Date: Sep 19, 2011 6:16 AM
Author: Jose Luis S. Blaquera
Subject: Re: Fundamental and trivial question on triangle inequality.

If A>0, B>0, C>0, A+B>C, A+C>B, C+B>A then a triangle can be constructed with sides A, B, C.
1. |A-B|<C<|A+B| Triangle Segment Range (?)
2. C>0 Deduction (The absolute value can only be a non-negative integer)
3. C<A+B Statement 1
4. |A-C|<B<|A+C| Triangle Segment Range (?)
5. B>0 Deduction (The absolute value can only be a non-negative integer)
6. B<A+C Statement 4
7. |B-C|<A<|B+C| Triangle Segment Range (?)
8. A>0 Deduction (The absolute value can only be a non-negative integer)
9. A<B+C Statement 7
The given sides satisfy the triangle segment range because they can be derived from it. Therefore, the given segment conditions can form a triangle.
QED

I think this is not quite clear. Ask me more if you did not understand.