1 Introduction
The classic “Analytic Inequalities” by Mitrinovic [1] has been hailed all over the world since it was published in 1970. The influence of this book on the various branches of mathematics cannot be overestimated and will last forever. As the author says in the introduction of his book, “The greater part of the results included have been checked, although this could not, of course, be done for all the results which appear in the book. We hope, however, that there are not many errors, but the very nature of this book is such that it seems impossible to expect it to be entirely free of them.” I find some errors in “Analytic Inequalities” and announce the specific contents.
The following conclusions are on p. 241 in [1].
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Proposition 1
([1, Theorem 3.4.20])
For every
\(t>0\),
$$\begin{aligned}& t-\frac{3\sin t}{2+\cos t}< \frac{1}{180}t^{5}, \end{aligned}$$
(1.1)
$$\begin{aligned}& t-\frac{3\sin t}{2+\cos t} \biggl( 1+ \frac{(1-\cos t)^{2}}{9(3+2\cos t)} \biggr) < \frac{t^{7}}{2100}. \end{aligned}$$
(1.2)
It is not difficult to find that these two inequalities above on the common interval \((0,\pi /2)\) are wrong. I read carefully the only one citation [2] by Frame in [1] for Theorem 3.4.20, which was published in the 1944 issue of “The American Mathematical Monthly” in the form of the report of the Mathematical Seminar. We judge that the object of Frame [2] is a right triangle, so the t must be the other two acute angles of a right triangle, that is, \(t\in (0,\pi /2)\). We can only find the related contents of (1.1) in [2] is the item “(7)”, but the one (1.2) at least did not appear in [2].
By using the analytic method, this paper has come to the corresponding conclusions of (1.1) and (1.2); specifically, these are, in the form of (1.1) and (1.2), the first of two inequalities holds for hyperbolic functions, while the second one must be reconstructed, and reversely for circular functions on the interval \((0,\pi )\).
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Theorem 1
Theorem 2
Let
\(x>0\). Then
and
hold, where
\(1/180\)
is the best constant in (1.5).
$$ x-\frac{3\sinh x}{2+\cosh x}< \frac{1}{180}x^{5} $$
(1.5)
$$ x-\frac{3\sinh x}{2+\cosh x} \biggl( 1+\frac{(1-\cosh x)^{2}}{9(3+2 \cosh x)} \biggr) < - \bigl( 9.537610179\cdot 10^{-5} \bigr) x^{7} $$
(1.6)
2 Lemmas
Lemma 1
(Mitrinovic–Adamovic inequality [3])
The inequality
holds for all
\(t\in (0,\pi /2)\), and the exponent 3 is best possible.
$$ \biggl( \frac{\sin t}{t} \biggr) ^{3}>\cos t $$
(2.1)
Lemma 2
(Lazarevic’s inequality [4])
Let
\(t\neq 0\). Then
holds, and the exponent 3 is best possible.
$$ \biggl( \frac{\sinh t}{t} \biggr) ^{3}>\cosh t $$
(2.2)
Lemma 3
([5])
Let
\(\vert x\vert <\pi \), and
\(B_{2n}\)
be the even-indexed Bernoulli numbers (see [6]), we have the following power series expansion:
$$ \frac{1}{\sin x}=\frac{1}{x}+\sum_{n=1}^{\infty } \frac{2^{2n}-2}{ ( 2n ) !}\vert B_{2n} \vert x^{2n-1}.$$
(2.3)
Lemma 4
([7])
Let
\(a_{n}\)
and
\(b_{n}\) (\(n=0,1,2,\ldots\)) be real numbers, and let the power series
\(A(t)=\sum_{n=0}^{\infty }a_{n}t^{n}\)
and
\(B(t)=\sum_{n=0}^{\infty }b_{n}t^{n}\)
be convergent for
\(\vert t\vert < R\) (\(R\leq +\infty \)). If
\(b_{n}>0\)
for
\(n=0,1,2,\ldots\) , and if
\(\varepsilon_{n}=a_{n}/b_{n}\)
is strictly increasing (or decreasing) for
\(n=0,1,2,\ldots\) , then the function
\(A(t)/B(t)\)
is strictly increasing (or decreasing) on
\((0,R)\) (\(R\leq +\infty \)).
Lemma 5
([8])
Let
\(B_{2n}\)
be the even-indexed Bernoulli numbers. Then the double inequality
holds for
\(n=1,2,\ldots \) .
$$ \frac{\pi^{2}(2^{2n+2}-1)}{(2n+2)(2n+1)(2^{2n}-1)}< \frac{\vert B _{2n}\vert }{\vert B_{2n+2}\vert }< \frac{\pi^{2}(2^{2n+1}-1)}{(2n+2)(2n+1)(2^{2n-1}-1)} $$
(2.4)
Lemma 6
Let
\(B_{2n}\)
be the even-indexed Bernoulli numbers. Then the series
is increasing for
\(n\geq 1\).
$$ \{ \delta_{n} \} = \biggl\{ \frac{n(2^{2n+2}-2)\vert B_{2n+2} \vert }{(2n+2)(2^{2n}-2)\vert B_{2n} \vert } \biggr\} $$
(2.5)
Proof
By Lemma 5 we have
In order to prove \(c_{n-1}< c_{n}\) for \(n\geq 2\) it suffices to show
that is,
Let \(2n=m\). Then, for \(m\geq 4\), we show
$$\begin{aligned}& \begin{aligned} \delta_{n-1} &= \frac{(n-1)(2^{2n}-2)\vert B_{2n} \vert }{(2n)(2^{2n-2}-2)\vert B_{2n-2} \vert } \\ &< \frac{(n-1)(2^{2n}-2)}{(2n)(2^{2n-2}-2)}\frac{(2n)(2n-1)(2^{2n-2}-1)}{\pi^{2}(2^{2n}-1)}, \end{aligned} \\& \begin{aligned} \delta_{n} &= \frac{n(2^{2n+2}-2)\vert B_{2n+2} \vert }{(2n+2)(2^{2n}-2)\vert B_{2n} \vert } \\ &>\frac{n(2^{2n+2}-2)}{(2n+2)(2^{2n}-2)}\frac{(2n+2)(2n+1)(2^{2n-1}-1)}{ \pi^{2}(2^{2n+1}-1)}. \end{aligned} \end{aligned}$$
$$\begin{aligned} &\frac{n(2^{2n+2}-2)}{(2n+2)(2^{2n}-2)}\frac{(2n+2)(2n+1)(2^{2n-1}-1)}{ \pi^{2}(2^{2n+1}-1)} \\ &\quad >\frac{(n-1)(2^{2n}-2)}{(2n)(2^{2n-2}-2)}\frac{(2n)(2n-1)(2^{2n-2}-1)}{ \pi^{2}(2^{2n}-1)}, \end{aligned}$$
$$ 2n(2n+1) \bigl(2^{2n}-8\bigr) \bigl(2^{2n}-1\bigr)>(2n-2) (2n-1) \bigl(2^{2n}-4\bigr) \bigl(2^{2n}-2\bigr). $$
(2.6)
$$ m(m+1) \bigl(2^{m}-8\bigr) \bigl(2^{m}-1\bigr)>(m-2) (m-1) \bigl(2^{m}-4\bigr) \bigl(2^{m}-2\bigr). $$
(2.7)
Let
Then
and
for \(m\geq 4\).
$$\begin{aligned}& u_{m} =m(m+1) \bigl(2^{m}-8\bigr) \bigl(2^{m}-1 \bigr)-(m-2) (m-1) \bigl(2^{m}-4\bigr) \bigl(2^{m}-2\bigr) \\& \hphantom{u_{m}}=2 (2m-1)2^{2m}-\bigl(3m^{2}+27m-12 \bigr)2^{m}+32m- 16, \\& v_{m} =2\cdot 2^{m}-\frac{3m^{2}+27m-12}{2m-1}. \end{aligned}$$
$$\begin{aligned} u_{m} &>2 (2m-1)2^{2m}-\bigl(3m^{2}+27m-12 \bigr)2^{m} \\ &= (2m-1)2^{m}v_{m} \end{aligned}$$
(2.8)
$$ v_{m+1}-2v_{m}= 3\cdot \frac{17m^{2}+2m^{3}+m-2}{ ( 2m-1 ) ( 2m+1 ) }>0 $$
(2.9)
Lemma 7
The function
is increasing on
\((0,\pi )\). In particular, we have
$$ K(t)=\frac{t^{3}-\sin^{3}t}{t^{3}\sin^{2}t} $$
(i)
The double inequality
holds for all
\(t\in (0,\pi /2)\), the constants
\((\pi^{3}-8)/\pi^{3}\)
and
\(1/2\)
are best possible.
$$ 1-\frac{\pi^{3}-8}{\pi^{3}}\sin^{2}t< \biggl( \frac{\sin t}{t} \biggr) ^{3}< 1-\frac{1}{2}\sin^{2}t $$
(2.10)
(ii)
The inequality
holds for all
\(t\in (\pi /2,\pi )\), and the constant
\((\pi^{3}-8)/\pi ^{3}\)
is best possible.
$$ \biggl( \frac{\sin t}{t} \biggr) ^{3}< 1-\frac{\pi^{3}-8}{\pi^{3}}\sin ^{2}t $$
(2.11)
Proof
Let
Then by (2.3) we obtain
and
where
Since
we know \(a_{0}/b_{0}< a_{1}/b_{1}\), and \(\{ a_{n}/b_{n} \} _{n\geq 1}\) is increasing by Lemma 6. So \(\{ a_{n}/b_{n} \} _{n\geq 0}\) is increasing, and \(K(t)=A(t)/B(t)\) is increasing on \((0,\pi )\) by Lemma 4. In view of
this completes the proof of Lemma 6. □
$$ K(t)=\frac{t^{3}-\sin^{3}t}{t^{3}\sin^{2}t}=\frac{ ( \frac{t}{ \sin t} ) ^{3}-1}{\frac{t^{3}}{\sin t}}:=\frac{A(t)}{B(t)},\quad 0< t< \pi . $$
$$\begin{aligned} \frac{1}{\sin^{3}t} &=\frac{1}{\sin t}+\frac{1}{2} \biggl( \frac{1}{ \sin t} \biggr) ^{\prime \prime } \\ &=\frac{1}{t}+\sum_{n=1}^{\infty } \frac{2^{2n}-2}{ ( 2n ) !}\vert B_{2n} \vert t^{2n-1} \\ &\quad {}+\frac{1}{t^{3}}+\sum_{n=2}^{\infty } \frac{(2^{2n}-2)(2n-1)(n-1)}{ ( 2n ) !}\vert B_{2n} \vert t^{2n-3} \end{aligned}$$
$$\begin{aligned}& \begin{aligned} A(t) &=t^{3} \Biggl( \frac{1}{t}+\sum _{n=1}^{\infty }\frac{2^{2n}-2}{ ( 2n ) !}\vert B_{2n} \vert t^{2n-1} \Biggr) -1 \\ &\quad {}+t^{3} \Biggl( \frac{1}{t^{3}}+\sum_{n=2}^{\infty } \frac{(2^{2n}-2)(2n-1)(n-1)}{ ( 2n ) !}\vert B_{2n} \vert t^{2n-3} \Biggr) \\ &=t^{2}+\sum_{n=1}^{\infty } \biggl[ \frac{(2^{2n}-2)\vert B_{2n} \vert }{ ( 2n ) !}+ \frac{(2^{2n+2}-2)(2n+1)n\vert B_{2n+2} \vert }{ ( 2n+2 ) !} \biggr] t ^{2n+2} \\ &:=\sum_{n=1}^{\infty }a_{n}t^{2n+2}, \end{aligned} \\& \begin{aligned} B(t) &=t^{3} \Biggl( \frac{1}{t}+\sum _{n=1}^{\infty }\frac{2^{2n}-2}{ ( 2n ) !}\vert B_{2n} \vert t^{2n-1} \Biggr) =t^{2}+\sum _{n=1}^{\infty }\frac{(2^{2n}-2)\vert B_{2n} \vert }{ ( 2n ) !}t^{2n+2} \\ &:=\sum_{n=0}^{\infty }b_{n}t^{2n+2}, \end{aligned} \end{aligned}$$
$$\begin{aligned}& a_{0} = 1,\qquad a_{n}=\frac{(2^{2n}-2)\vert B_{2n} \vert }{ ( 2n ) !}+\frac{(2^{2n+2}-2)(2n+1)n\vert B_{2n+2} \vert }{ ( 2n+2 ) !}, \\& b_{0} = 1,\qquad b_{n}=\frac{(2^{2n}-2)\vert B_{2n} \vert }{ ( 2n ) !}>0,\quad n \geq 1. \end{aligned}$$
$$ \frac{a_{0}}{b_{0}}=1,\qquad \frac{a_{n}}{b_{n}}=1+\frac{n(2^{2n+2}-2)\vert B_{2n+2} \vert }{(2n+2)(2^{2n}-2)\vert B_{2n} \vert }=1+ \delta_{n},\quad n\geq 1, $$
$$ K\bigl(0^{+}\bigr)=\frac{1}{2},\qquad K \biggl( \frac{\pi }{2} \biggr) =\frac{\pi^{3}-8}{ \pi^{3}},\qquad K\bigl(\pi^{-}\bigr)=+\infty, $$
In order to prove (1.6), we need the following lemmas. We introduce a useful auxiliary function \(H_{f,g}\). For \(-\infty \leq a< b\leq \infty \), let f and g be differentiable on \((a,b)\) and \(g^{\prime }\neq 0\) on \((a,b)\). Then the function \(H_{f,g}\) is defined by
The function \(H_{f,g}\) has some good properties and plays an important role in the proof of a monotonicity criterion for the quotient of power series.
$$ H_{f,g}:=\frac{f^{\prime }}{g^{\prime }}g-f. $$
Lemma 8
([9])
Let
\(C ( t ) =\sum_{k=0}^{\infty }c_{k}t^{k}\)
and
\(D ( t ) =\sum_{k=0}^{\infty }d_{k}t^{k}\)
be two real power series converging on
\(( -r,r ) \) (\(r\leq +\infty \)) and
\(d_{k}>0\)
for all k. Suppose that, for certain
\(m\in \mathbb{N} \), the non-constant sequence
\(\{ a_{k}/b_{k} \} \)
is increasing (resp. decreasing) for
\(0\leq k\leq m\)
and decreasing (resp. increasing) for
\(k\geq m\). Then the function
\(C/D\)
is strictly increasing (resp. decreasing) on
\(( 0,r ) \)
if and only if
\(H_{C,D} ( r^{-} ) \geq \) (resp. ≤) 0. Moreover, if
\(H_{C,D} ( r^{-} ) <\) (resp. >) 0, then there exists
\(t_{0}\in ( 0,r ) \)
such that the function
\(C/D\)
is strictly increasing (resp. decreasing) on
\(( 0,t_{0} ) \)
and strictly decreasing (resp. increasing) on
\(( t_{0},r ) \).
Lemma 9
Let
Then the function
\(L(t)\)
has a minimum point
\(t_{0}=2.72078\ldots \) , and
In particular, we see that the double inequality
holds for all
\(t\in (0,+\infty )\), the constant
θ
is best possible.
$$ L(t)=\frac{\sinh^{3}t-t^{3}}{t^{3}\sinh^{2}t},\quad t>0, $$
$$ \frac{\sinh^{3}t-t^{3}}{t^{3}\sinh^{2}t}\geq L(t_{0})=0.35803\ldots :=\theta . $$
$$ \biggl( \frac{\sinh t}{t} \biggr) ^{3}\geq 1+\theta \sinh^{2}t $$
(2.12)
Proof
Let
Then by using the infinite series of sinhx and coshx we obtain
and
where
$$ L(t)=\frac{ ( \frac{\sinh t}{t} ) ^{3}-1}{\sinh^{2}t}=\frac{ \sinh^{3}t-t^{3}}{t^{3}\sinh^{2}t}:=\frac{C(t)}{D(t)},\quad 0< t< +\infty . $$
$$\begin{aligned}& C(t) =\sinh^{3}t-t^{3}=\frac{1}{4} ( \sinh 3t-3 \sinh t ) -t ^{3}=\sum_{n=2}^{\infty }\frac{3^{2n+1}-3}{4 ( 2n+1 ) !}t^{2n+1}:= \sum_{n=2}^{\infty }c_{n}t^{2n+1} \end{aligned}$$
$$\begin{aligned} D(t) =&t^{3}\sinh^{2}t=\frac{1}{2} ( \cosh 2t-1 ) t^{3} \\ =&\sum_{n=2}^{\infty }\frac{2^{2n-3}}{ ( 2n-2 ) !}t^{2n+1}:= \sum_{n=2}^{\infty }d_{n}t^{2n+1}, \end{aligned}$$
$$ c_{n}=\frac{3^{2n+1}-3}{4 ( 2n+1 ) !},\qquad d_{n}=\frac{2^{2n-3}}{ ( 2n-2 ) !}>0,\quad n\geq 2. $$
Setting
we have
and \(\{ \zeta_{n} \} _{n\geq 4}\) is increasing since
for \(n\geq 4\). So
$$ \zeta_{n}:=\frac{c_{n}}{d_{n}}=\frac{3^{2n+1}-3}{ ( 2n+1 ) (2n)(2n-1)2^{2n-1}},\quad n \geq 2, $$
$$ \zeta_{2}=\frac{1}{2}>\zeta_{3}=\frac{13}{40}>\zeta_{4}= \frac{205}{672}, $$
$$ \zeta_{n+1}-\zeta_{n}=\frac{3}{4}\frac{ ( 10n^{2}-29n-12 ) 3^{2n}+6n ^{2}+21n+12}{2^{2n}n ( 2n+3 ) ( 2n+1 ) ( 2n-1 ) ( n+1 ) } >0 $$
$$ \zeta_{2}>\zeta_{3}>\zeta_{4}< \zeta_{5}< \zeta_{6}< \cdots . $$
We compute
and we find that there exists \(t_{0}\in ( 0,+\infty ) \) such that the function \(C/D\) is strictly decreasing on \(( 0,t_{0} ) \) and strictly increasing on \(( t_{0},+\infty ) \) by Lemma 8. Let \(t_{1}=2.72078,t_{2}=2.72079\). We calculate
and see that there exists \(t_{0}\in ( t_{1},t_{2} ) = ( 2.72078,2.72079 ) \subset ( 0,+\infty ) \) such that \(L^{\prime }(t_{0})=0\). So
and
$$ H_{C,D} ( +\infty ) =\lim_{x\rightarrow +\infty } \biggl( \frac{C ^{\prime }}{D^{\prime }}D-C \biggr) =+\infty, $$
$$\begin{aligned}& L^{\prime }(t_{1})=-3.9522\times10^{-7} < 0, \\& L^{\prime }(t_{2})=3.7644\times 10^{-7}>0, \end{aligned}$$
$$ L(t_{0})=\min_{t\in (0,+\infty )}L(t)=\min_{t\in (2.72078,2.72079)}L(t), $$
$$ L(t_{0})\geq L(t_{2})+(t_{0}-t_{2})L^{\prime }(t_{2})= 0.35803\ldots. $$
Obviously, \(L(t)\geq L(t_{0})\) implies (2.12). □
3 The proof of Theorem 1
The proof of the inequality (1.3)
Let
Then
In order to prove \(F^{\prime }(x)>0\) holds for \(x\in (0,\pi )\), it suffices to show
$$ F(x)= x-\frac{3\sin x}{2+\cos x}- \frac{1}{180}x^{5},\quad 0< x< \pi . $$
$$\begin{aligned} F^{\prime }(x) &=-\frac{1}{36 ( \cos x+2 ) ^{2}} \bigl( x^{4} ( \cos x+2 ) ^{2}-36 ( 1-\cos x ) ^{2} \bigr) \\ &=-\frac{x^{2} ( \cos x+2 ) +6 ( 1-\cos x ) }{36 ( \cos x+2 ) ^{2}} \bigl[ x^{2} ( \cos x+2 ) -6 ( 1-\cos x ) \bigr] . \end{aligned}$$
$$ x^{2} ( \cos x+2 ) -6 ( 1-\cos x ) < 0,\quad 0< x< \pi . $$
(3.1)
Since
we have
Let \(x/2=t\). Then (3.3) is equivalent to
In fact, when letting
then
By Lemma 1, we have
which implies
So \(F^{\prime }(x)>0\) holds for \(x\in (0,\pi )\), and
Since
this completes the proof of the inequality (1.3). □
$$ \cos x=-\frac{\tan^{2}\frac{x}{2}-1}{\tan^{2}\frac{x}{2}+1} $$
(3.2)
$$ x^{2} ( \cos x+2 ) -6 ( 1-\cos x ) < 0\quad \Longleftrightarrow \quad 1+ \frac{3}{ ( \tan \frac{x}{2} ) ^{2}}< \frac{3}{ ( \frac{x}{2} ) ^{2}}. $$
(3.3)
$$ 1+\frac{3}{\tan^{2}t}< \frac{3}{t^{2}},\quad 0< t< \frac{\pi }{2}. $$
(3.4)
$$ f(t)=\frac{3}{t^{2}}-1-\frac{3}{\tan^{2}t},\quad 0< t< \frac{\pi }{2}, $$
(3.5)
$$ f\bigl(0^{+}\bigr)=1,f \biggl( \biggl( \frac{\pi }{2} \biggr) ^{-} \biggr) = \frac{12}{\pi^{2}}-1\thickapprox 0.21585>0. $$
(3.6)
$$ f^{\prime }(t)=-\frac{6}{\sin^{3}t} \biggl[ \biggl( \frac{\sin t}{t} \biggr) ^{3}-\cos t \biggr] < 0,\quad 0< t< \frac{\pi }{2}, $$
(3.7)
$$ f(t)\geq \min_{t\in (0,\pi /2)}f(t)=f \biggl( \biggl( \frac{\pi }{2} \biggr) ^{-} \biggr) =\frac{12}{\pi^{2}}-1>0,\quad 0< t< \frac{\pi }{2}. $$
(3.8)
$$ F(x)= x-\frac{3\sin x}{2+\cos x}- \frac{1}{180}x^{5}>F(0)=0,\quad 0< x< \pi . $$
(3.9)
$$ \lim_{x\rightarrow 0^{+}}\frac{x-\frac{3\sin x}{2+\cos x}}{x^{5}}= \frac{1}{180}, $$
The proof of the inequality (1.4)
Let
Then \(G(0^{+})=0\), and
In order to prove \(G^{\prime }(x)>0\) holds for \(x\in (0,\pi )\), it suffices to prove
$$ G(x)=x-\frac{3\sin x}{2+\cos x} \biggl( 1+\frac{(1-\cos x)^{2}}{9(3+2 \cos x)} \biggr) - \frac{x^{7}}{2100},\quad 0< x< \pi . $$
$$ G^{\prime }(x)=-\frac{1}{300 ( 2\cos x+3 ) ^{2}} \bigl[ x^{6} ( 2\cos x+3 ) ^{2}- 200 ( 1-\cos x ) ^{3} \bigr] . $$
(3.10)
$$ x^{6}< 200\frac{ ( 1-\cos x ) ^{3}}{ ( 2\cos x+3 ) ^{2}},\quad 0< x< \pi . $$
(3.11)
Via (3.2) we have
and (3.11) is equivalent to
So (3.12) holds for \(x\in (0,\pi )\) when proving
or
Let \(x/2=t\). Then \(t\in (0,\pi /2)\), and (3.14) is equivalent to
or
In fact, by Lemma 7 we have
for all \(t\in (0,\pi /2)\) due to \(4/5>(\pi^{3}-8)/\pi^{3}=0.74199\ldots\) .
$$\begin{aligned}& 2\cos x+3=\frac{\tan^{2}\frac{x}{2}+5}{\tan^{2}\frac{x}{2}+1},\qquad 1- \cos x= 2 \frac{\tan^{2}\frac{x}{2}}{\tan^{2}\frac{x}{2}+1}, \end{aligned}$$
$$ x^{6}< \biggl( \frac{40\tan^{3}\frac{x}{2}}{(\sec \frac{x}{2})(\sec^{2} \frac{x}{2}+4)} \biggr) ^{2}. $$
(3.12)
$$ x^{3}< \frac{40\tan^{3}\frac{x}{2}}{(\sec \frac{x}{2})(\sec^{2} \frac{x}{2}+4)},\quad 0< x< \pi, $$
(3.13)
$$ \biggl( \frac{x}{2} \biggr) ^{3}< \frac{5\tan^{3}\frac{x}{2}}{(\sec \frac{x}{2})(\sec^{2}\frac{x}{2}+4)},\quad 0< x< \pi . $$
(3.14)
$$ t^{3}< \frac{5\tan^{3}t}{(\sec t)(\sec^{2}t+4)}= \frac{5\sin^{3}t}{4\cos^{2}t+1}, $$
(3.15)
$$ \biggl( \frac{\sin t}{t} \biggr) ^{3}>\frac{4\cos^{2}t+1}{5}=1- \frac{4}{5}\sin^{2}t. $$
(3.16)
$$ \biggl( \frac{\sin t}{t} \biggr) ^{3}>1-\frac{\pi^{3}-8}{\pi^{3}} \sin^{2}t>1- \frac{4}{5}\sin^{2}t $$
(3.17)
Since
this completes the proof of the inequality (1.4). □
$$ \lim_{x\rightarrow 0^{+}}\frac{x-\frac{3\sin x}{2+\cos x} ( 1+\frac{(1- \cos x)^{2}}{9(3+2\cos x)} ) }{x^{7}}= \frac{1}{2100}, $$
So the proof of Theorem 1 is complete.
4 The proof of Theorem 2
The proof of the inequality (1.5)
Let
Then
In order to prove that \(S^{\prime }(x)<0\) holds for \(x\in (0,+\infty )\), it suffices to show
$$ S(x)=x-\frac{3\sinh x}{2+\cosh x}- \frac{1}{180}x^{5},\quad 0< x< + \infty . $$
$$\begin{aligned} S^{\prime }(x)&=-\frac{x^{4} ( \cosh x+2 ) ^{2}-36 ( \cosh x-1 ) ^{2}}{36 ( \cosh x+2 ) ^{2}} \\ &=-\frac{x^{2} ( \cosh x+2 ) +6 ( \cosh x-1 ) }{36 ( \cos x+2 ) ^{2}} \bigl[ x^{2} ( \cosh x+2 ) -6 ( \cosh x-1 ) \bigr]. \end{aligned}$$
$$ x^{2} ( \cosh x+2 ) -6 ( \cosh x-1 ) >0,\quad 0< x< + \infty . $$
(4.1)
Since
we have
Let \(x/2=t\). Then (4.3) is equivalent to
In fact, when letting
we have
By Lemma 2 we can obtain
which implies
So \(S^{\prime }(x)<0\) holds for \(x\in (0,+\infty )\), and
Since
this completes the proof of the inequality (1.5). □
$$ \cosh x=-\frac{\tanh^{2}\frac{x}{2}+1}{\tanh^{2}\frac{x}{2}-1} $$
(4.2)
$$ x^{2} ( \cosh x+2 ) -6 ( \cosh x-1 ) >0\quad \Longleftrightarrow \quad 1+ \frac{3}{ ( \tanh \frac{x}{2} ) ^{2}}>\frac{3}{ ( \frac{x}{2} ) ^{2}}. $$
(4.3)
$$ 1+\frac{3}{\tanh^{2}t}>\frac{3}{t^{2}},\quad 0< t< +\infty . $$
(4.4)
$$ s(t)=\frac{3}{\tanh^{2}t}-1-\frac{3}{t^{2}},\quad 0< t< +\infty, $$
(4.5)
$$ s\bigl(0^{+}\bigr)=1,\qquad s ( +\infty ) =2. $$
(4.6)
$$ s^{\prime }(t)=\frac{6}{\sinh^{3}t} \biggl( \biggl( \frac{\sinh t}{t} \biggr) ^{3}-\cosh t \biggr) >0,\quad 0< t< +\infty, $$
(4.7)
$$ s(t)\geq \min_{t\in (0,+\infty )}s(t)=s\bigl(0^{+} \bigr)=1>0,\quad 0< t< +\infty . $$
(4.8)
$$ S(x)= x-\frac{3\sinh x}{2+\cosh x}- \frac{1}{180}x^{5}< S(0)=0,\quad 0< x< + \infty . $$
(4.9)
$$ \lim_{x\rightarrow 0^{+}}\frac{x-\frac{3\sinh x}{2+\cosh x}}{x^{5}}= \frac{1}{180}, $$
The proof of the inequality (1.6)
Let \(p=9.537610179\cdot 10^{-5}\), and
Then \(H(0^{+})=0\), and
We have
where \(t=x/2>0\). In fact, by (2.12) in Lemma 9 we have
The last inequality holds for \(t\in ( 0,+\infty ) \) due to
$$ H(x)=x-\frac{3\sinh x}{2+\cosh x} \biggl( 1+\frac{(1-\cosh x)^{2}}{9(3+2 \cosh x)} \biggr) +px^{7},\quad 0< x< +\infty . $$
$$ H^{\prime }(x)=\frac{21px^{6} ( 2\cosh x+3 ) ^{2}-2 ( \cosh x-1 ) ^{3}}{3 ( 2\cosh x+3 ) ^{2}}. $$
$$\begin{aligned}& H^{\prime }(x)>0 \quad \Longleftrightarrow \quad p< \frac{2 ( \cosh x-1 ) ^{3}}{21x^{6} ( 2\cosh x+3 ) ^{2}} \\& \hphantom{H^{\prime }(x)>0 }\quad \Longleftrightarrow \quad x^{6}< \frac{998.553028 ( \cosh x-1 ) ^{3}}{ ( 2\cosh x+3 ) ^{2}} \\& \hphantom{H^{\prime }(x)>0 \quad \Longleftrightarrow \quad x^{6}}=\frac{998.553028 ( 2\frac{\tanh^{2}\frac{x}{2}}{1-\tanh^{2}\frac{x}{2}} ) ^{3}}{ ( \frac{5-\tanh^{2}\frac{x}{2}}{1-\tanh^{2}\frac{x}{2}} ) ^{2}} =\frac{7988.424224\cosh^{2}\frac{x}{2}\tanh^{6}\frac{x}{2}}{ ( 5-\tanh^{2} \frac{x}{2} ) ^{2}} \\& \hphantom{H^{\prime }(x)>0 } \quad \Longleftrightarrow \quad x^{3}< 89.37798512 \frac{\cosh \frac{x}{2}\tanh^{3}\frac{x}{2}}{5-\tanh^{2}\frac{x}{2}} = 89.37798512\frac{\sinh^{3}\frac{x}{2}}{5\cosh^{2}x-\sinh^{2}\frac{x}{2}} \\& \hphantom{H^{\prime }(x)>0 }\quad \Longleftrightarrow \quad \biggl( \frac{\sin t}{t} \biggr) ^{3}> \frac{5+4\sinh^{2}t}{11.17224814}, \end{aligned}$$
$$ \biggl( \frac{\sinh t}{t} \biggr) ^{3}>1+0.35803\sinh^{2}t> \frac{5+4\sinh^{2}t}{11.17224814}. $$
$$\begin{aligned} & 11.17224814\bigl(1+0.35803\sinh^{2}t\bigr)-\bigl(5+4 \sinh^{2}t\bigr) \\ &\quad = 1.5642\times 10^{-9}\cosh 2t+ 6.17224814>0. \end{aligned}$$
Therefore \(H^{\prime }(x)>0\), and \(H(x)>H(0^{+})=0\) holds for \(x\in ( 0,+\infty ) \).
So the proof of Theorem 2 is complete. □
5 Remarks
Remark 1
Remark 2
Mortici [14] strengthened (5.1) to
It is in Frame [2] that the following double inequality was also given:
or
In order to compare the three inequalities (1.4), (5.2), and the right hand side of (5.4), we rewrite (1.4) as
$$ \frac{\sin x}{x}< \frac{2+\cos x}{3}-\frac{1}{180}x^{4}+ \frac{1}{3780}x ^{6},\quad 0< x< \frac{\pi }{2}. $$
(5.2)
$$ \frac{2+\cos x-\frac{x^{2}}{\pi^{2}}}{3-\frac{x^{2}}{\pi^{2}}}< \frac{ \sin x}{x}< \frac{2+\cos x-\frac{x^{2}}{10}}{3-\frac{x^{2}}{10}},\quad 0< x< \pi, $$
(5.3)
$$ \frac{2+\cos x}{3}-\frac{x(x-\sin x)}{3\pi^{2}}< \frac{\sin x}{x}< \frac{2+ \cos x}{3}- \frac{x(x-\sin x)}{30},\quad 0< x< \pi . $$
(5.4)
$$ \frac{\sin x}{x}< \frac{1-\frac{x^{6}}{2100}}{1+\frac{(1-\cos x)^{2}}{9(3+2 \cos x)}}\frac{2+\cos x}{3}. $$
(5.5)
(i) We first compare two inequalities (5.5) and (5.2) on the same interval \((0,\pi /2)\). We compute
where
Numerical results show that \(i(x)>0\) for all \(x\in (0, 0.0040)\) and \(i(x)<0\) for all \(x\in (0.0040,\pi /2)\). That is, the upper estimate in (5.5) is smaller than the one in (5.2) on the interval \((0,0.0040)\), meanwhile the upper estimate in (5.2) is smaller than the one in (5.5) on the interval \((0.0040,\pi /2)\). So these two inequalities (1.4) and (5.2) are not included in each other.
$$\begin{aligned}& \biggl( 1+\frac{(1-\cos x)^{2}}{9(3+2\cos x)} \biggr) \biggl( \frac{2+ \cos x}{3}- \frac{1}{180}x^{4}+\frac{1}{3780}x^{6} \biggr) - \biggl( 1-\frac{x ^{6}}{2100} \biggr) \frac{2+\cos x}{3} \\& \quad = \frac{1}{170\text{,}100}\frac{\cos x+2}{2\cos x+3}i(x), \end{aligned}$$
$$ i(x)=-12\text{,}600\cos x-105x^{4}\cos x+59x^{6}\cos x-1470x^{4}+151x^{6}+6300 \cos^{2}x+6300. $$
(ii) Then we compare the two inequalities (5.5) and the right hand side of (5.4) on the same interval \((0,\pi )\). Let us check the function
where
Numerical results show that \(j(x)>0\) for all \(x\in (0, 0.4878)\) and \(i(x)<0\) for all \(x\in (0.4878,\pi )\). That is, the upper estimate in (5.5) is smaller than the one in the right hand side of (5.4) on the interval \((0,0.4878)\), meanwhile the upper estimate in the right hand side of (5.4) is smaller than the one in (5.5) on the interval \((0.4878,\pi )\). So these two inequalities (1.4) and the right hand side of (5.4) are not included in each other.
$$\begin{aligned} \begin{aligned} &\frac{x^{6}}{2100}\frac{2+\cos x}{3}+\frac{(1-\cos x)^{2}}{9(3+2 \cos x)}\frac{2+\cos x}{3}- \frac{x(x-\sin x)}{30} \biggl( 1+\frac{(1- \cos x)^{2}}{9(3+2\cos x)} \biggr) \\ &\quad = \frac{1}{18\text{,}900}\frac{\cos x+2}{2\cos x+3}j(x), \end{aligned} \end{aligned}$$
$$\begin{aligned} j(x) &=-1400\cos x-70x^{2}\cos x+6x^{6}\cos x+980x\sin x-980x^{2}+9x ^{6} \\ &\quad {}+700\cos^{2}x+70x\cos x\sin x+700. \end{aligned}$$
6 Conclusions
In the present study, we find that there are two wrong inequalities for circular functions in the famous monograph “Analytic Inequalities” by Mitrinovic, and we reestablish two inequalities on this topic and create two corresponding inequalities for hyperbolic functions. These new inequalities are the generalization of the famous Cusa–Huygens inequality, one of them is not contained in other improved Cusa–Huygens inequalities showed in [14] and [2] and is stronger than the ones near \(x=0\).
Competing interests
The author declares that he has no competing interests.
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