# The range of X in the following expression is . abs(abs(x+1)+1)>=1 ?

Sep 27, 2017

All $x$ or

$\left\{x \in \mathbb{R}\right\}$

#### Explanation:

We don't need to try and remove absolute bars to solve this problem.

Notice in $| | x + 1 | + 1 | \ge 1$ that the value of $| x + 1 | \ge 0$ for any real $x$ since the absolute value is always positive.

So even at the minimum value of $0$

$| | 0 | + 1 | \ge 1$