5. Solve the inequality

$$
\left(\log _{\frac{1}{6} x^{2}+\frac{x}{3}+\frac{19}{27}}\left(1+\frac{x^{2}}{4}\right) \cdot \log _{\frac{1}{6} x^{2}+\frac{x}{3}+\frac{19}{27}}\left(1-\frac{x^{2}}{4}\right)+1\right) \log _{1-\frac{1}{16} x^{4}}\left(\frac{x^{2}}{6}+\frac{x}{3}+\frac{19}{27}\right) \geqslant 1
$$
Answer: $x \in\left[-\frac{4}{3} ; 0\right) \cup\left(0 ; \frac{8}{15}\right] \cup\left(\frac{2}{3} ; \frac{4}{3}\right]$.

Solution. The domain of definition of the logarithms in the inequality is determined by the conditions

$$
\begin{gathered}
1-\frac{x^{2}}{4}>0 \Leftrightarrow-20 \Leftrightarrow x \in \mathbb{R}, \\
1+\frac{x^{2}}{4} \neq 1 \Leftrightarrow x \neq 0, \\
\frac{x^{2}}{6}+\frac{x}{3}+\frac{19}{27}>0 \Leftrightarrow x \in \mathbb{R}, \\
\frac{x^{2}}{6}+\frac{x}{3}+\frac{19}{27} \neq 1 \Leftrightarrow\left\{\begin{array}{l}
x \neq-\frac{8}{3}, \\
x \neq \frac{2}{3} .
\end{array}\right.
\end{gathered}
$$

In the end, we get $x \in(-2 ; 2)$ and $x \neq 0, x \neq \frac{2}{3}$.

Let $\frac{x^{2}}{6}+\frac{x}{3}+\frac{19}{27}=u, 1+\frac{x^{2}}{4}=v, 1-\frac{x^{2}}{4}=w$. We write and transform the inequality:

$$
\begin{aligned}
& \left(\log _{u} v \cdot \log _{u} w+1\right) \log _{v w} u \geqslant 1 \Leftrightarrow \frac{\log _{u} v \cdot \log _{u} w+1}{\log _{u} v w}-1 \geqslant 0 \Leftrightarrow \\
& \Leftrightarrow \frac{\log _{u} v \cdot \log _{u} w+1-\log _{u} v-\log _{u} w}{\log _{u} v w} \geqslant 1 \Leftrightarrow \frac{\left(\log _{u} v-1\right)\left(\log _{u} w-1\right)}{\log _{u} v w} \geqslant 0
\end{aligned}
$$

To solve this inequality, we further apply the method of rationalization: the sign of the difference $\log _{a} b- \log _{a} c$ on the domain of admissible values coincides with the sign of the expression $\frac{b-c}{a-1}$; in particular (when $c=1$), the sign of the logarithm $\log _{a} b$ coincides with the sign of the expression $\frac{b-1}{a-1}$. Then from the last inequality we get $\frac{(u-1)(v-u)(u-1)(w-u)}{(u-1)(v w-1)} \geqslant 0 \Leftrightarrow \frac{(v-u)(w-u)}{(u-1)(v w-1)} \geqslant 0$. Substituting the expressions for $u, v, w$ here and solving the resulting inequality:

$$
\begin{aligned}
& \frac{\left(\frac{x^{2}}{12} \frac{x}{3}+\frac{8}{27}\right)\left(-\frac{5 x^{2}}{12}-\frac{x}{3}+\frac{8}{27}\right)}{\left(\frac{x^{2}}{6}+\frac{x}{3}-\frac{8}{27}\right)\left(-\frac{x^{4}}{16}\right)} \geqslant 0 \Leftrightarrow \frac{\left(\frac{3 x^{2}}{4}-3 x+\frac{8}{3}\right)\left(\frac{15 x^{2}}{4}+3 x-\frac{8}{3}\right)}{x^{2}\left(\frac{3 x^{2}}{2}+3 x-\frac{8}{3}\right)} \geqslant 0 \Leftrightarrow \\
& \Leftrightarrow \frac{\frac{3}{4}\left(x-\frac{8}{3}\right)\left(x-\frac{4}{3}\right) \frac{15}{4}\left(x-\frac{8}{15}\right)\left(x+\frac{4}{3}\right)}{\frac{3 x^{2}}{2}\left(x+\frac{8}{3}\right)\left(x-\frac{2}{3}\right)} \geqslant 0 \Leftrightarrow x \in\left(-\infty ;-\frac{8}{3}\right) \cup\left[-\frac{4}{3} ; 0\right) \cup\left(0 ; \frac{8}{15}\right] \cup\left(\frac{2}{3} ; \frac{4}{3}\right] \cup\left[\frac{8}{3} ;+\infty\right)
\end{aligned}
$$

Considering the domain of definition, we are left with $x \in\left[-\frac{4}{3} ; 0\right) \cup\left(0 ; \frac{8}{15}\right] \cup\left(\frac{2}{3} ; \frac{4}{3}\right]$.