Modulus inequality some different concepts

Consider, 

means, we have to find those values for x for which the difference of distance from -1 and 0 is less than 2

By distance method discussed in http://iitmathseasy.blogspot.in/2013/03/modulus-equality-solving-concept-ii.html

"d" denotes distance

from figure, it is clear that if we take any value less than -1 , consider 
 ( because nearer to -1 than 0)

therefore , 
means, for any point less than -1


e.g. take 
  ( put x = -3, |x + 1| )
 and,  ( put x = -3 , |x| )

therefore,  ( difference will always be -1, see figure )
i.e. 

therefore , for all values less than -1 inequality satisfies.

consider , any point between -1 and 0 including -1 and 0

since,  , 
therefore for no points the difference between difference from -1 and 0 will exceed 2.

implies that for all values between and including -1 and 0 inequality satisfies.

i.e. 

 , ( -1 gives -1 and 0 gives 1, )

consider, any point greater than 0,

clearly from figure difference will always be 1


therefore, for all values greater than 0 inequality satisfies.

considering all the statements, inequality satisfies for all Real Numbers.

or 

Second method,


from graph,

there are two curves, 

 and 

and we have to find those values of x for which  is less than  i.e.  will below the line 

consider, 
two critical points -1 and 0
for, 



for, 

 ,

we need two points to draw a line.
put x = -1
y = -1
and for x = 0
y = 1

for, 




plot the graph as mentioned above, 

as cleared from the above graph,  is always less than or below to ,

 therefore for  ,