If a, b, c are in AP, prove that b + c, c + a, and a + b are also in AP.

The conventional approach is the simplest.
As a, b, c are in AP, we can say that 2b = a + c.

Now, 
(b + c) + (a + b) 
= 2b + a + c
= a + c + a + c                          (we have just found out that 2b = a + c)
= 2(a + c)

=> (b + c), (c + a) and (a + b) are in AP

One might also think of simpler and smarter solutions. Such as

a, b, c are in AP. If we add b to all these terms, the resulting terms should also b in AP because the terms would still have the same common difference.

=> a + b, 2b, b + c are in AP
=> (a + b), (c + a), (b + c) are in AP.                    (because 2b = a + c)