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2015-04-16T10:45:14+05:30

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Since a^2,b^2,c^2 are in A.P, the difference between consecutive terms is same :
b^2-a^2=c^2-b^2\\(b+a)(b-a)=(c+b)(c-b)

Divide (a+b)(b+c)(c+a) each side and get
\dfrac{b-a}{(c+a)(c+b)}=\dfrac{c-b}{(a+b)(c+a)}

In view of splitting into two partial fractions, rewrite numerators as
\dfrac{(b+c)-(c+a)}{(c+a)(c+b)}=\dfrac{(c+a)-(a+b)}{(a+b)(c+a)}
\dfrac{1}{c+a}-\dfrac{1}{b+c}=\dfrac{1}{a+b}-\dfrac{1}{c+a}
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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