2022/048) Given $\frac{a^2}{b+c} = \frac{b^2}{c+a} = \frac{c^2}{a+b} = 1$ Find $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}$

 We are given

$\frac{a^2}{b+c} = 1$

or $a^2 = b + c$

or $a^2 + a = a + b + c$

or $a(a+1) = a + b + c$

or $\frac{1}{a+1} = \frac{ a}{a + b+ c}\cdots(1)$

similarly  $\frac{1}{b+1} = \frac{a}{a + b+ c}\cdots(2)$

and  $\frac{1}{c+1} = \frac{a}{a + b+ c}\cdots(3)$

adding all 3 above we get  $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{a+b+c}{a+b+c} = 1$