Let's solve the inequality given in the image:

$\log \left( \frac{\sqrt{36x^2 + 12x + 1}}{(6x + 1)^6} \right) - \log_3 \left( 27 \cdot (6x + 1)^2 \right) \geq -6$

Step-by-step solution:

1. Simplify the logarithmic terms:

   • The first part involves the logarithmic expression $\log \left( \frac{\sqrt{36x^2 + 12x + 1}}{(6x + 1)^6} \right)$. Using the logarithmic rule $\log \left( \frac{a}{b} \right) = \log(a) - \log(b)$, we split it: $\log \left( \sqrt{36x^2 + 12x + 1} \right) - \log \left( (6x + 1)^6 \right)$

   • Now, simplify both parts: [ \log \left( \sqrt{36x^2 + 12x + 1} \right) = \frac{1}{