We need to solve the inequality $\frac{4}{|x-1|+1} - \frac{5}{|x-1|-2} \geq 0$.

1. Problem Description

We need to solve the inequality $\frac{4}{|x-1|+1} - \frac{5}{|x-1|-2} \geq 0$.

2. Solution Steps

First, we rewrite the inequality:

$\frac{4}{|x-1|+1} - \frac{5}{|x-1|-2} \geq 0$

$\frac{4(|x-1|-2) - 5(|x-1|+1)}{(|x-1|+1)(|x-1|-2)} \geq 0$

$\frac{4|x-1| - 8 - 5|x-1| - 5}{(|x-1|+1)(|x-1|-2)} \geq 0$

$\frac{-|x-1| - 13}{(|x-1|+1)(|x-1|-2)} \geq 0$

$\frac{|x-1| + 13}{(|x-1|+1)(|x-1|-2)} \leq 0$

Since $|x-1|+13 > 0$ and $|x-1|+1 > 0$ for all $x$, we only need to consider the sign of $|x-1|-2$.

Thus, we have $|x-1|-2 < 0$, which means $|x-1| < 2$.

$-2 < x-1 < 2$

$-2+1 < x < 2+1$

$-1 < x < 3$

However, we need to consider the case where the denominator is zero.

$|x-1|-2 = 0$ implies $|x-1| = 2$.

$x-1 = 2$ or $x-1 = -2$

$x = 3$ or $x = -1$

Since $|x-1|-2$ is in the denominator, we must have $|x-1|-2 \neq 0$, which means $x \neq 3$ and $x \neq -1$.

Therefore, the solution to the inequality is $-1 < x < 3$.

3. Final Answer

$-1 < x < 3$