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

Correct option is D

Given: ​

​$$x^2=y+z, y^2=z+x \text { and } z^2=x+y$$ 

To find: $$\sqrt{\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}}$$ 

Solution: ​

Multiplying and dividing by x , y and z in $$\frac{1}{1+x}, \ \frac{1}{1+y}, \ \frac{1}{1+z}$$ respectively;

​$$= \sqrt{\frac{x}{x+x^2}+\frac{y}{y+y^2}+\frac{z}{z+z^2}}$$ 

Putting value of x² , y² and z² 

​$$= \sqrt{\frac{x}{x+y+z}+\frac{y}{y+x+z}+\frac{z}{z+y+x}} \\ \ \\ = \sqrt{\frac{x+y+z}{x+y+z}} \\ \ \\ = 1$$ 

Alternate Solution: 

By putting value of x, y and z = 2 

​$$=\sqrt{\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}} \\ \ \\ =\sqrt{\frac{1}{1+2}+\frac{1}{1+2}+\frac{1}{1+2}} \\ \ \\ =\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}} \\ \ \\ = \sqrt{\frac{3}{3}} \\ \ \\ = \sqrt{1} = 1$$​​