Solve for $x$ in the following inequality: $\frac{4}{x+1} \leq 3 \leq \frac{6}{x+1}$ where $x > 0$.

(N/A) Consider the first part of the inequality: $\frac{4}{x+1} \leq 3$.
Since $x > 0$,$x+1 > 0$,so we can multiply by $(x+1)$ without changing the inequality sign:
$4 \leq 3(x+1)$
$4 \leq 3x + 3$
$1 \leq 3x$
$x \geq \frac{1}{3} \quad (i)$
Consider the second part of the inequality: $3 \leq \frac{6}{x+1}$.
Since $x+1 > 0$,we have:
$3(x+1) \leq 6$
$3x + 3 \leq 6$
$3x \leq 3$
$x \leq 1 \quad (ii)$
Combining $(i)$ and $(ii)$,we get:
$\frac{1}{3} \leq x \leq 1$
Thus,the solution set is $x \in [\frac{1}{3}, 1]$.