Question #11f91

1 Answer
Feb 12, 2015

#| 1 - ( |x|/(1 + |x|) )| >= 1/2 for x epsilon [-1, +1]#

First note that #( 1 + |x| )# will always be a positive value greater than (the positive value) #|x|#.
Therefore #( |x|/(1 + |x|) )|# will always be less than #1#
and
#1 - ( |x|/(1 + |x|) )# will always be greater than #0#
(so we can ignore the "outside" absolute value bars).

# 1 - ( |x|/(1 + |x|) ) >= 1/2# can be simplified as

#1/2 >= |x|/(1+|x|)#

#rarr# #1+|x| >= 2|x|#

#rarr# #1 >= |x|#

#rarr# #x# must fall within the closed range #-1# to #+1#