# 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  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)

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 .

For $$q\neq0$$, the mean $$M_{x^{q},0}(a, b)$$ seems to be mysterious, Toader  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 $$I_{\nu}(z)$$  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.  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)

\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 )

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 )

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 )

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 )

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 )

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 )

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 )

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)

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

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

\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:

1. (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 $$q\geq1$$.

2. (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)

\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}

(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:

1. (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 $$p\geq \sqrt{6}/4$$.

2. (ii)

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

3. (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  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)

\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

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

(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)

\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 .

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 $$t>0$$.

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

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 

$$T(a,b)< I_{9/4}(a,b).$$
(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 , 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  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 $$n\in\mathbb{N}$$.

## References

1. Toader, G: Some mean values related to the arithmetic-geometric mean. J. Math. Anal. Appl. 218(2), 358-368 (1998)

2. Borwein, JM, Borwein, PB: Inequalities for compound mean iterations with logarithmic asymptotes. J. Math. Anal. Appl. 177(2), 572-582 (1993)

3. Vamanamurthy, MK, Vuorinen, M: Inequalities for means. J. Math. Anal. Appl. 183(1), 155-166 (1994)

4. Alzer, H: Sharp inequalities for the complete elliptic integral of the first kind. Math. Proc. Camb. Philos. Soc. 124(2), 309-314 (1998)

5. Barnard, RW, Pearce, K, Richards, KC: An inequality involving the generalized hypergeometric function and the arc length of an ellipse. SIAM J. Math. Anal. 31(3), 693-699 (2000)

6. Bracken, P: An arithmetic-geometric mean inequality. Expo. Math. 19(3), 273-279 (2001)

7. Alzer, H, Qiu, S-L: Monotonicity theorems and inequalities for the complete elliptic integrals. J. Comput. Appl. Math. 172(2), 289-312 (2004)

8. András, S, Baricz, Á: Bounds for complete elliptic integrals of the first kind. Expo. Math. 28(4), 357-364 (2010)

9. Chu, Y-M, Wang, M-K, Qiu, S-L, Qiu, Y-F: Sharp generalized Seiffert mean bounds for Toader mean. Abstr. Appl. Anal. 2011, Article ID 605259 (2011)

10. Chu, Y-M, Wang, M-K: Inequalities between arithmetic-geometric, Gini, and Toader means. Abstr. Appl. Anal. 2012, Article ID 830585 (2012)

11. Chu, Y-M, Wang, M-K, Qiu, S-L: Optimal combinations bounds of root-square and arithmetic means for Toader mean. Proc. Indian Acad. Sci. Math. Sci. 122(1), 41-51 (2012)

12. Chu, Y-M, Wang, M-K: Optimal Lehmer mean bounds for the Toader mean. Results Math. 61(3-4), 223-229 (2012)

13. Chu, Y-M, Wang, M-K, Ma, X-Y: Sharp bounds for Toader mean in terms of contraharmonic mean with applications. J. Math. Inequal. 7(2), 161-166 (2013)

14. Chu, Y-M, Wang, M-K, Qiu, Y-F, Ma, X-Y: Sharp two parameter bounds for the logarithmic mean and the arithmetic-geometric mean of Gauss. J. Math. Inequal. 7(3), 349-355 (2013)

15. Song, Y-Q, Jiang, W-D, Chu, Y-M, Yan, D-D: Optimal bounds for Toader mean in terms of arithmetic and contraharmonic means. J. Math. Inequal. 7(4), 751-757 (2013)

16. Li, W-H, Zheng, M-M: Some inequalities for bounding Toader mean. J. Funct. Spaces Appl. 2013, Article ID 394194 (2013)

17. Chu, Y-M, Qiu, S-L, Wang, M-K: Sharp inequalities involving the power mean and complete elliptic integral of the first kind. Rocky Mt. J. Math. 43(5), 1489-1496 (2013)

18. Hua, Y, Qi, F: The best possible bounds for Toader mean in terms of the centroidal and arithmetic means. Filomat 28(4), 775-780 (2014)

19. Hua, Y, Qi, F: A double inequality for bounding Toader mean by the centroidal mean. Proc. Indian Acad. Sci. Math. Sci. 124(4), 527-531 (2014)

20. Yang, Z-H, Song, Y-Q, Chu, Y-M: Sharp bounds for the arithmetic-geometric mean. J. Inequal. Appl. 2014, Article ID 192 (2014)

21. Abramowitz, M, Stegun, IA: Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. U.S. Government Printing Office, Washington (1964)

22. Qi, F, Shi, X-T, Liu, F-F, Yang, Z-H: A double inequality for an integral mean in terms of the exponential and logarithmic means. ResearchGate research. http://www.researchgate.net/publication/278968439. doi:10.13140/RG.2.1.2353.6800

23. Pólya, G, Szegő, G: Problems and Theorems in Analysis I. Springer, Berlin (1998)

24. Qi, F: Bounds for the ratio of two gamma functions. J. Inequal. Appl. 2010, Article ID 493058 (2010)

25. Biernacki, M, Krzyż, J: On the monotonity of certain functionals in the theory of analytic functions. Ann. Univ. Mariae Curie-Skłodowska, Sect. A 9, 135-147 (1955)

26. Yang, Z-H, Chu, Y-M, Wang, M-K: Monotonicity criterion for the quotient of power series with applications. J. Math. Anal. Appl. 428(1), 587-604 (2015)

27. Mitrinović, DS, Pečarić, JE, Fink, AM: Classical and New Inequalities in Analysis. Kluwer Academic, Dordrecht (1993)

28. Lupaş, A: An integral inequality for convex functions. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 381-409, 17-19 (1972)

29. Yang, Z-H: New sharp bounds for logarithmic mean and identric mean. J. Inequal. Appl. 2013, Article ID 116 (2013)

30. Pittenger, AO: Inequalities between arithmetic and logarithmic means. Publ. Elektroteh. Fak. Univ. Beogr., Ser. Mat. Fiz. 678-715, 15-18 (1980)

31. Stolarsky, KB: The power and generalized logarithmic means. Am. Math. Mon. 87(7), 545-548 (1980)

32. Yang, Z-H, Chu, Y-M: An optimal inequalities chain for bivariate means. J. Math. Inequal. 9(2), 331-343 (2015)

33. Yang, Z-H: Very accurate approximations for the elliptic integrals of the second kind in terms of Stolarsky mean (2015). arXiv:1508.05513 [math.CA]

34. Carlson, BC: The logarithmic mean. Am. Math. Mon. 79, 615-618 (1972)

35. Carlson, BC, Vuorinen, M: Inequality of the AGM and the logarithmic mean. SIAM Rev. 33(4), 653-654 (1991)

36. Alzer, H: Ungleichungen für Mittelwerte. Arch. Math. 47(5), 422-426 (1986)

## Acknowledgements

The authors would like to thank the anonymous referee for his/her valuable comments and suggestions. The research was supported by the Natural Science Foundation of China under Grants 11371125, 11401191, and 61374086.

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