Solve the inequality $3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}$ I tried to group the summands so that I could decompose them into multipliers, but nothing worked...
$$3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}\Leftrightarrow 3^{x^2+6x+9}+\frac{1}{9}\leq 3^{x^2-2}+3^{6x+9}$$
$$3^{x^2+6x+7}+1\leq 3^{x^2-4}+3^{6x+7}$$
How to solve it further?

Multiplying both sides by $9=3^2$,
\begin{align*}
3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}&\Longleftrightarrow 3^{(x+3)^2\color{red}{+2}}+1\leq 3^{x^2-2\color{red}{+2}}+3^{3(2x+3)\color{red}{+2}}\\
&\Longleftrightarrow 3^{x^2+6x+11}+1\leq 3^{x^2}+3^{6x+11}\\
&\Longleftrightarrow \left(3^{x^2}-1\right)\left(3^{6x+11}-1\right)\leq 0.
\end{align*}
Since $x^2\geq 0$, we have $3^{x^2}-1\geq 0$, so the inequality is equivalent to
$$3^{x^2}-1=0\ \ \text{ or }\ \ 3^{6x+11}-1\leq 0\Longleftrightarrow x=0 \ \ \text{ or }\ \ 6x+11\leq 0\Longleftrightarrow x=0 \ \ \text{ or }\ \  x\leq-\frac{11}6.$$
So the solutions of the inequality are
$$x\in\{0\}\bigcup \left(-\infty,-\frac{11}6\right].$$
