How to solve an inequality with absolute values on both sides?

Question

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.

Answer 1

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 $(-\infty, -3)$.

Answer 2

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 $$(x+3)^2\geq(x-1)^2$$

Answer 3

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.