Proving Bounds on |1+x| when |x-1|<d

[ImAnEngineer]

Homework Statement

Show the following for every d>0:
For every real number x with |x-1|<d it follows that |1+x|<2+d

2. The attempt at a solution
If x-1>0, then |x-1|=x-1<d. Hence x+2 = |x+2| < 2+d.
If x-1<0, then |x-1|=-(x-1)<d. Hence x-1>-d => x+1 > 2-d ...??

Is this really possible to do without putting any other restrictions on d?

 

[CompuChip]

Let me answer your last question first.

|x - 1| < d means, that the distance of x to 1 is no more than d. Now what is the distance of x to -1? Well, if x is "to the left" of 1, then it is definitely closer to -1 than d. For example, if x is at the maximal distance from 1, namely 1 - d, then it has distance 1 - (1 - d) = d to -1.
If x is on the other side of 1, its distance to -1 is equal to two units plus the distance to 1.

If you use the idea that you can measure the distance of x to -1 by going "to 1" first, you hopefully see how the triangle inequality comes in useful.

 

[ImAnEngineer]

Ah of course, the triangle inequality in one dimension :) (not really a triangle)
|a-c|\leq |a-b|+|b-c|

If I choose: a=1, b=-1, c=x

|1-x|\leq|1+1|+|-1-x|=2+|1+x|&lt;d
So surely:
|1+x|&lt;d+2

Thank you!