1. The problem is to solve the inequality $$\left|\frac{x+4}{3x+1}\right| \geq 0$$. 2. Recall that the absolute value of any real number or expression is always greater than or equal to zero. That is, for any expression $A$, $$|A| \geq 0$$ is always true. 3. Therefore, $$\left|\frac{x+4}{3x+1}\right| \geq 0$$ holds for all values of $x$ where the expression inside the absolute value is defined. 4. The expression $$\frac{x+4}{3x+1}$$ is undefined when the denominator is zero, so we find where $$3x+1=0$$. 5. Solve for $x$: $$3x+1=0 \implies 3x = -1 \implies x = -\frac{1}{3}$$. 6. So the expression is undefined at $x = -\frac{1}{3}$. 7. Hence, the solution to the inequality is all real numbers except $x = -\frac{1}{3}$. Final answer: $$x \in \mathbb{R} \setminus \left\{-\frac{1}{3}\right\}$$.