We are given the system of equations:

• $x^2 = y+z$ (1)
• $y^2 = z+x$ (2)
• $z^2 = x+y$ (3)

We need to find the value of the expression $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1}$.

Algebraic Manipulation of Equations

Add $x$ to both sides of equation (1):

$x^2 + x = x + y + z$

Factor the left side:

$x(x+1) = x+y+z$

Similarly, from equations (2) and (3), we get:

$y(y+1) = x+y+z$

$z(z+1) = x+y+z$

Let the sum $S = x+y+z$. Then:

$x(x+1) = S \quad \implies x+1 = \frac{S}{x} \quad (\text{if } x \ne 0)$

$y(y+1) = S \quad \implies y+1 = \frac{S}{y} \quad (\text{if } y \ne 0)$

$z(z+1) = S \quad \implies z+1 = \frac{S}{z} \quad (\text{if } z \ne 0)$

Evaluating the Expression

Consider the case where $S = 0$. This implies $x(x+1)=0, y(y+1)=0, z(z+1)=0$. Thus, $x, y, z$ must be either $0$ or $-1$. The only solution that satisfies the original equations when $S=0$ is $x=y=z=0$. In this case, the expression is:

$\frac{1}{0+1} + \frac{1}{0+1} + \frac{1}{0+1} = 1 + 1 + 1 = 3$

Now, consider the case where $S \ne 0$. This implies $x, y, z$ must be non-zero. We can rewrite the terms in the expression:

$\frac{1}{x+1} = \frac{1}{S/x} = \frac{x}{S}$

$\frac{1}{y+1} = \frac{1}{S/y} = \frac{y}{S}$

$\frac{1}{z+1} = \frac{1}{S/z} = \frac{z}{S}$

Substitute these back into the expression:

$\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \frac{x}{S} + \frac{y}{S} + \frac{z}{S}$

Combine the terms:

$= \frac{x+y+z}{S}$

Since $S = x+y+z$:

$= \frac{S}{S} = 1$

This case is valid for solutions like $x=y=z=2$, where $S=6 \ne 0$.

Given the options, the intended non-trivial solution yields the value 1.