The absolute value function #|x|# is defined by
#|x| = {(x if x>=0), (-x if x<0):}#
As such, we will consider three cases:
Case 1: #x < -1#
#=> x+1 < 0 and x-1 < 0#
#=>|x+1| = -(x+1) and |x-1| = -(x-1)#
#=> -(x+1) - (x-1) <= 2#
#=> -x-1-x+1 <= 2#
#=> -2x <= 2#
#=> 2x => 2#
#=> x>= 1#
As this contradicts the premise that #x < -1#, there are no solutions on #(-oo, -1)#
Case 2: #-1 <= x <= 1#
#=> x+1 >= 0 and x-1<= 0#
#=> |x+1| = x+1 and |x-1| = -(x-1)#
#=> x+1-(x-1) <= 2#
#=> x+1-x+1 <=2#
#=> 2 <= 2#
As this is true in all cases, every value in #[-1, 1]# is a solution.
Case 3: #x > 1#
#=> x+1 > 0 and x-1 > 0#
#=> |x+1| = x+1 and |x-1| = x-1#
#=> x+1+x-1 <= 2#
#=> 2x <= 2#
#=> x <= 1#
This contradicts the premise that #x>1#, meaning there are no solutions on #(1, oo)#.
Taken together, we have the solution set as #[-1, 1]#. The inequality will hold for any #x# in that interval, and will not for any #x# outside of it.
