An algebra problem by A Former Brilliant Member

$x^2-y^2=(y+z)-(z+x) \Rightarrow (x+y)(x-y)=-(x-y)$

So we have two cases, $x=y$ or $x+y=-1$ .

Case 1: $x=y$ $x^2=x+z \Rightarrow z=(x^2-x)$

$z^2=2x \Rightarrow (x^2-x)^2=2x \Rightarrow x^4-2x^3+x^2-2x=0 \Rightarrow x(x-2)(x^2+1)=0$

If $x$ is real then $x^2+1>0$ so we have $x=y=2$ as $x \neq 0$ .

$\boxed{x=y=z=2}$

Case 2: $x+y=-1$

$z^2=x+y \Rightarrow z^2=-1$

As we are assuming $z$ is real then this gives no real solutions.

Therefore:

$x=y=z=2 \Rightarrow \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\boxed{1}$

Note: I've assumed $x,y,z$ are real because otherwise there would be complex solutions for the answers.

$\dfrac{1}{x+1} = \dfrac{x}{x^{2}+x}=\dfrac{x}{x+y+z}$ ; since $x^{2}=y+z$

Similarly , $\dfrac{1}{y+1} = \dfrac{y}{x+y+z}$ and $\dfrac{1}{z+1} =\dfrac{z}{x+y+z}$

Therefore $\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1} = \dfrac{x+y+z}{x+y+z} = \boxed{\boxed{1}}$