How to solve an inequality with absolute values on both sides? I have the following inequality:
$$|x+3| \geq |x-1| $$
Following this answer I get:
$$
|x+3|=\left\{ \begin{align}
x+3 & \text{   , if }x\geq -3 \\
-x-3 & \text{   , if }x <-3 
\end{align}
\right\}
$$
$$
|x-1|=\left\{ \begin{align}
x-1 & \text{   , if }x\geq 1 \\
-x+1 & \text{   , if }x < 1 
\end{align}
\right\}
$$
Putting those together I get 3 sets of equations:
For $x<-3$:
$$-x-3\ge-x+1$$
For $-3 \le x< 1$:
$$x+3\ge-x+1$$
For $x\ge1$:
$$x+3\ge x-1$$
The first inequality however gives me: 
$$-3 \ge 1$$
What am I doing wrong here? The answer to the problem is $x \ge -1$ by the way.
 A: You haven't done anything wrong. Getting $-3\ge 1$ in the first case means that the first case cannot occur, so there is no solution in the interval $(-\infty, -3)$.
A: As you are no doubt aware, the first statement leads to no solution, the second gives you the solution, and the third, being a statement of fact, does not contradict that, so you do arrive at the correct answer.
As a matter of practicality, however, it would be easier to solve the inequality by solving $$(x+3)^2\geq(x-1)^2$$
A: You didn't do anything wrong.  You just misinterpreted what you were trying to do and what the information tells you:
You have 3 ranges to consider and in each range you have a set of inequalities to interpret.
If $x \le -3$ (which it might or might not be).
We have $-x - 3 \ge -x + 1$ so $-3 \ge 1$.  This is impossible so this is not an option.
So we know $x > -3$.
If $-3 < x \le 1$ (which it might or might not be).
We have $x + 3 \ge -x + 1$ so $x \ge -1$.
So if $-3< x \le 1 \implies x \ge -1 \implies -1 \le x \le 1$.  IF $-3 < x \le -1$ which it might or might not be.
And if $x > 1$ (which it might or might not be)
We have $x + 3 \ge x -1$ so $3 \ge -1$ which is trivially true.
So if $x > 1$ then SOMETHING TRUE.  So $x > 1$ is possible.  But it might or might not be true.  But it's possible and if it is true ... we conclude nothing further.
Putting those together we conclude:
$-1 \le x \le 1$  OR $x > 1$ or 
$x \in [-1,1] \cup [1,\infty) = [-1,\infty)$ or
$x \ge -1$.
.....
But another way, maybe better, to do it is:
$|x+3| \geq |x-1|$
$-|x+3| \le x-1 \le |x+3|$
$\min(x+3,-x-3) \le x-1 \le \max (x+3, -x - 3)$
$\min(x+4, -x -2) \le x \le \max (x+4, -x-2)$
So either $x+4 \le x \le -x-2$ or $-x-2 \le x \le x+4$.
The first one is clearly impossible so
$-x -2 \le x \le x+4$ ($x \le x + 4$ is redundant.)
so $-x-2 \le x$ so $x \ge -1$.
.....
I don't know.  Maybe that is harder.
====
Oh.
David Quinn offers a third option that is probably the easiest to solve, although intuitively I would have assumed it'd give the most extraneous impossible ranges.  As it turns out, it gives the least.
$|x + 3| \ge |x-1|$
$(x+3)^2 \ge (x-1)^2$
$x^2 + 6x + 9 \ge x^2 - 2x + 1$
$6x + 9 \ge -2x + 1$
which will give one and only one unambiguous solution range.
