Không mất tính tổng quát giả sử $y=\min \{x,y,z\}.$
$\frac{x^2-z^2}{y+z}+\frac{z^2-y^2}{x+y}+\frac{y^2-x^2}{z+x}$
$=\frac{x^2-z^2}{y+z}-\frac{x^2-z^2+y^2-x^2}{x+y}+\frac{y^2-x^2}{z+x}$
$=(x^2-z^2)\left ( \frac{1}{y+z}-\frac{1}{x+y} \right )-(y^2-x^2)\left ( \frac{1}{x+y}-\frac{1}{z+x} \right )$
$=(x^2-z^2)\frac{x-z}{(x+y)(y+z)}-(y^2-x^2)\frac{z-y}{(x+y)(x+z)}$
$=(x+z)\frac{(x-z)^2}{(x+y)(y+z)}+(x^2-y^2)\frac{z-y}{(x+y)(x+z)}$
$ \ge 0+0=0.$