If $x,y,z$ are real numbers satisfying$x/(y+z) +y/(z+x) +z/(x+y) =1$ then $x^2/(y+z) +y^2/(z+x)+z^2/(x+y)=$

Question

I have tried it a few times but I am not making any progress. Please help.

Answer 1

$$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$$

$$=\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z-(x+y+z)$$

$$=(x+y+z)\left(\dfrac x{y+z}+\cdots\right)-(x+y+z)$$

$$=?$$

Answer 2

Your first equation simplifies to $$x^3+y^3+z^3+xyz=0$$. The left-hand side of your given term is (factorized) $${\frac { \left( x+y+z \right) \left( {x}^{3}+xyz+{y}^{3}+{z}^{3} \right) }{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }} $$ Using your equation above so we get $$0$$