Solution. Let $1+x^{2}=u, 1-3 x^{2}+36 x^{4}=v, 1-27 x^{5}=w$. Then the inequality takes the form $\log _{u} w+\log _{v} u-\log _{v} w-1 \leqslant 0$. It can be transformed as follows:

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

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

$$
\frac{(w-u)(v-u)}{(u-1)(v-1)} \leqslant 0
$$

The domain of admissible values (DAV) of the original inequality is defined by the conditions $u>0, v>0, w>0, u \neq 1, v \neq 1$. The last two restrictions are automatically satisfied for any solution (??), since at $u=1$ or $v=1$ the denominator of the fraction turns to zero. In addition, $u=1+x^{2}$ and $v=1-3 x^{2}+36 x^{4}$ are positive for all values of $x$. Therefore, the only restriction from the DAV that needs to be taken into account is the inequality $1-27 x^{5}>0$, from which $x<\frac{1}{\sqrt[5]{27}}$. We solve the inequality $(?):$

$$
\begin{aligned}
\frac{\left(1-27 x^{5}-x^{2}-1\right)\left(1-3 x^{2}+36 x^{4}-1-x^{2}\right)}{\left(1+x^{2}-1\right)\left(1-3 x^{2}+36 x^{4}-1\right)} \leqslant 0 & \Leftrightarrow \frac{4 x^{4}\left(-27 x^{3}-1\right)\left(9 x^{2}-1\right)}{3 x^{4}\left(12 x^{2}-1\right)} \leqslant 0 \Leftrightarrow \\
\Leftrightarrow\left\{\begin{array}{l}
x \neq 0, \\
\frac{(3 x+1)\left(9 x^{2}-3 x+1\right)(3 x+1)(3 x-1)}{\left(12 x^{2}-1\right)} \geqslant 0
\end{array}\right. & \Leftrightarrow\left\{\begin{array}{l}
x \neq 0, \\
\frac{(3 x-1)(3 x+1)^{2}}{\left(x-\frac{1}{2 \sqrt{3}}\right)\left(x+\frac{1}{2 \sqrt{3}}\right)} \geqslant 0
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x \neq 0, \\
x \in\left\{-\frac{1}{3}\right\} \bigcup\left(-\frac{1}{2 \sqrt{3}} ; \frac{1}{2 \sqrt{3}}\right) \cup\left[\frac{1}{3} ;+\infty\right) . \\
\text { Taking into account the DAV, we finally get } x \in\left\{-\frac{1}{3}\right\} \bigcup\left(-\frac{1}{2 \sqrt{3}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{3}}\right) \bigcup\left[\frac{1}{3} ; \frac{1}{\sqrt[5]{27}}\right) .
\end{array}\right.
\end{aligned}
$$
5. Solve the inequality $\log _{1+x^{2}}\left(1-27 x^{5}\right)+\log _{1-3 x^{2}+36 x^{4}}\left(1+x^{2}\right) \leqslant 1+\log _{1-3 x^{2}+36 x^{4}}\left(1-27 x^{5}\right)$. Answer: $x \in\left\{-\frac{1}{3}\right\} \bigcup\left(-\frac{1}{2 \sqrt{3}} ; 0\right) \cup\left(0 ; \frac{1}{2 \sqrt{3}}\right) \cup\left[\frac{1}{3} ; \frac{1}{\sqrt[5]{27}}\right)$.