Ta có:
$\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$
$\to (x+y+z)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=1.(x+y+z)$
$\to (x+y+x)(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y})=x+y+z$(*)
Mà ta có:
$(x+y+z)(\dfrac{x}{y+z}+\dfrac{x^2}{y+z}+\dfrac{xy}{y+z}+\dfrac{xz}{y+z}$
$\to = \dfrac{x^2}{y+z}+\dfrac{x(y+z)}{y+z}$(**)
$\to = \dfrac{x^2}{y+z}+x$
Tương tự:
$(x+y+z)\dfrac{y}{z+x}=\dfrac{y^2}{z+x}+y$(***)
$(x+y+x)\dfrac{x}{x+y}=\dfrac{z^2}{x+y}+z$(****)
Từ (*),(**),(***),(****),suy ra:
$\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}=x+y+z$
$\to \dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0$
Mà:
$S=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}$
$\to S=0$
