Here is problem 16 from section 1.1 in the book "Calculus A Rigorous First Course", by Daniel J. Velleman:
\begin{align} \text{Solve} \; 2|x|-3 \ge |x-1| \; \text{by using theorem 1.1.2.} \end{align} Where theorem 1.1.2 states:
For any numbers $x$ and $y$, the following statements are true:
$|x|<y$ iff $-y<x<y$
$|x|\le y$ iff $-y\le x \le y$
$|x| \ge y$ iff either $x\le -y$ or $x\ge y$
$|x| > y$ iff either $x<-y$ or $x > y$
Edit: Here is what I've tried so far:
By the second statement, we get:
\begin{align} -2|x|+4\le x\le2|x|-2 \end{align}
We can split into two inequalities, and arrive at the following;
\begin{align} -\frac{x-4}{2}\ge|x| \; \; \text{and} \; \; \frac{x+2}{2}\le|x| \end{align}
Applying the second statement to the left inequality and the third statement to the other inequality gives us
\begin{align} \frac{x-4}{2}\le x \le \frac{4-x}{2} \; \; \text{and} \; \; x \le -\frac{x+2}{2} \; \text{or} \; x \ge \frac{x+2}{2} \end{align}
By this, our solution set is defined as: $\{ x : \text{either}\; -4 \le x \le \frac{4}{3} \; \text{or} \; x\ge 2 \} $. But, this is not the result I arrived at when reasoning by cases.
Case #1: Consider whenever $x<0$, then $-2x-3 \ge 1-x$, therefore true if $x\le-4$.
Case #2: Consider whenever $0\le x<1$, then $2x-3\ge x-1$, therefore $x\ge \frac{4}{3}$, but since this exceeds our initial condition, the interval $[0,1)$ is not included in our solution set.
Case #3: Consider whenever $x \ge 0$, then $2x-3 \ge x-1$, therefore true if $x \ge 2$.
By this, our solution set is defined as: $\{ x : \text{either}\; x \le -4 \; \text{or} \; x\ge 2 \} $.