)!}-24\cdot\frac{4^{2n}}{(2n)!}-26\cdot\frac{2^{2n}}{(2n)!}\\ &+2\cdot\frac{8^{2n+1}}{(2n+1)!}+8\cdot\frac{6^{2n+1}}{(2n+1)!}+12\cdot\frac{4^{2n+1}}{(2n+1)!}+8\cdot\frac{2^{2n+1}}{(2n+1)!}\\ &-8\cdot\frac{6^{2n+2}}{(2n+2)!}+16\cdot\frac{4^{2n+2}}{(2n+2)!}-24\cdot\frac{2^{2n+2}}{(2n+2)!}. \end{align*}$

It is easy to check that

$\begin{equation} a_{4} = 0, \; a_6 = \frac{2816}{9}, \; a_8 = \frac{4224}{5}. \end{equation}$

(2.3)

When $n\geqslant 4$ ,

$\begin{equation} \begin{split} a_{2n+2} > & \frac{8^{2n}}{(2n)!}\left[1-6\cdot\left(\frac{3}{4}\right)^{2n}-24\cdot\left(\frac{1}{2}\right)^{2n}-26\cdot\left(\frac{1}{4}\right)^{2n}\right]\\ & +\frac{6^{2n+1}}{(2n+1)!}\left(1-\frac{3}{n+1}\right)+8\cdot\frac{2^{2n+1}}{(2n+1)!}\left(1-\frac{3}{n+1}\right)\\ > & \frac{8^{2n}}{(2n)!}\left[1-6\cdot\left(\frac{3}{4}\right)^{8}-24\cdot\left(\frac{1}{2}\right)^{8}-26\cdot\left(\frac{1}{4}\right)^{8}\right]\\ = & \frac{8^{2n}}{(2n)!}\cdot\frac{625}{2^{11}}\\ > & 0. \end{split} \end{equation}$

(2.4)

Combining (2.2)–(2.4), we obtain that $\varphi(t) > 0$ for any $t\in(0, +\infty)$ , which implies

$F^{\prime}(t) > 0\ \ \text{for any}\ \ t\in(0, +\infty).$

So, $F(t)$ is strictly increasing on $(0, +\infty)$ . The proof of Lemma 2.1 is completed. □

Lemma 2.2. Let

$g(x) = \arcsin x-\frac{\left[\tanh^{-1}(x)\right]^2\sqrt{1-x^2}}{x-\frac{1}{45}x^5},$

then $g(x)$ is strictly increasing on $(0, 1)$ .

Proof. Let $t = \tanh^{-1}(x)\in(0, +\infty)$ , then $x = \tanh(t)$ . Define

$\begin{align*} G(t): = g\left(\tanh(t)\right)& = \arcsin\left(\tanh(t)\right)-\frac{\frac{t^2}{\cosh(t)}}{\tanh(t)-\frac{1}{45}\tanh^5(t)}\\ & = \arcsin\left(\tanh(t)\right)-\frac{45t^2\cosh^4t}{\left(45\cosh^4t-\sinh^4t\right)\sinh t}. \end{align*}$

In order to prove that $g(x)$ is strictly increasing on $(0, 1)$ , we only need to prove $G(t)$ is strictly increasing on $(0, +\infty)$ . Define

$\begin{equation} \begin{split} \psi(t): = &\left[\left(45\cosh^4t-\sinh^4t\right)^2\cosh t\sinh^2 t\right]G^{\prime}(t)\\ = &t^2\left(1980\cosh^6t-90\cosh^4t+315\cosh^2t-180\right)\cosh^4t\\ &-90t\left(45\cosh^4t-\sinh^4t\right)\sinh t\cosh^5t+\left(45\cosh^4t-\sinh^4t\right)^2\sinh^2t. \end{split} \end{equation}$

(2.5)

Since

$\begin{align*} A_2: = &\left(1980\cosh^6t-90\cosh^4t+315\cosh^2t-180\right)\cosh^4t\\ = &\frac{1}{2^7}\left[495\cosh(10t)+4860\cosh(8t)+22815\cosh(6t)+61560\cosh(4t)\right.\\ & \quad \left.+106290\cosh(2t)+63180\right]\\ = &\frac{1}{2^7}\left[495\cdot\sum\limits_{n = 0}^{\infty}\frac{(10t)^{2n}}{(2n)!}+4860\cdot\sum\limits_{n = 0}^{\infty}\frac{(8t)^{2n}}{(2n)!}+22815\cdot\sum\limits_{n = 0}^{\infty}\frac{(6t)^{2n}}{(2n)!} +61560\cdot\sum\limits_{n = 0}^{\infty}\frac{(4t)^{2n}}{(2n)!}\right.\\ & \quad \left.+106290\cdot\sum\limits_{n = 0}^{\infty}\frac{(2t)^{2n}}{(2n)!}+63180\right], \end{align*}$

$\begin{align*} B_2: = &-90\left(45\cosh^4t-\sinh^4t\right)\sinh t\cosh^5t\\ = &\frac{1}{2^7}\left[-990\sinh(10t)-8100\sinh(8t)-27450\sinh(6t)-48600\sinh(4t)-42300\sinh(2t)\right]\\ = &\frac{1}{2^7}\left[-990\cdot\sum\limits_{n = 0}^{\infty}\frac{(10t)^{2n+1}}{(2n+1)!}-8100\cdot\sum\limits_{n = 0}^{\infty}\frac{(8t)^{2n+1}}{(2n+1)!}-27450\cdot\sum\limits_{n = 0}^{\infty}\frac{(6t)^{2n+1}}{(2n+1)!} \right.\\ & \quad \left.-48600\cdot\sum\limits_{n = 0}^{\infty}\frac{(4t)^{2n+1}}{(2n+1)!}-42300\cdot\sum\limits_{n = 0}^{\infty}\frac{(2t)^{2n+1}}{(2n+1)!}\right], \end{align*}$

and

$\begin{align*} C_2: = &\left(45\cosh^4t-\sinh^4t\right)^2\sinh^2t\\ = &\frac{1}{2^7}\left[484\cosh(10t)+3080\cosh(8t)+6660\cosh(6t)+3840\cosh(4t)-7080\cosh(2t)-6984\right]\\ = &\frac{1}{2^7}\left[484\cdot\sum\limits_{n = 0}^{\infty}\frac{(10t)^{2n}}{(2n)!}+3080\cdot\sum\limits_{n = 0}^{\infty}\frac{(8t)^{2n}}{(2n)!}+6660\cdot\sum\limits_{n = 0}^{\infty}\frac{(6t)^{2n}}{(2n)!} +3840\cdot\sum\limits_{n = 0}^{\infty}\frac{(4t)^{2n}}{(2n)!}\right.\\ & \quad \left.-7080\cdot\sum\limits_{n = 0}^{\infty}\frac{(2t)^{2n}}{(2n)!}-6984\right], \end{align*}$

we have

$\begin{equation} \psi(t) = A_2t^2+B_2t+C_2 = \frac{1}{2^7}\sum\limits_{n = 1}^{\infty}b_{2n+2}t^{2n+2}, \end{equation}$

(2.6)

where

$\begin{align*} b_{2n+2} = &495\cdot\frac{10^{2n}}{(2n)!}+4860\cdot\frac{8^{2n}}{(2n)!}+22815\cdot\frac{6^{2n}}{(2n)!}+61560\cdot\frac{4^{2n}}{(2n)!}+106290\cdot\frac{2^{2n}}{(2n)!}\\ &-990\cdot\frac{10^{2n+1}}{(2n+1)!}-8100\cdot\frac{8^{2n+1}}{(2n+1)!}-27450\cdot\frac{6^{2n+1}}{(2n+1)!}-48600\cdot\frac{4^{2n+1}}{(2n+1)!}\\ &-42300\cdot\frac{2^{2n+1}}{(2n+1)!}+484\cdot\frac{10^{2n+2}}{(2n+2)!}+3080\cdot\frac{8^{2n+2}}{(2n+2)!}+6660\cdot\frac{6^{2n+2}}{(2n+2)!} \\ &+3840\cdot\frac{4^{2n+2}}{(2n+2)!}-7080\cdot\frac{2^{2n+2}}{(2n+2)!}. \end{align*}$

It is easy to check that

$\begin{equation} b_{2n+2} > 0, \ \ 1\leqslant n\leqslant 9. \end{equation}$

(2.7)

When $n\geqslant10$ ,

$\begin{equation} \begin{split} b_{2n+2} > &\frac{10^{2n}}{(2n)!}\cdot\left(495-\frac{990\times10}{2n+1}\right)+\frac{8^{2n}}{(2n)!}\cdot\left(4860-\frac{8100\times 8}{2n+1}\right)\\ &+\frac{6^{2n}}{(2n)!}\cdot\left(22815-\frac{27450\times 6}{2n+1}\right)+\frac{4^{2n}}{(2n)!}\cdot\left(61560-\frac{48600\times 4}{2n+1}\right)\\ &+\frac{2^{2n}}{(2n)!}\cdot\left(106290-\frac{42300\times 2}{2n+1}\right)+\frac{2^{2n+2}}{(2n+2)!}\cdot\left(3840\cdot 2^{2n+2}-7840\right)\\ > &0. \end{split} \end{equation}$

(2.8)

Combining (2.5)–(2.8), we obtain that $\psi(t) > 0$ for any $t\in(0, +\infty)$ , which implies

$G^{\prime}(t) > 0\ \ \text{for any}\ \ t\in(0, +\infty).$

So, $G(t)$ is strictly increasing on $(0, +\infty)$ . The proof of Lemma 2.2 is completed. □

Proof of Theorem 1.1. By Lemma 2.1 and

$\lim\limits_{x\to0^{+}}f(x) = 0,$

we get $f(x) > 0$ for any $x\in(0, 1)$ , which implies

$\begin{equation} \left[\tanh^{-1}(x)\right]^2 > \frac{\left(x-x^5\right)\arcsin x}{\sqrt{1-x^2}}, \ \ \ x\in(0, 1). \end{equation}$

(2.9)

By Lemma 2.2 and

$\lim\limits_{x\to0^{+}}g(x) = 0,$

we get $g(x) > 0$ for any $x\in(0, 1)$ , which implies

$\begin{equation} \left[\tanh^{-1}(x)\right]^2 < \frac{\left(x-\frac{1}{45}x^5\right)\arcsin x}{\sqrt{1-x^2}}, \ \ \ x\in(0, 1). \end{equation}$

(2.10)

Since

$\begin{align} \lim\limits_{x\to0^+}\frac{\left[\tanh^{-1}(x)\right]^2\cdot \sqrt{1-x^2}-x\arcsin x}{x^5\arcsin x}& = -1, \end{align}$

(2.11)

$\begin{align} \lim\limits_{x\to1^-}\frac{\left[\tanh^{-1}(x)\right]^2\cdot \sqrt{1-x^2}-x\arcsin x}{x^5\arcsin x}& = -\frac{1}{45}, \end{align}$

(2.12)

Theorem 1.1 follows from (2.9)–(2.12). □

3. Conclusions

In this paper, we give a new simple proof of a double inequality of Masjed-Jamei type proved by Zhu ^[1]. We believe that the technique used in this paper can be used to obtain other interesting analytic inequalities.

Based on numerical experiments and (1.9), we propose the following conjectures:

Conjecture 1.

$\phi(x) = \frac{\sqrt{1-x^2}\left[\tanh^{-1}(x)\right]^2-x\arcsin x}{x^5\arcsin x}$

is strictly increasing on $(0, 1)$ .

Conjecture 2.

$h(x) = x-\frac{\sqrt{1-x^2}\left[\tanh^{-1}(x)\right]^2}{\arcsin x}$

is absolutely monotonic on $(0, 1)$ .

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author was supported by the Foundation of Hubei Provincial Department of Eduction (No. Q20233003), the Scientific Research Fund of Hubei Provincial Department of Eduction (No. B2022207) and Hubei University of Education, Bigdata Modeling and Intelligent Computing Research Institute.

Conflict of interest

The author declares no conflict of interest in this paper.

References

[1]	L. Zhu, New double inequality of Masjed-Jamei-type, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat., 117 (2023), 41. https://doi.org/10.1007/s13398-022-01375-6 doi: 10.1007/s13398-022-01375-6
[2]	M. Masjed-Jamei, A main inequality for several special functions, Comput. Math. Appl., 60 (2010), 1280–1289. https://doi.org/10.1016/j.camwa.2010.06.007 doi: 10.1016/j.camwa.2010.06.007
[3]	L. Zhu, B. Male $\breve{\rm{s}}$ ević, Inequalities between the inverse hyperbolic tangent and the inverse sine and the analogue for corresponding functions, J. Inequal. Appl., 2019 (2019), 1–10. https://doi.org/10.1186/s13660-019-2046-2 doi: 10.1186/s13660-019-2046-2
[4]	L. Zhu, B. Male $\breve{\rm{s}}$ ević, Natural approximation of Masjed-Jamei's inequality, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat., 114 (2020), 25. https://doi.org/10.1007/s13398-019-00735-z doi: 10.1007/s13398-019-00735-z
[5]	C. P. Chen, B. Male $\breve{\rm{s}}$ ević, Inequalities related to certain inverse trigonometric and inverse hyperbolic functions, Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat., 114 (2020), 105. https://doi.org/10.1007/s13398-020-00836-0 doi: 10.1007/s13398-020-00836-0
[6]	C. Chesneau, , Y. J. Bagul, On a reverse trigonometric Masjed-Jamei inequality, Asia Pac. J. Math., 8 (2021), 1–5. https://doi.org/10.28924/APJM/8-13 doi: 10.28924/APJM/8-13
[7]	X. D. Chen, L. Nie, W. K. Huang, New inequalities between the inverse hyperbolic tangent and the analogue for corresponding functions, J. Inequal. Appl., 2020 (2020), 1–8. https://doi.org/10.1186/s13660-020-02396-8 doi: 10.1186/s13660-020-02396-8

This article has been cited by:

Fen Wang, Hai-Yan Xiao, A proof of Chen-Males̆ević’s conjecture, 2024, 118, 1578-7303, 10.1007/s13398-024-01637-5

Reader Comments

Your name:*

Email:*
© 2024 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

AIMS Mathematics

1.8 3.4

Metrics

Article views(995) PDF downloads(67) Cited by(1)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

AIMS Mathematics

A new proof of a double inequality of Masjed-Jamei type

Related Papers:

Abstract

1. Introduction

2. A new proof of Theorem 1.1

3. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Catalog

AIMS Mathematics

A new proof of a double inequality of Masjed-Jamei type

Related Papers:

Abstract

1. Introduction

2. A new proof of Theorem 1.1

3. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

This article has been cited by:

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog