己知x,y,z>0,x+y+z=1.求证x^3/(x^2+y^2)+y^3/(y^2+z^2)+z^3/(z^2+x^2)>=1/2

Question

己知x,y,z>0,x+y+z=1.求证
x^3/(x^2+y^2)+y^3/(y^2+z^2)+z^3/(z^2+x^2)>=1/2

Answer 1

x^3=(x^2+y^2-y^2)x=x(x^2+y^2)-xy^2,同理
y^3=y(y^2+z^2)-yz^2,
z^3=z(z^2+x^2)-zx^2,
所以x^3/(x^2+y^2)+y^3/(y^2+z^2)+z^3/(z^2+x^2)
=x-xy^2/(x^2+y^2)+y-yz^2/(y^2+z^2)+z-zx^2/(z^2+x^2)
=x+y+z-xy^2/(x^2+y^2)-yz^2/(y^2+z^2)-zx^2/(z^2+x^2)
≥1-xy^2/(2xy)-yz^2/(2yz)-zx^2/(2zx)
=1-y/2-z/2-x/2=1-1/2=1/2