The inequality given is:

$\frac{12}{x + 3} \leq 4$

Step-by-Step Solution:

1. Isolate the fraction: We start by eliminating the fraction. Multiply both sides of the inequality by $x + 3$, but be cautious: if $x + 3$ is negative, we need to reverse the inequality sign.

   First, assume $x + 3 > 0$ (so no sign change). Then:

   $12 \leq 4(x + 3)$

2. Simplify the inequality: Expand the right-hand side:

   $12 \leq 4x + 12$

3. Subtract 12 from both sides:

   $0 \leq 4x$

4. Solve for $x$:

   $0 \leq x$

   Or equivalently:

   $x \geq 0$

5. Consider the case where $x + 3 < 0$: If $x + 3$ is negative, multiplying by it would flip the inequality sign. The inequality would become:

   $12 \geq 4(x + 3)$

   Simplifying this:

   $12 \geq 4x + 12$

   Subtracting 12 from both sides:

   $0 \geq 4x$

   Dividing by 4:

   $0 \geq x$

   Or:

   $x \leq 0$

6. Combine both cases: From both cases, we get that $x \geq 0$ from the first case and $x \leq 0$ from the second. Therefore, the solution is:

   $x = 0$

Final Answer in Interval Notation:

$\{ 0 \}$

Let me know if you need any further clarification or a detailed explanation of any of the steps!

Would you like more details or have any questions? Here are some related questions:

1. What if the inequality was strict ($<$) instead of $\leq$?
2. How do we handle inequalities with absolute values?
3. What happens if the inequality involves a quadratic expression?
4. Can this method be applied to systems of inequalities?
5. How would the solution change if we had $\frac{12}{x+3} \geq 4$?

Tip: Always check for cases when multiplying or dividing inequalities by variables, as the inequality sign may change depending on whether the variable is positive or negative.