This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Since a^2,b^2,c^2 are in A.P, the difference between consecutive terms is same :

Divide (a+b)(b+c)(c+a) each side and get

In view of splitting into two partial fractions, rewrite numerators as
That means \dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b} are in A.P.

For part 2, we may start with the result from part 1 :
\dfrac{1}{b+c},\dfrac{1}{c+a},\dfrac{1}{a+b} are in A.P.

Multiplying each term by a+b+c gives another A.P :
\dfrac{a+b+c}{b+c},\dfrac{a+b+c}{c+a},\dfrac{a+b+c}{a+b} are in A.P.
1+\dfrac{a}{b+c},1+\dfrac{b}{c+a},1+\dfrac{c}{a+b} are in A.P.

Subtracting 1 from each term still gives another A.P :
\dfrac{a}{b+c},\dfrac{b}{c+a},\dfrac{c}{a+b} are in A.P.
as desired.

1 5 1