we can determine the real numerical factorization of the polynomial $$\label{Ineq_66}
\begin{split}
Q_6^{''}(\!s)=\alpha(s - s_1)(s - s_2)(s^2+p s+q),
\end{split}$$ with values $\alpha= 8.853\ldots 10^{10}$, $s_1=-3.45\ldots $, $s_2=5.381\ldots $, $p=-9.49\ldots $, $q=53.32\ldots $ Also, holds that inequality $p^2-4q<0$ is true. The polynomial $Q_6^{''}(s)$ has exactly two simple real roots with a symbolic radical representation and the corresponding numerical values: $s_1$, $s_2$. Since we have that $Q_6^{''}(0)<0$ that follows $Q_6^{''}(s)<0$ for $s\!\in\!(0, 2.3409 ] \subset(s_1, s_2)$.

Further, the function $Q_6^{'}(s)$ is monotonically decreasing function for $s\!\in\!(0,2.3409 ]$, $Q_6^{'}(1.53)>0$ and has the first positive root for $s=2.472\ldots $ which follows $Q_6^{'}(s)>0$ for $s\!\in\!(0,2.3409 ]$.

The function $Q_6(s)$ is monotonically increasing for $s\!\in\!(0,2.3409 ]$, has the first positive root $\mbox{\boldmath $b$}=2.358\ldots$ and holds $Q_6(1.53)<0$, which follows: $$\label{Ineq_67}
\begin{array}{rccl}
& Q_6(s)<0 & \mbox {for} & s\!\in\!(0,2.3409 ] \subset(0,\mbox{\boldmath $b$}) \\[1.0 ex]
\Longrightarrow\; & Q_{12}(x)>0 & \mbox {for} & x\!\in\!(0,1.53 ]              \\[1.0 ex]
\Longrightarrow\; & Q_{22}(x)>0 & \mbox {for} & x\!\in\!(0,1.53 ]              \\[1.0 ex]
\Longrightarrow\; & f(x)>0 & \mbox {for} & x\!\in\!(0, 1.53].
\end{array}$$

We can easily calculate the real root $\mbox{\boldmath $b$}$ of the polynomial $Q_6(s)$, and with arbitrary accuracy because of the monotonous increasing of the polynomial function. This also applies to $x^*=\sqrt{\mbox{\boldmath $b$}}=1.53579 