If 1/a, 1/b, 1/c are in AP, prove that (i) b + c / a, c +a / b, a + b / c are in AP (ii) a(b +c), b (c +a), c ( a + b) are in AP

given: 1/a , 1/b and 1/c are in AP

multiplying each term by (a+b+c) will also result as an AP.

(a+b+c) / a , (a+b+c)/b and (a+b+c)/c must form an AP

subtracting 1 from each term is also an AP

therefore

(a+b+c) / a -1 , (a+b+c)/b -1 and (a+b+c)/c -1  is also an AP.

therefore (b+c)/a , (a+c)/b and (a+b)/c is an AP.

given :

1/a , 1/b and 1/c are in AP

multiplying each term by -abc is also an AP

therefore 

-bc , -ca, -ab is also an AP

adding ab+bc+ca to each term, will also an AP

therefore

bc+ca , ab+bc, bc+ca is also an AP

i.e.

c(b+a) , b(a+c) and c(a+b) is also an AP

reverse of the sequence is also an AP

therefore

c(a+b) , b(a+c) and c(a+b) is also an AP