Yes.
The absolute value function is continuous and so is any combination of absolute value functions. And since such a function is continuous, the only place it can change from ">" to "<" is where it is "=". That is, a very good way to solve inequalities involving absolute values (or any complicated but continuous functions) is to solve the associated equation to separate the real line into intervals. You can then determine which of the intervals give ">" and which "<" by evaluating at one point in each interval.
For example, the inequality |||x - 1| - 1| - 1| >= 4 has the associated equation |||x-1|- 1|- 1|= 4. Removing the "outer" absolute value, we must have ||x-1|- 1|- 1= 4 or ||x-1|-1|- 1= -4.
a) ||x-1|- 1|- 1= 4 is the same as ||x-1|- 1|= 5. Removing the next absolute value, either |x-1|- 1= 5 or |x-1|- 1= -5.
a1) |x- 1|- 1= 5 is the same as |x-1|= 6. Removing the last absolute value, either x- 1= 6 or x- 1= -6.
a1I) x- 1=6 gives x= 7.
a1II) x- 1= -6 gives x= -5.
a2) |x- 1|- 1= -5 is the same as |x- 1|= -4. That's impossible, of course.
b) ||x-1|-1|- 1= -4 is the same as ||x-1|- 1|= -3. That's impossible.
So the only solutions to the equation are x= 7 and x= -5. Those two points divide the real number line into three intervals:
i) x< -5. x=-6 is in that set. |||-6-1|- 1|- 1|= |||-7|- 1|- 1|= ||7-1|- 1|= |6- 1|= 5 which is larger than 4 so the inequality is true for all x less than or equal to -5.
ii) -5< x< 7. x= 0 is in that set. |||0- 1|- 1|- 1|= |||-1|- 1|- 1|= ||1- 1|- 1|= |0 - 1|= -1 which is less than 4 so the inequality is false for all x between -5 and 7.
iii) x> 7. x= 8 is in that set. |||8- 1|- 1|- 1|= ||7- 1|- 1|= |6- 1|= 5 which is larger than 4. The inequality is true for all x larger than or equal to 7.
||x-1|-1|- 1= -4.
This thread has algebraic derivations - and I showed before a solution using distances.
Here's yet another way to do it:
Think of the set of x satisfying |||x-1|-1|-1|>=4 as the complement of the set of x satisfying |||x-1|-1|-1|<4. (Every real x will satisfy exactly one of those two inequalities.)
The point of this is that it's algebraically easier to work with |U| < V than it is to work with |U| > V.
That's because of this (you should know this by heart!): |Z - W| < K if and only if -K < Z - W < K.
Since it will be used twice in what follows, also convince yourself of this trivially true but perhaps odd looking statement:
"Given a and b positive reals, then for any real Y: -a < |Y| < b if and only if |Y| < b."
Now, verify that each line is true if and only if the line above it is true:
|||x-1|-1|-1|<4
-4 < ||x-1|-1|-1 < 4
-3 < ||x-1|-1| < 5
||x-1|-1| < 5
-5 < |x-1|-1 < 5
-4 < |x-1| < 6
|x-1| < 6
-6 < x-1 < 6
-5 < x < 7
Therefore the set of x such that |||x-1|-1|-1|<4 is the set: -5 < x < 7, or, in interval notation (-5, 7).
Therefore the set of x such that |||x-1|-1|-1|>=4 is the complement of that set, which is: x<=-5 or 7<= x. In interval notation (-infinity, -5] Union [7, infinity).