Question

If a²,b²,c² are in A.P, then prove that the following are in A.P.
1. 1/b+c,1/c+a,1/a+b
2. a/b+c,b/c+a,c/a+b

Answer 1

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.