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?

Hint: You can rewrite the inequality as follows: $(3^{6x+11}-1)(3^{x^2}-1) \le 0$. Can you take it from here?
