On approximating the modified Bessel function of the first kind and Toader-Qi mean

Abstract

In the article, we present several sharp bounds for the modified Bessel function of the first kind $I_{0}(t)=\sum_{n=0}^{\infty}\frac{t^{2n}}{2^{2n}(n!)^{2}}$ and the Toader-Qi mean $TQ(a,b)=\frac{2}{\pi}\int_{0}^{\pi/2}a^{\cos^{2}\theta }b^{\sin^{2}\theta}\,d\theta$ for all $t>0$ and $a, b>0$ with $a\neq b$.

1 Introduction

Let $a, b>0$, $p: (0, \infty)\rightarrow\mathbb{R^{+}}$ be a strictly monotone real function, $\theta\in(0, 2\pi)$ and

$$ r_{n}(\theta)=\left \{ \textstyle\begin{array}{l@{\quad}l} (a^{n}\cos^{2}\theta+b^{n}\sin^{2}\theta )^{1/n}, & n\neq 0, \\ a^{\cos^{2}\theta}b^{\sin^{2}\theta}, & n=0. \end{array}\displaystyle \right . $$

(1.1)

Then the mean $M_{p,n}(a,b)$ was first introduced by Toader in [1] as follows:

$$ M_{p,n}(a,b)=p^{-1} \biggl(\frac{1}{2\pi} \int_{0}^{2\pi }p\bigl( r_{n}(\theta)\bigr) \,d\theta \biggr)=p^{-1} \biggl(\frac{2}{\pi} \int_{0}^{\pi /2}p\bigl(r_{n}(\theta)\bigr)\,d \theta \biggr), $$

(1.2)

where $p^{-1}$ is the inverse function of p.

From (1.1) and (1.2) we clearly see that

$$ M_{1/x,2}(a, b)=\frac{\pi}{2\int_{0}^{\pi/2}\frac{d\theta}{\sqrt {a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}}}=\operatorname{AGM}(a,b) $$

is the classical arithmetic-geometric mean, which is related to the complete elliptic integral of the first kind ${\mathcal{K}}(r)=\int _{0}^{\pi/2} (1-r^{2}\sin^{2}\theta )^{-1/2}\,d\theta$. The Toader mean

$$ M_{x,2}(a, b)=\frac{2}{\pi} \int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2} \theta +b^{2}\sin^{2}\theta}\,d\theta=T(a,b) $$

is related to the complete elliptic integral of the second kind ${\mathcal{E}}(r)=\int_{0}^{\pi/2} (1-r^{2}\sin^{2}\theta )^{1/2}\,d\theta$. We have

$$ M_{x^{q},0}(a, b)= \biggl(\frac{2}{\pi} \int_{0}^{\pi/2}a^{q\cos^{2}\theta }b^{q\sin^{2}\theta}\,d \theta \biggr)^{1/q} \quad (q\neq0). $$

In particular,

$$ M_{x,0}(a, b)=\frac{2}{\pi} \int_{0}^{\pi/2}a^{\cos^{2}\theta}b^{\sin ^{2}\theta}\,d \theta=TQ(a,b) $$

(1.3)

is the Toader-Qi mean.

Recently, the arithmetic-geometric mean $\operatorname{AGM}(a,b)$ and the Toader mean $T(a,b)$ have attracted the attention of many researchers. In particular, many remarkable inequalities for $\operatorname{AGM}(a,b)$ and $T(a,b)$ can be found in the literature [2–20].

For $q\neq0$, the mean $M_{x^{q},0}(a, b)$ seems to be mysterious, Toader [1] said that he did not know how to determine any sense for this mean.

Let $z\in\mathbb{C}$, $\nu\in\mathbb{R}\backslash\{-1, -2, -3,\ldots\}$ and $\Gamma(z)=\lim_{n\rightarrow\infty}n!n^{z}/ [\Pi _{k=0}^{\infty}(z+k) ]$ be the classical gamma function. Then the modified Bessel function of the first kind $I_{\nu}(z)$ [21] is given by

$$ I_{\nu}(z)=\sum_{n=0}^{\infty} \frac{z^{2n+\nu}}{n!2^{2n+\nu}\Gamma(\nu+n+1)}. $$

(1.4)

Very recently, Qi et al. [22] proved the identity

$$ M_{x^{q},0}(a, b)= \biggl(\frac{2}{\pi} \int_{0}^{\pi/2}a^{q\cos^{2}\theta }b^{q\sin^{2}\theta}\,d \theta \biggr)^{1/q}=\sqrt{ab}I_{0}^{1/q} \biggl( \frac {q}{2}\log\frac{a}{b} \biggr) $$

(1.5)

and inequalities

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

(1.6)

for all $q\neq0$ and $a, b>0$ with $a\neq b$, where $L(a,b)=(b-a)/(\log b-\log a)$, $A(a,b)=(a+b)/2$, $G(a,b)=\sqrt{ab}$, and $I(a,b)=(b^{b}/a^{a})^{1/(b-a)}/e$ are, respectively, the logarithmic, arithmetic, geometric, and identric means of a and b.

Let $b>a>0$, $p\in\mathbb{R}$, $t=(\log b-\log a)/2>0$, and the pth power mean $A_{p}(a,b)$ be defined by

$$ A_{p}(a,b)= \biggl(\frac{a^{p}+b^{p}}{2} \biggr)^{1/p} \quad (p \neq0),\qquad A_{0}(a,b)=\sqrt{ab}=G(a,b). $$

Then the logarithmic mean $L(a,b)$, the identric mean $I(a,b)$, and the pth power mean $A_{p}(a,b)$ can be expressed as

$$ \begin{aligned} &L(a,b)=\sqrt{ab}\frac{\sinh t}{t}, \qquad I(a,b)= \sqrt{ab}e^{t/\tanh t-1}, \\ &A_{p}(a,b)=\sqrt{ab} \cosh^{1/p}(pt) \quad (p\neq0) \end{aligned} $$

(1.7)

and (1.3)-(1.5) lead to

$$\begin{aligned} \frac{TQ(a,b)}{\sqrt{ab}} =&\frac{M_{x, 0}(a,b)}{\sqrt{ab}}=\frac{2}{\pi } \int_{0}^{\pi/2}e^{t\cos(2\theta)}\,d\theta=I_{0}(t) \\ =&\frac{2}{\pi} \int_{0}^{\pi/2}\cosh(t\cos\theta)\,d\theta= \frac{2}{\pi } \int_{0}^{\pi/2}\cosh(t\sin\theta)\,d\theta. \end{aligned}$$

(1.8)

The main purpose of this paper is to present several sharp bounds for the modified Bessel function of the first kind $I_{0}(t)$ and the Toader-Qi mean $TQ(a, b)$.

2 Lemmas

In order to establish our main results we need several lemmas, which we present in this section.

Lemma 2.1

(See [23])

Let $\binom{n}{k}$ be the number of combinations of n objects taken k at a time, that is,

$$ \binom{n}{k}=\frac{n!}{k!(n-k)!}. $$

Then

$$ \sum_{k=0}^{\infty}\binom{n}{k}^{2}= \binom{2n}{n}. $$

Lemma 2.2

(See [23])

Let $\{a_{n}\}_{n=0}^{\infty}$ and $\{b_{n}\}_{n=0}^{\infty }$ be two real sequences with $b_{n}>0$ and $\lim_{n\rightarrow\infty }{a_{n}}/{b_{n}}=s$. Then the power series $\sum_{n=0}^{\infty }a_{n}t^{n}$ is convergent for all $t\in\mathbb{R}$ and

$$ \lim_{t\rightarrow\infty}\frac{\sum_{n=0}^{\infty}a_{n}t^{n}}{\sum_{n=0}^{\infty}b_{n}t^{n}}=s $$

if the power series $\sum_{n=0}^{\infty}b_{n}t^{n}$ is convergent for all $t\in\mathbb{R}$.

Lemma 2.3

The Wallis ratio

$$ W_{n}=\frac{\Gamma (n+\frac{1}{2} )}{\Gamma (\frac {1}{2} )\Gamma(n+1)} $$

(2.1)

is strictly decreasing with respect to all integers $n\geq0$ and strictly log-convex with respect to all real numbers $n\geq0$.

Proof

It follows from (2.1) that

$$ \frac{W_{n+1}}{W_{n}}=1-\frac{1}{2(n+1)}< 1 $$

(2.2)

for all integers $n\geq0$.

Therefore, $W_{n}$ is strictly decreasing with respect to all integers $n\geq0$ follows from (2.2).

Let $f(x)=\Gamma(x+1/2)/\Gamma(x+1)$ and $\psi(x)=\Gamma^{\prime }(x)/\Gamma(x)$ be the psi function. Then it follows from the monotonicity of $\psi^{\prime}(x)$ that

$$ \bigl[\log f(x)\bigr]^{\prime\prime}=\psi^{\prime} \biggl(x+ \frac{1}{2} \biggr)-\psi ^{\prime}(x+1)>0 $$

(2.3)

for all $x\geq0$.

Therefore, $W_{n}$ is strictly log-convex with respect to all real numbers $n\geq0$ follows from (2.1) and (2.3). □

Lemma 2.4

(See [24])

The double inequality

$$ \frac{1}{(x+a)^{1-a}}< \frac{\Gamma(x+a)}{\Gamma(x+1)}< \frac{1}{x^{1-a}} $$

holds for all $x>0$ and $a\in(0, 1)$.

Lemma 2.5

Let $s_{n}=(2n)!(2n+1)!/[2^{4n}(n!)^{4}]$. Then the sequence $\{s_{n}\} _{n=0}^{\infty}$ is strictly decreasing and

$$ \lim_{n\rightarrow\infty}s_{n}=\frac{2}{\pi}. $$

(2.4)

Proof

The monotonicity of the sequence $\{s_{n}\}_{n=0}^{\infty}$ follows from

$$ \frac{s_{n+1}}{s_{n}}=\frac{(2n+1)(2n+3)}{4(n+1)^{2}}< 1. $$

To prove (2.4), we rewrite $s_{n}$ as

$$\begin{aligned} s_{n} =&(2n+1) \biggl[\frac{(2n-1)!!}{2^{n}n!} \biggr]^{2}=\frac{2n+1}{ \Gamma^{2} (\frac{1}{2} )} \biggl[\frac{\Gamma (n+\frac {1}{2} )}{\Gamma(n+1)} \biggr]^{2} \\ =&\frac{2 (n+\frac{1}{2} )}{\pi} \biggl[\frac{\Gamma (n+\frac {1}{2} )}{\Gamma(n+1)} \biggr]^{2}. \end{aligned}$$

(2.5)

It follows from Lemma 2.4 and (2.5) that

$$ \frac{2}{\pi}=\frac{2}{\pi}\frac{n+\frac{1}{2}}{n+\frac {1}{2}}< s_{n}< \frac{2}{\pi}\frac{n+\frac{1}{2}}{n}. $$

(2.6)

Therefore, equation (2.4) follows from (2.6). □

Lemma 2.6

(See [25])

Let $A(t)=\sum_{k=0}^{\infty}a_{k}t^{k}$ and $B(t)=\sum_{k=0}^{\infty}b_{k}t^{k}$ be two real power series converging on $(-r, r)$ ($r>0$) with $b_{k}>0$ for all k. If the non-constant sequence $\{a_{k}/b_{k}\}$ is increasing (decreasing) for all k, then the function $A(t)/B(t)$ is strictly increasing (decreasing) on $(0, r)$.

Lemma 2.7

(See [26])

Let $A(t)=\sum_{k=0}^{\infty}a_{k}t^{k}$ and $B(t)=\sum_{k=0}^{\infty}b_{k}t^{k}$ be two real power series converging on $\mathbb{R}$ with $b_{k}>0$ for all k. If there exists $m\in\mathbb{N}$ such that the non-constant sequence $\{a_{k}/b_{k}\}$ is increasing (decreasing) for $0\leq k\leq m$ and decreasing (increasing) for $k\geq m$, then there exists $t_{0}\in(0,\infty)$ such that the function $A(t)/B(t)$ is strictly increasing (decreasing) on $(0,t_{0})$ and strictly decreasing (increasing) on $(t_{0},\infty)$.

Lemma 2.8

The identity

$$ I^{2}_{0}(t)=\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^{4}}t^{2n} $$

holds for all $t\in\mathbb{R}$.

Proof

From (1.4) and Lemma 2.1 together with the Cauchy product we have

$$\begin{aligned} I^{2}_{0}(t) =&\sum_{n=0}^{\infty} \Biggl(\sum_{k=0}^{n}\frac {1}{2^{2k}(k!)^{2}} \frac{1}{2^{2(n-k)}[(n-k)!]^{2}} \Biggr) t^{2n} \\ =&\sum_{n=0}^{\infty} \Biggl(\frac{1}{2^{2n}(n!)^{2}} \sum_{k=0}^{n}\frac {(n!)^{2}}{(k!)^{2}[(n-k)!]^{2}} \Biggr)t^{2n} =\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^{4}}t^{2n}. \end{aligned}$$

 □

Lemma 2.9

(See [27])

Let $-\infty< a< b<\infty$ and $f, g: [a, b]\rightarrow\mathbb {R}$. Then

$$ \int_{a}^{b}f(x)g(x)\,dx\geq\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \int_{a}^{b}g(x)\,dx $$

if both f and g are increasing or decreasing on $(a, b)$.

Lemma 2.10

(See [28])

Let $-\infty< a< b<\infty$ and $f, g: (a, b)\rightarrow\mathbb {R}$. Then

$$\begin{aligned}& \int_{a}^{b}f(x)g(x)\,dx-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \int_{a}^{b}g(x)\,dx \\& \quad \geq\frac{12}{(b-a)^{3}} \int_{a}^{b} \biggl(x-\frac{a+b}{2} \biggr)f(x)\,dx \int_{a}^{b} \biggl(x-\frac{a+b}{2} \biggr)g(x)\,dx \end{aligned}$$

(2.7)

if both f and g are convex on the interval $(a, b)$, and inequality (2.7) becomes an equality if and only if f or g is a linear function on $(a,b)$.

3 Main results

Theorem 3.1

The double inequalities

$$ \frac{e^{t}}{1+2t}< I_{0}(t)< \frac{e^{t}}{\sqrt{1+2t}} $$

(3.1)

and

$$ \frac{b}{1+\log(b/a)}< TQ(a,b)< \frac{b}{\sqrt{1+\log(b/a)}} $$

(3.2)

hold for all $t>0$ and $b>a>0$.

Proof

From (1.8) we have

$$ I_{0}(t)=\frac{2}{\pi} \int_{0}^{\pi/2}\cosh(t\sin\theta)\,d\theta= \frac {2}{\pi} \int_{0}^{1}\frac{\cosh(tx)}{\sqrt{1-x^{2}}}\,dx $$

(3.3)

and

$$\begin{aligned} e^{-t}I_{0}(t) =&\frac{2}{\pi} \int_{0}^{\pi/2}e^{t[\cos(2\theta )-1]}\,d\theta= \frac{2}{\pi} \int_{0}^{\pi/2}\frac{d\theta}{e^{2t\sin ^{2}\theta}} \\ < &\frac{2}{\pi} \int_{0}^{\pi/2}\frac{d\theta}{1+2t\sin^{2}\theta}=\frac {1}{\sqrt{1+2t}}. \end{aligned}$$

(3.4)

We clearly see that both $\cosh(tx)$ and $1/\sqrt{1-x^{2}}$ are increasing with respect to x on $(0, 1)$. Then Lemma 2.9 and (3.3) lead to

$$\begin{aligned} I_{0}(t) \geq&\frac{2}{\pi} \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}} \int _{0}^{1}\cosh(tx)\,dx=\frac{\sinh t}{t} \\ =&\frac{e^{t}}{2t} \biggl(1-\frac{1}{e^{2t}} \biggr)>\frac{e^{t}}{2t} \biggl(1-\frac{1}{1+2t} \biggr)=\frac{e^{t}}{1+2t}. \end{aligned}$$

(3.5)

Therefore, inequality (3.1) follows from (3.4) and (3.5).

Let $t=\log(b/a)/2$. Then it follows from (1.8) and (3.1) that

$$ \frac{\sqrt{b/a}}{1+\log(b/a)}< \frac{TQ(a,b)}{\sqrt{ab}}< \frac{\sqrt {b/a}}{\sqrt{1+\log(b/a)}}. $$

(3.6)

Therefore, inequality (3.2) follows from (3.6). □

Remark 3.1

From Theorem 3.1 we clearly see that

$$ \lim_{t\rightarrow\infty}e^{-t}I_{0}(t)=\lim _{x\rightarrow0^{+}}TQ(x, 1)=0. $$

Theorem 3.2

The double inequalities

$$ \alpha_{1}\sqrt{\frac{\sinh(2t)}{t}}< I_{0}(t)< \beta_{1}\sqrt{\frac{\sinh (2t)}{t}} $$

(3.7)

and

$$ \alpha_{2}\sqrt{L(a,b)A(a,b)}< TQ(a,b)< \beta_{2}\sqrt{L(a,b)A(a,b)} $$

(3.8)

hold for all $t>0$ and $a, b>0$ with $a\neq b$ if and only if $\alpha _{1}\leq1/\sqrt{\pi}$, $\beta_{1}\geq\sqrt{2}/2$, $\alpha_{2}\leq\sqrt{2/\pi}$ and $\beta_{2}\geq1$.

Proof

Let

$$\begin{aligned}& R_{0}(t)=\frac{I^{2}_{0}(t)}{\sinh(2t)/(2t)}, \end{aligned}$$

(3.9)

$$\begin{aligned}& a_{n}=\frac{(2n)!}{2^{2n}(n!)^{4}}, \qquad b_{n}=\frac{2^{2n}}{(2n+1)!}. \end{aligned}$$

(3.10)

Then simple computation leads to

$$ \frac{a_{n}}{b_{n}}=\frac{(2n)!(2n+1)!}{2^{4n}(n!)^{4}}. $$

(3.11)

It follows from Lemma 2.5 and (3.11) that the sequence $\{a_{n}/b_{n}\} _{n=0}^{\infty}$ is strictly decreasing and

$$ \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{2}{\pi}. $$

(3.12)

From Lemma 2.8 we have

$$ R_{0}(t)=\frac{\sum_{n=0}^{\infty}a_{n}t^{2n}}{\sum_{n=0}^{\infty}b_{n}t^{2n}}. $$

(3.13)

Lemma 2.6 and (3.13) together with the monotonicity of the sequence $\{ a_{n}/b_{n}\}_{n=0}^{\infty}$ lead to the conclusion that $R_{0}(t)$ is strictly decreasing on the interval $(0, \infty)$. Therefore, we have

$$ \lim_{t\rightarrow\infty}R_{0}(t)< R_{0}(t)< \lim_{t\rightarrow 0^{+}}R_{0}(t)=\frac{a_{0}}{b_{0}}=1. $$

(3.14)

From Lemma 2.2, (3.12), and (3.13) we know that

$$ \lim_{t\rightarrow\infty}R_{0}(t)= \frac{2}{\pi}. $$

(3.15)

Therefore, inequality (3.7) holds for all $t>0$ if and only if $\alpha _{1}\leq1/\sqrt{\pi}$ and $\beta_{1}\geq\sqrt{2}/2$ follows easily from (3.9), (3.14), and (3.15) together with the monotonicity of $R_{0}(t)$.

Let $b>a>0$ and $t=\log(b/a)/2$. Then inequality (3.8) holds for $a, b>0$ with $a\neq b$ if and only if $\alpha_{2}\leq\sqrt{2/\pi}$, and $\beta_{2}\geq1$ follows from (1.7) and (1.8) together with inequality (3.7) for all $t>0$ if and only if $\alpha_{1}\leq1/\sqrt{\pi}$ and $\beta_{1}\geq\sqrt{2}/2$. □

Remark 3.2

Equations (3.9) and (3.15) imply that

$$ \lim_{t\rightarrow\infty}e^{-t}\sqrt{t}I_{0}(t)= \frac{1}{\sqrt{2\pi}} $$

or we have the asymptotic formula

$$ I_{0}(t)\sim\frac{e^{t}}{\sqrt{2\pi t}} \quad (t\rightarrow\infty). $$

Theorem 3.3

Let $\lambda_{1}, \lambda_{2}>0$, $t_{0}=2.7113\ldots$ be the unique solution of the equation

$$ \frac{d}{dt} \biggl[\frac{tI^{2}_{0}(t)-\sinh t}{(\cosh t-1)\sinh t} \biggr]=0 $$

(3.16)

on $(0, \infty)$ and

$$ \lambda_{0}=\frac{t_{0}I^{2}_{0}(t_{0})-\sinh t_{0}}{(\cosh t_{0}-1)\sinh t_{0}}=0.6766\ldots. $$

(3.17)

Then the double inequality

$$ \sqrt{(\lambda_{1}\cosh t+1-\lambda_{1}) \frac{\sinh t}{t}}< I_{0}(t)< \sqrt{(\lambda_{2}\cosh t+1- \lambda_{2})\frac{\sinh t}{t}} $$

(3.18)

or

$$ \sqrt{\bigl[\lambda_{1}A(a,b)+(1-\lambda_{1})G(a,b) \bigr]L(a,b)}< TQ(a,b)< \sqrt {\bigl[\lambda_{2}A(a,b)+(1- \lambda_{2})G(a,b)\bigr]L(a,b)} $$

holds for all $t>0$ or $a,b>0$ with $a\neq b$ if and only if $\lambda _{1}\leq2/\pi$, $\lambda_{2}>\lambda_{0}$.

Proof

Let

$$\begin{aligned}& R_{1}(t)=\frac{I^{2}_{0}(t)-\frac{\sinh t}{t}}{\frac{(\cosh t-1)\sinh t}{t}}, \end{aligned}$$

(3.19)

$$\begin{aligned}& c_{n}=\frac{(2n)!}{2^{2n}(n!)^{4}}-\frac{1}{(2n+1)!}, \qquad d_{n}=\frac {2^{2n}-1}{(2n+1)!}, \qquad s_{n}=\frac{(2n)!(2n+1)!}{2^{4n}(n!)^{4}}, \end{aligned}$$

(3.20)

and

$$ s^{\prime}_{n}= \bigl(2^{2n}+3n^{2}+6n+2 \bigr)s_{n}- \bigl(3n^{2}+6n+3 \bigr). $$

(3.21)

Then it follows from Lemma 2.2, Lemma 2.5, Lemma 2.8, and (3.19)-(3.21) that

$$\begin{aligned}& R_{1}(t)=\frac{\sum_{n=1}^{\infty}c_{n}t^{2n}}{\sum_{n=1}^{\infty}d_{n}t^{2n}}, \end{aligned}$$

(3.22)

$$\begin{aligned}& \lim_{t\rightarrow\infty}R_{1}(t)=\lim_{n\rightarrow\infty} \frac {c_{n}}{d_{n}}=\lim_{n\rightarrow\infty}\frac {2^{2n}s_{n}-1}{2^{2n}-1}= \frac{2}{\pi}, \end{aligned}$$

(3.23)

$$\begin{aligned}& \frac{c_{1}}{d_{1}}=\frac{2}{3}< \frac{c_{2}}{d_{2}}=\frac{41}{60}> \frac {c_{3}}{d_{3}}=\frac{19}{28}, \end{aligned}$$

(3.24)

$$\begin{aligned}& \frac{c_{n+1}}{d_{n+1}}-\frac{c_{n}}{d_{n}}=-\frac{2^{2n}s^{\prime }_{n}}{(n+1)^{2} (2^{2n+2}-1 ) (2^{2n}-1 )}, \end{aligned}$$

(3.25)

and we have the inequality

$$\begin{aligned} s^{\prime}_{n} >&\frac{2}{\pi} \bigl(2^{2n}+3n^{2}+6n+2 \bigr)- \bigl(3n^{2}+6n+3 \bigr) \\ >&\frac{3}{5} \bigl(2^{2n}+3n^{2}+6n+2 \bigr)- \bigl(3n^{2}+6n+3 \bigr)=\frac{3}{5} \bigl[2^{2n}- \bigl(2n^{2}+4n+3 \bigr) \bigr]>0 \end{aligned}$$

(3.26)

for all $n\geq3$.

From (3.24)-(3.26) we know that the sequence $\{c_{n}/d_{n}\} _{n=1}^{\infty}$ is strictly increasing for $1\leq n\leq2$ and strictly decreasing for $n\geq2$. Then Lemma 2.7 and (3.22) lead to the conclusion that there exists $t_{0}\in(0, \infty)$ such that $R_{1}(t)$ is strictly increasing on $(0, t_{0})$ and decreasing on $(t_{0}, \infty)$. Therefore, we have

$$ \min\Bigl\{ R_{1}\bigl(0^{+}\bigr), \lim _{t\rightarrow\infty}R_{1}(t)\Bigr\} < R_{1}(t)\leq R_{1}(t_{0}) $$

(3.27)

for all $t>0$, and $t_{0}$ is the unique solution of equation (3.16) on $(0, \infty)$.

Note that

$$ R_{1}\bigl(0^{+}\bigr)=\frac{c_{1}}{d_{1}}= \frac{2}{3}. $$

(3.28)

From (3.17), (3.19), (3.23), (3.27), and (3.28) we get

$$ \frac{2}{\pi}< R_{1}(t)\leq R_{1}(t_{0})= \lambda_{0}. $$

(3.29)

Therefore, inequality (3.18) holds for all $t>0$ if and only if $\lambda _{1}\leq2/\pi$, $\lambda_{2}\geq\lambda_{0}$ follows from (3.19) and (3.29) together with the piecewise monotonicity of $R_{1}(t)$ on $(0, \infty)$. Numerical computations show that $t_{0}=2.7113\ldots$ and $\lambda_{0}=0.6766\ldots$ . □

Theorem 3.4

Let $p, q\in\mathbb{R}$. Then the double inequality

$$ \cosh^{1-p} t \biggl(\frac{\sinh t}{t} \biggr)^{p}< I_{0}(t)< q\frac{\sinh t}{t}+(1-q)\cosh t $$

(3.30)

or

$$ L^{p}(a,b)A^{1-p}(a,b)< TQ(a,b)< qL(a,b)+(1-q)A(a,b) $$

holds for all $t>0$ or $a, b>0$ with $a\neq b$ if and only if $p\geq 3/4$ and $q\leq3/4$.

Proof

If the first inequality of (3.30) holds for all $t>0$, then

$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-\cosh^{1-p} t (\frac{\sinh t}{t} )^{p}}{t^{2}}=\frac{1}{3} \biggl(p- \frac{3}{4} \biggr)\geq0, $$

which implies that $p\geq3/4$.

It is not difficult to verify that the function $\cosh^{1-p} t(\sinh t/t)^{p}$ is strictly decreasing with respect to $p\in\mathbb{R}$ for any fixed $t>0$, hence we only need to prove the first inequality of (3.30) for all $t>0$ and $p=3/4$, that is,

$$ I^{4}_{0}(t)> \biggl(\frac{\sinh t}{t} \biggr)^{3}\cosh t. $$

(3.31)

Making use of the power series and Cauchy product formulas together with Lemma 2.8 we have

$$\begin{aligned}& I^{4}_{0}(t)- \biggl(\frac{\sinh t}{t} \biggr)^{3}\cosh t \\& \quad =\sum_{n=0}^{\infty} \Biggl[\sum _{k=0}^{n} \biggl(\frac {(2k)!}{2^{2k}(k!)^{4}} \frac{(2(n-k))!}{ 2^{2(n-k)}((n-k)!)^{4}} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \Biggr]t^{2n}. \end{aligned}$$

(3.32)

Let $W_{n}$ and $s_{n}$ be, respectively, defined by Lemma 2.3 and Lemma 2.5, and

$$ u_{n}=\sum_{k=0}^{n} \biggl(\frac{(2k)!}{2^{2k}(k!)^{4}}\frac{(2(n-k))!}{ 2^{2(n-k)}((n-k)!)^{4}} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!}. $$

(3.33)

Then simple computations lead to

$$ u_{0}=u_{1}=0,\qquad u_{2}= \frac{3}{80}, \qquad u_{3}=\frac{4}{189}. $$

(3.34)

It follows from Lemma 2.1 and Lemmas 2.3-2.5 together with (3.33) that

$$\begin{aligned} u_{n} =&\sum_{k=0}^{n} \biggl(\frac{1}{(k!)^{2}((n-k)!)^{2}}\frac {(2k)!}{2^{2k}(k!)^{2}}\frac{(2(n-k))!}{ 2^{2(n-k)}((n-k)!)^{2}} \biggr)- \frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\sum_{k=0}^{n} \biggl( \frac{1}{(k!)^{2}((n-k)!)^{2}}W_{k}W_{n-k} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ >&\sum_{k=0}^{n} \biggl( \frac{1}{(k!)^{2}((n-k)!)^{2}}W^{2}_{n/2} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\sum_{k=0}^{n} \biggl[ \frac{1}{(k!)^{2}((n-k)!)^{2}} \biggl(\frac{\Gamma (n/2+1/2)}{\Gamma(1/2)\Gamma(n/2+1)} \biggr)^{2} \biggr]- \frac {2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ >&\sum_{k=0}^{n} \biggl( \frac{1}{\pi(n/2+1/2)(k!)^{2}((n-k)!)^{2}} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\frac{2}{\pi(n+1)(n!)^{2}}\sum_{k=0}^{n} \frac {(n!)^{2}}{(k!)^{2}((n-k)!)^{2}}-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\frac{2}{\pi(n+1)(n!)^{2}}\frac{(2n)!}{(n!)^{2}}-\frac {2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\frac{2^{2n+1} (2^{2n+2}-1 )}{\pi(2n+3)!} \biggl[\frac {2^{2n+2}}{2^{2n+2}-1} \biggl(n+\frac{3}{2} \biggr)s_{n}-\pi \biggr] \\ >&\frac{2^{2n+1} (2^{2n+2}-1 )}{\pi(2n+3)!} \biggl[ \biggl(4+\frac {3}{2} \biggr) \frac{2}{\pi}-\pi \biggr]>0 \end{aligned}$$

(3.35)

for all $n\geq4$.

Therefore, inequality (3.31) follows from (3.32)-(3.35).

If the second inequality of (3.30) holds for all $t>0$, then we have

$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-q\frac{\sinh t}{t}-(1-q)\cosh t}{t^{2}}=\frac{1}{3} \biggl(q- \frac{3}{4} \biggr)\leq0, $$

which implies that $q\leq3/4$.

Since $\cosh t>\sinh t/t$, we only need to prove that the second inequality of (3.3) holds for all $t>0$ and $q=3/4$, that is,

$$ \frac{\cosh t-I_{0}(t)}{\cosh t-\sinh t/t}>\frac{3}{4}. $$

(3.36)

Let

$$ \alpha_{n}=\frac{2^{n}n!-(2n-1)!!}{2^{n}n!(2n)!}, \qquad \beta_{n}= \frac {2n}{(2n+1)!}, \qquad \gamma_{n}=\frac{(n+2)(2n+1)}{2(n+1)}W_{n}, $$

and $W_{n}$ be defined by (2.1).

Then simple computations lead to

$$\begin{aligned}& \frac{\cosh t-I_{0}(t)}{\cosh t-\sinh t/t}=\frac{\sum_{n=1}^{\infty }\alpha_{n}t^{2n}}{\sum_{n=1}^{\infty}\beta_{n}t^{2n}}, \end{aligned}$$

(3.37)

$$\begin{aligned}& \frac{\alpha_{n+1}}{\beta_{n+1}}-\frac{\alpha_{n}}{\beta_{n}}=\frac {2n+3}{2n+2}(1-W_{n+1})- \frac{2n+1}{2n}(1-W_{n})=\frac{\gamma_{n}-1}{2n(n+1)}, \end{aligned}$$

(3.38)

$$\begin{aligned}& \frac{\gamma_{n+1}}{\gamma_{n}}=1+\frac{n+1}{2(n+2)^{2}}>1, \qquad \gamma _{1}= \frac{9}{8}>1. \end{aligned}$$

(3.39)

From (3.38) and (3.39) we clearly see that the sequence $\{\alpha _{n}/\beta_{n}\}_{n=1}^{\infty}$ is strictly increasing, then Lemma 2.6 and (3.37) lead to the conclusion that the function $(\cosh t-I_{0}(t))/[\cosh t-\sinh t/t]$ is strictly increasing on the interval $(0, \infty)$. Therefore, inequality (3.36) follows from the monotonicity of $(\cosh t-I_{0}(t))/[\cosh t-\sinh t/t]$ and the fact that

$$ \lim_{t\rightarrow0^{+}}\frac{\cosh t-I_{0}(t)}{\cosh t-\sinh t/t}=\frac{\alpha_{1}}{\beta_{1}}= \frac{3}{4}. $$

 □

Theorem 3.5

Let $p, q>0$, $t_{0}$ be the unique solution of the equation

$$ \frac{d [\frac{p^{2}(I_{0}(t)-1)}{\cosh(pt)-1} ]}{dt}=0 $$

(3.40)

and

$$ \mu_{0}=\frac{p^{2}(I_{0}(t_{0})-1)}{\cosh(pt_{0})-1}. $$

(3.41)

Then the following statements are true:

• (i) The double inequality

  $$ 1-\frac{1}{2p^{2}}+\frac{1}{2p^{2}}\cosh(pt)< I_{0}(t)< 1- \frac {1}{2q^{2}}+\frac{1}{2q^{2}}\cosh(qt) $$

  (3.42)

  or

  $$\begin{aligned}& \biggl(1-\frac{1}{2p^{2}} \biggr)G(a,b)+\frac {1}{2p^{2}}A^{p}_{p}(a,b)G^{1-p}(a,b) \\& \quad < TQ(a,b) < \biggl(1-\frac{1}{2q^{2}} \biggr)G(a,b)+ \frac {1}{2q^{2}}A^{q}_{q}(a,b)G^{1-q}(a,b) \end{aligned}$$

  holds for all $t>0$ or $a, b>0$ with $a\neq b$ if and only if $p\leq \sqrt{3}/2$ and $q\geq1$.

• (ii) The inequality

  $$ I_{0}(t)\geq1-\frac{\mu_{0}}{p^{2}}+\frac{\mu_{0}}{p^{2}} \cosh(pt) $$

  (3.43)

  or

  $$ TQ(a,b)\geq \biggl(1-\frac{\mu_{0}}{p^{2}} \biggr)G(a,b)+\frac{\mu _{0}}{p^{2}}A^{p}_{p}(a,b)G^{1-p}(a,b) $$

  holds for all $t>0$ or $a, b>0$ with $a\neq b$ if $p\in(\sqrt{3}/2, 1)$.

Proof

(i) Let

$$\begin{aligned}& R_{2}(t)=\frac{p^{2}(I_{0}(t)-1)}{\cosh(pt)-1}, \\& u_{n}=\frac{1}{2^{2n}(n!)^{2}}, \qquad v_{n}=\frac{p^{2n-2}}{(2n)!}. \end{aligned}$$

(3.44)

Then simple computations lead to

$$\begin{aligned}& R_{2}(t)=\frac{\sum_{n=1}^{\infty}u_{n}t^{2n}}{\sum_{n=1}^{\infty}v_{n}t^{2n}}, \end{aligned}$$

(3.45)

$$\begin{aligned}& \frac{u_{n+1}}{v_{n+1}}-\frac{u_{n}}{v_{n}}=-\frac {(2n)!}{(2p)^{2n}(n!)^{2}} \biggl(p^{2}-\frac{2n+1}{2n+2} \biggr). \end{aligned}$$

(3.46)

From (3.46) we clearly see that the sequence $\{u_{n}/v_{n}\} _{n=1}^{\infty}$ is strictly decreasing if $p\geq1$ and strictly increasing if $p\leq\sqrt{3}/2$. Then Lemma 2.6 and (3.45) lead to the conclusion that the function $R_{2}(t)$ is strictly decreasing if $p\geq1$ and strictly increasing if $p\leq\sqrt{3}/2$. Hence, we have

$$ R_{2}(t)< \lim_{t\rightarrow0^{+}}R_{2}(t)= \frac{u_{1}}{v_{1}}=\frac{1}{2} $$

(3.47)

for all $t>0$ if $p\geq1$ and

$$ R_{2}(t)>\lim_{t\rightarrow0^{+}}R_{2}(t)= \frac{u_{1}}{v_{1}}=\frac{1}{2} $$

(3.48)

for all $t>0$ if $p\leq\sqrt{3}/2$.

Therefore, inequality (3.42) holds for all $t>0$ if $p\leq\sqrt{3}/2$ and $q\geq1$ follows easily from (3.44) and (3.47) together with (3.48).

If the first inequality (3.42) holds for all $t>0$, then we have

$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)- (1-\frac{1}{2p^{2}}+\frac {1}{2p^{2}}\cosh(pt) )}{ t^{4}}=\frac{1}{48} \biggl( \frac{3}{4}-p^{2} \biggr)\geq0, $$

which implies that $p\leq\sqrt{3}/2$.

If there exists $q_{0}\in(\sqrt{3}/2, 1)$ such that the second inequality of (3.42) holds for all $t>0$, then we have

$$ \lim_{t\rightarrow\infty}\frac{I_{0}(t)- (1-\frac {1}{2q_{0}^{2}}+\frac{1}{2q_{0}^{2}}\cosh(q_{0}t) )}{ e^{q_{0}t}}\leq0. $$

(3.49)

But the first inequality of (3.1) leads to

$$\begin{aligned}& \frac{I_{0}(t)- (1-\frac{1}{2q_{0}^{2}}+\frac{1}{2q_{0}^{2}}\cosh (q_{0}t) )}{e^{q_{0}t}} \\& \quad >\frac{e^{(1-q_{0})t}}{1+2t}- \biggl(1-\frac{1}{2q_{0}^{2}} \biggr)e^{-q_{0}t}- \frac{1+e^{-2q_{0}t}}{4q_{0}^{2}}\rightarrow\infty \quad (t\rightarrow\infty), \end{aligned}$$

which contradicts inequality (3.49).

(ii) If $p\in(\sqrt{3}/2, 1)$, then from (3.46) we know that there exists $n_{0}\in\mathbb{N}$ such that the sequence $\{u_{n}/v_{n}\} _{n=1}^{\infty}$ is strictly decreasing for $n\leq n_{0}$ and strictly increasing for $n\geq n_{0}$. Then (3.45) and Lemma 2.7 lead to the conclusion that there exists $t_{0}\in(0, \infty)$ such that the function $R_{2}(t)$ is strictly decreasing on $(0, t_{0}]$ and strictly increasing on $[t_{0}, \infty)$. We clearly see that $t_{0}$ satisfies equation (3.40). It follows from (3.41) and (3.44) together with the piecewise monotonicity of $R_{2}(t)$ that

$$ R_{2}(t)\geq R_{2}(t_{0})= \mu_{0}. $$

(3.50)

Therefore, inequality (3.43) holds for all $t>0$ follows from (3.44) and (3.50). □

It is not difficult to verify that the function

$$ 1-\frac{1}{2p^{2}}+\frac{1}{2p^{2}}\cosh(pt) $$

is strictly increasing with respect to p on the interval $(0, \infty )$ and

$$ 2\cosh \biggl(\frac{t}{2} \biggr)-1>\cosh^{1/2} t $$

for $t>0$.

Letting $p=\sqrt{3}/2, 3/4, \sqrt{2}/2, 2/3, 1/2$ and $q=1$ in Theorem 3.5(i), then we get Corollary 3.1 immediately.

Corollary 3.1

The inequalities

$$\begin{aligned} \cosh^{1/2}(t) < &2\cosh \biggl(\frac{t}{2} \biggr)-1< \frac{9}{8}\cosh \biggl(\frac{2t}{3} \biggr)-\frac{1}{8}< \cosh \biggl(\frac{\sqrt{2}t}{2} \biggr) \\ < &\frac{8}{9}\cosh \biggl(\frac{3t}{4} \biggr)+ \frac{1}{9}< \frac{2}{3}\cosh \biggl(\frac{\sqrt{3}t}{2} \biggr)+ \frac{1}{3}< I_{0}(t)< \frac{1+\cosh t}{2} \end{aligned}$$

or

$$\begin{aligned} G^{1/2}(a,b)A^{1/2}(a,b) < &2A^{1/2}_{1/2}(a,b)G^{1/2}(a,b)-G(a,b) \\ < & \frac {9}{8}A^{2/3}_{2/3}(a,b)G^{1/3}(a,b)- \frac{1}{8}G(a,b) \\ < &A^{\sqrt{2}/2}_{\sqrt{2}/2}(a,b)G^{1-\sqrt{2}/2}(a,b) \\ < & \frac {8}{9}A^{3/4}_{3/4}(a,b)G^{1/4}(a,b)+ \frac{1}{9}G(a,b) \\ < &\frac{2}{3}A^{\sqrt{3}/2}_{\sqrt{3}/2}(a,b)G^{1-\sqrt{3}/2}(a, b)+ \frac {1}{3}G(a,b) \\ < &TQ(a,b)< \frac{A(a,b)+G(a,b)}{2} \end{aligned}$$

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

Theorem 3.6

Let $p>0$. Then the following statements are true:

• (i) The inequality

  $$ I_{0}(t)>\bigl[\cosh(pt)\bigr]^{\frac{1}{2p^{2}}} $$

  (3.51)

  or

  $$ TQ(a,b)>G^{1-\frac{1}{2p}}(a,b)A_{p}^{\frac{1}{2p}}(a,b) $$

  (3.52)

  holds for all $t>0$ or $a, b>0$ with $a\neq b$ if and only if $p\geq \sqrt{6}/4$.

• (ii) The inequality (3.51) or (3.52) is reversed if and only if $p\leq1/2$.

• (iii) The inequalities

  $$ \cosh^{1/2} t< \cosh \biggl(\frac{\sqrt{2}t}{2} \biggr)< \biggl[\cosh \biggl(\frac{\sqrt{6}t}{4} \biggr) \biggr]^{4/3}< I_{0}(t)< \cosh^{2} \biggl(\frac {t}{2} \biggr)< e^{t^{2}/4} $$

  (3.53)

  or

  $$\begin{aligned} G^{1/2}(a,b)A^{1/2}(a,b) < &A^{\sqrt{2}/2}_{\sqrt{2}/2}(a,b)G^{1-\sqrt {2}/2}(a,b) \\ < &A^{\sqrt{6}/3}_{\sqrt{6}/4}(a,b)G^{1-\sqrt{6}/3}(a,b) \\ < &TQ(a,b)< \frac{A(a,b)+G(a,b)}{2} \\ < &G(a,b)e^{ (A^{2}(a,b)-G^{2}(a,b) )/ (4L^{2}(a,b) )} \end{aligned}$$

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

Proof

(i) If inequality (3.51) holds for all $t>0$, then we have

$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-[\cosh(pt)]^{\frac {1}{2p^{2}}}}{t^{4}}=\frac{1}{24} \biggl(p^{2}-\frac{3}{8} \biggr)\geq0, $$

which implies that $p\geq\sqrt{6}/4$.

It follows from Lemma 2 of [29] that the function $[\cosh (pt)]^{1/(2p^{2})}$ is strictly decreasing with respect to $p\in(0, \infty)$ for any fixed $t>0$, hence we only need to prove that inequality (3.51) holds for all $t>0$ and $p=\sqrt{6}/4$. From the sixth inequality of Corollary 3.1 we clearly see that it suffices to prove that

$$ \frac{2}{3}\cosh \biggl(\frac{\sqrt{3}t}{2} \biggr)+\frac{1}{3}> \biggl[\cosh \biggl(\frac{\sqrt{6}t}{4} \biggr) \biggr]^{4/3} $$

for all $t>0$, which is equivalent to

$$ \log \biggl[\frac{2}{3}\cosh(\sqrt{2}x)+\frac{1}{3} \biggr]>\frac{4}{3}\log (\cosh x) $$

(3.54)

for all $x>0$, where $x=\sqrt{6}t/4$.

Let

$$\begin{aligned}& f_{1}(x)=\log \biggl[\frac{2}{3}\cosh( \sqrt{2}x)+\frac{1}{3} \biggr]-\frac {4}{3}\log(\cosh x), \\& f_{2}(x)=6\sqrt{2}\sinh(\sqrt{2}x)\cosh x-8\cosh(\sqrt{2}x)\sinh x -4 \sinh x, \\& \xi_{n}=(3\sqrt{2}-4) (\sqrt{2}+1)^{2n-1}+(3\sqrt{2}+4) ( \sqrt{2}-1)^{2n-1}-4, \\& \eta_{n}=(\sqrt{2}+1)^{2n-1}. \end{aligned}$$

(3.55)

Then simple computations lead to

$$\begin{aligned}& f_{1}(0)=0, \end{aligned}$$

(3.56)

$$\begin{aligned}& f^{\prime}_{1}(x)=\frac{f_{2}(x)}{3\cosh x[2\cosh(\sqrt{2}x)+1]}, \end{aligned}$$

(3.57)

$$\begin{aligned}& f_{2}(x)=\sum_{n=1}^{\infty} \frac{\xi_{n}}{(2n-1)!}x^{2n-1}, \end{aligned}$$

(3.58)

$$\begin{aligned}& \xi_{1}=\xi_{2}=0, \end{aligned}$$

(3.59)

$$\begin{aligned}& \eta_{n}\xi_{n}=(3\sqrt{2}-4) (\eta_{n}- \eta_{1}) (\eta_{n}-\eta_{2}). \end{aligned}$$

(3.60)

From (3.58)-(3.60) and $\eta_{n}>\eta_{2}>\eta_{1}>0$ for $n\geq3$ we know that

$$ f_{2}(x)>0 $$

(3.61)

for all $x>0$.

Therefore, inequality (3.54) follows easily from (3.55)-(3.57) and (3.61).

(ii) The sufficiency follows easily from the monotonicity of the function $p\rightarrow [\cosh(pt)]^{1/(2p^{2})}$ and the last inequality in Corollary 3.1 together with the identity $(1+\cosh t)/2=\cosh^{2}(t/2)$.

Next, we prove the necessity. If there exists $p_{0}\in(1/2, \sqrt {6}/4)$ such that $I_{0}(t)< [\cosh(p_{0}t)]^{1/(2p_{0}^{2})}$ for all $t>0$, then we have

$$ \lim_{t\rightarrow\infty}\frac{I_{0}(t)-[\cosh (p_{0}t)]^{1/(2p_{0}^{2})}}{e^{t/(2p_{0})}}\leq0. $$

(3.62)

But the first inequality of (3.1) leads to

$$ \frac{I_{0}(t)-[\cosh(p_{0}t)]^{1/(2p_{0}^{2})}}{e^{t/(2p_{0})}}>\frac {1}{1+2t}\frac{e^{t}}{e^{t/(2p_{0})}} - \biggl( \frac{1+e^{-2p_{0}t}}{2} \biggr)^{1/(2p_{0}^{2})}\rightarrow \infty \quad (t\rightarrow \infty), $$

which contradicts (3.62).

(iii) Let $p=1, \sqrt{2}/2, \sqrt{6}/4, 1/2, 0^{+}$. Then parts (i) and (ii) together with the monotonicity of the function $p\rightarrow[\cosh(pt)]^{1/(2p^{2})}$ lead to (3.53). □

Theorem 3.7

Let $\theta\in[0, \pi/2]$. Then the inequality

$$ I_{0}(t)>\frac{\cosh(t\cos\theta)+\cosh(t\sin\theta)}{2} $$

(3.63)

or

$$ TQ(a,b)>\frac{A^{\cos\theta}_{\cos\theta}(a,b)G^{1-\cos\theta }(a,b)+A^{\sin\theta}_{\sin\theta}(a,b)G^{1-\sin\theta}(a,b)}{2} $$

holds for all $t>0$ or all $a, b>0$ with $a\neq b$ if and only if $\theta\in[\pi/8, 3\pi/8]$. In particular, the inequalities

$$\begin{aligned} I_{0}(t) >&\frac{1}{2} \biggl[\cosh \biggl( \frac{\sqrt{2-\sqrt{2}}}{2}t \biggr)+\cosh \biggl(\frac{\sqrt{2+\sqrt{2}}}{2}t \biggr) \biggr] \\ >&\frac{1}{2} \biggl[\cosh \biggl(\frac{\sqrt{3}}{2}t \biggr)+\cosh \biggl(\frac {1}{2}t \biggr) \biggr]>\cosh \biggl(\frac{\sqrt{2}}{2}t \biggr) \end{aligned}$$

(3.64)

or

$$\begin{aligned} TQ(a,b) >&\frac{A^{\sqrt{2-\sqrt{2}}/2}_{\sqrt{2-\sqrt {2}}/2}(a,b)G^{1-\sqrt{2-\sqrt{2}}/2}(a,b) +A^{\sqrt{2+\sqrt{2}}/2}_{\sqrt{2+\sqrt{2}}/2}(a,b)G^{1-\sqrt{2+\sqrt {2}}/2}(a,b)}{2} \\ >&\frac{A^{\sqrt{3}/2}_{\sqrt{3}/2}(a,b)G^{1-\sqrt {3}/2}+A^{1/2}_{1/2}(a,b)G^{1/2}(a,b)}{2}>A^{\sqrt{2}/2}_{\sqrt {2}/2}(a,b)G^{1-\sqrt{2}/2}(a,b) \end{aligned}$$

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

Proof

If inequality (3.63) holds for all $t>0$, then we have

$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-\frac{\cosh(t\cos\theta)+\cosh (t\sin\theta)}{2}}{t^{4}}=-\frac{1}{192}\cos(4 \theta)\geq0, $$

which implies that $\theta\in[\pi/8, 3\pi/8]$.

Next, we prove the sufficiency of inequality (3.63). Simple computations lead to

$$\begin{aligned}& \frac{\partial[\cosh(t\cos\theta)+\cosh(t\sin\theta)]}{\partial\theta }=\frac{t^{2}\sin(2\theta)}{2} \biggl[ \frac{\sinh(t\sin\theta)}{t\sin \theta}-\frac{\sinh(t\cos\theta)}{t\cos\theta} \biggr], \end{aligned}$$

(3.65)

$$\begin{aligned}& \biggl(\frac{\sinh x}{x} \biggr)^{\prime}=\frac{1}{x} \biggl( \cosh x-\frac {\sinh x}{x} \biggr)>0 \end{aligned}$$

(3.66)

for $x>0$.

Equation (3.65) and inequality (3.66) imply that the function $\theta \rightarrow[\cosh(t\cos\theta)+\cosh(t\sin\theta)]$ is decreasing on $[0, \pi/4]$ and increasing on $[\pi/4, \pi/2]$ for any fixed $t>0$. Hence, it suffices to prove that inequality (3.63) holds for all $t>0$ and $\theta=\theta_{0}=\pi/8$.

Let

$$\begin{aligned}& \rho_{n}=\frac{ (\frac{2-\sqrt{2}}{4} )^{n}+ (\frac{2+\sqrt {2}}{4} )^{n}}{(2n)!}, \qquad \sigma_{n}= \frac{2}{2^{2n}(n!)^{2}}, \\& R_{3}(t)=\frac{\cosh(t\cos\theta_{0})+\cosh(t\sin\theta_{0})}{2I_{0}(t)}. \end{aligned}$$

(3.67)

Then simple computations lead to

$$\begin{aligned}& R_{3}(t)=\frac{\sum_{n=0}^{\infty}\rho_{n}t^{2n}}{\sum_{n=0}^{\infty }\sigma_{n}t^{2n}}, \end{aligned}$$

(3.68)

$$\begin{aligned}& \frac{\rho_{0}}{\sigma_{0}}=\frac{\rho_{1}}{\sigma_{1}}=\frac{\rho _{2}}{\sigma_{2}}= \frac{\rho_{3}}{\sigma_{3}}=1, \end{aligned}$$

(3.69)

$$\begin{aligned}& \frac{\frac{\rho_{n+1}}{\sigma_{n+1}}}{\frac{\rho_{n}}{\sigma_{n}}}-1 =-\frac{\sqrt{2} [(n+\sqrt{2}-1)(\sqrt{2}-1)^{n-1}+ (n-\sqrt{2}-1)(\sqrt{2}+1)^{n-1} ]}{2(2n+1) [(\sqrt {2}-1)^{n}+(\sqrt{2}+1)^{n} ]}< 0 \end{aligned}$$

(3.70)

for $n\geq3$.

It follows from Lemma 2.6 and (3.68)-(3.70) that $R_{3}(t)$ is strictly decreasing on $(0, \infty)$. Therefore,

$$ I_{0}(t)>\frac{\cosh(t\cos\theta_{0})+\cosh(t\sin\theta_{0})}{2} $$

(3.71)

follows from (3.67) and the monotonicity of $R_{3}(t)$ together with $R_{3}(0^{+})=\rho_{0}/\sigma_{0}=1$.

Let $\theta=\pi/8, \pi/6, \pi/4$. Then inequality (3.64) follows easily from (3.63) and the monotonicity of the function $\theta\rightarrow [\cosh(t\cos\theta)+\cosh(t\sin\theta)]$. □

Theorem 3.8

The inequality

$$ I_{0}(t)>\frac{\sinh t}{t}+\frac{3(4-\pi)(t\sinh t-2\cosh t+2)}{\pi t^{2}} $$

(3.72)

holds for all $t>0$.

Proof

It is easy to verify that

$$ \frac{d^{2}}{dx^{2}} \biggl(\frac{1}{\sqrt{1-x^{2}}} \biggr)=\frac {1+2x^{2}}{ (1-x^{2} )^{5/2}}>0,\qquad \frac{\partial^{2}\cosh(tx)}{\partial x^{2}}=t^{2}\cosh(tx)>0 $$

for all $t>0$ and $x\in(0, 1)$, which implies that the two functions $1/\sqrt{1-x^{2}}$ and $\cosh(tx)$ are convex with respect to x on the interval $(0, 1)$. Then from Lemma 2.10 and (3.3) we have

$$\begin{aligned}& \frac{\pi}{2}I_{0}(t)-\frac{\pi}{2} \frac{\sinh t}{t} \\& \quad = \int_{0}^{1}\frac{\cosh(tx)}{\sqrt{1-x^{2}}}\,dx- \int_{0}^{1}\frac {dx}{\sqrt{1-x^{2}}} \int_{0}^{1}\cosh(tx)\,dx \\& \quad >12 \int_{0}^{1}\frac{x-\frac{1}{2}}{\sqrt{1-x^{2}}}\,dx \int_{0}^{1} \biggl(x-\frac{1}{2} \biggr) \cosh(tx)\,dx \\& \quad =\frac{3(4-\pi)(t\sinh t-2\cosh t+2)}{2t^{2}}. \end{aligned}$$

(3.73)

Therefore, inequality (3.72) follows from (3.73). □

Remark 3.3

The inequality $I_{0}(t)>\sinh(t)/t$ in (3.5) is equivalent to the first inequality $TQ(a,b)>L(a,b)$ in (1.6). Therefore, Theorem 3.8 is an improvement of the first inequality in (1.6).

Let $p\in\mathbb{R}$ and $M(a,b)$ be a bivariate mean of two positive a and b. Then the pth power-type mean $M_{p}(a,b)$ is defined by

$$ M_{p}(a,b)=M^{1/p} \bigl(a^{p}, b^{p} \bigr) \quad (p\neq0),\qquad M_{0}(a,b)=\sqrt{ab}. $$

We clearly see that

$$ M_{\lambda p}(a,b)=M_{p}^{1/\lambda} \bigl(a^{\lambda}, b^{\lambda} \bigr) $$

for all $\lambda, p\in\mathbb{R}$ and $a, b>0$ if M is a bivariate mean.

Theorem 3.9

The inequality

$$ TQ(a,b)< I_{p}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\geq3/4$.

Proof

The second inequality (1.6) can be rewritten as

$$ TQ(a,b)< A_{1/2}(a,b). $$

(3.74)

In [30, 31], the authors proved that the inequality

$$ I(a,b)>A_{2/3}(a,b) $$

(3.75)

holds for all distinct positive real numbers a and b with the best possible constant $2/3$.

Inequalities (3.74) and (3.75) lead to

$$ TQ(a,b)< A_{1/2}(a,b)=A_{2/3}^{4/3} \bigl(a^{3/4}, b^{3/4} \bigr) < I^{4/3} \bigl(a^{3/4}, b^{3/4} \bigr)=I_{3/4}(a,b) $$

(3.76)

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

If $p\geq3/4$, then $TQ(a,b)< I_{3/4}(a,b)\leq I_{p}(a,b)$ follows from (3.76) and the function $p\rightarrow I_{p}(a,b)$ is strictly increasing [32].

If $TQ(a,b)< I_{p}(a,b)$ for all $a, b>0$ with $a\neq b$. Then

$$ I_{0}(t)-e^{t/\tanh(pt)-1/p}< 0 $$

(3.77)

for all $t>0$.

Inequality (3.77) leads to

$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-e^{t/\tanh (pt)-1/p}}{t^{2}}=\frac{1}{3} \biggl( \frac{3}{4}-p \biggr)\leq0, $$

which implies that $p\geq3/4$. □

Remark 3.4

For all $a, b>0$ with $a\neq b$, the Toader mean $T(a,b)$ satisfies the double inequality [5, 7]

$$ A_{3/2}(a,b)< T(a,b)< A_{\log2/(\log\pi-\log2)}(a,b) $$

(3.78)

with the best possible constants $3/2$ and $\log2/(\log\pi-\log2)$, and the one-sided inequality [33]

$$ T(a,b)< I_{9/4}(a,b). $$

(3.79)

It follows from (3.78) and (3.79) that

$$\begin{aligned} A_{1/2}^{1/3}(a,b) =&A_{3/2} \bigl(a^{1/3}, b^{1/3} \bigr)< T \bigl(a^{1/3}, b^{1/3} \bigr) \\ =&T^{1/3}_{1/3}(a,b)< I_{9/4} \bigl(a^{1/3}, b^{1/3} \bigr)=I_{3/4}^{1/3}(a,b), \end{aligned}$$

which can be rewritten as

$$ A_{1/2}(a,b)< T_{1/3}(a,b)< I_{3/4}(a,b). $$

(3.80)

Inequalities (3.74) and (3.80) lead to the inequalities

$$ TQ(a,b)< A_{1/2}(a,b)< T_{1/3}(a,b)< I_{3/4}(a,b) $$

(3.81)

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

Remark 3.5

For all $a, b>0$ with $a\neq b$, Theorem 3.4 shows that

$$ L^{3/4}(a,b)A^{1/4}(a,b)< TQ(a,b)< \frac{3L(a,b)+A(a,b)}{4}. $$

(3.82)

It follows from $L(a,b)< A(a,b)/3+2G(a,b)/3$, given by Carlson in [34], and $A(a,b)>L(a,b)$ that

$$ L(a,b)< L^{3/4}(a,b)A^{1/4}(a,b), \qquad \frac{A(a,b)+G(a,b)}{2}> \frac {3L(a,b)+A(a,b)}{4}. $$

Therefore, inequality (3.82) is an improvement of the first and second inequalities of (1.6).

Remark 3.6

In [2, 20, 35], the authors proved that the inequalities

$$ L(a,b)< \operatorname{AGM}(a,b)< L^{3/4}(a,b)A^{1/4}(a,b)< L_{3/2}(a,b) $$

(3.83)

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

Inequalities (3.81)-(3.83) lead to the chain of inequalities

$$\begin{aligned} L(a,b) < &\operatorname{AGM}(a,b)< L^{3/4}(a,b)A^{1/4}(a,b) \\ < &TQ(a,b)< A_{1/2}(a,b)< T_{1/3}(a,b)< I_{3/4}(a,b) \end{aligned}$$

(3.84)

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

Motivated by the first inequality in (3.82) and the third inequality in (3.83), we propose Conjecture 3.1.

Conjecture 3.1

The inequality

$$ TQ(a,b)>L_{3/2}(a,b) $$

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

For all $a, b>0$ with $a\neq b$, inspired by the double inequality

$$ \sqrt{A(a,b)G(a,b)}< TQ(a,b)< \frac{A(a,b)+G(a,b)}{2} $$

given in Corollary 3.1 and the inequalities

$$ \sqrt{A(a,b)G(a,b)}< \sqrt{I(a,b)L(a,b)}< \frac{I(a,b)+L(a,b)}{2}< \frac {A(a,b)+G(a,b)}{2} $$

proved by Alzer in [36] we propose Conjecture 3.2.

Conjecture 3.2

The inequality

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

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

Remark 3.7

Let $W_{n}$ be the Wallis ratio defined by (2.1), and $c_{n}$, $d_{n}$, and $s_{n}$ be defined by (3.20). Then it follows from Lemma 2.5 and the proof of Theorem 3.3 that the sequence $\{s_{n}\}_{n=1}^{\infty}$ is strictly decreasing and $\lim_{n\rightarrow\infty}s_{n}=2/\pi$, and the sequence $\{ c_{n}/d_{n}\}_{n=1}^{\infty}$ is strictly increasing for $n=1, 2$ and strictly decreasing for $n\geq2$. Hence, we have

$$ \frac{2}{\pi}< s_{n}=(2n+1)W_{n}^{2} \leq s_{1}=\frac{3}{4} $$

(3.85)

and

$$ \frac{2}{\pi}=\min \biggl\{ \frac{c_{1}}{d_{1}}, \lim _{n\rightarrow\infty }\frac{c_{n}}{d_{n}} \biggr\} < \frac{c_{n}}{d_{n}}= \frac{2^{2n}s_{n}-1}{2^{2n}-1}\leq\frac {c_{2}}{d_{2}}=\frac{41}{60} $$

(3.86)

for all $n\in\mathbb{N}$.

Inequalities (3.85) and (3.86) lead to the Wallis ratio inequalities

$$ \frac{1}{\sqrt{\pi (n+\frac{1}{2} )}}< W_{n}\leq\frac{\sqrt {6}}{4\sqrt{n+\frac{1}{2}}} $$

and

$$ \sqrt{\frac{2^{-2n}(\pi-2)+2}{\pi(2n+1)}}< W_{n}\leq\sqrt{\frac {41+19\times2^{-2n}}{60(2n+1)}} $$

for all $n\in\mathbb{N}$.