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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.

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