2019/003) if $a^2=b+c$, $b^2=c+a$, $c^2 = a+b$ find $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}$

We have
$a^2=b+c$
Hence $a^2+a= a+b+c$
Hence $a(a+1) = a + b+ c$
or $\frac{1}{a+1} = \frac{a}{a+b+c}\cdots(1)$
Similarly $\frac{1}{b+1} = \frac{b}{a+b+c}\cdots(2)$
and $\frac{1}{c+1} = \frac{c}{a+b+c}\cdots(3)$

adding these 3 we get

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