$t= x^2$ we can transform the polynomial $Q_{10}(x)$ into polynomial $$\label{Ineq_31}
\begin{array}{rcl}
Q_5(t)\!\!&\!=\!&\!\!262144 t^5-5203625 t^4+69322260 t^3-665557650 t^2 \\[1.0 ex]
\!\!&\! \!&\!\!+3412527300 t-5237832600,
\end{array}$$ for $t\!\in\!(0,2.4649]$. The first derivate of the polynomial $Q_5(t)$ is the polynomial of $4^{th}$ degree $$\label{Ineq_32}
Q_5^ '(t)\!=\!1312070 t^4\!-\!20814500 t^3\!+\!207966780 t^2\!-\!1331115300 t\!+\!3412527300.$$ Using MATLAB software we can determine the real numerical factorization of the polynomial $$\label{Ineq_33}
Q_5'(t)=\alpha(t^2+p_1 t+q_1)(t^2+p_2 t+q_2),$$ where $\alpha=1310720$, $p_1=-11.655\ldots$, $q_1=34.966\ldots$, $p_2=-4.224\ldots $, $q_2=74.457\ldots\;$. Also, holds that inequalities $p_1^2-4q_1<0$ and $p_2^2-4q_2<0$ are true. The polynomial $Q_5'(t)$ has no real roots. Let us remark that roots and constants $p_1$, $q_1$, $p_2$, $ q_2$ can be represented in symbolic form. The polynomial $Q_5'(t)$ is positive function for $t\!\in\!R$ therefore the polynomial $Q_5(t)$ is monotonically increasing function for $t\!\in\!R$. Further, the function $Q_5(t)$ has real root in $\mbox{\boldmath $a_1$}=2.464993 \ldots > 2.4649$ and $Q_5(0)<0$, so we have that the function $Q_5(t)<0$ for $t\!\in\!(0,\mbox{\boldmath $a_1$})$ which follows that the function $Q_{10}(x)<0$ for $ x\!\in\!(0,1.57]$.

After all we can conclude following: $$\label{Ineq_34}
\begin{array}{rccl}
                & Q_{10}(x)<0 & \mbox {for} & x\!\in\!(0,1.57] \\[1.0 ex]
\Longrightarrow & Q_{20}(x)>0 & \mbox {for} & x\!\in\!(0,1.57] \\[1.0 ex]
\Longrightarrow & f(x)>0       & \mbox {for} & x\!\in\!(0,1.57].
\end{array}$$ Let us remark that we can easily calculate the 