證：不妨設(shè)x≥y≥z
則x^2≥y^2≥z^2
原式=[(z^2-y^2)+(y^2-x^2)]/(x+y)+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)
=(z^2-y^2)/(x+y)+(y^2-x^2)/(x+y))
+(x^2-y^2)/(y+z)+(y^2-z^2)/(z+x)
=[(x^2-y^2)/(y+z)+(y^2-x^2)/(x+y)]
+[(y^2-z^2)/(z+x)+(z^2-y^2)/(x+y)
=(x^2-y^2)[1/(y+z)-1/(x+y)]+(y^2-z^2)[1/(z+x)-1/(x+y)]
=(x^2-y^2)·(x-z)/(y+z)(x+y)+(y^2-z^2)·(y-z)/(z+x)((x+y)
=[(x-y)(x-z)]/(y+z)+[(y+z)(y-z)^2]/(z+x)(x+y)
∵ x-y≥0 ,x-z≥0
∴原式≥0