If x² = y + z, y² = z + x, z² = x + y, then the value of $$\frac{1}{{x + 1}}$$   + $$\frac{1}{{y + 1}}$$   + $$\frac{1}{{z + 1}} = \,?$$

Solution (By Examveda Team)

$$\eqalign{ & {x^2} = y + z \cr & {y^2} = z + x \cr & {z^2} = x + y \cr & \Rightarrow {x^2} + x = x + y + z \cr & {\text{Adding }}x{\text{ on both sides }} \cr & x\left( {x + 1} \right) = x + y + z \cr & \frac{1}{{\left( {x + 1} \right)}} = \frac{x}{{x + y + z}} \cr & {\text{Similarly,}} \cr & \frac{1}{{\left( {y + 1} \right)}} = \frac{y}{{x + y + z}} \cr & \frac{1}{{\left( {z + 1} \right)}} = \frac{z}{{x + y + z}} \cr & \therefore \frac{1}{{\left( {x + 1} \right)}} + \frac{1}{{\left( {y + 1} \right)}} + \frac{1}{{\left( {z + 1} \right)}} \cr & = \frac{x}{{x + y + z}} + \frac{y}{{x + y + z}} + \frac{z}{{x + y + z}} \cr & = \frac{{x + y + z}}{{x + y + z}} \cr & = 1 \cr} $$