1 Introduction

Masjed-Jamei [6] obtained the following inequality:

$$\begin{aligned} (\arctan x)^2\le \frac{x\ln (x+\sqrt{1+x^2})}{\sqrt{1+x^2}} \end{aligned}$$
(1.1)

for \(|x|<1\). By using Maple software, Masjed-Jamei [6] pointed out that the inequality (1.1) holds for \(x\in {\mathbb {R}}\). Zhu and Malešević [13, Theorem 1.1] proved that the inequality (1.1) holds for all \(x\in {\mathbb {R}}\), and the power number 2 is the best in (1.1). Inequality (1.1) gives the upper bound for the square of the inverse tangent function \(\arctan x\) by the inverse hyperbolic sine function \({{\,\mathrm{arcsinh}\,}}x=\ln (x+\sqrt{1+x^2})\).

Zhu and Malešević [14] obtained a general result on the natural approximation of the function \((\arctan x)^2- (x {{\,\mathrm{arcsinh}\,}}x)/\sqrt{1+x^2}\), and proved a conjecture raised by Zhu and Malešević [13].

Zhu and Malešević [13, Theorem 1.4] showed the analogue for inverse hyperbolic tangent function \({{\,\mathrm{arctanh}\,}}x = \frac{1}{2}\ln \frac{1+x}{1-x}\) and inverse sine function \(\arcsin x\). More precisely, these authors proved that the inequality

$$\begin{aligned} ({{\,\mathrm{arctanh}\,}}x)^2\le \frac{x\arcsin x}{\sqrt{1-x^2}} \end{aligned}$$
(1.2)

holds for all \(x\in (-1, 1)\), and the power number 2 is the best in (1.2).

The first aim of the present paper is to develop (1.1) to produce a sharp double inequality (Theorem 2.1). The second aim of the present paper is to provide a lower bound of \(({{\,\mathrm{arctanh}\,}}x)^2\) (Theorem 2.2).

The numerical values given have been calculated using the computer program MAPLE 11.

2 Results

Theorem 2.1 develops (1.1) to produce a sharp double inequality.

Theorem 2.1

For \(x>0\), we have

$$\begin{aligned} \frac{x{{\,\mathrm{arcsinh}\,}}x}{\sqrt{1+x^2+\alpha x^4}}<(\arctan x)^2<\frac{x{{\,\mathrm{arcsinh}\,}}x}{\sqrt{1+x^2+\beta x^4}}, \end{aligned}$$
(2.1)

with the best possible constants

$$\begin{aligned} \alpha =\frac{2}{45} \quad \text {and}\quad \beta =0. \end{aligned}$$
(2.2)

Proof

Zhu and Malešević [13, Theorem 1.1] have proved that the right-hand side of (2.1) with \(\beta =0\) is valid for \(x>0\).

We now prove that the left-hand side of (2.1) with \(\alpha =\frac{2}{45}\) is valid for \(x>0\), namely,

$$\begin{aligned} \frac{x{{\,\mathrm{arcsinh}\,}}x}{\sqrt{1+x^2+\frac{2}{45} x^4}}<(\arctan x)^2,\quad x>0. \end{aligned}$$
(2.3)

The inequality (2.3) is proved by considering the function F(x) defined, for \(x>0\), by

$$\begin{aligned} F(x)=(\arctan x)^2\frac{\sqrt{1+x^2+\frac{2}{45} x^4}}{x}-{{\,\mathrm{arcsinh}\,}}x. \end{aligned}$$

We consider two cases to prove \(F(x)>0\) for \(x>0\).

Case 1.\(0<x<1\).

From the continued fraction [7, p.122, Eq.(4.25.4)]

$$\begin{aligned} \arctan x=\frac{x}{1+\dfrac{x^2}{3+\dfrac{4x^2}{5+\dfrac{9x^2}{7+\dfrac{16x^2}{9 +\dfrac{25x^2}{11+\ddots }}}}}}, \end{aligned}$$

we find, for \(x>0\),

$$\begin{aligned} \frac{x}{1+\dfrac{x^2}{3+\dfrac{4x^2}{5+\dfrac{9x^2}{7+\dfrac{16x^2}{9 +\dfrac{25x^2}{11}}}}}}<\arctan x<\frac{x}{1+\dfrac{x^2}{3+\dfrac{4x^2}{5+\dfrac{9x^2}{7+\dfrac{16x^2}{9}}}}}, \end{aligned}$$

which can be written for \(x>0\) as

$$\begin{aligned} \frac{7x(165+170x^2+33x^4)}{5(231+315x^2+105x^4+5x^6)}<\arctan x< \frac{x(945+735x^2+64x^4)}{15(63+70x^2+15x^4)}. \end{aligned}$$
(2.4)

From the continued fraction [7, p.129, Eq.(4.39.2)]

$$\begin{aligned} \frac{{{\,\mathrm{arcsinh}\,}}x}{\sqrt{1+x^2}}=\frac{x}{1+\dfrac{1\cdot 2x^2}{3+\dfrac{1\cdot 2x^2}{5+\dfrac{3\cdot 4x^2}{7+\dfrac{3\cdot 4x^2}{9+\ddots }}}}}, \end{aligned}$$

we find, for \(x>0\),

$$\begin{aligned} \frac{x}{1+\dfrac{1\cdot 2x^2}{3+\dfrac{1\cdot 2x^2}{5+\dfrac{3\cdot 4x^2}{7}}}}<\frac{{{\,\mathrm{arcsinh}\,}}x}{\sqrt{1+x^2}}<\frac{x}{1+\dfrac{1\cdot 2x^2}{3+\dfrac{1\cdot 2x^2}{5+\dfrac{3\cdot 4x^2}{7+\dfrac{3\cdot 4x^2}{9}}}}}, \end{aligned}$$

which can be written for \(x>0\) as

$$\begin{aligned} \sqrt{1+x^2}\frac{5x(21+10x^2)}{3(35+40x^2+8x^4)}<{{\,\mathrm{arcsinh}\,}}x<\sqrt{1+x^2}\frac{15x(315+210x^2+8x^4)}{(21+28x^2+8x^4)}. \end{aligned}$$
(2.5)

Using the left-hand side of (2.4) and the right-hand side of (2.5), we have

$$\begin{aligned} F(x)>F_1(x)-F_2(x), \end{aligned}$$

where

$$\begin{aligned} F_1(x)=&\left( \frac{7x(165+170x^2+33x^4)}{5(231+315x^2+105x^4+5x^6)}\right) ^2 \frac{\sqrt{1+x^2+\frac{2}{45} x^4}}{x} \end{aligned}$$

and

$$\begin{aligned} F_2(x)=\sqrt{1+x^2}\frac{15x(315+210x^2+8x^4)}{(21+28x^2+8x^4)}. \end{aligned}$$

For \(0<x<1\), we have

$$\begin{aligned}&\left( F_1(x)\right) ^2-\left( F_2(x)\right) ^2=\frac{x^8F_3(x)}{28125(231 +315x^2+105x^4+5x^6)^4(21+28x^2+8x^4)^2}, \end{aligned}$$

where

$$\begin{aligned} F_3(x)&=74744153426250-687500000x^{26}-5000000x^{28}\\&\quad +x^2(579606935205375 -344451926144788x^{12})\\&\quad +x^4(1824891791790375-226558849074978x^{12})\\&\quad +x^6(3069233024942700 -70320861146025x^{12})\\&\quad +x^8(2949115869590100-11855324999049x^{12}) \\&\quad +x^{10}(1480818426676610 -1034978520912x^{12})\\&\quad +x^{12}(103550407764650-41061562500x^{12}). \end{aligned}$$

Noting that \(F_3(x)>0\) for \(0<x<1\), we obtain, for \(0<x<1\),

$$\begin{aligned} F(x)&>F_1(x)-F_2(x)>0. \end{aligned}$$

Case 2.\(x\ge 1\).

Shafer [9] proved that for \(x>0\),

$$\begin{aligned} \frac{8x}{3+\sqrt{25+\frac{80}{3}x^2}}<\arctan x. \end{aligned}$$
(2.6)

The inequality (2.6) can also be found in [8, 10,11,12]. We have, by (2.6),

$$\begin{aligned} F(x)>\left( \frac{8x}{3+\sqrt{25+\frac{80}{3}x^2}}\right) ^2\frac{\sqrt{1+x^2 +\frac{2}{45} x^4}}{x}-{{\,\mathrm{arcsinh}\,}}x=:F_4(x). \end{aligned}$$

Differentiation yields

$$\begin{aligned} F_4'(x)=F_5(x)-\frac{1}{\sqrt{1+x^2}}, \end{aligned}$$

where

$$\begin{aligned} F_5(x)&=\frac{576\Big ((135+270x^2+18x^4)\sqrt{225+240x^2}+3375+3150x^2+450x^4 +160x^6\Big )}{(9+\sqrt{225+240x^2})^3\sqrt{225+240x^2}\sqrt{225+225x^2+10x^4}}. \end{aligned}$$

For \(x\ge 1\), we have

$$\begin{aligned}&\big (F_5(x)\big )^2-\left( \frac{1}{\sqrt{1+x^2}}\right) ^2=\frac{864F_6(x)}{5(9 +\sqrt{225+240x^2})^6(15+16x^2)(45+45x^2+2x^4)(1+x^2)}, \end{aligned}$$

where

$$\begin{aligned} F_6(x)=F_7(x)+F_8(x)-F_9(x), \end{aligned}$$
(2.7)

and

$$\begin{aligned} F_7(x)&= 23236875-46018125x^2-80566650x^4+80740800x^6\\&{+}107019360x^8+18091776x^{10}+3772416x^{12}+655360x^{14}, \end{aligned}$$
$$\begin{aligned} F_8(x)=(147456x^{12}+2198016x^{10}+3073950x^4+3894075x^2)\sqrt{225+240x^2}, \end{aligned}$$
$$\begin{aligned} F_9(x)=(2371680x^8+6793200x^6+1549125)\sqrt{225+240x^2}. \end{aligned}$$

Write (2.7) as

$$\begin{aligned} F_6(x)=F_7(x)-0.5F_8(x)+1.5F_8(x)-F_9(x), \end{aligned}$$

We find, for \(x\ge 1\),

$$\begin{aligned}&\big (F_7(x)\big )^2-\big (0.5F_8(x)\big )^2=P_{28}(x-1)\quad \text {and}\quad \big (1.5F_8(x)\big )^2-\big (F_9(x)\big )^2=P_{26}(x-1), \end{aligned}$$

where

$$\begin{aligned} P_{28}(x)&=\frac{5402979408824191}{4}+143110860826405993 x+\cdots +429496729600 x^{28} \end{aligned}$$

and

$$\begin{aligned} P_{26}(x)&=\frac{149502812513856165}{4}+297654306451453215 x +\cdots +11741366845440x^{26}. \end{aligned}$$

The polynomials \(P_{28}(x)\) and \(P_{26}(x)\) have all coefficients positive, so \(F_6(x)>0\) for \(x\ge 1\). We then obtain, for \(x\ge 1\),

$$\begin{aligned} F_5(x)>\frac{1}{\sqrt{1+x^2}} \Longrightarrow F_4'(x)>0. \end{aligned}$$

Hence, \(F_4(x)\) is strictly increasing for \(x\ge 1\), and we have, for \(x\ge 1\),

$$\begin{aligned} F(x)>F_4(x)\ge F_4(1)&=\frac{64\sqrt{115}-(455+15\sqrt{465}) \ln (1+\sqrt{2})}{(455+15\sqrt{465})}\\&=0.0002714659399\ldots >0. \end{aligned}$$

If we write (2.1) as

$$\begin{aligned} \beta<\frac{1}{x^4}\left[ \left( \frac{x{{\,\mathrm{arcsinh}\,}}x}{(\arctan x)^2}\right) ^2 -1-x^2\right] <\alpha ,\quad x>0, \end{aligned}$$

we find

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1}{x^4}\left[ \left( \frac{x{{\,\mathrm{arcsinh}\,}}x}{(\arctan x)^2}\right) ^2 -1-x^2\right] =\frac{2}{45} \end{aligned}$$

and

$$\begin{aligned} \lim _{x\rightarrow \infty }\frac{1}{x^4}\left[ \left( \frac{x{{\,\mathrm{arcsinh}\,}}x}{(\arctan x)^2} \right) ^2-1-x^2\right] =0. \end{aligned}$$

Hence, the double inequality (2.1) holds for \(x>0\), with the best possible constants \(\alpha =\frac{2}{45}\) and \( \beta =0\). The proof of Theorem 2.1 is complete.

We provide another proof of (2.3) in the Appendix.

Theorem 2.2 provides a lower bound of \(({{\,\mathrm{arctanh}\,}}x)^2\). \(\square \)

Theorem 2.2

For \(0<x<1\), we have

$$\begin{aligned} \frac{x\arcsin x}{1-\frac{1}{2}x^2}<({{\,\mathrm{arctanh}\,}}x)^2, \end{aligned}$$
(2.8)

and the constant \(\frac{1}{2}\) in the lower bound is the best possible.

Proof

Let \(\arcsin x = t, x \in (0,1)\). Then \(x = \sin t, t \in (0,\pi /2)\). We see that

$$\begin{aligned} {{\,\mathrm{arctanh}\,}}(\sin t)= \frac{1}{2}\ln \frac{1+\sin t}{1-\sin t}= \frac{1}{2}\ln \frac{(1+\sin t)^2}{(1-\sin t)(1+\sin t)}=\ln \frac{1+\sin t}{\cos t}, \end{aligned}$$

and (2.8) is equivalent to

$$\begin{aligned} \frac{t\sin t}{1-\frac{1}{2}\sin ^2t}<\left( \ln \frac{1+\sin t}{\cos t}\right) ^2,\qquad 0<t<\frac{\pi }{2}. \end{aligned}$$
(2.9)

In order to prove (2.9), it suffices to show that for \(0<t<\pi /2\),

$$\begin{aligned} \frac{t\sin t}{1-\frac{1}{2}\sin ^2t}<t^2+\frac{1}{3} t^4+\frac{1}{9} t^6<\left( \ln \frac{1+\sin t}{\cos t}\right) ^2. \end{aligned}$$
(2.10)

The left-hand side of (2.10) can be written for \(0<t<\pi /2\) as

$$\begin{aligned} \frac{1}{t+\frac{1}{3} t^3+\frac{1}{9} t^5}<\csc t-\frac{1}{2}\sin t. \end{aligned}$$

Using the power series expansions for \(\csc t\) and \(\sin t\), we have

$$\begin{aligned}&\csc t-\frac{1}{2}\sin t-\frac{1}{t+\frac{1}{3} t^3+\frac{1}{9} t^5}\nonumber \\&\quad =\frac{1}{t}+\sum _{n=1}^{\infty }\frac{(-1)^{n-1}2(2^{2n-1}-1)B_{2n}}{(2n)!} t^{2n-1}-\frac{1}{2}\sum _{n=0}^{\infty }\frac{(-1)^n}{(2n+1)!}t^{2n+1} -\frac{1}{t+\frac{1}{3} t^3+\frac{1}{9} t^5}\nonumber \\&\quad =\frac{1}{t}-\frac{1}{3}t+\frac{37}{360}t^3-\frac{2}{945}t^5 +\sum _{n=4}^{\infty }\left\{ \frac{2(2^{2n-1}-1)|B_{2n}|}{(2n)!}-\frac{(-1)^{n-1}}{2\cdot (2n-1)!}\right\} t^{2n-1}\nonumber \\&\qquad -\frac{1}{t+\frac{1}{3} t^3+\frac{1}{9} t^5}, \end{aligned}$$
(2.11)

where \(B_n\)\((n\in {\mathbb {N}}_0)\) are the Bernoulli numbers defined by

$$\begin{aligned} \frac{z}{e^z-1}=\sum _{n=0}^{\infty }\,B_n\,\frac{z^n}{n!}, \qquad |z|<2\pi . \end{aligned}$$

By the inequality (see [1, p. 805])

$$\begin{aligned} |B_{2n}|>\frac{2\cdot (2n)!}{(2\pi )^{2n}},\qquad n=1,2,\ldots , \end{aligned}$$

we find, for \(n\ge 4\),

$$\begin{aligned} \frac{2(2^{2n-1}-1)|B_{2n}|}{(2n)!}>\frac{4(2^{2n-1}-1)}{(2\pi )^{2n}} >\frac{1}{2\cdot (2n-1)!}. \end{aligned}$$
(2.12)

By induction on n, the second inequality in (2.12) can be proved (we here omit the proof). We then obtain from (2.11) that, for \(0<t<\pi /2\),

$$\begin{aligned} \csc t-\frac{1}{2}\sin t-\frac{1}{t+\frac{1}{3} t^3+\frac{1}{9} t^5}&>\frac{1}{t}-\frac{1}{3}t+\frac{37}{360}t^3-\frac{2}{945}t^5 -\frac{1}{t+\frac{1}{3} t^3+\frac{1}{9} t^5}\\&=\frac{t^3\Big ((6993-333t^2)+t^4(729-16t^2)\Big )}{7560(9+3t^2+t^4)}>0. \end{aligned}$$

Hence, the left-hand side of (2.10) holds for \(0<t<\pi /2\).

We now prove the right-hand side of (2.10). For \(0<t<\pi /2\), let

$$\begin{aligned} G(t)=\left( \ln \frac{1+\sin t}{\cos t}\right) ^2-\left( t^2+\frac{1}{3} t^4+\frac{1}{9} t^6\right) . \end{aligned}$$

Differentiation yields

$$\begin{aligned} \frac{\cos t}{2}G'(t)&=\ln \frac{1+\sin t}{\cos t}-\left( t+\frac{2}{3}t^3+\frac{1}{3}t^5\right) \cos t=:H(t), \end{aligned}$$

and

$$\begin{aligned} H'(t)&=\frac{(t^5+2t^3+3t)\sin t\cos t+3\sin ^2t-(5t^4+6t^2)\cos ^2t}{3\cos t}, \end{aligned}$$

which can be written as

$$\begin{aligned} \frac{3}{\cos t} H'(t)&=(t^5+2t^3+3t)\tan t+3\tan ^2t-5t^4-6t^2. \end{aligned}$$

Using the power series expansion of \(\tan t\)

$$\begin{aligned} \tan t =\sum _{j=1}^{\infty }\frac{2^{2j}(2^{2j}-1){|B_{2j}|}}{(2j)!}t^{2j-1}=t+\frac{1}{3} t^3+\frac{2}{15} t^5+\ldots , \end{aligned}$$

we obtain, for \(0<t<\pi /2\),

$$\begin{aligned} \frac{3}{\cos t} H'(t)&=(t^5+2t^3+3t)\left( t+\frac{1}{3} t^3 +\frac{2}{15} t^5+\ldots \right) +3\left( t+\frac{1}{3} t^3 +\frac{2}{15} t^5+\ldots \right) ^2\\&\quad -5t^4-6t^2\\&=\frac{16}{5} t^6+\frac{142}{105} t^8+\ldots >0. \end{aligned}$$

Hence, H(t) is strictly increasing for \(0<t<\pi /2\), and we have

$$\begin{aligned} H(t)>H(0)=0 \quad \text {and}\quad G'(t)>0 \quad \text {for}\quad 0<t<\frac{\pi }{2}. \end{aligned}$$

Therefore, G(t) is strictly increasing for \(0<t<\pi /2\), and we have

$$\begin{aligned} G(t)>G(0)=0 \quad \text {for}\quad 0<t<\frac{\pi }{2}. \end{aligned}$$

This means that the right-hand side of (2.10) holds for \(0< t< \pi /2\).

If we write (2.8) as

$$\begin{aligned} \frac{1}{x^2}\left[ 1-\frac{x\arcsin x}{({{\,\mathrm{arctanh}\,}}x)^2}\right] >\frac{1}{2}, \end{aligned}$$

we find

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1}{x^2}\left[ 1-\frac{x\arcsin x}{({{\,\mathrm{arctanh}\,}}x)^2}\right] =\frac{1}{2}. \end{aligned}$$

Hence, the inequality (2.8) holds for \(0<x<1\), and the constant \(\frac{1}{2}\) in the lower bound is the best possible. The proof of Theorem 2.2 is complete. \(\square \)

Conjecture 2.1

For \(0<x<1\), we have

$$\begin{aligned} \frac{x\arcsin x}{\Big (1-\frac{41}{45}x^2\Big )^{\frac{45}{82}}}<({{\,\mathrm{arctanh}\,}}x)^2. \end{aligned}$$
(2.13)

Remark 2.1

The lower bound in (2.13) is better than the lower bound in (2.8). By using Maple software, we find the following approximation formulas near the origin:

$$\begin{aligned}&({{\,\mathrm{arctanh}\,}}x)^2-\frac{x\arcsin x}{\Big (1-\frac{41}{45}x^2\Big )^{\frac{45}{82}}}=O\left( x^8\right) , \end{aligned}$$
(2.14)
$$\begin{aligned}&({{\,\mathrm{arctanh}\,}}x)^2-\frac{x\arcsin x}{1-\frac{1}{2}x^2}=O\left( x^6\right) , \end{aligned}$$
(2.15)
$$\begin{aligned}&({{\,\mathrm{arctanh}\,}}x)^2-\frac{x\arcsin x}{\sqrt{1-x^2}}=O\left( x^6\right) . \end{aligned}$$
(2.16)

This shows that, among approximation formulas (2.14)–(2.16), the formula (2.14) would be the best one.

Appendix: Another proof of (2.3)

By an elementary change of variable \(x = \tan t\, (0< t < \pi /2)\), the inequality (2.3) is equivalent to

$$\begin{aligned} \ln (\tan t+\sec t)<\frac{t^2}{\tan t}\sqrt{\sec ^2t+\frac{2}{45}\tan ^4t},\quad 0<t<\frac{\pi }{2}. \end{aligned}$$

For \(0\le t<\pi /2\), let

$$\begin{aligned} f(t)=\left( \frac{t^2}{\tan t}\sqrt{\sec ^2t+\frac{2}{45}\tan ^4t}\right) ^2-\Big (\ln (\tan t+\sec t)\Big )^2,\quad t\not =0 \end{aligned}$$

and

$$\begin{aligned} f(0)=\lim _{t\rightarrow 0}f(t)=0. \end{aligned}$$

Differentiation yields

$$\begin{aligned} \frac{\cos t}{2} f'(t)&=\frac{t^3}{45\cos ^2t\sin ^3t}\Big (4\sin t\cos t+82\sin t\cos ^3t+4\sin t\cos ^5t\\&\qquad \qquad -4t\cos ^2t-43t\cos ^4t+2t\Big )-\ln \frac{1+\sin t}{\cos t}=:g(t) \end{aligned}$$

and

$$\begin{aligned} g'(t)&=\frac{h(t)}{45(1-\cos t)^2(1+\cos t)^2\cos ^3t}, \end{aligned}$$

where

$$\begin{aligned} h(t)&=-12t^2\cos ^8t+4t^3\cos ^7t\sin t-45\cos ^6t-234t^2\cos ^6t+43t^4\cos ^6t\\&\quad \,\,-266t^3\cos ^5t\sin t+90\cos ^4t+234t^2\cos ^4t+98t^4\cos ^4t-110t^3\cos ^3t\sin t\\&\quad \,\,+12t^2\cos ^2t-10t^4\cos ^2t-45\cos ^2t+12t^3\cos t\sin t+4t^4. \end{aligned}$$

We now use the method from paper [4] to prove \(h(t)>0\) for \(t \in (0,\pi /2)\). Let us start from the function h(t) in the form of multiple angles

$$\begin{aligned} \begin{array}{rcl} h(t) \!&{}\! = \!&{}\! \underbrace{{\left( \displaystyle \frac{2053}{32} t^4 + \displaystyle \frac{129}{16} t^2 + \displaystyle \frac{45}{32} \right) }}_{(>\,0)} \cos 2t + \underbrace{{\left( \displaystyle \frac{325}{16} t^4 + \displaystyle \frac{45}{16} \right) }}_{(>\,0)} \cos 4t - \displaystyle \frac{69}{4} t^2 \cos 4t \\ \!&{}\! \!&{}\! + \underbrace{{\left( \displaystyle \frac{43}{32} t^4 - \displaystyle \frac{129}{16} t^2 - \displaystyle \frac{45}{32} \right) }}_{(<\,0)} \cos 6t - \displaystyle \frac{3}{32} \, t^2 \cos 8t - \displaystyle \frac{501}{8} \, t^3 \sin 2t \\ \!&{}\! \!&{}\! - \displaystyle \frac{745}{16} \, t^3 \sin 4t - \displaystyle \frac{65}{8} \, t^3 \sin 6t + \displaystyle \frac{1}{32} \, t^3 \sin 8t + \displaystyle \frac{787}{16} \, t^4 + \displaystyle \frac{555}{32} \, t^2 - \displaystyle \frac{45}{16}, \end{array} \end{aligned}$$

for \(t \in (0,\pi /2)\). Let us denote with \(T_{m}^{\varphi ,a}(x)\) Taylor development of function \(\varphi (x)\) in the point \(x=a\) of degree m [4]. The following inequality is true

$$\begin{aligned} \begin{array}{rcl} h(t) \!&{}\! \ge \!&{}\! P_{n_1,n_2,n_3,n_4,n_5,n_6,n_7,n_8,n_9}(t) \\ \!&{}\! = \!&{} \left( \displaystyle \frac{2053}{32} t^4 + \displaystyle \frac{129}{16} t^2 + \displaystyle \frac{45}{32} \right) T_{4n_1+2}^{\cos , 0}(2t) + \left( \displaystyle \frac{325}{16} t^4 + \displaystyle \frac{45}{16} \right) T_{4n_2+2}^{\cos , 0}(4t) - \displaystyle \frac{69}{4} t^2 T_{4n_3+0}^{\cos , 0}(4t) \\ \!&{}\! \!&{}\!\!\!\!\! \quad + \left( \displaystyle \frac{43}{32} t^4 - \displaystyle \frac{129}{16} t^2 - \displaystyle \frac{45}{32} \right) T_{4n_4+0}^{\cos , 0}(6t) - \displaystyle \frac{3}{32} \, t^2 T_{4n_5+0}^{\cos , 0}(8t) - \displaystyle \frac{501}{8} \, t^3 T_{4n_6+1}^{\sin , 0}(2t) \\ \!&{}\! \!&{}\!\!\!\! \quad -\, \displaystyle \frac{745}{16} \, t^3 T_{4n_7+1}^{\sin , 0}(4t) - \displaystyle \frac{65}{8} \, t^3 T_{4n_8+1}^{\sin , 0}(6t) + \displaystyle \frac{1}{32} \, t^3 T_{4n_9+3}^{\sin , 0}(8t) + \displaystyle \frac{787}{16} \, t^4 + \displaystyle \frac{555}{32} \, t^2 - \displaystyle \frac{45}{16}, \end{array} \end{aligned}$$

for each \(n_1,n_2,n_3,n_4,n_5,n_6,n_7,n_8,n_9 \in {\mathbb {N}}_0: = {\mathbb {N}} \cup \{0\}\) (\({\mathbb {N}}=\{1,2,\ldots \}\)) and \(t \in (0,\pi /2)\). For the following choice

$$\begin{aligned} n_1\!=\!n_2\!=\!n_3\!=\!n_4\!=\!n_5\!=\!n_6\!=\!n_7\!=\!n_8\!=\!n_9\!=\!6 \end{aligned}$$

we obtain the polynomial with rational coefficients

$$\begin{aligned} Q(t)&=P_{6,6,6,6,6,6,6,6,6}(t) \\&=-\frac{36029974779582599}{5192217564782310562500}t^{30} +\frac{2291243636599537}{32050725708532781250}t^{28}\\&\quad \,\,-\frac{31610419202861}{22610741240587500}t^{26} +\frac{27106191652}{2406129350625}t^{24}-\frac{16899377227}{227949096375}t^{22}\\&\quad \,\, +\frac{81737813}{212837625}t^{20}-\frac{63051259}{42567525}t^{18} + \frac{604216}{155925}t^{16}-\frac{359}{63}t^{14}+\frac{149}{63}t^{12}+\frac{8}{3}t^{10}. \end{aligned}$$

We now prove

$$\begin{aligned} Q(t) > 0 \quad \text {for}\,\, t \in (0,\pi /2). \end{aligned}$$

One proof previously polynomial inequality, which we give here, is based on the Sturm’s algorithm [3, Section 4 in Chapter 6]. Namely, using Sturm’s algorithm it is possible verify that polynomial Q(t) not have zeros in interval (0, b), using for the right bound the rational number \(b = (22/7) / 2 > \pi /2\). Let us remark for example \(Q(1) > 0\) is true and therefore we obtain the following conclusion

$$\begin{aligned} h(t)> Q(t) > 0 \quad \text {for}\,\, t \in (0,\pi /2). \end{aligned}$$

On this way we present one proof of inequality \(h(t) > 0\), for \(t \in (0,\pi /2)\). Let us emphasize that previous proof in all steps is available and via system simthep—simple theprover for automatic proving of inequalities of mixed trigonometric polynomial functions class [2].

We then obtain \(g'(t)>0\) for \(0<t<\pi /2\). Hence, g(t) is strictly increasing for \(0<t<\pi /2\), and we have

$$\begin{aligned} g(t)>\lim _{u\rightarrow 0}g(u)=0 \quad \text {and}\quad f'(t)>0 \quad \text {for}\,\, 0<t<\frac{\pi }{2}. \end{aligned}$$

Therefore, f(t) is strictly increasing for \(0<t<\pi /2\), and we have

$$\begin{aligned} f(t)>f(0)=0 \quad \text {for}\,\, 0<t<\frac{\pi }{2}. \end{aligned}$$

This means that (2.3) holds for \(x>0\).

Remark 2.2

The inequality (1.1) can be traced back by the generalization of the famous Cauchy-Schwarz inequality, which can be found in [5] and the references cited therein.