1 Introduction
Let \(a, b>0\), \(p: (0, \infty)\rightarrow\mathbb{R^{+}}\) be a strictly monotone real function, \(\theta\in(0, 2\pi)\) and Then the mean \(M_{p,n}(a,b)\) was first introduced by Toader in [1] as follows: where \(p^{-1}\) is the inverse function of p.
$$ r_{n}(\theta)=\left \{ \textstyle\begin{array}{l@{\quad}l} (a^{n}\cos^{2}\theta+b^{n}\sin^{2}\theta )^{1/n}, & n\neq 0, \\ a^{\cos^{2}\theta}b^{\sin^{2}\theta}, & n=0. \end{array}\displaystyle \right . $$
(1.1)
$$ M_{p,n}(a,b)=p^{-1} \biggl(\frac{1}{2\pi} \int_{0}^{2\pi }p\bigl( r_{n}(\theta)\bigr) \,d\theta \biggr)=p^{-1} \biggl(\frac{2}{\pi} \int_{0}^{\pi /2}p\bigl(r_{n}(\theta)\bigr)\,d \theta \biggr), $$
(1.2)
From (1.1) and (1.2) we clearly see that 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 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 In particular, is the Toader-Qi mean.
$$ M_{1/x,2}(a, b)=\frac{\pi}{2\int_{0}^{\pi/2}\frac{d\theta}{\sqrt {a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta}}}=\operatorname{AGM}(a,b) $$
$$ M_{x,2}(a, b)=\frac{2}{\pi} \int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2} \theta +b^{2}\sin^{2}\theta}\,d\theta=T(a,b) $$
$$ M_{x^{q},0}(a, b)= \biggl(\frac{2}{\pi} \int_{0}^{\pi/2}a^{q\cos^{2}\theta }b^{q\sin^{2}\theta}\,d \theta \biggr)^{1/q} \quad (q\neq0). $$
$$ M_{x,0}(a, b)=\frac{2}{\pi} \int_{0}^{\pi/2}a^{\cos^{2}\theta}b^{\sin ^{2}\theta}\,d \theta=TQ(a,b) $$
(1.3)
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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 and inequalities 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.
$$ M_{x^{q},0}(a, b)= \biggl(\frac{2}{\pi} \int_{0}^{\pi/2}a^{q\cos^{2}\theta }b^{q\sin^{2}\theta}\,d \theta \biggr)^{1/q}=\sqrt{ab}I_{0}^{1/q} \biggl( \frac {q}{2}\log\frac{a}{b} \biggr) $$
(1.5)
$$ L(a,b)< TQ(a,b)< \frac{A(a,b)+G(a,b)}{2}< \frac{2A(a,b)+G(a,b)}{3}< I(a,b) $$
(1.6)
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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 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 and (1.3)-(1.5) lead to
$$ A_{p}(a,b)= \biggl(\frac{a^{p}+b^{p}}{2} \biggr)^{1/p} \quad (p \neq0),\qquad A_{0}(a,b)=\sqrt{ab}=G(a,b). $$
$$ \begin{aligned} &L(a,b)=\sqrt{ab}\frac{\sinh t}{t}, \qquad I(a,b)= \sqrt{ab}e^{t/\tanh t-1}, \\ &A_{p}(a,b)=\sqrt{ab} \cosh^{1/p}(pt) \quad (p\neq0) \end{aligned} $$
(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 [23])
Let
\(\binom{n}{k}\)
be the number of combinations of
n
objects taken
k
at a time, that is,
Then
$$ \binom{n}{k}=\frac{n!}{k!(n-k)!}. $$
$$ \sum_{k=0}^{\infty}\binom{n}{k}^{2}= \binom{2n}{n}. $$
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
if the power series
\(\sum_{n=0}^{\infty}b_{n}t^{n}\)
is convergent for all
\(t\in\mathbb{R}\).
$$ \lim_{t\rightarrow\infty}\frac{\sum_{n=0}^{\infty}a_{n}t^{n}}{\sum_{n=0}^{\infty}b_{n}t^{n}}=s $$
Lemma 2.3
The Wallis ratio
is strictly decreasing with respect to all integers
\(n\geq0\)
and strictly log-convex with respect to all real numbers
\(n\geq0\).
$$ W_{n}=\frac{\Gamma (n+\frac{1}{2} )}{\Gamma (\frac {1}{2} )\Gamma(n+1)} $$
(2.1)
Proof
It follows from (2.1) that for all integers \(n\geq0\).
$$ \frac{W_{n+1}}{W_{n}}=1-\frac{1}{2(n+1)}< 1 $$
(2.2)
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 for all \(x\geq0\).
$$ \bigl[\log f(x)\bigr]^{\prime\prime}=\psi^{\prime} \biggl(x+ \frac{1}{2} \biggr)-\psi ^{\prime}(x+1)>0 $$
(2.3)
Lemma 2.4
(See [24])
The double inequality
holds for all
\(x>0\)
and
\(a\in(0, 1)\).
$$ \frac{1}{(x+a)^{1-a}}< \frac{\Gamma(x+a)}{\Gamma(x+1)}< \frac{1}{x^{1-a}} $$
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)
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
holds for all
\(t\in\mathbb{R}\).
$$ I^{2}_{0}(t)=\sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^{4}}t^{2n} $$
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
if both
f
and
g
are increasing or decreasing on
\((a, b)\).
$$ \int_{a}^{b}f(x)g(x)\,dx\geq\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \int_{a}^{b}g(x)\,dx $$
Lemma 2.10
(See [28])
Let
\(-\infty< a< b<\infty\)
and
\(f, g: (a, b)\rightarrow\mathbb {R}\). Then
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
\((a,b)\).
$$\begin{aligned}& \int_{a}^{b}f(x)g(x)\,dx-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \int_{a}^{b}g(x)\,dx \\& \quad \geq\frac{12}{(b-a)^{3}} \int_{a}^{b} \biggl(x-\frac{a+b}{2} \biggr)f(x)\,dx \int_{a}^{b} \biggl(x-\frac{a+b}{2} \biggr)g(x)\,dx \end{aligned}$$
(2.7)
3 Main results
Theorem 3.1
The double inequalities
and
hold for all
\(t>0\)
and
\(b>a>0\).
$$ \frac{e^{t}}{1+2t}< I_{0}(t)< \frac{e^{t}}{\sqrt{1+2t}} $$
(3.1)
$$ \frac{b}{1+\log(b/a)}< TQ(a,b)< \frac{b}{\sqrt{1+\log(b/a)}} $$
(3.2)
Proof
From (1.8) we have and
$$ I_{0}(t)=\frac{2}{\pi} \int_{0}^{\pi/2}\cosh(t\sin\theta)\,d\theta= \frac {2}{\pi} \int_{0}^{1}\frac{\cosh(tx)}{\sqrt{1-x^{2}}}\,dx $$
(3.3)
$$\begin{aligned} e^{-t}I_{0}(t) =&\frac{2}{\pi} \int_{0}^{\pi/2}e^{t[\cos(2\theta )-1]}\,d\theta= \frac{2}{\pi} \int_{0}^{\pi/2}\frac{d\theta}{e^{2t\sin ^{2}\theta}} \\ < &\frac{2}{\pi} \int_{0}^{\pi/2}\frac{d\theta}{1+2t\sin^{2}\theta}=\frac {1}{\sqrt{1+2t}}. \end{aligned}$$
(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)
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
and
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
\(\beta_{2}\geq1\).
$$ \alpha_{1}\sqrt{\frac{\sinh(2t)}{t}}< I_{0}(t)< \beta_{1}\sqrt{\frac{\sinh (2t)}{t}} $$
(3.7)
$$ \alpha_{2}\sqrt{L(a,b)A(a,b)}< TQ(a,b)< \beta_{2}\sqrt{L(a,b)A(a,b)} $$
(3.8)
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
Theorem 3.3
Let
\(\lambda_{1}, \lambda_{2}>0\), \(t_{0}=2.7113\ldots\)
be the unique solution of the equation
on
\((0, \infty)\)
and
Then the double inequality
or
holds for all
\(t>0\)
or
\(a,b>0\)
with
\(a\neq b\)
if and only if
\(\lambda _{1}\leq2/\pi\), \(\lambda_{2}>\lambda_{0}\).
$$ \frac{d}{dt} \biggl[\frac{tI^{2}_{0}(t)-\sinh t}{(\cosh t-1)\sinh t} \biggr]=0 $$
(3.16)
$$ \lambda_{0}=\frac{t_{0}I^{2}_{0}(t_{0})-\sinh t_{0}}{(\cosh t_{0}-1)\sinh t_{0}}=0.6766\ldots. $$
(3.17)
$$ \sqrt{(\lambda_{1}\cosh t+1-\lambda_{1}) \frac{\sinh t}{t}}< I_{0}(t)< \sqrt{(\lambda_{2}\cosh t+1- \lambda_{2})\frac{\sinh t}{t}} $$
(3.18)
$$ \sqrt{\bigl[\lambda_{1}A(a,b)+(1-\lambda_{1})G(a,b) \bigr]L(a,b)}< TQ(a,b)< \sqrt {\bigl[\lambda_{2}A(a,b)+(1- \lambda_{2})G(a,b)\bigr]L(a,b)} $$
Proof
Let
and
$$\begin{aligned}& R_{1}(t)=\frac{I^{2}_{0}(t)-\frac{\sinh t}{t}}{\frac{(\cosh t-1)\sinh t}{t}}, \end{aligned}$$
(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)
$$ s^{\prime}_{n}= \bigl(2^{2n}+3n^{2}+6n+2 \bigr)s_{n}- \bigl(3n^{2}+6n+3 \bigr). $$
(3.21)
Then it follows from Lemma 2.2, Lemma 2.5, Lemma 2.8, and (3.19)-(3.21) that
and we have the inequality for all \(n\geq3\).
$$\begin{aligned}& R_{1}(t)=\frac{\sum_{n=1}^{\infty}c_{n}t^{2n}}{\sum_{n=1}^{\infty}d_{n}t^{2n}}, \end{aligned}$$
(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)
$$\begin{aligned} s^{\prime}_{n} >&\frac{2}{\pi} \bigl(2^{2n}+3n^{2}+6n+2 \bigr)- \bigl(3n^{2}+6n+3 \bigr) \\ >&\frac{3}{5} \bigl(2^{2n}+3n^{2}+6n+2 \bigr)- \bigl(3n^{2}+6n+3 \bigr)=\frac{3}{5} \bigl[2^{2n}- \bigl(2n^{2}+4n+3 \bigr) \bigr]>0 \end{aligned}$$
(3.26)
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 for all \(t>0\), and \(t_{0}\) is the unique solution of equation (3.16) on \((0, \infty)\).
$$ \min\Bigl\{ R_{1}\bigl(0^{+}\bigr), \lim _{t\rightarrow\infty}R_{1}(t)\Bigr\} < R_{1}(t)\leq R_{1}(t_{0}) $$
(3.27)
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
or
holds for all
\(t>0\)
or
\(a, b>0\)
with
\(a\neq b\)
if and only if
\(p\geq 3/4\)
and
\(q\leq3/4\).
$$ \cosh^{1-p} t \biggl(\frac{\sinh t}{t} \biggr)^{p}< I_{0}(t)< q\frac{\sinh t}{t}+(1-q)\cosh t $$
(3.30)
$$ L^{p}(a,b)A^{1-p}(a,b)< TQ(a,b)< qL(a,b)+(1-q)A(a,b) $$
Proof
If the first inequality of (3.30) holds for all \(t>0\), then which implies that \(p\geq3/4\).
$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-\cosh^{1-p} t (\frac{\sinh t}{t} )^{p}}{t^{2}}=\frac{1}{3} \biggl(p- \frac{3}{4} \biggr)\geq0, $$
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 for all \(n\geq4\).
$$\begin{aligned} u_{n} =&\sum_{k=0}^{n} \biggl(\frac{1}{(k!)^{2}((n-k)!)^{2}}\frac {(2k)!}{2^{2k}(k!)^{2}}\frac{(2(n-k))!}{ 2^{2(n-k)}((n-k)!)^{2}} \biggr)- \frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\sum_{k=0}^{n} \biggl( \frac{1}{(k!)^{2}((n-k)!)^{2}}W_{k}W_{n-k} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ >&\sum_{k=0}^{n} \biggl( \frac{1}{(k!)^{2}((n-k)!)^{2}}W^{2}_{n/2} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\sum_{k=0}^{n} \biggl[ \frac{1}{(k!)^{2}((n-k)!)^{2}} \biggl(\frac{\Gamma (n/2+1/2)}{\Gamma(1/2)\Gamma(n/2+1)} \biggr)^{2} \biggr]- \frac {2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ >&\sum_{k=0}^{n} \biggl( \frac{1}{\pi(n/2+1/2)(k!)^{2}((n-k)!)^{2}} \biggr)-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\frac{2}{\pi(n+1)(n!)^{2}}\sum_{k=0}^{n} \frac {(n!)^{2}}{(k!)^{2}((n-k)!)^{2}}-\frac{2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\frac{2}{\pi(n+1)(n!)^{2}}\frac{(2n)!}{(n!)^{2}}-\frac {2^{4n+3}-2^{2n+1}}{(2n+3)!} \\ =&\frac{2^{2n+1} (2^{2n+2}-1 )}{\pi(2n+3)!} \biggl[\frac {2^{2n+2}}{2^{2n+2}-1} \biggl(n+\frac{3}{2} \biggr)s_{n}-\pi \biggr] \\ >&\frac{2^{2n+1} (2^{2n+2}-1 )}{\pi(2n+3)!} \biggl[ \biggl(4+\frac {3}{2} \biggr) \frac{2}{\pi}-\pi \biggr]>0 \end{aligned}$$
(3.35)
If the second inequality of (3.30) holds for all \(t>0\), then we have which implies that \(q\leq3/4\).
$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-q\frac{\sinh t}{t}-(1-q)\cosh t}{t^{2}}=\frac{1}{3} \biggl(q- \frac{3}{4} \biggr)\leq0, $$
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 and \(W_{n}\) be defined by (2.1).
$$ \alpha_{n}=\frac{2^{n}n!-(2n-1)!!}{2^{n}n!(2n)!}, \qquad \beta_{n}= \frac {2n}{(2n+1)!}, \qquad \gamma_{n}=\frac{(n+2)(2n+1)}{2(n+1)}W_{n}, $$
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
and
$$ \frac{d [\frac{p^{2}(I_{0}(t)-1)}{\cosh(pt)-1} ]}{dt}=0 $$
(3.40)
$$ \mu_{0}=\frac{p^{2}(I_{0}(t_{0})-1)}{\cosh(pt_{0})-1}. $$
(3.41)
Then the following statements are true:
(i)
The double inequality
or
holds for all
\(t>0\)
or
\(a, b>0\)
with
\(a\neq b\)
if and only if
\(p\leq \sqrt{3}/2\)
and
\(q\geq1\).
$$ 1-\frac{1}{2p^{2}}+\frac{1}{2p^{2}}\cosh(pt)< I_{0}(t)< 1- \frac {1}{2q^{2}}+\frac{1}{2q^{2}}\cosh(qt) $$
(3.42)
$$\begin{aligned}& \biggl(1-\frac{1}{2p^{2}} \biggr)G(a,b)+\frac {1}{2p^{2}}A^{p}_{p}(a,b)G^{1-p}(a,b) \\& \quad < TQ(a,b) < \biggl(1-\frac{1}{2q^{2}} \biggr)G(a,b)+ \frac {1}{2q^{2}}A^{q}_{q}(a,b)G^{1-q}(a,b) \end{aligned}$$
(ii)
The inequality
or
holds for all
\(t>0\)
or
\(a, b>0\)
with
\(a\neq b\)
if
\(p\in(\sqrt{3}/2, 1)\).
$$ I_{0}(t)\geq1-\frac{\mu_{0}}{p^{2}}+\frac{\mu_{0}}{p^{2}} \cosh(pt) $$
(3.43)
$$ TQ(a,b)\geq \biggl(1-\frac{\mu_{0}}{p^{2}} \biggr)G(a,b)+\frac{\mu _{0}}{p^{2}}A^{p}_{p}(a,b)G^{1-p}(a,b) $$
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 for all \(t>0\) if \(p\geq1\) and for all \(t>0\) if \(p\leq\sqrt{3}/2\).
$$ R_{2}(t)< \lim_{t\rightarrow0^{+}}R_{2}(t)= \frac{u_{1}}{v_{1}}=\frac{1}{2} $$
(3.47)
$$ R_{2}(t)>\lim_{t\rightarrow0^{+}}R_{2}(t)= \frac{u_{1}}{v_{1}}=\frac{1}{2} $$
(3.48)
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 which implies that \(p\leq\sqrt{3}/2\).
$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)- (1-\frac{1}{2p^{2}}+\frac {1}{2p^{2}}\cosh(pt) )}{ t^{4}}=\frac{1}{48} \biggl( \frac{3}{4}-p^{2} \biggr)\geq0, $$
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 which contradicts inequality (3.49).
$$\begin{aligned}& \frac{I_{0}(t)- (1-\frac{1}{2q_{0}^{2}}+\frac{1}{2q_{0}^{2}}\cosh (q_{0}t) )}{e^{q_{0}t}} \\& \quad >\frac{e^{(1-q_{0})t}}{1+2t}- \biggl(1-\frac{1}{2q_{0}^{2}} \biggr)e^{-q_{0}t}- \frac{1+e^{-2q_{0}t}}{4q_{0}^{2}}\rightarrow\infty \quad (t\rightarrow\infty), \end{aligned}$$
(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)
It is not difficult to verify that the function is strictly increasing with respect to p on the interval \((0, \infty )\) and for \(t>0\).
$$ 1-\frac{1}{2p^{2}}+\frac{1}{2p^{2}}\cosh(pt) $$
$$ 2\cosh \biggl(\frac{t}{2} \biggr)-1>\cosh^{1/2} t $$
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
or
hold for all
\(t>0\)
or all
\(a, b>0\)
with
\(a\neq b\).
$$\begin{aligned} \cosh^{1/2}(t) < &2\cosh \biggl(\frac{t}{2} \biggr)-1< \frac{9}{8}\cosh \biggl(\frac{2t}{3} \biggr)-\frac{1}{8}< \cosh \biggl(\frac{\sqrt{2}t}{2} \biggr) \\ < &\frac{8}{9}\cosh \biggl(\frac{3t}{4} \biggr)+ \frac{1}{9}< \frac{2}{3}\cosh \biggl(\frac{\sqrt{3}t}{2} \biggr)+ \frac{1}{3}< I_{0}(t)< \frac{1+\cosh t}{2} \end{aligned}$$
$$\begin{aligned} G^{1/2}(a,b)A^{1/2}(a,b) < &2A^{1/2}_{1/2}(a,b)G^{1/2}(a,b)-G(a,b) \\ < & \frac {9}{8}A^{2/3}_{2/3}(a,b)G^{1/3}(a,b)- \frac{1}{8}G(a,b) \\ < &A^{\sqrt{2}/2}_{\sqrt{2}/2}(a,b)G^{1-\sqrt{2}/2}(a,b) \\ < & \frac {8}{9}A^{3/4}_{3/4}(a,b)G^{1/4}(a,b)+ \frac{1}{9}G(a,b) \\ < &\frac{2}{3}A^{\sqrt{3}/2}_{\sqrt{3}/2}(a,b)G^{1-\sqrt{3}/2}(a, b)+ \frac {1}{3}G(a,b) \\ < &TQ(a,b)< \frac{A(a,b)+G(a,b)}{2} \end{aligned}$$
Theorem 3.6
Let
\(p>0\). Then the following statements are true:
(i)
The inequality
or
holds for all
\(t>0\)
or
\(a, b>0\)
with
\(a\neq b\)
if and only if
\(p\geq \sqrt{6}/4\).
$$ I_{0}(t)>\bigl[\cosh(pt)\bigr]^{\frac{1}{2p^{2}}} $$
(3.51)
$$ TQ(a,b)>G^{1-\frac{1}{2p}}(a,b)A_{p}^{\frac{1}{2p}}(a,b) $$
(3.52)
(iii)
The inequalities
or
hold for all
\(t>0\)
or all
\(a, b>0\)
with
\(a\neq b\).
$$ \cosh^{1/2} t< \cosh \biggl(\frac{\sqrt{2}t}{2} \biggr)< \biggl[\cosh \biggl(\frac{\sqrt{6}t}{4} \biggr) \biggr]^{4/3}< I_{0}(t)< \cosh^{2} \biggl(\frac {t}{2} \biggr)< e^{t^{2}/4} $$
(3.53)
$$\begin{aligned} G^{1/2}(a,b)A^{1/2}(a,b) < &A^{\sqrt{2}/2}_{\sqrt{2}/2}(a,b)G^{1-\sqrt {2}/2}(a,b) \\ < &A^{\sqrt{6}/3}_{\sqrt{6}/4}(a,b)G^{1-\sqrt{6}/3}(a,b) \\ < &TQ(a,b)< \frac{A(a,b)+G(a,b)}{2} \\ < &G(a,b)e^{ (A^{2}(a,b)-G^{2}(a,b) )/ (4L^{2}(a,b) )} \end{aligned}$$
Proof
(i) If inequality (3.51) holds for all \(t>0\), then we have which implies that \(p\geq\sqrt{6}/4\).
$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-[\cosh(pt)]^{\frac {1}{2p^{2}}}}{t^{4}}=\frac{1}{24} \biggl(p^{2}-\frac{3}{8} \biggr)\geq0, $$
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 for all \(t>0\), which is equivalent to for all \(x>0\), where \(x=\sqrt{6}t/4\).
$$ \frac{2}{3}\cosh \biggl(\frac{\sqrt{3}t}{2} \biggr)+\frac{1}{3}> \biggl[\cosh \biggl(\frac{\sqrt{6}t}{4} \biggr) \biggr]^{4/3} $$
$$ \log \biggl[\frac{2}{3}\cosh(\sqrt{2}x)+\frac{1}{3} \biggr]>\frac{4}{3}\log (\cosh x) $$
(3.54)
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 for all \(x>0\).
$$ f_{2}(x)>0 $$
(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 which contradicts (3.62).
$$ \frac{I_{0}(t)-[\cosh(p_{0}t)]^{1/(2p_{0}^{2})}}{e^{t/(2p_{0})}}>\frac {1}{1+2t}\frac{e^{t}}{e^{t/(2p_{0})}} - \biggl( \frac{1+e^{-2p_{0}t}}{2} \biggr)^{1/(2p_{0}^{2})}\rightarrow \infty \quad (t\rightarrow \infty), $$
(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
or
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
or
hold for all
\(t>0\)
or all
\(a, b>0\)
with
\(a\neq b\).
$$ I_{0}(t)>\frac{\cosh(t\cos\theta)+\cosh(t\sin\theta)}{2} $$
(3.63)
$$ TQ(a,b)>\frac{A^{\cos\theta}_{\cos\theta}(a,b)G^{1-\cos\theta }(a,b)+A^{\sin\theta}_{\sin\theta}(a,b)G^{1-\sin\theta}(a,b)}{2} $$
$$\begin{aligned} I_{0}(t) >&\frac{1}{2} \biggl[\cosh \biggl( \frac{\sqrt{2-\sqrt{2}}}{2}t \biggr)+\cosh \biggl(\frac{\sqrt{2+\sqrt{2}}}{2}t \biggr) \biggr] \\ >&\frac{1}{2} \biggl[\cosh \biggl(\frac{\sqrt{3}}{2}t \biggr)+\cosh \biggl(\frac {1}{2}t \biggr) \biggr]>\cosh \biggl(\frac{\sqrt{2}}{2}t \biggr) \end{aligned}$$
(3.64)
$$\begin{aligned} TQ(a,b) >&\frac{A^{\sqrt{2-\sqrt{2}}/2}_{\sqrt{2-\sqrt {2}}/2}(a,b)G^{1-\sqrt{2-\sqrt{2}}/2}(a,b) +A^{\sqrt{2+\sqrt{2}}/2}_{\sqrt{2+\sqrt{2}}/2}(a,b)G^{1-\sqrt{2+\sqrt {2}}/2}(a,b)}{2} \\ >&\frac{A^{\sqrt{3}/2}_{\sqrt{3}/2}(a,b)G^{1-\sqrt {3}/2}+A^{1/2}_{1/2}(a,b)G^{1/2}(a,b)}{2}>A^{\sqrt{2}/2}_{\sqrt {2}/2}(a,b)G^{1-\sqrt{2}/2}(a,b) \end{aligned}$$
Proof
If inequality (3.63) holds for all \(t>0\), then we have which implies that \(\theta\in[\pi/8, 3\pi/8]\).
$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-\frac{\cosh(t\cos\theta)+\cosh (t\sin\theta)}{2}}{t^{4}}=-\frac{1}{192}\cos(4 \theta)\geq0, $$
Next, we prove the sufficiency of inequality (3.63). Simple computations lead to
for \(x>0\).
$$\begin{aligned}& \frac{\partial[\cosh(t\cos\theta)+\cosh(t\sin\theta)]}{\partial\theta }=\frac{t^{2}\sin(2\theta)}{2} \biggl[ \frac{\sinh(t\sin\theta)}{t\sin \theta}-\frac{\sinh(t\cos\theta)}{t\cos\theta} \biggr], \end{aligned}$$
(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)
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
for \(n\geq3\).
$$\begin{aligned}& R_{3}(t)=\frac{\sum_{n=0}^{\infty}\rho_{n}t^{2n}}{\sum_{n=0}^{\infty }\sigma_{n}t^{2n}}, \end{aligned}$$
(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)
It follows from Lemma 2.6 and (3.68)-(3.70) that \(R_{3}(t)\) is strictly decreasing on \((0, \infty)\). Therefore, follows from (3.67) and the monotonicity of \(R_{3}(t)\) together with \(R_{3}(0^{+})=\rho_{0}/\sigma_{0}=1\).
$$ I_{0}(t)>\frac{\cosh(t\cos\theta_{0})+\cosh(t\sin\theta_{0})}{2} $$
(3.71)
Theorem 3.8
The inequality
holds for all
\(t>0\).
$$ I_{0}(t)>\frac{\sinh t}{t}+\frac{3(4-\pi)(t\sinh t-2\cosh t+2)}{\pi t^{2}} $$
(3.72)
Proof
It is easy to verify that 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
$$ \frac{d^{2}}{dx^{2}} \biggl(\frac{1}{\sqrt{1-x^{2}}} \biggr)=\frac {1+2x^{2}}{ (1-x^{2} )^{5/2}}>0,\qquad \frac{\partial^{2}\cosh(tx)}{\partial x^{2}}=t^{2}\cosh(tx)>0 $$
$$\begin{aligned}& \frac{\pi}{2}I_{0}(t)-\frac{\pi}{2} \frac{\sinh t}{t} \\& \quad = \int_{0}^{1}\frac{\cosh(tx)}{\sqrt{1-x^{2}}}\,dx- \int_{0}^{1}\frac {dx}{\sqrt{1-x^{2}}} \int_{0}^{1}\cosh(tx)\,dx \\& \quad >12 \int_{0}^{1}\frac{x-\frac{1}{2}}{\sqrt{1-x^{2}}}\,dx \int_{0}^{1} \biggl(x-\frac{1}{2} \biggr) \cosh(tx)\,dx \\& \quad =\frac{3(4-\pi)(t\sinh t-2\cosh t+2)}{2t^{2}}. \end{aligned}$$
(3.73)
Remark 3.3
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 for all \(\lambda, p\in\mathbb{R}\) and \(a, b>0\) if M is a bivariate mean.
$$ M_{\lambda p}(a,b)=M_{p}^{1/\lambda} \bigl(a^{\lambda}, b^{\lambda} \bigr) $$
Theorem 3.9
The inequality
holds for all
\(a, b>0\)
with
\(a\neq b\)
if and only if
\(p\geq3/4\).
$$ TQ(a,b)< I_{p}(a,b) $$
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 holds for all distinct positive real numbers a and b with the best possible constant \(2/3\).
$$ I(a,b)>A_{2/3}(a,b) $$
(3.75)
Inequalities (3.74) and (3.75) lead to for all \(a, b>0\) with \(a\neq b\).
$$ TQ(a,b)< A_{1/2}(a,b)=A_{2/3}^{4/3} \bigl(a^{3/4}, b^{3/4} \bigr) < I^{4/3} \bigl(a^{3/4}, b^{3/4} \bigr)=I_{3/4}(a,b) $$
(3.76)
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 for all \(t>0\).
$$ I_{0}(t)-e^{t/\tanh(pt)-1/p}< 0 $$
(3.77)
Inequality (3.77) leads to which implies that \(p\geq3/4\). □
$$ \lim_{t\rightarrow0^{+}}\frac{I_{0}(t)-e^{t/\tanh (pt)-1/p}}{t^{2}}=\frac{1}{3} \biggl( \frac{3}{4}-p \biggr)\leq0, $$
Remark 3.4
For all \(a, b>0\) with \(a\neq b\), the Toader mean \(T(a,b)\) satisfies the double inequality [5, 7] with the best possible constants \(3/2\) and \(\log2/(\log\pi-\log2)\), and the one-sided inequality [33]
$$ A_{3/2}(a,b)< T(a,b)< A_{\log2/(\log\pi-\log2)}(a,b) $$
(3.78)
$$ T(a,b)< I_{9/4}(a,b). $$
(3.79)
It follows from (3.78) and (3.79) that which can be rewritten as
$$\begin{aligned} A_{1/2}^{1/3}(a,b) =&A_{3/2} \bigl(a^{1/3}, b^{1/3} \bigr)< T \bigl(a^{1/3}, b^{1/3} \bigr) \\ =&T^{1/3}_{1/3}(a,b)< I_{9/4} \bigl(a^{1/3}, b^{1/3} \bigr)=I_{3/4}^{1/3}(a,b), \end{aligned}$$
$$ A_{1/2}(a,b)< T_{1/3}(a,b)< I_{3/4}(a,b). $$
(3.80)
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
$$ L(a,b)< L^{3/4}(a,b)A^{1/4}(a,b), \qquad \frac{A(a,b)+G(a,b)}{2}> \frac {3L(a,b)+A(a,b)}{4}. $$
Remark 3.6
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
holds for all
\(a, b>0\)
with
\(a\neq b\).
$$ TQ(a,b)>L_{3/2}(a,b) $$
For all \(a, b>0\) with \(a\neq b\), inspired by the double inequality given in Corollary 3.1 and the inequalities proved by Alzer in [36] we propose Conjecture 3.2.
$$ \sqrt{A(a,b)G(a,b)}< TQ(a,b)< \frac{A(a,b)+G(a,b)}{2} $$
$$ \sqrt{A(a,b)G(a,b)}< \sqrt{I(a,b)L(a,b)}< \frac{I(a,b)+L(a,b)}{2}< \frac {A(a,b)+G(a,b)}{2} $$
Conjecture 3.2
The inequality
holds for all
\(a, b>0\)
with
\(a\neq b\).
$$ TQ(a,b)< \sqrt{I(a,b)L(a,b)} $$
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 and for all \(n\in\mathbb{N}\).
$$ \frac{2}{\pi}< s_{n}=(2n+1)W_{n}^{2} \leq s_{1}=\frac{3}{4} $$
(3.85)
$$ \frac{2}{\pi}=\min \biggl\{ \frac{c_{1}}{d_{1}}, \lim _{n\rightarrow\infty }\frac{c_{n}}{d_{n}} \biggr\} < \frac{c_{n}}{d_{n}}= \frac{2^{2n}s_{n}-1}{2^{2n}-1}\leq\frac {c_{2}}{d_{2}}=\frac{41}{60} $$
(3.86)
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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Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.