interval $ x\!\in\!\left(0,\nfrac{\pi}{2}\right)$.

**The proof of Theorem 2.2.**

Transforming inequality (\[Ineq\_8\]) we have the following considerations.

Let us prove the left side of the inequality $$\label{Ineq_80}
2+\left(\nfrac{16}{\pi^4}+c(x)\right)x^3 \tan x<\left(\nfrac{\sin⁡ x}{x}\right)^{\!2}+ \nfrac{\tan x}{x},$$ for $x\!\in\! \left(0,\nfrac{\pi}{2}\right)$. The inequality (\[Ineq\_80\]) is equivalent to the mixed trigonometric inequality $$\label{Ineq_81}
\begin{array}{rcl}
f(x)\!\!&\!\!=\!\!&\!\!1-8 x^2 + h_1(x)\cos4 x+h_2(x)\cos2 x+h_3(x)\sin2 x\\ [1.0 ex]
\!\!&\!\!=\!\!&\!\! 1-8 x^2-\cos4 x-8 x^2 \cos2 x +\left(4 x-4 \left(\nfrac{16}{\pi^4}+c(x) \right) x^5 \right)\sin2 x>0,
\end{array}$$ for $x\!\in\! \left(0,\nfrac{\pi}{2} \right) $, and $h_1(x)=-1<0$, $h_2(x)=-8 x^2<0$, $h_3(x)=4 x-4 \left(\nfrac{16}{\pi^4} +c(x) \right) x^5$.

Now let us consider two cases:

[**(C/I)**]{} [$x\!\in\!(0,0.98]$]{} Let us determine sign of the polynomial $h_3(x)$. As we see, that polynomial is the polynomial of $6^{th}$ degree $$\label{Ineq_83}
\begin{array}{rcl}
h_3(x)\!\!&\!\!=\!\!&\!\!P_6(x)\!=\!4x\!-\!4\left(\nfrac{16}{\pi^4}+c(x)\right)x^5\!=\!\left(\nfrac{64}{\pi^5}\!-\!\nfrac{64}{\pi^3}\right)\!x^6+\!\left(\nfrac{\!-\!384}{\pi^4}\!+\!\nfrac{32}{\pi^2}\right)x^5\!+\!4x.
\end{array}$$ Using factorization of the polynomial $P_6(x)$ we have $$\label{Ineq_84}
\begin{array}{rcl}
P_6(x)\!\!&\!\!=\!\!&\!\!\nfrac{4 \left(-2 x+\pi \right)\!\left((8\pi^2-80)x^4+8\pi x^3+4\pi^2 x^2+2\pi^3 x+\pi^4 \right) x }{\pi^5} \\[1.5 ex]
\!\!&\!\!=\!\!&\!\!\nfrac{4 \left(-2 x+\pi \right) x }{\pi^5} P_4(x),
\end{array}$$ where $$\label{Ineq_85}
P_4(x)=(8\pi^2-80)x^4+8\pi x^3+4\pi^2 x^2+2\pi^3 x+\pi^4,$$ for $ x\!\in\!(0,0.98]$. Using MATLAB software we can determine the real numerical factorization of the polynomial $$\label{Ineq_86}
P_4(x)=\alpha(x- x_1)(x- x_2)(x^2+p x+q),$$ with values $\alpha= -1.043\ldots $, $ x_1=-1.524\ldots $, $ x_2=25.663\ldots $, $p=0.046\ldots $, $q=2.387\ldots $. Also, holds that inequality $p^2-4q<0$ is true. The polynomial $P_4(x)$ has 