Solve $|x - 1| + |x + 1| < 2$ algebraically.

Question

By drawing a number line and labeling $-1$ to $1$, it becomes apparent that there is no solution because there is no point whose distance from $-1$ and distance from $1$ sum to less than $2$. How can you use algebra to reach the same conclusion?

Answer 1

HINT

The standard method with absolute value is to distinguish the following cases

1) $x\ge1$

$x - 1 + x + 1 < 2\implies x<1$

2) $-1\le x<1$

$-x + 1+ x + 1 < 2\implies 2<2$

3) x<-1

$-x + 1 -x -1< 2\implies x>-1$

Answer 2

By the triangle inequality $$2>|x-1|+|x+1|=|x+1|+|1-x|\geq|x+1+1-x|=2,$$ which says that this inequality has no solutions.

Also, you can consider three cases:

1. $x\geq1$, which gives $x+1+x-1<2$;

2. $-1<x<1$, which gives $2<2$ and

3. $x\leq-1$, which gives $1-x-x-1<2.$

Answer 3

Draw a picture:

$B_1((-1,1),2) \cap \{ (x,x) \}_x = \emptyset$.

The $1$ subscript refers to the $1$-norm, $\| y\|_1 = |y_1|+|y_2|$.

Alternative solution: For a more analytic approach, note that $\phi(x) = |1+x|+|x-1|$ is convex and differentiable for $|x| <1$ and $\phi'(0) = 0$, hence $0$ is a minimiser of $\phi$. Since $\phi(0) = 2$, we see that $\phi(x) \ge 2$ for all $x$.

Answer 4

After squaring we get $$x^2-1+|x^2-1|<0$$ Let $t=x^2-1$, so we have $$t+|t|<0$$

So if $t>0$ we get $2t<0$. A contradiction.

If $t\leq 0$ we get $0<0$. A contradiction.

So given inequality doesn' t have a solution.

Answer 5

You can go to complex numbers, draw two circles with flexible radii centered on +1 and -1 and see that if the sum of those two radiuses is smaller than 2 they will never reach each other. In fact you will by using this method also see that the closest they will get for any sum of radii is along the real axis.