Determine all possible triplets (a,b,c) of positive integers, with a>b>c, such that:
- a+bc < b+ca < c+ab are three expressions in arithmetic sequence, and:
- a+b-c=22
It was not necessary to specify that (a,b,c) are positive integers, as it follows from the problem statement.
Consider (a+bc) < (c+ab)
Manipulating gives (a-c) < b(a-c)
Because (a-c) > 0, we can divide without reversing the sign.
Dividing by (a-c) gives 1 < b, so b must be positive
Similarly, (a+bc) < (b + ca)
leads to 1 < c
Similarly, (b + ca) < (c + ab)
leads to 1 < a
No negative solutions
