Question

$a^2,b^2,c^2$ are in A.P
Then prove that $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in A.P

Answer 1

Given that $a^2,b^2,c^2$ are in AP

$\Rightarrow a^2+ab+bc+ca,b^{2}ab+bc+ca,$

$c^2+ab+bc+ca$ are in AP

$\Rightarrow (a+b)(a+b),(b+c)(b+a),(c+a)(c+b)$ are in AP

$\Rightarrow \frac{(a+b)(a+c)(b+c)}{b+c},\frac{(b+c)(b+c)(c+a)}{(c+a)},$

$\frac{(c+a)(c+b)(a+b)}{a+b}$ are in AP

$\Rightarrow \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in AP

$\Rightarrow \frac{a+b+c}{b+c},\frac{a+b+c}{c+a},\frac{a+b+c}{a+b}$ are in AP

$\Rightarrow \frac{a+b+c}{b+c}-1,\frac{a+b+c}{c+a}-1,\frac{a+b+c}{a+b}-1$ are in AP

$\Rightarrow \frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}are$ in AP

Hence the proof.