x^3+y^3+z^3-3xyz
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]/2
≥0
x^3+y^3+z^3≥3xyz
设,a=x^3,b=y^3,c=z^3
代人上式有：
a+b+c大于等于3倍的3次根号下abc