x+y+z=1,why x(1-2x)(1-3x)+y(1-2y)(1-3y)+z(1-2z)(1-3z)>=0

构造函数f(t)=(t-x)(t-y)(t-z)
f(t)=t³-(x+t+z)t²+(xy+yz+zx)t-xyz=t³-t²+(xy+yz+zx)t-xyz
∵f(x)=0
∴x³-x²+(xy+yz+zx)x-xyz=0
同理
y³-y²+(xy+yz+zx)y-xyz=0
z³-z²+(xy+yz+zx)z-xyz=0
三式相加得
(x³+y³+z³)-(x²+y²+z²)+(xy+yz+zx)(x+y+z)-3xyz=0
即
x³+y³+z³= (x²+y²+z²)-(xy+yz+zx)+3xyz

原式左=1-5(x²+y²+z²)+6(x³+y³+z³)
= (x²+y²+z²) -6(xy+yz+zx)+18xyz+1

然后你可以试试x²+y²+z²>=xy+yz+zx (利用cauchy不等式)

然后就不会了，你自己再想想吧……