Question

证明 (a+b+c)/3大于等于三倍根号abc

Answer 1

x,y,z是非负数时

x^3+y^3+z^3-3xyz

=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)

=(x+y+z)[(x-y)^2+(y-z)^2+(x-z)^2]/2≥0

所以,

x^3+y^3+z^3≥3xyz

设x^3=a,y^3=b,z^3=c

则:(a+b+c)/3≥三次根号(abc)