If $$x^2=y+z,y^2=z+x$$​ and $$z^2=x+y$$​, then find the value of $$\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}.$$

Correct option is D

Given:
x² = y + z
y² = z + x
z² = x + y
Solution:
1) x² = y + z
 x² + x = y + z + x 
​$$\Rightarrow \frac{1}{x + 1} = \frac{x}{x + y + z}$$ ............(1)​
2) y² = z + x 
 y² + y = x + z + y = x + y + z
​$$\Rightarrow \frac{1}{y + 1} = \frac{y}{x + y + z}$$ ..........(2)
3) z² = x + y
 z² + z = x + y + z
$$\Rightarrow \frac{1}{z + 1} = \frac{z}{x + y + z}$$..........(3)
Add 1, 2, and 3
​$$\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \frac{x}{x + y + z} + \frac{y}{x + y + z} + \frac{z}{x + y + z}\\ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = \frac{x + y + z}{x + y + z} = 1$$​​