New Sharp Bounds for the Modified Bessel Function of the First Kind and Toader-Qi Mean ^†

Abstract

Let $I_v\left(x\right)$ be he modified Bessel function of the first kind of order v. We prove the double inequality $\sqrt{\frac{sinht}{t}{cosh}^{1/q}\left(qt\right)}<I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{1/p}\left(pt\right)}$ holds for $t>0$ if and only if $p\ge 2/3$ and $q\le \left(ln2\right)/ln\pi$. The corresponding inequalities for means improve already known results.

1. Introduction

The modified Bessel function of the first kind of order v, denoted by $I_v\left(x\right)$, is a particular solution of the second-order differential equation ([1], p. 77)

$$x^2{y}^{″}\left(x\right)+x{y}^{\prime }\left(x\right)-\left(x^2+v^2\right)y\left(x\right)=0,$$

(1)

which can be represented explicitly by the infinite series as

$$I_v\left(x\right)=\sum _{n=0}^{\infty }\frac{{\left(x/2\right)}^{2n+v}}{n!\mathsf{\Gamma }\left(v+n+1\right)},x\in \mathbb{R},v\in \mathbb{R}\setminus \{-1,-2,\dots \},$$

(2)

where $\mathsf{\Gamma }\left(x\right)$ is the gamma function [2,3,4]. There are many properties of $I_v\left(x\right)$, see for example, [5,6,7,8,9,10,11].

In this paper, we are interested in a special case of $I_v\left(x\right)$, that is, $I_0\left(x\right)$, which is related to Toader-Qi mean of positive numbers a and b defined by

$$TQ\left(a,b\right)=\frac{2}{\pi }{\int }_0^{\pi /2}{a}^{{cos}^2\theta }{b}^{{sin}^2\theta }d\theta =\sqrt{ab}{I}_0\left(ln\sqrt{\frac{a}{b}}\right)$$

(see [12,13,14]), where and in what follows $a,b>0$ with $a\ne b$. It is undoubted that Toader-Qi mean $TQ\left(a,b\right)$ is a new newcomer. Recall that some classical means including the arithmetic mean, geometric mean, logarithmic mean, exponential mean and power mean of order p defined by

$$\begin{array}{cc}A\equiv A\left(a,b\right)={\displaystyle \frac{a+b}{2}},\hfill & G\equiv G\left(a,b\right)=\sqrt{ab},\hfill \\ L\equiv L\left(a,b\right)={\displaystyle \frac{a-b}{lna-lnb}},\hfill & I\equiv I\left(a,b\right)=e^{-1}{\left({\displaystyle \frac{b^b}{a^a}}\right)}^{1/\left(b-a\right)},\hfill \end{array}$$

$$A_p\equiv A_p\left(a,b\right)={\left(\frac{a^p+b^p}{2}\right)}^{1/p}\mathrm{if}p\ne 0\mathrm{and}{A}_0\equiv A_0\left(a,b\right)=\sqrt{ab},$$

respectively. Clearly, $A\left(a,b\right)=A_1\left(a,b\right)$ and $G\left(a,b\right)=A_0\left(a,b\right)$. It is known that $p\mapsto A_p\left(a,b\right)$ is increasing on $\mathbb{R}$. A simple relation among these elementary means is the following inequalities:

$$G<L<A_{1/3}<\frac{A+2G}{3}<A_{1/2}<\frac{2A+G}{3}<A_{2/3}<I<A_{ln2}<A_1$$

(3)

(see [15,16,17,18,19,20,21]). Another interesting relation proven in [22] is that:

$$\sqrt{AG}<\sqrt{LI}<\frac{L+I}{2}<\frac{A+G}{2}.$$

(4)

Let $b>a>0$ and $t=ln\sqrt{a/b}$. Then those means mentioned above can be represented in terms of hyperbolic functions:

$$\begin{array}{ccc}\hfill \frac{L\left(a,b\right)}{\sqrt{ab}}& =& \frac{sinht}{t},\frac{I\left(a,b\right)}{\sqrt{ab}}=exp\left(\frac{t}{tanht}-1\right),\hfill \\ \hfill \frac{TQ\left(a,b\right)}{\sqrt{ab}}& =& I_0\left(t\right),\frac{A_p\left(a,b\right)}{\sqrt{ab}}={cosh}^{1/p}\left(pt\right)\mathrm{for}p\ne 0.\hfill \end{array}$$

Correspondingly, the inequalities mentioned above are equivalent to

$$\begin{array}{ccc}\hfill 1& <& \frac{sinht}{t}<{cosh}^3\left(\frac{t}{3}\right)<\frac{cosht+2}{3}<{cosh}^2\left(\frac{t}{2}\right)\hfill \\ & <& \frac{2cosht+1}{3}<{cosh}^{3/2}\left(\frac{2t}{3}\right)<exp\left(\frac{t}{tanht}-1\right)<{cosh}^{1/ln2}\left(tln2\right),\hfill \end{array}$$

$$\sqrt{cosht}<\sqrt{\frac{sinht}{t}exp\left(\frac{t}{tanht}-1\right)}<\frac{1}{2}\left[\frac{sinht}{t}+exp\left(\frac{t}{tanht}-1\right)\right]<\frac{cosht+1}{2}.$$

for $t>0$.

Let us return to Toader-Qi mean. In 2015, Qi, Shi, Liu and Yang [13] proved that the inequalities

$$L\left(a,b\right)<TQ\left(a,b\right)<\frac{A\left(a,b\right)+G\left(a,b\right)}{2}<\frac{2A\left(a,b\right)+G\left(a,b\right)}{3}<I\left(a,b\right)$$

(5)

hold. Yang and Chu (Theorem 3.3 of [23]) established a series of sharp inequalities for $TQ\left(a,b\right)$ and $I_0\left(t\right)$, for example, the inequalities

$$\sqrt{\frac{sinh\left(2t\right)}{\pi t}}<I_0\left(t\right)<\sqrt{\frac{sinh\left(2t\right)}{2t}},$$

(6)

$$\sqrt{\left(\frac{2}{\pi }cosht+1-\frac{2}{\pi }\right)\frac{sinht}{t}}<I_0\left(t\right)<\sqrt{\left({\lambda }_{0}cosht+1-{\lambda }_0\right)\frac{sinht}{t}},$$

(7)

$${\left(\frac{sinht}{t}\right)}^{3/4}{\left(cosht\right)}^{1/4}<I_0\left(t\right)<\frac{3}{4}\frac{sinht}{t}+\frac{1}{4}cosht,$$

(8)

hold for $t>0$ with ${\lambda }_0=0.6766\dots$. Inspired by the inequalities (3) and (4), Yang and Chu conjectured further that the inequality

$$TQ\left(a,b\right)<\sqrt{L\left(a,b\right)I\left(a,b\right)}$$

(9)

holds, which was proven in Theorem 3.1 of [24] by Yang, Chu and Song. In fact, they proved the following double inequality

$$\sqrt{\frac{e}{\pi }}\sqrt{L\left(a,b\right)I\left(a,b\right)}<TQ\left(a,b\right)<\sqrt{L\left(a,b\right)I\left(a,b\right)}$$

(10)

holds with the best coefficients $\sqrt{e/\pi }=0.930\dots$ and 1. More inequalities for $TQ\left(a,b\right)$ can be seen in [25,26].

Motivated by the inequalities (9) and $A_{2/3}<I$ listed in (3), the aim of this paper is to find the best constants p and q such that double inequality

$$\sqrt{L\left(a,b\right)A_q\left(a,b\right)}<TQ\left(a,b\right)<\sqrt{L\left(a,b\right)A_p\left(a,b\right)}$$

(11)

holds, or equivalently,

$$\sqrt{\frac{sinht}{t}{cosh}^{1/q}\left(qt\right)}<I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{1/p}\left(pt\right)}$$

(12)

for $t>0$. Our main results are as follows.

Theorem 1.

The function

$$F\left(t\right)=\frac{t{I}_0{\left(t\right)}^2}{{cosh}^{3/2}\left(2t/3\right)sinht}$$

is strictly decreasing from $\left(0,\infty \right)$ onto $\left(\sqrt{8}/\pi ,1\right)$. Therefore, the double inequality

$$\frac{2^{3/4}}{\sqrt{\pi }}\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}<I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}$$

holds for $t>0$, or equivalently,

$$\frac{2^{3/4}}{\sqrt{\pi }}\sqrt{L\left(a,b\right)A_{2/3}\left(a,b\right)}<TQ\left(a,b\right)<\sqrt{L\left(a,b\right)A_{2/3}\left(a,b\right)}$$

(13)

holds, where the coefficients $2^{3/4}/\sqrt{\pi }=0.94885\dots$ and 1 are the best.

Theorem 2.

The double inequality (12) holds for $t>0$, or equivalently, (11) holds for $a,b>0$ with $a\ne b$, if and only if $p\ge 2/3$ and $q\le p_0=\left(ln2\right)/ln\pi =0.605\dots$.

2. Tools and Lemmas

To prove our results, we need two tools. The first tool was due to Biernacki and Krzyz [27], which play an important role in dealing with the monotonicity of the ratio of power series.

Lemma 1

([27]). Let $A\left(t\right)={\sum }_{k=0}^{\infty }{a}_k{t}^k$ and $B\left(t\right)={\sum }_{k=0}^{\infty }{b}_k{t}^k$ be two real power series converging on $\left(-r,r\right)$ ($r>0$) with $b_k>0$ for all k. If the sequence $\{a_k/b_k\}$ is increasing (decreasing) for all k, then the function $t\mapsto A\left(t\right)/B\left(t\right)$ is also increasing (decreasing) on $\left(0,r\right)$.

Remark 1.

Recently, another monotonicity rule in the case when the sequence ${\{a_k/b_k\}}_{k\ge 0}$ is piecewise monotonic was presented in Theorem 1 of [28], which is now applied preliminarily, see for example, [29,30,31,32].

The second tool is the so-called “L’Hospital Monotone Rule” (or, for short, LMR), which is very effective in studying the monotonicity of ratios of two functions.

Lemma 2

([33], Theorem 2). Let $-\infty <a<b<\infty$, and let $f,g:[a,b]\to \mathbb{R}$ be continuous functions that are differentiable on $\left(a,b\right)$, with $f\left(a\right)=g\left(a\right)=0$ or $f\left(b\right)=g\left(b\right)=0$. Assume that $g^{\prime }\left(x\right)\ne 0$ for each x in $(a,b)$. If $f^{\prime }/g^{\prime }$ is increasing (decreasing) on $(a,b)$ then so is $f/g$.

The following two lemmas will be used to prove Proposition 1.

Lemma 3

([23], Lemma 2.8). We have

$$I_0{\left(t\right)}^2=\sum _{n=0}^{\infty }\frac{\left(2n\right)!}{2^{2n}n{!}^4}{t}^{2n}.$$

(14)

Lemma 4

([34], Problems 85, 94). The two given sequences ${\left\{a_n\right\}}_{n\ge 0}$ and ${\left\{b_n\right\}}_{n\ge 0}$ satisfy the conditions

$$b_n>0;\sum _{n=0}^{\infty }{b}_n{t}^{n}convergesforallvaluesoft;\underset{n\to \infty }{lim}\frac{a_n}{b_n}=s.$$

Then ${\sum }_{n=0}^{\infty }{a}_n{t}^n$ converges too for all values of t and in addition

$$\underset{t\to \infty }{lim}\frac{{\sum }_{n=0}^{\infty }{a}_n{t}^n}{{\sum }_{n=0}^{\infty }{b}_n{t}^n}=s.$$

3. Three Propositions

The proofs of Theorems 1 and 2 rely on the following propositions.

Proposition 1.

Let

$$f_0\left(t\right)=\theta \frac{2cosht+1}{3}\frac{sinht}{t}+\left(1-\theta \right)\left(1+\frac{1}{2}{t}^2+\frac{229}{6720}{t}^4\right),$$

(15)

where $\theta =11,009/10,449$. The function

$$F_0\left(t\right)=\frac{I_0{\left(t\right)}^2}{f_0\left(t\right)}$$

is strictly decreasing from $\left(0,\infty \right)$ onto $\left(3/\left(\theta \pi \right),1\right)$.

Proof. 

Expanding in power series yields

$$\begin{array}{ccc}\hfill f_0\left(t\right)& =& \theta \frac{sinh2t+sinht}{3t}+\left(1-\theta \right)\left(1+\frac{1}{2}{t}^2+\frac{229}{6720}{t}^4\right)\hfill \\ & =& \theta \sum _{n=0}^{\infty }\frac{2^{2n+1}+1}{3\left(2n+1\right)!}{t}^{2n}+\left(1-\theta \right)\left(1+\frac{1}{2}{t}^2+\frac{229}{6720}{t}^4\right)\hfill \\ & =& 1+\frac{1}{2}{t}^2+\frac{387\theta +229}{6720}{t}^4+\sum _{n=3}^{\infty }\frac{\theta \left(2^{2n+1}+1\right)}{3\left(2n+1\right)!}{t}^{2n}:=\sum _{n=0}^{\infty }{v}_n{t}^{2n},\hfill \end{array}$$

where $v_0=1$, $v_1=1/2$,

$$v_2={\displaystyle \frac{387\theta +229}{6720}}\mathrm{and}{v}_n={\displaystyle \frac{\theta \left(2^{2n+1}+1\right)}{3\left(2n+1\right)!}}\mathrm{for}n\ge 3.$$

By Lemma 3, we see that

$$I_0{\left(t\right)}^2=\sum _{n=0}^{\infty }\frac{\left(2n\right)!}{2^{2n}n{!}^4}{t}^{2n}:=\sum _{n=0}^{\infty }{u}_n{t}^{2n}.$$

Direct calculations gives

$$\begin{array}{ccc}\hfill \frac{u_0}{v_0}& =& \frac{u_1}{v_1}=1,\frac{u_2}{v_2}={\displaystyle \frac{630}{387\theta +229}}\hfill \\ \hfill \frac{u_n}{v_n}& =& {\displaystyle \frac{\left(2n\right)!}{2^{2n}n{!}^4}}/{\displaystyle \frac{\theta \left(2^{2n+1}+1\right)}{3\left(2n+1\right)!}}\mathrm{for}n\ge 3,\hfill \end{array}$$

then

$$\begin{array}{ccc}\hfill \frac{u_1}{v_1}-\frac{u_0}{v_0}& =& 0,\frac{u_2}{v_2}-\frac{u_1}{v_1}=-\frac{387\theta -401}{387\theta +229}<0,\hfill \\ \hfill \frac{u_3}{v_3}-\frac{u_2}{v_2}& =& -\frac{35}{172}\frac{1161\theta -1145}{\theta \left(387\theta +229\right)}<0,\hfill \end{array}$$

$$\frac{u_{n+1}}{v_{n+1}}/\frac{u_n}{v_n}-1=-\frac{2^{2n+1}-\left(3n^2+6n+2\right)}{{\left(n+1\right)}^2\left(2^{2n+3}+1\right)}<0\mathrm{for}n\ge 3,$$

where the last inequality holds due to

$$\begin{array}{ccc}\hfill 2^{2n+1}-\left(3n^2+6n+2\right)& >& 1+\left(2n+1\right)+\frac{\left(2n+1\right)\left(2n\right)}{2!}+\frac{\left(2n+1\right)\left(2n\right)\left(2n-1\right)}{3!}\hfill \\ \hfill -\left(3n^2+6n+2\right)& =& \frac{1}{3}n\left(4n+5\right)\left(n-2\right)>0\mathrm{for}n\ge 3.\hfill \end{array}$$

This shows that the sequence ${\left\{u_n/v_n\right\}}_{n\ge 0}$ is strictly decreasing, so is $I_0{\left(t\right)}^2/f_0\left(t\right)$ on $\left(0,\infty \right)$ by Lemma 1. It is easy to check that

$$\underset{t\to 0}{lim}\frac{I_0{\left(t\right)}^2}{f_0\left(t\right)}=\frac{u_0}{v_0}=1\mathrm{and}\underset{t\to \infty }{lim}\frac{I_0{\left(t\right)}^2}{f_0\left(t\right)}=\underset{n\to \infty }{lim}\frac{u_n}{v_n}=\frac{3}{\pi \theta },$$

where the second limits holds due to Lemma 4, thereby completing the proof. ☐

Proposition 2.

Let $f_0\left(t\right)$ be defined by (15). The function

$$F_1\left(t\right)=\frac{t{f}_0\left(t\right)}{{cosh}^{3/2}\left(2t/3\right)sinht}$$

is strictly decreasing from $\left(0,\infty \right)$ onto $\left(\sqrt{8}\theta /3,1\right)$, where $\theta =11,009/10,449$.

Proof. 

Let

$$\begin{array}{ccc}\hfill f_1\left(t\right)& =& ln{F}_1\left(t\right)=ln\left[\theta \frac{2cosht+1}{3}\frac{sinht}{t}+\left(1-\theta \right)\left(1+\frac{1}{2}{t}^2+\frac{229}{6720}{t}^4\right)\right]\hfill \\ & & -\frac{3}{2}ln\left(cosh\frac{2t}{3}\right)-ln\frac{sinht}{t}.\hfill \end{array}$$

Differentiation yields

$$f_1^{\prime }\left(t\right)=-\frac{1}{6tsinhtcosh\left(2t/3\right)}\frac{f_2\left(t\right)}{f_0\left(t\right)},$$

where

$$f_2\left(t\right)=t^5{f}_{25}\left(t\right)+t^4{f}_{24}\left(t\right)+t^3{f}_{23}\left(t\right)+t^2{f}_{22}\left(t\right)+t{f}_{21}\left(t\right)+f_{20}\left(t\right),$$

(16)

$$\begin{array}{ccc}\hfill f_{25}\left(t\right)& =& \frac{229}{1120}\left(1-\theta \right)\left(cosh\frac{2t}{3}cosht+3sinh\frac{2t}{3}sinht\right),\hfill \\ \hfill f_{24}\left(t\right)& =& \frac{229}{224}\left(\theta -1\right)cosh\frac{2t}{3}sinht,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill f_{23}\left(t\right)& =& \frac{3}{2}\left(1-\theta \right)\left(2cosh\frac{2t}{3}cosht+3sinh\frac{2t}{3}sinht\right)\hfill \\ \hfill f_{22}\left(t\right)& =& 9\left(\theta -1\right)cosh\frac{2t}{3}sinht,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill f_{21}\left(t\right)& =& 3\left(1-\theta \right)\left(2cosh\frac{2t}{3}cosht+3sinh\frac{2t}{3}sinht\right),\hfill \\ \hfill f_{20}\left(t\right)& =& 6\left(\theta -1\right)cosh\frac{2t}{3}sinht-4\theta cosh\frac{2t}{3}{sinh}^{3}t\hfill \\ & & +3\theta sinh\frac{2t}{3}{sinh}^{2}t+6\theta sinh\frac{2t}{3}cosht{sinh}^{2}t.\hfill \end{array}$$

Expanding in power series gives

$$f_{25}\left(3s\right)=-\frac{229}{4480}\left(\theta -1\right)\left(5cosh5s-coshs\right)=-\frac{229}{4480}\left(\theta -1\right)\sum _{n=2}^{\infty }\frac{5^{2n-3}-1}{\left(2n-4\right)!}{s}^{2n-4},$$

$$f_{24}\left(3s\right)=\frac{229}{448}\left(\theta -1\right)\left(sinh5s+sinhs\right)=\frac{229}{448}\left(\theta -1\right)\sum _{n=2}^{\infty }\frac{5^{2n-3}+1}{\left(2n-3\right)!}{s}^{2n-3},$$

$$f_{23}\left(3s\right)=-\frac{3}{4}\left(\theta -1\right)\left(5cosh5s-coshs\right)=-\frac{3}{4}\left(\theta -1\right)\sum _{n=1}^{\infty }\frac{5^{2n-1}-1}{\left(2n-2\right)!}{s}^{2n-2},$$

$$f_{22}\left(3s\right)=\frac{9}{2}\left(\theta -1\right)\left(sinh5s+sinhs\right)=\frac{9}{2}\left(\theta -1\right)\sum _{n=1}^{\infty }\frac{5^{2n-1}+1}{\left(2n-1\right)!}{s}^{2n-1},$$

$$f_{21}\left(3s\right)=-\frac{3}{2}\left(\theta -1\right)\left(5cosh5s-coshs\right)=-\frac{3}{2}\left(\theta -1\right)\sum _{n=0}^{\infty }\frac{5^{2n+1}-1}{\left(2n\right)!}{s}^{2n},$$

$$\begin{array}{ccc}\hfill f_{20}\left(3s\right)& =& \frac{1}{4}\theta sinh11s+\frac{3}{4}\theta sinh8s-\frac{5}{4}\theta sinh7s+\left(\frac{15}{4}\theta -3\right)sinh5s\hfill \\ & & -\frac{3}{4}\theta sinh4s-\frac{3}{2}\theta sinh2s+\left(\frac{21}{4}\theta -3\right)sinhs\hfill \end{array}$$

$$\begin{array}{cc}=& \sum _{n=0}^{\infty }\left[\frac{\theta }{4}{11}^{2n+1}+\frac{3\theta }{4}{8}^{2n+1}-\frac{5\theta }{4}{7}^{2n+1}+\left(\frac{15\theta }{4}-3\right)5^{2n+1}\right.\hfill \\ & \left.-\frac{3\theta }{4}{4}^{2n+1}-\frac{3\theta }{2}{2}^{2n+1}+\left(\frac{21}{4}\theta -3\right)\right]\frac{s^{2n+1}}{\left(2n+1\right)!}\hfill \end{array}$$

Then $f_2\left(3s\right)$ defined by (16) can be written as

$$f_2\left(3s\right)=243s^5{f}_{25}\left(3s\right)+81s^4{f}_{24}\left(3s\right)+27s^3{f}_{23}\left(3s\right)+9s^2{f}_{22}\left(3s\right)+3s{f}_{21}\left(3s\right)+f_{20}\left(3s\right)$$

$$=54s^3+\left(\theta -1\right)\sum _{n=2}^{\infty }{a}_n^{\left[1\right]}\frac{s^{2n+1}}{\left(2n+1\right)!}+\sum _{n=2}^{\infty }{a}_n^{\left[2\right]}{s}^{2n+1}+\frac{3\theta }{4}\sum _{n=2}^{\infty }{a}_n^{\left[3\right]}\frac{s^{2n+1}}{\left(2n+1\right)!},$$

where

$$\begin{array}{ccc}\hfill a_n^{\left[1\right]}& =& -\frac{55,647}{4480}\frac{\left(2n+1\right)!}{\left(2n-4\right)!}{5}^{2n-3}+\frac{18,549}{448}\frac{\left(2n+1\right)!}{\left(2n-3\right)!}{5}^{2n-3}\hfill \\ & & -\frac{81}{4}\frac{\left(2n+1\right)!}{\left(2n-2\right)!}{5}^{2n-1}+\frac{81}{2}\frac{\left(2n+1\right)!}{\left(2n-1\right)!}{5}^{2n-1}\hfill \\ & & -\frac{9}{2}\frac{\left(2n+1\right)!}{\left(2n\right)!}{5}^{2n+1}+\frac{3}{4}\frac{5\theta -4}{\theta -1}{5}^{2n+1}+\frac{1}{4}\frac{\theta }{\theta -1}{11}^{2n+1},\hfill \end{array}$$

$$\begin{array}{ccc}\hfill a_n^{\left[2\right]}& =& \frac{55,647}{4480}\frac{\theta -1}{\left(2n-4\right)!}+\frac{18,549}{448}\frac{\theta -1}{\left(2n-3\right)!}+\frac{81}{4}\frac{\theta -1}{\left(2n-2\right)!}\hfill \\ & & +\frac{81}{2}\frac{\theta -1}{\left(2n-1\right)!}+\frac{9}{2}\frac{\theta -1}{\left(2n\right)!}+\frac{1}{4}\frac{3\left(7\theta -4\right)}{\left(2n+1\right)!},\hfill \\ \hfill a_n^{\left[3\right]}& =& 8^{2n+1}-\frac{5}{3}\times 7^{2n+1}-4^{2n+1}-2^{2n+2}.\hfill \end{array}$$

It remains to prove $a_n^{\left[i\right]}>0$ for $i=1,2,3$ and $n\ge 2$. It is clear that $a_n^{\left[2\right]}>0$ due to $\theta =11,009/10,449>1$. For $a_n^{\left[3\right]}$, it is easy to check that

$$a_{n+1}^{\left[3\right]}-49a_n^{\left[3\right]}=12\left(10\times 2^{4n}+11\times 2^{2n}+15\right)\times 2^{2n}>0,$$

which together with $a_2^{\left[3\right]}=11,005>0$ yields $a_n^{\left[3\right]}>0$ for all $n\ge 2$. For $a_n^{\left[1\right]}$, since $\left(5\theta -4\right)>5\left(\theta -1\right)$ and

$$\frac{\theta }{\theta -1}=\frac{11,009}{560}=19.659\dots >18,$$

we have

$$\begin{array}{ccc}\hfill a_n^{\left[1\right]}& >& -\frac{55,647}{4480}\frac{\left(2n+1\right)!}{\left(2n-4\right)!}{5}^{2n-3}+\frac{18,549}{448}\frac{\left(2n+1\right)!}{\left(2n-3\right)!}{5}^{2n-3}\hfill \\ & & -\frac{81}{4}\frac{\left(2n+1\right)!}{\left(2n-2\right)!}{5}^{2n-1}+\frac{81}{2}\frac{\left(2n+1\right)!}{\left(2n-1\right)!}{5}^{2n-1}\hfill \\ & & -\frac{9}{2}\frac{\left(2n+1\right)!}{\left(2n\right)!}{5}^{2n+1}+\frac{15}{4}\times 5^{2n+1}+\frac{9}{2}\times {11}^{2n+1}\hfill \end{array}$$

$$\begin{array}{cc}=& \frac{9}{2}\times {11}^{2n+1}-3\left(\frac{18,549}{28}{n}^5-\frac{154,575}{56}{n}^4+\frac{138,915}{16}{n}^3\right.\hfill \\ & \left.-\frac{1,357,425}{224}{n}^2+\frac{848,523}{224}n+\frac{3125}{4}\right)\times 5^{2n-4}:=a_n^{\left[0\right]}.\hfill \end{array}$$

The sequence ${\left\{a_n^{\left[0\right]}\right\}}_{n\ge 2}$ satisfies the recurrence relation

$$\begin{array}{ccc}\hfill \frac{a_{n+1}^{\left[0\right]}-121a_n^{\left[0\right]}}{9\times 5^{2n-4}}& =& \frac{148,392}{7}{n}^5-\frac{463,725}{4}{n}^4+\frac{2,202,435}{7}{n}^3\hfill \\ & & -\frac{36,754,425}{112}{n}^2+\frac{3,895,809}{56}n-\frac{21,875}{2},\hfill \end{array}$$

which can be written as

$$\begin{array}{c}\frac{148,392}{7}{\left(n-2\right)}^5+\frac{2,689,605}{28}{\left(n-2\right)}^4+\frac{1,645,965}{7}{\left(n-2\right)}^3\hfill \\ +\frac{52,997,895}{112}{\left(n-2\right)}^2+\frac{29,042,799}{56}\left(n-2\right)+\frac{312,145}{2}>0\hfill \end{array}$$

for $n\ge 2$. This in combination with $a_2^{\left[0\right]}=10,126,407/16>0$ leads to $a_n^{\left[0\right]}>0$ for $n\ge 2$, and so is $a_n^{\left[1\right]}$. Therefore, $f_1\left(t\right)>0$ for $t>0$, so $f_0\left(t\right)$ is strictly increasing on $\left(0,\infty \right)$. An easy computation yields

$$\underset{t\to 0}{lim}{f}_1\left(t\right)=0\mathrm{and}\underset{t\to \infty }{lim}{f}_1\left(t\right)=ln\frac{\sqrt{8}\theta }{3},$$

which completes the proof. ☐

Using Lemma 2 we can prove the following lemma, which will be use to prove Theorem 2.

Proposition 3.

Let $q\ne 0,1/2,1$. The ratio

$$t\mapsto \frac{{cosh}^{1/q}\left(qt\right)-1}{cosht-1}$$

is strictly increasing on $\left(0,\infty \right)$ if $q\in \left(-\infty ,0\right)\cup \left(1/2,1\right)$ and strictly decreasing on $\left(0,\infty \right)$ if $q\in \left(0,1/2\right)\cup \left(1,\infty \right)$. Consequently, the double inequality

$$qcosht+1-q<{cosh}^{1/q}\left(qt\right)<c_{q}cosht+1-c_q$$

(17)

holds for $t>0$ if $q\in \left(-\infty ,0\right)\cup \left(1/2,1\right)$, where the weights q and $c_q=2^{1-1/q}$ if $q>0$ and $c_q=0$ if $q<0$ are the best possible. If $q\in \left(0,1/2\right)\cup \left(1,\infty \right)$, then the double inequality (17) is reversed.

Proof. 

Let

$$g_1\left(t\right)={cosh}^{1/q}\left(qt\right)-1\mathrm{and}{g}_2\left(t\right)=cosht-1.$$

Clearly, $g_1\left(0^{+}\right)=g_2\left(0^{+}\right)=0$, and

$$\underset{t\to 0}{lim}\frac{g_1\left(t\right)}{g_2\left(t\right)}=q\mathrm{and}\underset{t\to \infty }{lim}\frac{g_1\left(t\right)}{g_2\left(t\right)}=c_q=\left\{\begin{array}{cc}{2}^{1-1/q}& \mathrm{if}q>0,\\ 0& \mathrm{if}q<0.\end{array}\right.$$

Differentiation yields

$$\begin{array}{ccc}\hfill \frac{g_1^{\prime }\left(t\right)}{g_2^{\prime }\left(t\right)}& =& \frac{{cosh}^{1/q-1}\left(qt\right)sinh\left(qt\right)}{sinht},\hfill \\ \hfill {\left[\frac{g_1^{\prime }\left(t\right)}{g_2^{\prime }\left(t\right)}\right]}^{\prime }& =& \frac{\left(1-2q\right)t}{2{sinh}^{2}t{cosh}^{2-1/q}\left(qt\right)}\left(\frac{sinh\left|\left(1-2q\right)t\right|}{\left|\left(1-2q\right)t\right|}-\frac{sinht}{t}\right).\hfill \end{array}$$

Since the function $\left(sinhx\right)/x$ is strictly increasing on $\left(0,\infty \right)$, we find that

$${\left[\frac{g_1^{\prime }\left(t\right)}{g_2^{\prime }\left(t\right)}\right]}^{\prime }\left\{\begin{array}{cc}>0& \mathrm{if}\left(\left|1-2q\right|-1\right)\left(1-2q\right)>0,\mathrm{i}.\mathrm{e}.,q\in \left(-\infty ,0\right)\cup \left(\frac{1}{2},1\right),\\ <0& \mathrm{if}\left(\left|1-2q\right|-1\right)\left(1-2q\right)<0,\mathrm{i}.\mathrm{e}.,q\in \left(1,\infty \right)\cup \left(0,\frac{1}{2}\right).\end{array}\right.$$

By Lemma 2, the desired monotonicity follows. The double inequality (17) and its reverse follow from the monotonicity of $g_1\left(t\right)/g_2\left(t\right)$ on $\left(0,\infty \right)$. This completes the proof. ☐

Remark 2.

Taking $q=p_0=\left(ln2\right)/ln\pi$ in the double inequality (17) we obtain the double inequality

$$p_{0}cosht+1-p_0<{cosh}^{1/p_0}\left(p_{0}t\right)<\frac{2}{\pi }cosht+1-\frac{2}{\pi }$$

(18)

for $t>0$.

Remark 3.

The generalized Heronian mean [35] is defined by

$$H_w\left(a,b\right)=\frac{a+b+w\sqrt{ab}}{w+2}.$$

Let $t=ln\sqrt{a/b}$ with $b>a>0$ and $q=w/\left(w+2\right)>0$. Then Proposition 3 give a best approximation for $H_w\left(a,b\right)$ by power means:

$$\begin{array}{ccc}\hfill H_w\left(a,b\right)& <& A_{w/\left(w+2\right)}\left(a,b\right)\mathrm{if}w\in \left(2,\infty \right),\hfill \\ \hfill H_w\left(a,b\right)& >& A_{w/\left(w+2\right)}\left(a,b\right)\mathrm{if}q\in \left(0,2\right).\hfill \end{array}$$

Our proof is clearly concise than Li, Long and Chu’s given in [35].

4. Proofs of Theorem 1 and 2

We are now in a position to prove Theorems 1 and 2.

Proof of Theorem 1.

We have

$$F\left(t\right)=\frac{t{I}_0{\left(t\right)}^2}{{cosh}^{3/2}(2t/3)sinht}=\frac{I_0{\left(t\right)}^2}{f_0\left(t\right)}\times \frac{t{f}_0\left(t\right)}{{cosh}^{3/2}(2t/3)sinht}=F_0\left(t\right)\times F_1\left(t\right).$$

As shown in Propositions 1 and 2, the functions $F_0\left(t\right)$ and $F_1\left(t\right)$ are both strictly positive and decreasing on $\left(0,\infty \right)$, so is $F\left(t\right)$. And, we easily obtain

$$\begin{array}{ccc}\hfill \underset{t\to 0}{lim}F\left(t\right)& =& \underset{t\to 0}{lim}{F}_0\left(t\right)\times \underset{t\to 0}{lim}{F}_1\left(t\right)=1,\hfill \\ \hfill \underset{t\to \infty }{lim}F\left(t\right)& =& \underset{t\to \infty }{lim}{F}_0\left(t\right)\times \underset{t\to \infty }{lim}{F}_1\left(t\right)=\frac{3}{\pi \theta }\frac{\sqrt{8}\theta }{3}=\frac{\sqrt{8}}{\pi }.\hfill \end{array}$$

Using the monotonicity of $F\left(t\right)$, the desired double inequality follows. This completes the proof. ☐

Proof of Theorem 2.

(i) The necessary condition for the right hand side inequality of (12) to hold follows from the limit relation

$$\underset{t\to 0}{lim}\frac{I_0{\left(t\right)}^2-{cosh}^{1/p}\left(pt\right)\left(sinht\right)/t}{t^2}=-\frac{1}{6}\left(3p-2\right)\le 0.$$

The sufficiency follow from Theorem 1 and the increasing property of $p\mapsto {cosh}^{1/p}\left(pt\right)$ on $\mathbb{R}$.

(ii) The necessary condition for the left hand side inequality of (12) to hold follows from the limit relation

$$\underset{t\to \infty }{lim}\frac{{\left(cosh\left(qt\right)\right)}^{1/q}\left(sinht\right)/t}{I_0{\left(t\right)}^2}\le 1.$$

Since $I_0\left(t\right)\sim e^t/\sqrt{2\pi t}$ as $t\to \infty$ (see [36], 9.7.1) and

$$\begin{array}{ccc}\hfill {cosh}^{1/q}\left(qt\right)\frac{sinht}{t}& \le & \frac{sinht}{t}\sim \frac{e^t}{2t}\mathrm{if}q\le 0,\hfill \\ \hfill {cosh}^{1/q}\left(qt\right)\frac{sinht}{t}& =& e^t{\left(\frac{1+e^{-2qt}}{2}\right)}^{1/q}{e}^t\frac{1-e^{-2t}}{2t}\sim \frac{1}{2^{1/q}}\frac{e^{2t}}{2t},\hfill \end{array}$$

we have

$$\underset{t\to \infty }{lim}\frac{{cosh}^{1/q}\left(qt\right)\left(sinht\right)/t}{I_0{\left(t\right)}^2}=\left\{\begin{array}{cc}0& \mathrm{if}q\le 0,\\ {\displaystyle \frac{\pi }{2^{1/q}}}& \mathrm{if}q>0.\end{array}\right.$$

Therefore, the necessary condition is that $\pi /2^{1/q}\le 1$ if $q>0$ and $q\le 0$, that is, $q\le \left(ln2\right)/ln\pi =p_0$.

By the increasing property of $q\mapsto {cosh}^{1/q}\left(qt\right)$, to prove the sufficiency, it suffices to prove the left hand side inequality of (12) holds when $q=p_0$. From the first inequality of (7) and the second inequality of (18) it follows that

$$I_0\left(t\right)>\sqrt{\frac{sinht}{t}\left(\frac{2}{\pi }cosht+1-\frac{2}{\pi }\right)}>\sqrt{\frac{sinht}{t}{cosh}^{1/p_0}\left(p_{0}t\right)}$$

for $t>0$, which proves the sufficiency, and the proof is completed. ☐

5. Concluding Remarks

In this paper, we obtained the best constants p and q such that the double inequality (12) holds for $t>0$, or equivalently, (11) holds for $a,b>0$ with $a\ne b$. This improved the result in [24]. We close the paper by giving two remarks on our results.

Remark 4.

It was shown in ([20], 5.25) that

$$A_{2/3}\left(a,b\right)<I\left(a,b\right)<\frac{2\sqrt{2}}{e}{A}_{2/3}\left(a,b\right).$$

Then the double inequality (11) can be extended as

$$\begin{array}{ccc}\hfill \sqrt{\frac{e}{\pi }}\sqrt{L\left(a,b\right)I\left(a,b\right)}& <& \frac{2^{3/4}}{\sqrt{\pi }}\sqrt{L\left(a,b\right)A_{2/3}\left(a,b\right)}<TQ\left(a,b\right)\hfill \\ & <& \sqrt{L\left(a,b\right)A_{2/3}\left(a,b\right)}<\sqrt{L\left(a,b\right)I\left(a,b\right)}.\hfill \end{array}$$

Remark 5.

As a computable bound, the upper bound $\sqrt{t^{-1}sinht{cosh}^{3/2}\left(2t/3\right)}$ for $I_0\left(t\right)$ is superior to those given (6) and (8). In fact, we have

$$I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}<\sqrt{\frac{sinht}{t}cosht}=\sqrt{\frac{sinh\left(2t\right)}{2t}}$$

(19)

and

$$I_0\left(t\right)<\sqrt{\frac{sinht}{t}{cosh}^{3/2}\left(\frac{2t}{3}\right)}<\frac{3}{4}\frac{sinht}{t}+\frac{1}{4}cosht$$

(20)

for $t>0$. The inequalities (19) are clear, and we have to check (20). Let

$$h\left(t\right)=ln\sqrt{\frac{sinh\left(3t/2\right)}{3t/2}{cosh}^{3/2}\left(t\right)}-ln\left[\frac{3}{4}\frac{sinh\left(3t/2\right)}{3t/2}+\frac{1}{4}cosh\left(3t/2\right)\right].$$

Differentiation yields

$$h^{\prime }\left(t\right)=\frac{1}{6}\frac{h_1\left(t\right)}{t\left[sinh\left(3t\right)+tcosh\left(3t\right)\right]cosh\left(2t\right)sinh\left(3t\right)},$$

where

$$\begin{array}{ccc}\hfill h_1\left(t\right)& =& t^2\left(3cosh2t{cosh}^{2}3t+3sinh2tcosh3tsinh3t-6cosh2t{sinh}^{2}3t\right)\hfill \\ & & +\left(3sinh2t{sinh}^{2}3t-4cosh2tcosh3tsinh3t\right)t+cosh2t{sinh}^{2}3t.\hfill \end{array}$$

Using “product into sum" formulas for hyperbolic functions and expanding in power series give

$$\begin{array}{ccc}\hfill h_1\left(t\right)& =& \frac{9}{2}{t}^{2}cosh2t-\frac{3}{2}{t}^{2}cosh4t-\frac{3}{2}tsinh2t-\frac{7}{4}tsinh4t-\frac{1}{4}tsinh8t\hfill \\ & & +\frac{1}{4}cosh4t-\frac{1}{2}cosh2t+\frac{1}{4}cosh8t\hfill \end{array}$$

$$\begin{array}{ccc}\hfill h_1\left(t\right)& =& \frac{9}{2}\sum _{n=1}^{\infty }\frac{2^{2n-2}}{\left(2n-2\right)!}{t}^{2n}-\frac{3}{2}\sum _{n=1}^{\infty }\frac{4^{2n-2}}{\left(2n-2\right)!}{t}^{2n}\hfill \\ & & -\frac{3}{2}\sum _{n=1}^{\infty }\frac{2^{2n-1}}{\left(2n-1\right)!}{t}^{2n}-\frac{7}{4}\sum _{n=1}^{\infty }\frac{4^{2n-1}}{\left(2n-1\right)!}{t}^{2n}\hfill \\ & & -\frac{1}{4}\sum _{n=1}^{\infty }\frac{8^{2n-1}}{\left(2n-1\right)!}{t}^{2n}+\frac{1}{4}\sum _{n=0}^{\infty }\frac{4^{2n}}{\left(2n\right)!}{t}^{2n}\hfill \\ & & -\frac{1}{2}\sum _{n=0}^{\infty }\frac{2^{2n}}{\left(2n\right)!}{t}^{2n}+\frac{1}{4}\sum _{n=0}^{\infty }\frac{8^{2n}}{\left(2n\right)!}{t}^{2n}=-\frac{1}{16}\sum _{n=1}^{\infty }\frac{b_n{\left(2t\right)}^{2n}}{\left(2n\right)!},\hfill \end{array}$$

where

$$b_n=\left(n-4\right)4^{2n}+\left(6n^2+11n-4\right)2^{2n}-4\left(18n^2-15n-2\right).$$

Since $b_1=b_2=0$, $b_3=756$ and for $n\ge 4$,

$$\begin{array}{ccc}\hfill b_n& \ge & \left(6n^2+11n-4\right)2^8-4\left(18n^2-15n-2\right)\hfill \\ & =& 4\left(366n^2+719n-254\right)>0,\hfill \end{array}$$

we have $h_1\left(t\right)<0$ for $t>0$, so is $h^{\prime }\left(t\right)$. This leads to $h\left(t\right)<{lim}_{t\to 0}h\left(t\right)=0$, which proves the second inequality of (20) holds for $t>0$.

Remark 6.

Due to

$$\begin{array}{ccc}\hfill L\left(a,b\right)& =& {\displaystyle \frac{a-b}{lna-lnb}}={\int }_0^1{a}^s{b}^{1-s}ds,\hfill \\ \hfill TQ\left(a,b\right)& =& \frac{2}{\pi }{\int }_0^{\pi /2}{a}^{{cos}^2\theta }{b}^{{sin}^2\theta }d\theta =\frac{1}{\pi }{\int }_0^1{a}^s{b}^{1-s}{\left(s\left(1-s\right)\right)}^{-1/2}ds,\hfill \end{array}$$

the referee introduces a new family of means $L_{\alpha }\left(a,b\right)$ defined for $\alpha >0$ by

$$L_{\alpha }\left(a,b\right)=\frac{\mathsf{\Gamma }\left(2\alpha \right)}{\mathsf{\Gamma }{\left(\alpha \right)}^2}{\int }_0^1{a}^s{b}^{1-s}{\left(s\left(1-s\right)\right)}^{\alpha -1}ds.$$

The referee also gives an interesting relation between this new mean and the modified Bessel functions of the first kind:

$$\frac{L_{\alpha }\left(a,b\right)}{\sqrt{ab}}=\mathsf{\Gamma }\left(\alpha +\frac{1}{2}\right){\left(\frac{t}{2}\right)}^{1/2-\alpha }{I}_{\alpha -1/2}\left(t\right),t=ln\sqrt{\frac{a}{b}}.$$

It is easy to check that

$$\underset{\alpha \to 0}{lim}{L}_{\alpha }\left(a,b\right)=\frac{a+b}{2}$$

and

$$\begin{array}{ccc}\hfill L_{\alpha }\left(a,b\right)& =& \frac{\mathsf{\Gamma }\left(2\alpha \right)}{\mathsf{\Gamma }{\left(\alpha \right)}^2}{\int }_0^1{a}^s{b}^{1-s}{\left(s\left(1-s\right)\right)}^{\alpha -1}ds\hfill \\ & <& \frac{\mathsf{\Gamma }\left(2\alpha \right)}{\mathsf{\Gamma }{\left(\alpha \right)}^2}{\int }_0^1\left(sa+\left(1-s\right)b\right){\left(s\left(1-s\right)\right)}^{\alpha -1}ds=\frac{a+b}{2}.\hfill \end{array}$$

However, more problems remain to be researched on this new family of means, for example: (i) checking the monotonicity of this mean with respect to the parameter α; (ii) finding the lower and upper bounds for this mean in terms of elementary means; (iii) comparing this new mean with others.