Question

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

Answer 1

$a^2,b^2,c^2$ are in A.P.
$\therefore 2b^2=a^2+c^2\Rightarrow b^2-a^2=c^2-b^2\Rightarrow (b+a)(b-a)=(c-b)(c+b)\Rightarrow \frac{b-a}{c+b}=\frac{c-b}{b+a}\Rightarrow \frac{b-a}{(c+a)(c+b)}=\frac{c-b}{(b+a)(c+a)}$
[Multiplying both the sides by $\frac{1}{c+a}$]
$\Rightarrow \frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$
$\therefore \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in A.P.
Thus, $\frac{a}{b+c}+1,\frac{b}{c+a}+1,\frac{c}{a+b}+1$ are in A.P.
Hence, $\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in A.P.