Solve the following

$a^{2}, b^{2}, c^{2}$ are in A.P.

$\Rightarrow b^{2}-a^{2}=c^{2}-b^{2}$

$\Rightarrow(b+a)(b-a)=(c-b)(c+b)$

$\Rightarrow \frac{b-a}{c+b}=\frac{c-b}{b+a}$

$\Rightarrow \frac{b-a}{(c+a)(c+b)}=\frac{c-b}{(b+a)(c+a)} \quad\left[\right.$ Multiplying both the sides by $\left.\frac{1}{c+a}\right]$

$\Rightarrow \frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$

$\therefore \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in A. P.

Multiplying each term by $(a+b+c)$ :

$\frac{a+b+c}{b+c}, \frac{a+b+c}{c+a}, \frac{a+b+c}{a+b}$ 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.