{for} & x\!\in\! \left(1.57,\nfrac{\pi}{2} \right).
\end{array}$$ Hence we proved that the function $f(x)$ is positive on interval $ x\!\in\!(0,1.57]$ we conclude that the function $f(x)$ is positive on whole interval $x\!\in\!\left(0, \nfrac{\pi}{2}\right)$.

[**(B)**]{} Let us now prove the right side of inequality. If we write inequality in the following form $$\label{Ineq_51}
\left(\nfrac{\sin x }{ x} \right)^{\!2}+\frac{\tan x}{ x}<2+\left(\frac{8}{45}-b_1(x) \right) x^3 \tan x \quad \mbox {for} \quad x\!\in\! \left(0,\frac{\pi}{2} \right),$$ where $$\label{Ineq_52}
\begin{split}
b_1(x) = \frac{8x^2}{945}-\frac{\mbox{\boldmath $a$} x^4}{14175},\;\mbox{\boldmath $a$} = \frac{480\pi^6-40320\pi^4+3628800}{\pi^8} =17.15041 \ldots \; .
\end{split}$$ The inequality (\[Ineq\_51\]) is equivalent to the mixed trigonometric inequality $$\label{Ineq_53}
\begin{array}{rcl}
f(x)\!\!&\!=\!&\!\! 8x^2\!-\!1+h_1(x)\cos4 x+h_2(x)\cos2 x+h_3(x)\sin2 x \\[1.0 ex]
\!\!&\!=\!&\!\! 8x^2\!-\!1\!+\!\cos4 x\!+\!8x^2\cos2 x\!+\!\left(4\left(\frac{8}{45}\!-\!b_1(x)\right)x^5\!-\!4 x\right)\sin2 x>0,
\end{array}$$ for $x\!\in\!\left(0,\frac{\pi}{2}\right)$, and $h_1(x)=1>0$, $h_2(x)=8 x^2>0$, $h_3(x)=4\left(\frac{8}{45}-b_1(x) \right) x^5-4 x$.

Now let us consider two cases:

[**(B/I)**]{} [$ x\!\in\!(0,1.53]$]{} Let us determine sign of the polynomial $h_3(x)$. As we see, that polynomial is the polynomial of $9^{th}$ degree $$\label{Ineq_55}
\begin{split}
h_3(x)\!=\!P_9(x)=4\left(\nfrac{8}{45}\!-\!b_1(x)\right)x^5\!-\!4x\!=4\nfrac{\mbox{\boldmath $a$}}{14175}x^9\!-\!\nfrac{32}{945}x^7\!+\!\!\nfrac{32}{45}x^5\!-\!4x.
\end{split}$$ Using factorization of the polynomial $P_9(x)$ we have $$\label{Ineq_56}
\begin{array}{rcl}
P_9(x)\!\!&\!=\!&\!\!-\nfrac{4}{945}\nfrac{x}{\pi^8}\left(-2x+\pi \right)\!\left(\pi+2 x \right)\left (\left(8\pi^6-672\pi^4+60480 \right) x^6 \right.\\[2.5 ex]
\!\!&\! \!&\!\! \left. +\left(-168\pi^6+15120\pi^2 \right) x^4+3780\pi^4 x^2+945\pi^6\right) \\[1.0 ex]
\!\!&\!=\!&\!\! -\nfrac{4}{945} \nfrac{ x}{\pi^8} \left(-2 x+\pi \right)\!\left(\pi+2 x \right)P_6(x),
\end{array}$$ where $$\label{Ineq_57}
P_6(x)=(8\pi^6\!-\!672\pi^4\!+\!60480)x^6\!+\!(\!-\!168\pi^6\!+\!15120\pi^2)x^4\!+\!3780\pi^4x^2\!+\!945\pi^6,$$ for $x\!\in\!(0,1.53]$. Introducing the substitution 