In this section we will formulate and prove the refinements of the aforesaid inverse trigonometric inequalities.
Theorem 3.1
For every
\(0<\vert x\vert <1\), the following inequality holds:
$$ \biggl( \frac{\arcsin x}{x} \biggr) ^{2}+ \frac{\arctan x}{x}>\frac{A ( x ) }{3\text{,}600 ( 5x^{2}+7 ) ( 25x^{2}-42 ) ^{2}}>2, $$
(3.1)
where
\(A ( x ) =18\text{,}605x^{10}+84\text{,}847x^{8}+13\text{,}233\text{,}120x^{6}-27\text{,}306\text{,}720x^{4}-42\text{,}336\text{,}000x^{2}+88\text{,}905\text{,}600\).
Proof
We remark that if the above inequality is true for \(x\in ( 0,1 ) \), then it holds clearly for \(x\in ( -1,0 ) \). Therefore, it is sufficient to prove only for \(x\in ( 0,1 ) \).
By using Lemmas 2.1 and 2.2, we have
$$\begin{aligned} \biggl( \frac{\arcsin x}{x} \biggr) ^{2}+\frac{\arctan x}{x} >& \biggl( \frac{61x^{4}+1\text{,}080x^{2}-2\text{,}520}{1\text{,}500x^{2}-2\text{,}520} \biggr) ^{2}+\frac {-4x^{4}+40x^{2}+105}{75x^{2}+105} \\ =&\frac{A ( x ) }{3\text{,}600 ( 5x^{2}+7 ) ( 25x^{2}-42 ) ^{2}}. \end{aligned}$$
Additionally,
$$\begin{aligned}& \frac{A ( x ) }{3\text{,}600 ( 5x^{2}+7 ) ( 25x^{2}-42 ) ^{2}}-2 \\& \quad =\frac{x^{4} ( 18\text{,}605x^{6}+84\text{,}847x^{4} -9\text{,}266\text{,}880x^{2}+16\text{,}793\text{,}280 )}{3\text{,}600 ( 5x^{2}+7 ) ( 25x^{2}-42 )^{2}}>0, \end{aligned}$$
which implies the desired inequality (3.1). The proof of Theorem (3.1) is completed. □
Remark 1
By making use of the Taylor expansion, we have
$$ \biggl( \frac{\arcsin x}{x} \biggr) ^{2}+ \frac{\arctan x}{x}=2+\frac{17}{45}x^{4}-\frac{1}{35}x^{6}+O \bigl(x^{8}\bigr). $$
(3.2)
We note that
$$\begin{aligned}& \frac{A ( x ) }{3\text{,}600 ( 5x^{2}+7 ) ( 25x^{2}-42 ) ^{2}}- \biggl( 2+\frac{17}{45}x^{4}- \frac{1}{35}x^{6} \biggr) \\& \quad = \frac{x^{8} ( 250\text{,}000x^{4}-3\text{,}781\text{,}085x^{2}+6\text{,}074\text{,}481 ) }{2\text{,}800 ( 42-25x^{2} ) ^{2} ( 5x^{2}+7 ) }>0 \end{aligned}$$
for \(0<\vert x\vert <1\). Hence the Padé approximation provides a better inequality than the inequality below
$$ \biggl( \frac{\arcsin x}{x} \biggr) ^{2}+ \frac{\arctan x}{x}> 2+\frac{17}{45}x^{4}-\frac{1}{35}x^{6}. $$
(3.3)
The Taylor’s approximation prompts to consider another method for proving the inequality asserted in Theorem 3.1, that is, we may add more terms to the Taylor polynomial (3.2) in order to prove the inequality (3.1). However, in this way, we will face complicated calculations on the high-order derivatives of \(((\arcsin x)/{x})^{2}+{(\arctan x)}/{x}\) and some cumbersome inequalities resulting from it.
Theorem 3.2
Let
\(0\leq a\leq\frac{1}{2}\). Then for
\(0< x<1\)
we have the following inequality:
$$ \frac{ ( 1+a ) x}{a+\sqrt{1+x^{2}}}< \frac{105x+55x^{3}}{105+90x^{2}+9x^{4}}< \arctan x< \frac{15x+4x^{3}}{15+9x^{2}}< \frac{ ( \frac{\pi }{2} ) x}{a+\sqrt{1+x^{2}}}. $$
(3.4)
Proof
The first inequality can be rewritten as
$$ 105+ ( 90+35a ) x^{2}+9 ( 1+a ) x^{4}< \sqrt {1+x^{2}} \bigl( 105+55x^{2} \bigr) . $$
Since \(0\leq a\leq\frac{1}{2}\), it is easy to find that
$$105+ ( 90+35a ) x^{2}+9 ( 1+a ) x^{4}\leq \frac {210+215x^{2}+27x^{4}}{2}. $$
Thus we need to prove the following inequality:
$$ \frac{210+215x^{2}+27x^{4}}{2}< \sqrt{1+x^{2}} \bigl( 105+55x^{2} \bigr) $$
or, equivalently,
$$\bigl(210+215x^{2}+27x^{4}\bigr)^{2}< 4 \bigl(1+x^{2}\bigr) \bigl( 105+55x^{2} \bigr) ^{2}, $$
which can be deduced from a simple calculation:
$$\begin{aligned} & \bigl(210+215x^{2}+27x^{4}\bigr)^{2}-4 \bigl(1+x^{2}\bigr) \bigl( 105+55x^{2} \bigr) ^{2} \\ &\quad =x^{4} \bigl( 729x^{4}-490x^{2}-735 \bigr) \\ &\quad =x^{4} \bigl[ - \bigl( 729x^{2}+239 \bigr) ( 1-x ) (1+x)-496 \bigr] \\ &\quad < 0. \end{aligned}$$
Therefore, we obtain
$$ \frac{ ( 1+a ) x}{a+\sqrt{1+x^{2}}}< \frac{105x+55x^{3}}{105+90x^{2}+9x^{4}}. $$
(3.5)
The last inequality can be rewritten as
$$ \bigl( 15+4x^{2} \bigr) \sqrt{1+x^{2}}< \biggl( \frac{15\pi}{2}-15a \biggr) + \biggl( \frac{9\pi}{2}-4a \biggr) x^{2}. $$
Since \(0\leq a\leq\frac{1}{2}\), it is easy to observe that
$$\biggl(\frac{15\pi}{2}-15a \biggr) + \biggl( \frac{9\pi}{2}-4a \biggr) x^{2}\geq\frac{15\pi-15}{2}+\frac{9\pi-4}{2}x^{2}. $$
Hence we need to prove the following inequality:
$$\frac{15\pi-15}{2}+\frac{9\pi-4}{2}x^{2}> \bigl( 15+4x^{2} \bigr) \sqrt{1+x^{2}}$$
or, equivalently,
$$\bigl(15\pi-15+(9\pi-4)x^{2}\bigr)^{2}>4 \bigl( 15+4x^{2} \bigr) ^{2}\bigl(1+x^{2}\bigr). $$
Now, from
$$\begin{aligned} & \bigl(15\pi-15+(9\pi-4)x^{2}\bigr)^{2}-4 \bigl( 15+4x^{2} \bigr) ^{2}\bigl(1+x^{2}\bigr) \\ &\quad = \bigl( 1-x^{2} \bigr) \bigl[64x^{4}+ \bigl( 72 \pi-81\pi ^{2}+592 \bigr) \bigl(1+x^{2}\bigr) \bigr] \\ &\qquad {} + \bigl( 270\pi^{2}-390\pi-1\text{,}260 \bigr) x^{2}+ \bigl( 306\pi^{2}-522\pi-1267 \bigr), \end{aligned}$$
together with
$$72\pi-81\pi^{2}+592>0, \qquad 270\pi^{2}-390\pi-1 \text{,}260>0,\qquad 306\pi^{2}-522\pi-1\text{,}267>0, $$
we conclude that
$$\bigl(15\pi-15+(9\pi-4)x^{2}\bigr)^{2}-4 \bigl( 15+4x^{2} \bigr) ^{2}\bigl(1+x^{2}\bigr)>0, $$
which leads to the inequality
$$ \frac{15x+4x^{3}}{15+9x^{2}}< \frac{ ( \frac{\pi}{2} ) x}{a+\sqrt{1+x^{2}}}. $$
(3.6)
In conclusion, the inequality (3.4) follows immediately from inequalities (3.5), (3.6), and the inequality (2.3) given by Lemma 2.3. This completes the proof of Theorem 3.2. □
Finally, we deal with the improved version of the Shafer-Fink inequality (1.5).
Using the well-known trigonometric identity
$$ 2\arctan x=\arcsin\frac{2x}{1+x^{2}},\quad x\in ( 0,1 ), $$
and the double inequality from the Lemma 2.3, we obtain
$$\frac{210x+110x^{3}}{105+90x^{2}+9x^{4}}< \arcsin\frac {2x}{1+x^{2}}< \frac{30x+8x^{3}}{15+9x^{2}}. $$
Substituting the expression \(\frac{2x}{1+x^{2}}\) by y (\(0< y<1\)), we get
$$\begin{aligned}& \frac{210 (\frac{1-\sqrt{1-y^{2}}}{y} )+110 ( \frac{1-\sqrt {1-y^{2}}}{y} ) ^{3}}{105+90 ( \frac{1-\sqrt{1-y^{2}}}{y} )^{2} +9 ( \frac{1-\sqrt{1-y^{2}}}{y} ) ^{4}} \\& \quad < \arcsin y< \frac{30 (\frac{1-\sqrt{1-y^{2}}}{y} )+8 ( \frac{1-\sqrt{1-y^{2}}}{y} ) ^{3}}{15+9 ( \frac{1-\sqrt{1-y^{2}}}{y} ) ^{2}}. \end{aligned}$$
After some elementary calculations, the above inequality can be transformed to the following refinement of Shafer’s inequality for the arcsine function.
Theorem 3.3
For every
\(x\in ( 0,1 ) \), one has
$$ \frac{x ( 80+25\sqrt{1-x^{2}} ) }{57-6x^{2}+48\sqrt{1-x^{2}}}< \arcsin x< \frac{x ( 19+11\sqrt{1-x^{2}} ) }{3 ( 1+\sqrt{1-x^{2}} ) ( 4+\sqrt{1-x^{2}} ) } . $$
(3.7)
Proof
For the sake of completeness, besides the solution above, we will provide another elementary solution in regard to the inequalities (3.7).
Putting \(\arcsin x=t\), \(t\in ( 0,\frac{\pi}{2} ) \) in (3.7) gives
$$ \frac{(\sin t) ( 80+25\cos t ) }{57-6\sin^{2}t+48\cos t}< t< \frac{(\sin t) ( 19+11\cos t ) }{3 ( 1+\cos t ) ( 4+\cos t ) }. $$
(3.8)
The left-hand side inequality of (3.8) can be rewritten as
$$ 108t+96t\cos t+6t\cos2t-160\sin t-25\sin2t>0. $$
Let us consider the function \(u: ( 0,\frac{\pi}{2} ) \rightarrow\mathbb{R} \),
$$u ( t ) =108t+96t\cos t+6t\cos2t-160\sin t-25\sin2t. $$
Its derivatives are
$$\begin{aligned}& u^{\prime} ( t ) =108-64\cos t-96t\sin t-12t\sin2t-44\cos 2t, \\& u^{\prime\prime} ( t ) =-32\sin t+76\sin2t-96t\cos t-24t\cos2t, \end{aligned}$$
and
$$\begin{aligned} u^{ ( 3 ) } ( t ) =&128 ( -\cos t+\cos2t ) +96t\sin t+96t\sin t\cos t \\ =&64 \biggl(\sin\frac{t}{2} \biggr) \biggl( -4\sin \frac{3t}{2}+3t \cos\frac{t}{2}+3t\cos t\cos\frac{t}{2} \biggr) \\ =&64 \biggl(\sin\frac{t}{2} \biggr)v(t). \end{aligned}$$
Using the formula
$$ \sin3\alpha=3\sin\alpha-4\sin^{3}\alpha, $$
the function \(v(t)\) can be further rearranged as
$$ v ( t ) =4\sin\frac{t}{2}-16\sin\frac{t}{2}\cos^{2} \frac {t}{2}+6t\cos^{3}\frac{t}{2}. $$
Then
$$\begin{aligned} v^{\prime} ( t ) =&2\cos\frac{t}{2}-2\cos^{3} \frac{t}{2}+16\cos\frac{t}{2}\sin^{2} \frac{t}{2}-9t\cos^{2}\frac{t}{2}\sin\frac {t}{2} \\ =&9 \biggl(\sin\frac{t}{2}\cos\frac{t}{2} \biggr) \biggl( 2\sin \frac {t}{2}-t\cos\frac{t}{2} \biggr) \\ =&9 \biggl(\sin\frac{t}{2}\cos\frac{t}{2} \biggr)w(t). \end{aligned}$$
Note that
$$w^{\prime} ( t ) =\frac{t}{2}\sin\frac{t}{2}>0,\quad t\in \biggl( 0,\frac{\pi}{2} \biggr). $$
Therefore \(w(t)\) is strictly increasing on \(( 0,\frac{\pi }{2} ) \). As \(w ( 0 ) =0\), it follows that \(w(t)>0\) for \(t \in ( 0,\frac{\pi}{2} ) \). Hence \(v^{\prime } ( t ) >0\) for \(t\in ( 0,\frac{\pi}{2} ) \).
Using similar arguments, we have
$$v(t)>0,\qquad u^{ ( 3 ) }(t)>0,\qquad u^{\prime\prime}(t) >0,\qquad u^{\prime }(t)>0,\qquad u(t)>0 $$
for \(t \in ( 0,\frac{\pi}{2} ) \). The left-hand side inequality of (3.8) is proved.
In order to prove the right-hand side inequality of (3.8), we observe that
$$ \frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}. $$
It is easy to find that the right-hand side inequality of (3.8) is equivalent to the following inequality:
$$ 24t\sin t+3t\sin2t+16\cos t+11\cos2t-27< 0. $$
Define a function \(s: ( 0,\frac{\pi}{2} ) \rightarrow \mathbb{R}\) by
$$s ( t ) =24t\sin t+3t\sin2t+16\cos t+11\cos2t-27. $$
Differentiating \(s(t)\) with respect to t gives
$$\begin{aligned} \begin{aligned} &s^{\prime} ( t ) = 8\sin t+24t\cos t-19\sin2t+6t\cos 2t, \\ &s^{\prime\prime } ( t ) = 32 ( \cos t-\cos 2t ) -24t(\sin t) ( 1+\cos t ) \\ &\hphantom{s^{\prime\prime } ( t )}= 32 \biggl(\sin\frac{t}{2} \biggr) \biggl( 2\sin\frac {3t}{2}-3t \cos^{3}\frac{t}{2} \biggr) \\ &\hphantom{s^{\prime\prime } ( t )} = 32 \biggl(\sin\frac{t}{2} \biggr) \biggl( 6\sin\frac{t}{2}-8 \sin ^{3}t-3t\cos^{3}\frac{t}{2} \biggr) \\ &\hphantom{s^{\prime\prime } ( t )} = 32 \biggl(\sin\frac{t}{2} \biggr)r(t). \end{aligned} \end{aligned}$$
The function \(r ( t ) \) has the derivative
$$\begin{aligned} r^{\prime} ( t ) =&9 \biggl(\sin\frac{t}{2}\cos\frac {t}{2} \biggr) \biggl( \frac{1}{2}t\cos\frac{t}{2}-\sin\frac{t}{2} \biggr) \\ =&9 \biggl(\sin\frac{t}{2}\cos\frac{t}{2} \biggr)p(t). \end{aligned}$$
Also, we have
$$p^{\prime} ( t ) =-\frac{t}{4}\sin\frac{t}{2}< 0, \quad t\in \biggl( 0,\frac{\pi}{2} \biggr). $$
Since \(p ( 0 ) =0\), it follows that \(p(t)<0\) for \(t \in ( 0,\frac{\pi}{2} ) \). Therefore \(r^{\prime}(t)<0\) for \(t\in ( 0,\frac{\pi}{2} ) \).
Using the same arguments, we have
$$r(t)< 0,\qquad s^{\prime\prime}(t)< 0,\qquad s^{\prime}(t)< 0, \qquad s(t)< 0 $$
for \(t \in ( 0,\frac{\pi}{2} )\), the right-hand side inequality of (3.8) is proved. The proof of Theorem 3.3 is completed. □
Remark 2
It is easy to observe that
$$\begin{aligned} & \frac{3x}{2+\sqrt{1-x^{2}}}-\frac{x(80+25\sqrt{1-x^{2}})}{57-6x^{2}+48\sqrt{1-x^{2}}} \\ &\quad =\frac{ 7x ( 2\sqrt{1-x^{2}}+x^{2}-2 ) }{(2+\sqrt{1-x^{2}})(57-6x^{2}+48\sqrt{1-x^{2}})} \\ & \quad =\frac{-7x^{5}}{(2+\sqrt{1-x^{2}})(57-6x^{2}+48\sqrt{1-x^{2}}) ( 2\sqrt{1-x^{2}}+2-x^{2} ) } \\ &\quad < 0. \end{aligned}$$
Thus, the left-hand side inequality of (3.7) is stronger than the left-hand side inequality of (1.5).
Also, we have
$$\begin{aligned} & \frac{x(19+11\sqrt{1-x^{2}})}{3(1+\sqrt{1-x^{2}})(4+\sqrt{1-x^{2}})}-\frac{\pi x}{2+\sqrt{1-x^{2}}} \\ &\quad =\frac{x ( 3\pi x^{2}-11x^{2}+49-15\pi-\sqrt{1-x^{2}} ( 15\pi-41 ) ) }{3(1+\sqrt{1-x^{2}})(4+\sqrt{1-x^{2}})(2+\sqrt{1-x^{2}})} \end{aligned}$$
and
$$\begin{aligned} & \bigl(3\pi x^{2}-11x^{2}+49-15\pi\bigr)^{2}- \bigl(1-x^{2}\bigr) ( 15\pi-41 ) ^{2} \\ &\quad = \bigl( 9\pi^{2}-66\pi+121 \bigr) x^{4}+ \bigl( 135 \pi ^{2}-606\pi+603 \bigr) x^{2}+720-240\pi \\ &\quad =(x-C_{1}) (x+C_{1}) \bigl(x^{2}+C_{2} \bigr), \end{aligned}$$
where
$$\begin{aligned}& C_{1} =\sqrt{ \frac{606\pi-135\pi^{2}-603+ ( 15\pi-41 ) \sqrt{ 81\pi^{2}-246\pi+9}}{ 2 ( 3\pi-11 ) ^{2}}}\approx 0.99876, \\& C_{2} =\frac{606\pi-135\pi^{2}-603- ( 15\pi-41 ) \sqrt { 81\pi^{2}-246\pi+9}}{ 2 ( 3\pi-11 ) ^{2}}\approx13.729. \end{aligned}$$
Hence we conclude that
$$\frac{x(19+11\sqrt{1-x^{2}})}{3(1+\sqrt{1-x^{2}})(4+\sqrt{1-x^{2}})}\leq\frac{\pi x}{2+\sqrt{1-x^{2}}}$$
holds for \(0< x<0.99876\), which implies that the right-hand side inequality of (3.7) is stronger than the right-hand side inequality of (1.5) when \(0< x<0.99876\).