1 Introduction
The well-known Lazarević inequality [1, 2] states that for all \(x>0\) if and only if \(p\geq3\).
$$ \biggl(\frac{\sinh(x)}{x} \biggr)^{p}>\cosh(x) $$
(1.1)
Anzeige
Let \(p\in(-\infty, 8/15]\cup(1, \infty)\). Then the inequality holds for \(x>0\) if and only if \(q\geq3(1-p)\).
$$ \biggl(\frac{\sinh(x)}{x} \biggr)^{q}>p+(1-p)\cosh(x) $$
For \(p>0\), Yang [4] proved that the inequality holds for all \(x>0\) if and only if \(p\geq\sqrt{5}/5\), and inequality (1.2) is reversed if and only if \(p\leq1/3\).
$$ \biggl(\frac{\sinh(x)}{x} \biggr)^{3p^{2}}>\cosh(px) $$
(1.2)
Neuman and Sándor [5] proved that the Cusa type inequality holds for all \(x>0\).
$$ \frac{\sinh(x)}{x}< \frac{2+\cosh(x)}{3} $$
(1.3)
In [6], Zhu presented a more general result which contains Lazarević inequality (1.1) and Cusa type inequality (1.3) as follows.
Anzeige
Theorem A
The following statements are true:
(i)
If
\(p\geq4/5\), then the double inequality
holds for all
\(x>0\)
if and only if
\(\lambda\leq0\)
and
\(\eta\geq1/3\).
$$ 1-\lambda+\lambda\cosh^{p}(x)< \biggl(\frac{\sinh(x)}{x} \biggr)^{p}< 1-\eta +\eta\cosh^{p}(x) $$
(ii)
If
\(p<0\), then the inequality
holds for all
\(x>0\)
if and only if
\(\eta\leq1/3\).
$$ \biggl(\frac{\sinh(x)}{x} \biggr)^{p}< 1-\eta+\eta\cosh^{p}(x) $$
More inequalities for the hyperbolic sine and cosine functions can be found in the literature [7‐19].
Let \(p, q\in\mathbb{R}\) and \(H_{p,q}(x)\) be defined on \((0, \infty)\) by where the function \(U_{p}(t)\) is defined on \((1, \infty)\) by
$$ H_{p, q}(x)=\frac{U_{p} (\frac{\sinh(x)}{x} )}{U_{q}(\cosh(x))}, $$
(1.4)
$$ U_{p}(t)=\frac{t^{p}-1}{p} \quad (p\neq0),\qquad U_{0}(t)=\lim_{p\rightarrow 0}\frac{t^{p}-1}{p}=\log t. $$
(1.5)
The main purpose of this paper is to deal with the monotonicity of \(H_{p, q}(x)\) on \((0, \infty)\), generalize and improve the Lazarević and Cusa type inequalities, and present the new bounds for certain bivariate means.
2 Monotonicity
Lemma 2.1
Let
\(p\in\mathbb{R}\)
and
\(U_{p}(t)\)
be defined on
\((1, \infty)\)
by (1.5). Then the function
\(p\mapsto U_{p}(t)\)
is increasing on
\(\mathbb{R}\)
and
\(U_{p}(t)>0\)
for all
\(t\in(1, \infty)\).
Proof
Let \(p\neq0\), then the monotonicity of the function \(p\mapsto U_{p}(t)\) follows easily from
$$ \frac{\partial U_{p}(t)}{\partial p}=\frac{t^{p}}{p^{2}} \bigl[ \bigl(t^{-p}-1 \bigr)-\log \bigl(t^{-p} \bigr) \bigr]>0. $$
It follows from the monotonicity of the function \(p\mapsto U_{p}(t)\) that □
$$ U_{p}(t)>\lim_{p\rightarrow-\infty}U_{p}(t)=\lim _{p\rightarrow-\infty }\frac{t^{p}-1}{p}=0. $$
Lemma 2.2
Let
\(f, g: [a, b]\mapsto\mathbb{R}\)
be continuous on
\([a,b]\)
and differentiable on
\((a,b)\), and
\(g^{\prime}\neq0\)
on
\((a,b)\). If
\(f^{\prime}/g^{\prime}\)
is increasing (decreasing) on
\((a,b)\), then the functions
are also increasing (decreasing) on
\((a,b)\).
$$ \frac{f(x)-f(a)}{g(x)-g(a)}, \qquad \frac{f(x)-f(b)}{g(x)-g(b)} $$
Lemma 2.3
(See [21])
Let
\(\{a_{n}\}_{n=0}^{\infty}\)
and
\(\{b_{n}\}_{n=0}^{\infty}\)
be two real sequences with
\(b_{n}>0\)
for
\(n=0, 1, 2, \ldots \) , and the power series
\(A(t)=\sum_{n=0}^{\infty}a_{n}t^{n}\)
and
\(B(t)=\sum_{n=0}^{\infty}b_{n}t^{n}\)
have the radius of convergence
\(r>0\). If the non-constant sequence
\(\{a_{n}/b_{n}\}_{n=0}^{\infty}\)
is increasing (decreasing), then the function
\(A(t)/B(t)\)
is also increasing (decreasing) on
\((0, r)\).
Let \(\operatorname{Sh}_{p}(x)\) and \(\operatorname{Ch}_{p}(x)\) be defined on \((0, \infty)\) by and respectively, where the function \(U_{p}\) is defined by (1.5). Then from (1.4) and (1.5) we clearly see that the function \(H_{p,q}(x)\) can be rewritten as
$$ \operatorname{Sh}_{p}(x)=U_{p} \biggl( \frac{\sinh(x)}{x} \biggr) $$
(2.1)
$$ \operatorname{Ch}_{p}(x)=U_{p}\bigl(\cosh(x)\bigr), $$
(2.2)
$$ H_{p,q}(x)=\frac{\operatorname{Sh}_{p}(x)}{\operatorname{Ch}_{p}(x)}=\frac {\operatorname{Sh}_{p}(x)-\operatorname{Sh}_{p}(0^{+})}{\operatorname{Ch}_{p}(x)-\operatorname{Ch}_{p}(0^{+})}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{q}{p}\frac{ (\frac{\sinh(x)}{x} )^{p}-1}{\cosh ^{q}(x)-1}, & pq\neq0, \\ \frac{1}{p}\frac{ (\frac{\sinh(x)}{x} )^{p}-1}{\log(\cosh (x))}, & p\neq0, q=0, \\ q\frac{\log\frac{\sinh(x)}{x}}{\cosh^{q}(x)-1}, & p=0, q\neq0, \\ \frac{\log\frac{\sinh(x)}{x}}{\log(\cosh(x))}, & p=q=0. \end{array}\displaystyle \right . $$
(2.3)
Let \(pq\neq0\). Then it follows from (1.5), (2.1), and (2.2) that
where with
where \(C(x)>0\) due to by the Wilker type inequality given in [15].
$$\begin{aligned}& \frac{\operatorname{Sh}^{\prime}_{p}(x)}{\operatorname{Ch}^{\prime}_{q}(x)}=\frac{\cosh ^{1-q}(x)}{x^{2}\sinh(x)} \biggl(\frac{\sinh(x)}{x} \biggr)^{p-1}\bigl[x\cosh (x)-\sinh(x)\bigr]=:f_{1}(x), \end{aligned}$$
(2.4)
$$\begin{aligned}& f^{\prime}_{1}(x)=\frac{f_{2}(x)}{x^{2}\sinh^{3}(x)\cosh^{q}(x)} \biggl(\frac{\sinh(x)}{x} \biggr)^{p}, \end{aligned}$$
(2.5)
$$ f_{2}(x)=pA(x)-qB(x)+C(x) $$
(2.6)
$$\begin{aligned}& A(x)=\bigl[x\cosh(x)-\sinh(x)\bigr]^{2}\cosh(x)>0, \end{aligned}$$
(2.7)
$$\begin{aligned}& B(x)=x\bigl[x\cosh(x)-\sinh(x)\bigr]\sinh^{2}(x)>0, \end{aligned}$$
(2.8)
$$\begin{aligned}& C(x)=-2x^{2}\cosh(x)+x\sinh(x)+\cosh(x)\sinh^{2}(x)>0, \end{aligned}$$
(2.9)
$$ C(x)=x^{2}\cosh(x) \biggl[\frac{\sinh^{2}(x)}{x^{2}}+\frac{\tanh (x)}{x}-2 \biggr]>0 $$
Let
$$\begin{aligned}& a_{n}= \bigl(4n^{2}-14n+9 \bigr)9^{n-1}+12n^{2}-10n-1, \end{aligned}$$
(2.10)
$$\begin{aligned}& b_{n}=4n(n-2)9^{n-1}-4n(n-2), \end{aligned}$$
(2.11)
$$\begin{aligned}& c_{n}=9^{n}-32n^{2}+24n-1. \end{aligned}$$
(2.12)
Then making use of (2.7)-(2.9) together with the power series formulas \(\sinh(x)=\sum_{n=0}^{\infty}x^{2n+1}/(2n+1)!\) and \(\cosh(x)=\sum_{n=0}^{\infty}x^{2n}/(2n)!\) we have
$$\begin{aligned}& A(x)=\bigl[x\cosh(x)-\sinh(x)\bigr]^{2}\cosh(x) \\& \hphantom{A(x)}=\frac{1}{4}x^{2}\cosh(3x)+\frac{3}{4}x^{2} \cosh(x)-\frac{1}{2}x\sinh (3x) \\& \hphantom{A(x)={}}{}-\frac{1}{2}x\sinh(x)+ \frac{1}{4}\cosh(3x)-\frac{1}{4}\cosh(x) \\& \hphantom{A(x)}=\sum_{n=3}^{\infty} \frac{ (4n^{2}-14n+9 )9^{n-1}+12n^{2}-10n-1}{4(2n)!}x^{2n}=\sum_{n=3}^{\infty} \frac {a_{n}}{4(2n)!}x^{2n}, \end{aligned}$$
(2.13)
$$\begin{aligned}& B(x)=x\bigl[x\cosh(x)-\sinh(x)\bigr]\sinh^{2}(x) \\& \hphantom{B(x)}=\frac{1}{4}x^{2}\cosh(3x)-\frac{1}{4}x^{2} \cosh(x)-\frac{1}{4}x\sinh (3x)+\frac{3}{4}x\sinh(x) \\& \hphantom{B(x)}=\sum_{n=3}^{\infty} \frac{4n(n-2)9^{n-1}-4n(n-2)}{4(2n)!}x^{2n}=\sum_{n=3}^{\infty} \frac{b_{n}}{4(2n)!}x^{2n}, \end{aligned}$$
(2.14)
$$\begin{aligned}& C(x)=-2x^{2}\cosh(x)+x\sinh(x)+\cosh(x)\sinh^{2}(x) \\& \hphantom{C(x)}=-2x^{2}\cosh(x)+x\sinh(x)+\frac{1}{4}\cosh(3x)- \frac{1}{4}\cosh(x) \\& \hphantom{C(x)}=\sum_{n=3}^{\infty} \frac{9^{n}-32n^{2}+24n-1}{4(2n)!}x^{2n}=\sum_{n=3}^{\infty} \frac{c_{n}}{4(2n)!}x^{2n}. \end{aligned}$$
(2.15)
In order to investigate the monotonicity of the function \(H_{p,q}\), we need Lemma 2.4.
Lemma 2.4
Let
\(A(x)\), \(B(x)\), and
\(C(x)\)
be, respectively, defined by (2.7), (2.8), and (2.9), \(f_{3}\)
be defined on
\((0, \infty)\)
by
and
Then the following statements are true:
$$\begin{aligned}& f_{3}(x)=\frac{pA(x)-qB(x)}{C(x)}+1, \end{aligned}$$
(2.16)
$$\begin{aligned}& I_{1}= \{q=0, p>0 \}\cup \biggl\{ q>0, \frac{p}{q}\geq \frac {23}{17} \biggr\} \cup \biggl\{ q< 0, \frac{p}{q}\leq1 \biggr\} , \end{aligned}$$
(2.17)
$$ I_{2}= \{q=0, p< 0 \}\cup \biggl\{ q>0, \frac{p}{q} \leq1 \biggr\} \cup \biggl\{ q< 0, \frac{p}{q}\geq\frac{23}{17} \biggr\} . $$
(2.18)
(i)
\(f_{3}\)
is increasing from
\((0, \infty)\)
onto
\(((5p-15q)/8+1, \infty )\)
if
\((p, q)\in I_{1}\);
(ii)
\(f_{3}\)
is decreasing from
\((0, \infty)\)
onto
\((-\infty, (5p-15q)/8+1 )\)
if
\((p, q)\in I_{2}\).
Proof
Let \(a_{n}\), \(b_{n}\), and \(c_{n}\) be, respectively, defined by (2.10), (2.11), and (2.12). Then it follows from (2.13)-(2.16) that
where
$$\begin{aligned}& f_{3}(x)-1=\frac{\sum_{n=3}^{\infty}\frac {pa_{n}-qb_{n}}{4(2n)!}x^{2n}}{\sum_{n=3}^{\infty}\frac{c_{n}}{4(2n)!}x^{2n}}, \end{aligned}$$
(2.19)
$$\begin{aligned}& \frac{\frac{pa_{n}-qb_{n}}{4(2n)!}}{\frac{c_{n}}{4(2n)!}}=\frac {pa_{n}-qb_{n}}{c_{n}}, \end{aligned}$$
(2.20)
$$\begin{aligned}& \frac{pa_{n+1}-qb_{n+1}}{c_{n+1}}-\frac{pa_{n}-qb_{n}}{c_{n}} \\& \quad =\frac{p ( a_{n+1}c_{n}-a_{n}c_{n+1} ) -q ( b_{n+1}c_{n}-b_{n}c_{n+1} ) }{c_{n}c_{n+1}} \\& \quad =\frac{pv_{n}-qu_{n}}{c_{n}c_{n+1}}=\left \{ \textstyle\begin{array}{l@{\quad}l} p\frac{v_{n}}{c_{n}c_{n+1}}, & q=0, \\ \frac{v_{n}}{c_{n}c_{n+1}}q (\frac{p}{q}-\frac{u_{n}}{v_{n}} ), & q\neq0, \end{array}\displaystyle \right . \end{aligned}$$
(2.21)
$$\begin{aligned}& u_{n}=b_{n+1}c_{n}-b_{n}c_{n+1} \\& \hphantom{u_{n}}=2\times9^{n-1} \bigl[(36n-18)9^{n}- \bigl( 512n^{4}-384n^{3}-560n^{2}+792n-36 \bigr) \bigr] \\& \hphantom{\hphantom{u_{n}}={}}{}+4 \bigl(42n+40n^{2}-1 \bigr), \end{aligned}$$
(2.22)
$$\begin{aligned}& v_{n}=a_{n+1}c_{n}-a_{n}c_{n+1} \\& \hphantom{v_{n}}=2\times9^{n-1} \bigl[(36n-45)9^{n}- \bigl(512n^{4}-1\text{,}152n^{3}+1\text{,}072n^{2} \bigr) \bigr] \\& \hphantom{v_{n}={}}{}+2\times \bigl[2(28n+5)9^{n}- \bigl(16n^{2}+60n+5 \bigr) \bigr]. \end{aligned}$$
(2.23)
We claim that and for \(n\geq3\).
$$ c_{n}>0 $$
(2.24)
$$ v_{n}>0 $$
(2.25)
Indeed, making use of the binomial expansion we have for \(n\geq3\).
$$\begin{aligned}& c_{n}=9^{n}-32n^{2}+24n-1=(1+8)^{n}-32n^{2}+24n-1 \\& \hphantom{c_{n}}>1+8n+\frac{n(n-1)}{2}8^{2}-32n^{2}+24n-1=0, \\& (36n-45)9^{n}- \bigl(512n^{4}-1\text{,}152n^{3}+1 \text{,}072n^{2} \bigr) \\& \quad >(36n-45) \biggl(1+8n+\frac{n(n-1)}{2}8^{2}+ \frac {n(n-1)(n-2)}{6}8^{3} \biggr) \\& \qquad {}- \bigl(512n^{4}-1 \text{,}152n^{3}+1\text{,}072n^{2} \bigr) \\& \quad =2\text{,}560(n-3)^{4}+19\text{,}968(n-3)^{3}+55 \text{,}760(n-3)^{2}+65\text{,}340(n-3)+25\text{,}991>0, \\& 2(28n+5)9^{n}- \bigl(16n^{2}+60n+5 \bigr) \\& \quad >2(28n+5) (1+8n)- \bigl(16n^{2}+60n+5 \bigr) \\& \quad =432n^{2}+76n+5>0 \end{aligned}$$
We divide the proof into two cases.
Case 1. \(q=0\). Then (2.19)-(2.21), (2.24), (2.25), and Lemma 2.3 lead to the conclusion that \(f_{3}(x)\) is increasing on \((0, \infty )\) if \(p>0\) and decreasing on \((0, \infty)\) if \(p<0\). Therefore, we have \(5p/8+1<\lim_{x\rightarrow0^{+}}f_{3}(x)<f_{3}(x)<\lim_{x\rightarrow\infty}f_{3}(x)=\infty\) for \(p>0\), \(f_{3}(x)=1\) for \(p=0\), and \(-\infty<\lim_{x\rightarrow\infty}f_{3}(x)<f_{3}(x)<\lim_{x\rightarrow0^{+}}f_{3}(x)=5p/8+1\) for \(p<0\).
Case 2. \(q\neq0\). We first prove that \(u_{n}/v_{n}\) is decreasing for \(n\geq3\). From (2.25) we know that it suffices to show that \(u_{n}v_{n+1}-u_{n+1}v_{n}>0\) for \(n\geq3\). It follows from (2.22) and (2.23) that
$$\begin{aligned}& {u_{n}v_{n+1}-u_{n+1}v_{n}} \\& \quad =\frac{16c_{n+1}}{3}\bigl[9^{3n+2}- \bigl(1\text{,}024n^{4}-2 \text{,}560n^{3}+2\text{,}752n^{2}+243 \bigr)9^{2n} \\& \qquad {}+ \bigl(1\text{,}024n^{4}+2\text{,}560n^{3}+2 \text{,}752n^{2}+243 \bigr)9^{n}-81\bigr]. \end{aligned}$$
(2.26)
Note that for \(n\geq3\).
$$\begin{aligned}& 9^{n+2}- \bigl(1\text{,}024n^{4}-2 \text{,}560n^{3}+2\text{,}752n^{2}+243 \bigr) \\& \quad >1+8(n+2)+\frac{(n+2)(n+1)}{2}8^{2}+\frac{(n+2)(n+1)n}{6}8^{3} \\& \qquad {}+ \frac {(n+2)(n+1)n(n-1)}{24}8^{4}+\frac{(n+2)(n+1)n(n-1)(n-2)}{120}8^{5} \\& \qquad {}- \bigl(1\text{,}024n^{4}-2 \text{,}560n^{3}+2\text{,}752n^{2}+243 \bigr) \\& \quad =2\text{,}048(n-3)^{5}+24\text{,}320(n-3)^{4}+119 \text{,}680(n-3)^{3} \\& \qquad {}+297\text{,}040(n-3)^{2}+355 \text{,}692(n-3)+151\text{,}605>0 \end{aligned}$$
(2.27)
From (2.21), (2.24), (2.25), and the monotonicity of \(u_{n}/v_{n}\) we clearly see that
$$\begin{aligned}& 1=\lim_{n\rightarrow\infty}\frac{u_{n}}{v_{n}}< \frac{u_{n}}{v_{n}}\leq \frac{u_{3}}{v_{3}}=\frac{23}{17}, \\& \frac{pa_{n+1}-qb_{n+1}}{c_{n+1}}-\frac{pa_{n}-qb_{n}}{c_{n}}=\left \{ \textstyle\begin{array}{l@{\quad}l} >0, & q>0,\frac{p}{q}\geq\frac{23}{17}, \\ < 0, & q< 0,\frac{p}{q}\geq\frac{23}{17}, \\ < 0, & q>0,\frac{p}{q}\leq1, \\ >0, & q< 0,\frac{p}{q}\leq1. \end{array}\displaystyle \right . \end{aligned}$$
(2.28)
Therefore, the desired results follows easily from (2.19), (2.20), (2.28), and Lemma 2.3 together with the facts that □
$$\begin{aligned}& \lim_{x\rightarrow0^{+}}f_{3}(x)=\frac{5p-15q}{8}+1, \\& \lim_{x\rightarrow\infty}f_{3}(x)=\left \{ \textstyle\begin{array}{l@{\quad}l} \infty, & q>0,\frac{p}{q}\geq\frac{23}{17} \text{ or } q< 0,\frac {p}{q}\leq1, \\ -\infty, & q< 0,\frac{p}{q}\geq\frac{23}{17} \text{ or } q>0,\frac {p}{q}\leq1. \end{array}\displaystyle \right . \end{aligned}$$
From Lemma 2.4, we get the monotonicity of \(H_{p,q}\) as follows.
Proposition 2.1
Let
\(I_{1}\)
and
\(I_{2}\)
be defined, respectively, by (2.17) and (2.18), and
\(H_{p,q}(x)\)
be defined on
\((0, \infty)\)
by (2.3). Then the following statements are true:
(i)
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
if
\((p,q)\in I_{1}\cup (0, 0)\cap\{(5p-15q)/8+1\geq0\}\);
(ii)
\(H_{p,q}(x)\)
is decreasing on
\((0, \infty)\)
if
\((p,q)\in I_{2}\cap\{(5p-15q)/8+1\leq0\}\).
Proof
It is easily to verify that \((p,q)\in I_{1}\cup(0, 0)\cap\{ (5p-15q)/8+1\geq0\}\) is equivalent to or and \((p,q)\in I_{2}\cap\{(5p-15q)/8+1\leq0\}\) is equivalent to or
$$ p\geq \left \{ \textstyle\begin{array}{l@{\quad}l} 3q-8/5, & q\in[34/35,\infty), \\ 23q/17, & q\in[0,34/35), \\ q,& q\in(-\infty,0) \end{array}\displaystyle \right . $$
$$ q\leq \left \{ \textstyle\begin{array}{l@{\quad}l} p/3+8/15, & p\in[46/35,\infty), \\ 17p/23, & p\in[0,46/35), \\ p, & p\in(-\infty,0), \end{array}\displaystyle \right . $$
$$ p\leq \left \{ \textstyle\begin{array}{l@{\quad}l} q, & q\in[4/5,\infty), \\ 3q-8/5, & q\in(-\infty,4/5) \end{array}\displaystyle \right . $$
$$ q\geq \left \{ \textstyle\begin{array}{l@{\quad}l} p, & p\in[4/5,\infty), \\ p/3+8/15,& p\in(-\infty,4/5). \end{array}\displaystyle \right . $$
Proposition 2.2
Let
\(H_{p,q}(x)\)
be defined on
\((0, \infty)\)
by (2.3). Then the following statements are true:
(i)
If
\(q\in[34/35, \infty)\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(p\geq3q-8/5\)
and decreasing for
\(p\leq q\).
(ii)
If
\(q\in[4/5, 34/35)\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(p\geq23q/17\)
and decreasing for
\(p\leq q\).
(iii)
If
\(q\in(0, 4/5)\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(p\geq23q/17\)
and decreasing for
\(p\leq3q-8/5\).
(iv)
If
\(q\in(-\infty, 0]\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(p\geq q\)
and decreasing for
\(p\leq3q-8/5\).
Proposition 2.3
Let
\(H_{p,q}(x)\)
be defined on
\((0, \infty)\)
by (2.3). Then the following statements are true:
(i)
If
\(p\in[46/35, \infty)\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q\leq p/3+8/15\)
and decreasing for
\(q\geq p\).
(ii)
If
\(p\in[4/5, 46/35)\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q\leq17p/23\)
and decreasing for
\(q\geq p\).
(iii)
If
\(p\in(0, 4/5)\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q\leq17p/23\)
and decreasing for
\(q\geq p/3+8/15\).
(iv)
If
\(p\in(-\infty, 0]\), then
\(H_{p,q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q\leq p\)
and decreasing for
\(q\geq p/3+8/15\).
Let \(p=kq\), then Proposition 2.1 leads to the following corollary.
Corollary 2.1
Let
\(H_{p,q}(x)\)
be defined on
\((0, \infty)\)
by (2.3). Then the following statements are true:
(i)
If
\(k\in(3, \infty)\), then
\(H_{kq, q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q\geq0\)
and decreasing for
\(q\leq8/[5(3-k)]\).
(ii)
If
\(k=3\), then
\(H_{kq, q}(x)\)
is increasing on
\((0, \infty)\)
for all
\(q\in\mathbb{R}\).
(iii)
If
\(k\in[23/17, 3)\), then
\(H_{kq, q}(x)\)
is increasing on
\((0, \infty)\)
for
\(0\leq q \leq8/[5(3-k)]\).
(iv)
If
\(k\in(1, 23/17)\), then
\(H_{kq, q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q=0\).
(v)
If
\(k\in(-\infty, 1]\), then
\(H_{kq, q}(x)\)
is increasing on
\((0, \infty)\)
for
\(q\leq0\)
and decreasing for
\(q \geq8/[5(3-k)]\).
Let \(I_{1}\) and \(I_{2}\) be defined by (2.17) and (2.18), respectively. If \((5p-15q)/8+1=0\), then we clearly see that and Proposition 2.1 leads to the following corollary.
$$\begin{aligned}& I_{1}\cup\{0, 0\}\cap \biggl\{ \frac{5p-15q}{8}+1=0 \biggr\} = \biggl\{ q\geq \frac{34}{35}, p=3q-\frac{8}{5} \biggr\} = \biggl\{ p\geq \frac{46}{35}, q=\frac{5p+8}{15} \biggr\} , \\& I_{2}\cap \biggl\{ \frac{5p-15q}{8}+1=0 \biggr\} = \biggl\{ q\leq \frac{4}{5}, p=3q-\frac{8}{5} \biggr\} = \biggl\{ p\leq \frac{4}{5}, q=\frac {5p+8}{15} \biggr\} , \end{aligned}$$
Corollary 2.2
Let
\(H_{p,q}(x)\)
be defined on
\((0, \infty)\)
by (2.3). Then
\(H_{3q-8/5, q}(x)\)
is increasing on
\((0, \infty)\)
if
\(q\geq34/35\)
and decreasing if
\(q\leq4/5\). In other words, \(H_{p, (5p+8)/15}(x)\)
is increasing on
\((0, \infty)\)
if
\(p\geq46/35\)
and decreasing if
\(p\leq4/5\).
3 Main results
In this section, we present several Lazarević and Cusa type inequalities involving the hyperbolic sine and cosine functions with two parameters.
Let \(\operatorname{Sh}_{p}(x)\), \(\operatorname{Ch}_{p}(x)\), \(H_{p, q}(x)\), \(I_{1}\), and \(I_{2}\) be, respectively, defined by (2.1), (2.2), (2.3), (2.17), and (2.18). Then it is not difficult to verify that
$$\begin{aligned}& I_{1}\cup\{0, 0\}\cap \biggl\{ \frac{5p-15q}{8}+1\geq0 \biggr\} \\& \quad = \biggl\{ 0\leq q\leq\min \biggl(\frac{17p}{23}, \frac{5p+8}{15} \biggr) \biggr\} \cup \biggl\{ q\leq\min \biggl(0, p, \frac{5p+8}{15} \biggr) \biggr\} , \\& I_{2}\cap \biggl\{ \frac{5p-15q}{8}+1\leq0 \biggr\} \\& \quad = \biggl\{ \max \biggl(\frac{17p}{23}, \frac{5p+8}{15} \biggr)\leq q< 0 \biggr\} \cup \biggl\{ q\geq\max \biggl(0, p, \frac{5p+8}{15} \biggr) \biggr\} , \\& H_{p, q}\bigl(0^{+}\bigr)=\frac{1}{3}. \end{aligned}$$
Theorem 3.1
(i)
If
\(0\leq q\leq\min\{17p/23, (5p+8)/15\}\)
or
\(q\leq\min\{0, p, (5p+8)/15\}\), then the inequalities
hold for
\(x\in(0, \infty)\)
with the best possible constant
\(1/3\).
$$\begin{aligned}& \frac{ (\frac{\sinh(x)}{x} )^{p}-1}{p}>\frac{1}{3}\frac{\cosh ^{q}(x)-1}{q}\quad (pq \neq0), \end{aligned}$$
(3.1)
$$\begin{aligned}& \log \biggl(\frac{\sinh(x)}{x} \biggr)>\frac{1}{3}\frac{\cosh^{q}(x)-1}{q} \quad (p=0, q\neq0), \end{aligned}$$
(3.2)
$$\begin{aligned}& \frac{ (\frac{\sinh(x)}{x} )^{p}-1}{p}>\frac{1}{3}\log\bigl[\cosh (x)\bigr]\quad (p\neq0, q= 0), \end{aligned}$$
(3.3)
$$\begin{aligned}& \log \biggl(\frac{\sinh(x)}{x} \biggr)>\frac{1}{3}\log\bigl[\cosh(x)\bigr] \quad (p=q=0), \end{aligned}$$
(3.4)
For clarity of expression, in the following we directly write \(\operatorname{Sh}_{p}(x)\), \(\operatorname{Ch}_{p}(x)\), and \(H_{p,q}(x)\), and so on. For their general formulas, if \(pq=0\), then we regard them as limit at \(p=0\) or \(q=0\), unless otherwise specified.
Lemma 3.1 will be used to establish sharp inequalities for hyperbolic functions.
Lemma 3.1
Let
\(\operatorname{Sh}_{p}\)
and
\(\operatorname{Ch}_{q}\)
be, respectively, defined on
\((0, \infty)\)
by (2.1) and (2.2), and
\(D_{p,q}\)
be defined on
\((0, \infty)\)
by
Then we have
$$ D_{p, q}(x)=\operatorname{Sh}_{p}(x)- \frac{1}{3}\operatorname{Ch}_{q}(x)=\frac{ (\frac{\sinh (x)}{x} )^{p}-1}{p}- \frac{\cosh^{q}(x)-1}{3q}. $$
(3.5)
$$\begin{aligned}& \lim_{x\rightarrow0^{+}}\frac{D_{p, q}(x)}{x^{4}}=\frac{1}{72} \biggl(p-3q+ \frac{8}{5} \biggr), \end{aligned}$$
(3.6)
$$\begin{aligned}& \lim_{x\rightarrow0^{+}}\frac{D_{3q-8/5, q}(x)}{x^{6}}=\frac {1}{270} \biggl(q- \frac{34}{35} \biggr), \end{aligned}$$
(3.7)
$$\begin{aligned}& \lim_{x\rightarrow\infty} \bigl[e^{-qx}D_{p,q}(x) \bigr]= \left \{ \textstyle\begin{array}{l@{\quad}l} \infty, & p>q\geq0, \\ \infty, & p\geq q=0, \\ -\frac{2^{-q}}{3q}, & q\geq p>0, \\ -\frac{2^{-q}}{3q}, & q>p=0, \end{array}\displaystyle \right . \end{aligned}$$
(3.8)
$$\begin{aligned}& \lim_{x\rightarrow\infty}D_{p,q}(x) =\left \{ \textstyle\begin{array}{l@{\quad}l} \infty, & p\geq0,q< 0, \\ -\infty, & p< 0,q\geq0, \\ \frac{1}{3q}-\frac{1}{p}, & p< 0, q< 0. \end{array}\displaystyle \right . \end{aligned}$$
(3.9)
Proof
Let \(x\rightarrow0^{+}\), then making use of power series formulas and (3.5) we get
$$ D_{p, q}(x)=\frac{5p-15q+8}{360}x^{4}+ \frac {35p^{2}-42p-315q^{2}+630q-320}{45\text{,}360}x^{6}+o\bigl(x^{6}\bigr). $$
(3.10)
We divide the proof of (3.8) into four cases.
Case 1. \(p>0\), \(q>0\). Then (3.5) leads to
$$\begin{aligned}& \lim_{x\rightarrow\infty} \bigl[e^{-qx}D_{p,q}(x) \bigr] \\& \quad =\lim_{x\rightarrow\infty} \biggl[\frac{1}{p}\frac {e^{(p-q)x}}{x^{p}} \biggl(\frac{1-e^{-2x}}{2} \biggr)^{p}- \frac{1}{3q} \biggl( \frac{1+e^{-2x}}{2} \biggr)^{q}- \biggl(\frac{1}{p}- \frac {1}{3q} \biggr)e^{-qx} \biggr] \\& \quad =\left \{ \textstyle\begin{array}{l@{\quad}l} \infty, & p>q>0, \\ -\frac{2^{-q}}{3q},& q\geq p>0. \end{array}\displaystyle \right . \end{aligned}$$
Case 2. \(p=0\), \(q>0\). Then it follows from (3.5) that
$$\begin{aligned}& \lim_{x\rightarrow\infty} \bigl[e^{-qx}D_{0,q}(x) \bigr] \\& \quad =\lim_{x\rightarrow\infty} \biggl[xe^{-qx}+e^{-qx} \log \biggl(\frac {1-e^{-2x}}{2} \biggr)-e^{-qx}\log x-\frac{1}{3q} \biggl(\frac {1+e^{-2x}}{2} \biggr)^{q}+\frac{e^{-qx}}{3q} \biggr] \\& \quad =-\frac{2^{-q}}{3q}. \end{aligned}$$
Case 3. \(p=0\), \(q=0\). Then (3.5) leads to
$$ \lim_{x\rightarrow\infty}D_{0,0}(x)=\lim_{x\rightarrow\infty} \biggl[x \biggl(\frac{2}{3}-\frac{\log x}{x} \biggr)+\log \biggl( \frac {1-e^{-2x}}{2} \biggr)-\frac{1+e^{-2x}}{6} \biggr]=\infty. $$
Case 4. \(p>0\), \(q=0\). Then it follows from Lemma 2.1 and (3.5) that for \(x\in(0, \infty)\).
$$ D_{p, 0}(x)>D_{0, 0}(x) $$
(3.11)
Therefore, follows from Case 3 and (3.11).
$$ \lim_{x\rightarrow\infty}D_{p, 0}(x)=\infty $$
Theorem 3.2
The following statements are true:
(i)
If
\(q\in[34/35, \infty)\), then the double inequality
holds for
\(x\in(0, \infty)\)
if and only if
\(p_{1}\geq3q-8/5\)
and
\(p_{2}\leq q\).
$$ \frac{ (\frac{\sinh(x)}{x} )^{p_{2}}-1}{p_{2}}< \frac{\cosh ^{q}(x)-1}{3q}< \frac{ (\frac{\sinh(x)}{x} )^{p_{1}}-1}{p_{1}} $$
(3.12)
(ii)
(iii)
Proof
All the sufficiencies in (i)-(iv) follow from Proposition 2.2 and \(H_{p, q}(0^{+})=1/3\). Next, we prove the necessities in (i)-(iv).
(i) If \(q\in[34/35, \infty)\) and the second inequality of (3.12) holds for all \(x\in(0, \infty)\), then (3.5) and (3.6) lead to the conclusion that \(p_{1}\geq3q-8/5\). If \(q\in[34/35, \infty)\) and the first inequality of (3.12) holds for all \(x\in(0, \infty)\), then we claim that \(p_{2}\leq q\), otherwise, \(p_{2}>q\in[34/35, \infty)\) and the first inequality (3.12) imply that \(D_{p_{2}, q}(x)<0\) for all \(x\in(0, \infty)\), which contradicts with (3.8).
(ii) If \(q\in[4/5, 34/25)\) and the first inequality of (3.12) holds for \(x\in(0, \infty)\), then the proof of \(p_{2}\leq q\) is similar to part two of (i).
(iii) If \(q\in(0, 4/5)\) and the first inequality of (3.12) holds for \(x\in(0, \infty)\), then the proof of \(p_{2}\leq3q-8/5\) is similar to part one of (i).
Theorem 3.3
The following statements are true:
(i)
If
\(p\in[46/35, \infty)\), then the double inequality
holds for all
\(x\in(0, \infty)\)
if and only if
\(q_{1}\leq(5p+8)/15\)
and
\(q_{2}\geq p\).
$$ \frac{\cosh^{q_{1}}(x)-1}{3q_{1}}< \frac{ (\frac{\sinh(x)}{x} )^{p}-1}{p} < \frac{\cosh^{q_{2}}(x)-1}{3q_{2}} $$
(3.13)
(ii)
(iii)
Remark 3.1
Remark 3.2
Let \(t\in(1, \infty)\), and \(\Omega_{p, q}\) and \(M(t; p, q)\) be, respectively, defined by and
$$ \Omega_{p, q}=\bigl\{ (p, q): p\geq0\bigr\} \cup\bigl\{ (p, q): 3q\leq p< 0\bigr\} $$
(3.15)
$$ M(t;p,q)=\left \{ \textstyle\begin{array}{l@{\quad}l} (1-\frac{p}{3q}+\frac{p}{3q}t^{q} )^{1/p}, & pq\neq0, (p, q)\in\Omega_{p,q}, \\ e^{ (t^{q}-1 )/(3q)}, & p=0, q\neq0, \\ (\frac{p}{3}\log t+1 )^{1/p}, & p>0,q=0, \\ t^{1/3},& p=q=0. \end{array}\displaystyle \right . $$
(3.16)
Then we clearly see that \(H_{p,q}(x)<(>)\,H_{p,q}(0^{+})=1/3\) for all \(x\in(0, \infty)\) is equivalent to \(\sinh(x)/x> (<)\,M(\cosh(x); p, q)\). Moreover, it is not difficult to verify that \(M(t; p, q)\) is decreasing with respect to p and increasing with respect to q if \((p, q)\in \Omega_{p, q}\).
Remark 3.3
From Remark 3.2 we know that Theorems 3.2 and 3.3 are also true if \((p, q)\in\Omega_{p,q}\) and replacing (3.12) and (3.13), respectively, with and
$$ \biggl(1-\frac{p_{1}}{3q}+\frac{p_{1}}{3q}\cosh^{q}(x) \biggr)^{1/p_{1}}< \frac{\sinh(x)}{x} < \biggl(1-\frac{p_{2}}{3q}+ \frac{p_{2}}{3q}\cosh^{q}(x) \biggr)^{1/p_{2}} $$
$$ \biggl(1-\frac{p}{3q_{1}}+\frac{p}{3q_{1}}\cosh^{q_{1}}(x) \biggr)^{1/p}< \frac{\sinh(x)}{x} < \biggl(1-\frac{p}{3q_{2}}+ \frac{p}{3q_{2}}\cosh^{q_{2}}(x) \biggr)^{1/p}. $$
Corollary 3.1
The double inequality
holds for all
\(x\in(0, \infty)\)
if and only if
\(p_{1}\geq7/5\)
and
\(0\leq p_{2}\leq1\).
$$ \biggl(1-\frac{p_{1}}{3}+\frac{p_{1}}{3}\cosh(x) \biggr)^{1/p_{1}}< \frac {\sinh(x)}{x} < \biggl(1-\frac{p_{2}}{3}+\frac{p_{2}}{3}\cosh(x) \biggr)^{1/p_{2}} $$
Remark 3.4
Letting \(p_{1}=7/5, 3/2, 2, 3\) and making use of the monotonicity of \(M(\cosh(x); p, q)\) with respect to p, then Corollary 3.1 leads to the inequalities for \(x\in(0, \infty)\), which is better than the inequalities given in (3.23) of [7].
$$\begin{aligned} \cosh^{1/3}(x) < & \biggl(\frac{1}{3}+\frac{2}{3}\cosh(x) \biggr)^{1/2}< \biggl(\frac{1}{2}+\frac{1}{2}\cosh(x) \biggr)^{2/3} \\ < & \biggl(\frac{8}{15}+\frac{7}{15}\cosh(x) \biggr)^{5/7}< \frac{\sinh (x)}{x}< \frac{2}{3}+\frac{1}{3}\cosh(x) \end{aligned}$$
Corollary 3.2
Let
\(x\in(0, \infty)\). Then the double inequality
holds if and only if
\(q_{1}\leq0\)
and
\(q_{2}\geq8/5\), and the double inequality
holds if and only if
\(q_{1}\leq17/23\)
and
\(q_{2}\geq1\).
$$ e^{(\cosh^{q_{1}}(x)-1)/(3q_{1})}< \frac{\sinh(x)}{x}< e^{(\cosh ^{q_{2}}(x)-1)/(3q_{2})} $$
$$ 1-\frac{1}{3q_{1}}+\frac{1}{3q_{1}}\cosh^{q_{1}}(x)< \frac{\sinh (x)}{x}< 1-\frac{1}{3q_{2}}+\frac{1}{3q_{2}}\cosh^{q_{2}}(x) $$
(3.17)
Remark 3.5
Letting \(q_{1}=17/23, 2/3, 1/2, 1/3, 1/6, 0\) and making use of the monotonicity of \(M(\cosh(x); p, q)\) with respect to q, then inequality (3.17) leads to for \(x\in(0, \infty)\).
$$\begin{aligned} \frac{1}{3}\log\bigl[\cosh(x)\bigr]+1 < &2\cosh^{1/6}(x)-1< \cosh^{1/3}(x)< \frac {1}{3}+\frac{2}{3}\cosh^{1/2}(x) \\ < &\frac{1}{2}+\frac{1}{2}\cosh^{2/3}(x)< \frac{28}{51}+\frac{23}{51}\cosh ^{17/23}(x) \\ < &\frac{\sinh(x)}{x}< \frac{2}{3}+\frac{1}{3}\cosh(x) \end{aligned}$$
Let \(p=kq\), then (3.15) and (3.16) become
$$\begin{aligned}& \Omega_{kq, q}=\bigl\{ (k,q): k, q\geq0\bigr\} \cup\bigl\{ (k,q): k, q< 0 \bigr\} \cup\bigl\{ (k,q): 0< k\leq3, q\leq0\bigr\} , \\& M(t;kq,q) =\left \{ \textstyle\begin{array}{l@{\quad}l} (1-\frac{k}{3}+\frac{k}{3}t^{q} )^{1/(kqt)}, & kq\neq0, (k,q)\in\Omega_{kq,q}, \\ e^{\frac{t^{q}-1}{3q}}, & q\neq0, k=0, \\ t^{1/3},& q=0. \end{array}\displaystyle \right . \end{aligned}$$
Remark 3.6
It is not difficult to verify that \(M(t; kq, q)\) is decreasing (increasing) with respect to q if \(k>(<)\,3\), and \(M(t; kq, q)\) is decreasing (increasing) with respect to k if \(q>(<)\, 0\).
Theorem 3.4
Let
\(x\in(0, \infty)\)
and
\(k\in[0, 3)\). Then the following statements are true:
(i)
If
\(k\in[23/17, 3)\), then the inequality
holds if and only if
\(q\leq8/[5(3-k)]\).
$$ \frac{\sinh(x)}{x}> \biggl(1-\frac{k}{3}+\frac{k}{3} \cosh^{q}(x) \biggr)^{1/(kq)} $$
(3.18)
(ii)
If
\(k\in(0, 1]\), then the double inequality
holds if and only if
\(q_{1}\leq0\)
and
\(q_{2}\geq8/[5(3-k)]\).
$$ \biggl(1-\frac{k}{3}+\frac{k}{3} \cosh^{q_{1}}(x) \biggr)^{1/(kq_{1})}< \frac{\sinh(x)}{x} < \biggl(1- \frac{k}{3}+\frac{k}{3}\cosh^{q_{2}}(x) \biggr)^{1/(kq_{2})} $$
(3.19)
Proof
(i) If \(k\in[23/17, 3)\), then it follows from Corollary 2.1(iii) that \(H_{kq, q}(x)>H_{kq, q}(0^{+})=1/3\) and (3.18) holds for \(0\leq q\leq8/[5(3-k)]\). For \(q<0\), we have \(M^{k}(\cosh(x); kq, q)< M^{k}(\cosh(x); 0, 0)\) due to \(M^{k}(\cosh(x); kq, q)\) is a weighted qth power mean of \(\cosh(x)\) and 1, so (3.18) still holds.
If (3.18) holds, then \(D_{kq, q}(x)>0\) and (3.6) leads to and \(q\leq8/[5(3-k)]\).
$$ \lim_{x\rightarrow0^{+}}\frac{D_{kq, q}(x)}{x^{4}}=\frac{1}{72} \biggl(kq-3q+ \frac{8}{5} \biggr)\geq0 $$
(ii) If \(k\in(0, 1]\), then it follows from Corollary 2.1(v) that \(H_{kq_{1}, q_{1}}(x)>H_{kq_{1}, q_{1}}(0^{+})=1/3\), \(H_{kq_{2}, q_{2}}(x)< H_{kq_{2}, q_{2}}(0^{+})=1/3\), and (3.19) holds for \(q_{1}\leq0\) and \(q\geq8/[5(3-k)]\).
If the second inequality of (3.19) holds, then \(D_{kq_{2}, q_{2}}(x)<0\) and (3.6) leads to \(q_{2}\geq8/[5(3-k)]\).
Next, we prove that the condition \(q_{1}\leq0\) is necessary such that the first inequality of (3.19) holds for all \(x\in(0, \infty)\). Indeed, the first inequality of (3.19) leads to \(D_{kq_{1}, q_{1}}(x)<0\) for all \(x\in(0, \infty)\). If \(q_{1}>0\) and \(k\in(0, 1]\), then \(0< kq_{1}\leq q_{1}\) and (3.8) leads to which implies that there exists large enough \(X>0\) such that \(D_{kq_{1}, q_{1}}(x)<0\) for \(x\in(X, \infty)\). □
$$ \lim_{x\rightarrow\infty} \bigl[e^{-q_{1}x}D_{kq_{1}, q_{1}}(x) \bigr]=- \frac{2^{-q_{1}}}{3q_{1}}< 0, $$
Corollary 3.3
The inequalities
and
hold for all
\(x\in(0, \infty)\)
if and only if
\(p_{1}\leq0\), \(p_{2}\geq4/5\), \(p\leq16/15\), and
\(q\leq8/5\).
$$\begin{aligned}& \biggl(\frac{2}{3}+\frac{1}{3}\cosh^{p_{1}}(x) \biggr)^{1/p_{1}}< \frac {\sinh(x)}{x}< \biggl(\frac{2}{3}+ \frac{1}{3}\cosh^{p_{2}}(x) \biggr)^{1/p_{2}}, \end{aligned}$$
(3.20)
$$\begin{aligned}& \frac{\sinh(x)}{x}> \biggl(\frac{1}{2}+\frac{1}{2} \cosh^{p}(x) \biggr)^{2/(3p)}, \end{aligned}$$
(3.21)
$$ \frac{\sinh(x)}{x}> \biggl(\frac{1}{3}+\frac{2}{3} \cosh^{q}(x) \biggr)^{1/(2q)} $$
(3.22)
Let \(p=3q-8/5\), then (3.15) and (3.16) become
$$\begin{aligned}& \Omega_{3q-8/5, q}= \biggl\{ q: q\geq\frac{8}{15} \biggr\} , \\& M \biggl(t;3q-\frac{8}{5},q \biggr)=\left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{8}{15q}+ (1-\frac{8}{15q} )t^{q} )^{5/(15q-8)}, & q>\frac{8}{15}, \\ e^{5 (t^{8/15}-1 )/8},& q=\frac{8}{15}. \end{array}\displaystyle \right . \end{aligned}$$
It is easy to prove that \(M(t; 3q-8/5, q)\) is decreasing with respect to q on the interval \([8/15, \infty)\) and
$$ \lim_{q\rightarrow\infty}M \biggl(t;3q-\frac{8}{5},q \biggr)=t^{1/3}. $$
Theorem 3.5
Let
\(q>8/15\). Then the inequality
holds for all
\(x\in(0, \infty)\)
if and only if
\(q\geq34/35\), and inequality (3.23) is reversed if and only if
\(q\leq4/5\).
$$ \frac{\sinh(x)}{x}> \biggl[\frac{8}{15q}+ \biggl(1- \frac{8}{15q} \biggr)\cosh ^{q}(x) \biggr]^{5/(15q-8)} $$
(3.23)
Proof
The sufficiency can be derived from Corollary 2.2. If inequality (3.23) holds, then \(D_{3q-8/5, q}(x)>0\) and (3.7) leads to the conclusion that \(q\geq34/35\).
Next, we prove that \(q\leq4/5\) if the reversed inequality (3.23) holds.
Let \(q=34/35, 1, 16/15, 6/5, 8/5, 2, \infty\mbox{ and }4/5, 7/10, 2/3, 3/5, 8/15^{+}\). Then Theorem 3.5 leads to Corollary 3.4.
Corollary 3.4
The inequalities
hold for all
\(x\in(0, \infty)\).
$$\begin{aligned} \cosh^{1/3}(x) < & \biggl(\frac{11}{15}\cosh^{2}(x)+ \frac{4}{15} \biggr)^{5/22}< \biggl(\frac{2}{3} \cosh^{8/5}(x)+\frac{1}{3} \biggr)^{5/16} \\ < & \biggl(\frac{5}{9}\cosh^{6/5}(x)+\frac{4}{9} \biggr)^{1/2}< \biggl(\frac {1}{2}\cosh^{16/15}(x)+ \frac{1}{2} \biggr)^{5/8} < \biggl(\frac{7}{15}\cosh(x)+ \frac{8}{15} \biggr)^{5/7} \\ < & \biggl(\frac{23}{51}\cosh^{34/35}(x)+\frac{28}{51} \biggr)^{35/46}< \frac {\sinh(x)}{x}< \biggl(\frac{1}{3} \cosh^{4/5}(x)+\frac{2}{3} \biggr)^{5/4} \\ < & \biggl(\frac{5}{21}\cosh^{7/10}(x)+\frac{16}{21} \biggr)^{2}< \biggl(\frac {1}{5}\cosh^{2/3}(x)+ \frac{4}{5} \biggr)^{5/2} < e^{5(\cosh^{8/15}(x)-1)/8} \end{aligned}$$
4 Applications
Let \(a, b>0\). Then the geometric mean \(G(a,b)\), arithmetic mean \(A(a,b)\), quadratic mean \(Q(a,b)\), logarithmic mean \(L(a,b)\), Neuman-Sándor mean \(M(a,b)\) [22] and second Yang mean \(V(a,b)\) [23] are defined by and respectively.
$$ G(a,b)=\sqrt{ab}, \qquad A(a,b)=\frac{a+b}{2}, \qquad Q(a,b)=\sqrt{ \frac {a^{2}+b^{2}}{2}}, $$
$$\begin{aligned}& L(a,b)=\frac{b-a}{\log b-\log a} \quad (a\neq b), \qquad L(a,a)=a, \\& M(a,b)=\frac{b-a}{2\sinh^{-1} (\frac{b-a}{b+a} )}\quad (a\neq b), \qquad M(a,a)=a, \\& V(a,b)=\frac{b-a}{\sqrt{2}\sinh^{-1} (\frac{b-a}{\sqrt{2ab}} )}=V(a,b)\quad (a\neq b), \qquad V(a,a)=a, \end{aligned}$$
The Schwab-Borchardt mean \(\operatorname{SB}(a,b)\) [22, 24, 25] of \(a\geq0\) and \(b>0\) is given by where \(\cosh^{-1}(x)=\log(x+\sqrt{x^{2}-1})\) is the inverse hyperbolic cosine function.
$$ \operatorname{SB}(a,b)=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{\sqrt{b^{2}-a^{2}}}{\arccos(a/b)}, & a< b, \\ a, & a=b, \\ \frac{\sqrt{a^{2}-b^{2}}}{\cosh^{-1}(a/b)}, & a>b, \end{array}\displaystyle \right . $$
Let \(p, q\in\mathbb{R}\) and the function \(t\rightarrow \operatorname{Sh}(p, q, t)\) be defined on \((0, \infty)\) by
$$ \operatorname{Sh}(p, q, t) =\left \{ \textstyle\begin{array}{l@{\quad}l} (\frac{q}{p}\frac{\sinh(pt)}{\sinh(qt)} )^{1/ (p-q)},& pq(p-q)\neq0, \\ (\frac{\sinh(pt)}{pt} )^{1/p}, & p\neq0, q=0, \\ (\frac{\sinh(qt)}{qt} )^{1/q}, & p=0, q\neq0, \\ e^{t\coth(pt)-1/p}, & p=q, pq\neq0, \\ 1, & p=q=0. \end{array}\displaystyle \right . $$
Recently, Yang [23] proved that \(\operatorname{Sh}_{p, q}(b, a)\), defined by is a mean of a and b for all \(b\geq a>0\) if \((p ,q)\in\{(p,q): p>0, q>0, p+q\leq3, L(p, q)\leq1/\log2\} \cup\{(p,q): p<0, 0\leq p+q\leq3\}\cup\{(p,q): q<0, 0\leq p+q\leq3\}\).
$$ \operatorname{Sh}_{p, q}(b, a) =\left \{ \textstyle\begin{array}{l@{\quad}l} a\times \operatorname{Sh} [p, q,\cosh^{-1}(b/a) ], & a< b, \\ a, & a=b, \end{array}\displaystyle \right . $$
In particular, \(\operatorname{Sh}_{1, 0}(b,a)=a\sinh[\cosh^{-1}(b/a)]/\cosh ^{-1}(b/a)=\operatorname{SB}(b,a)\) for all \(b\geq a>0\). Let \(t=\cosh^{-1}(b/a)\), then Theorems 3.2-3.5 lead to Theorems 4.1-4.4.
Theorem 4.1
Let
\(b\geq a>0\)
and
\((p,q)\in\{(p,q): p\geq0\}\cup\{(p,q): 3q\leq p<0\}\). Then the following statements are true:
(i)
If
\(q\in[34/35, \infty)\), then the double inequality
holds if and only if
\(p_{1}\geq3q-8/5\)
and
\(p_{2}\leq q\).
$$\begin{aligned}& \biggl[ \biggl(1-\frac{p_{1}}{3q} \biggr)a^{q}+ \frac{p_{1}}{3q}b^{q} \biggr]^{1/p_{1}}a^{1-q/p_{1}} \\& \quad < \operatorname{SB}(b,a) < \biggl[ \biggl(1-\frac{p_{2}}{3q} \biggr)a^{q}+ \frac{p_{2}}{3q}b^{q} \biggr]^{1/p_{2}}a^{1-q/p_{2}} \end{aligned}$$
(4.1)
(ii)
(iii)
Theorem 4.2
Let
\(b\geq a>0\)
and
\((p,q)\in\{(p,q): p\geq0\}\cup\{(p,q): 3q\leq p<0\}\). Then the following statements are true:
(i)
If
\(p\in[46/35, \infty)\), then the double inequality
holds if and only if
\(q_{1}\leq(5p+8)/15\)
and
\(q_{2}\geq p\).
$$\begin{aligned}& \biggl[ \biggl(1-\frac{p}{3q_{1}} \biggr)a^{q_{1}}+ \frac {p}{3q_{1}}b^{q_{1}} \biggr]^{1/p}a^{1-q_{1}/p} \\& \quad < \operatorname{SB}(b,a) < \biggl[ \biggl(1-\frac{p}{3q_{2}} \biggr)a^{q_{2}}+ \frac {p}{3q_{2}}b^{q_{1}} \biggr]^{1/p}a^{1-q_{2}/p} \end{aligned}$$
(4.2)
(ii)
(iii)
Theorem 4.3
Let
\(b\geq a>0\)
and
\(k\in[0, 3)\). Then the following statements are true:
(i)
If
\(k\in[23/17, 3)\), then the inequality
holds if and only if
\(q\leq8/[5(3-k)]\).
$$ \operatorname{SB}(b,a)> \biggl[ \biggl(1-\frac{k}{3} \biggr)a^{q}+\frac{k}{3}b^{q} \biggr]^{1/(kq)}a^{1-1/k} $$
(ii)
If
\(k\in[0, 1]\), then the double inequality
holds if and only if
\(q_{1}\leq0\)
and
\(q_{2}\geq8/[5(3-k)]\).
$$\begin{aligned}& \biggl[ \biggl(1-\frac{k}{3} \biggr)a^{q_{1}}+ \frac{k}{3}b^{q_{1}} \biggr]^{1/(kq_{1})}a^{1-1/k} \\& \quad < \operatorname{SB}(b,a) < \biggl[ \biggl(1-\frac{k}{3} \biggr)a^{q_{2}}+ \frac{k}{3}b^{q_{2}} \biggr]^{1/(kq_{2})}a^{1-1/k} \end{aligned}$$
Theorem 4.4
Let
\(b\geq a>0\)
and
\(p, q>8/15\). Then the double inequality
holds if and only if
\(p\geq34/35\)
and
\(q\leq4/5\).
$$\begin{aligned}& \biggl[\frac{8}{15p}a^{p}+ \biggl(1-\frac{8}{15p} \biggr)b^{p} \biggr]^{5/(15p-8)}a^{-p/(15p-8)} \\& \quad < \operatorname{SB}(b,a) < \biggl[\frac{8}{15q}a^{q}+ \biggl(1-\frac{8}{15q} \biggr)b^{q} \biggr]^{5/(15q-8)}a^{-q/(15q-8)} \end{aligned}$$
Remark 4.1
Let \(a, b>0\) with \(a\neq b\). Then we clearly see that and Theorems 4.1-4.4 still hold true if we replace \((b, a, \operatorname{SB}(a,b))\) with \((A(a,b), G(a,b), L(a,b))\), \((Q(a,b), A(a,b), M(a,b))\) and \((Q(a,b), G(a,b), V(a,b))\).
$$\begin{aligned}& \operatorname{Sh}_{1, 0}\bigl[A(a,b), G(a,b)\bigr]=\operatorname{SB} \bigl[A(a,b), G(a,b)\bigr]=\frac{b-a}{\log b-\log a}=L(a,b), \\& \operatorname{Sh}_{1, 0}\bigl[Q(a,b), A(a,b)\bigr]=\operatorname{SB} \bigl[Q(a,b), A(a,b)\bigr]=\frac{b-a}{2\sinh^{-1} (\frac{b-a}{b+a} )}=M(a,b), \\& \operatorname{Sh}_{1, 0}\bigl[Q(a,b), G(a,b)\bigr]=\operatorname{SB} \bigl[Q(a,b), G(a,b)\bigr]=\frac{b-a}{\sqrt{2}\sinh ^{-1} (\frac{b-a}{\sqrt{2ab}} )}=V(a,b), \end{aligned}$$
Acknowledgements
The research was supported by the Natural Science Foundation of China under Grant 61374086 and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.
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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.