Question

If $a^2,b^2,c^2$ are in AP, prove that

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

Answer 1

$\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ will be in AP

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

[on adding 1 to each term]

i.e. if $\frac{a+b+c}{b+c},\frac{a+b+c}{c+a},\frac{a+b+c}{a+b}$ are in AP

i.e., if $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in AP

[on dividing each term by (a + b + c)]

i.e., if $\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$

i.e., if $\frac{(b+c)-(c+a)}{(c+a)(b+c)}=\frac{(c+a)-(a+b)}{(a+b)(c+a)}$

i.e., if $\frac{(b-a)}{(b+c)}=\frac{(c-b)}{(a+b)}$

i.e., if (b - a) (b + a) = (c - b) (c + b)

i.e., if $b^2-a^2=c^2-b^2$

i.e., if $a^2,b^2,c^2$ are in AP.

Hence, $a^2,b^2,c^2$ are in $AP\Rightarrow \frac{a}{(b+c)},\frac{b}{(c+a)},\frac{c}{(a+b)}$ are in AP.