Question

设x,y,z为正实数,且x+y+z>=xyz,求x^2+y^2+z^2/xyz的最小值

Answer 1

化成齐次式

((x^2+y^2+z^2)/xyz)^2 >= (xx+yy+zz)^2 /((x+y+z)xyz)

xx+yy+zz>=1/3*(x+y+z)^2

x+y+z >= 3(xyz)^(1/3)

xx+yy+zz >= 3(xyz)^(2/3)

三式相乘：

(xx+yy+zz)^2 >= 3(x+y+z)xyz

=>

((x^2+y^2+z^2)/xyz)^2 >=3

=>

(x^2+y^2+z^2)/xyz >= 3^(1/2）