急....F.6 pure ineq.

1
(a+b+c)/3>=(abc)^1/3
or 8/27>=abc...(1)
Also
{[(1-a)+(1-b)+(1-c)]/3}^3>=(1-a)(1-b)(1-c)
1/27>=(1-a)(1-b)(1-c)...(2)
Combine (1 ) and (2)
So abc/(1-a)(1-b)(1-c)>=8
2
我覺得條題目有問題﹐因為我可以證到。
(xy+yz+xz)-3xyz<=2/9
Consider
(1-3x)(1-3y)(1-3z)
=(1-3x-3y+9xy)(1-3z)
=(1-3z-3x+9xz-3y+9yz+9xy-27xyz)
=(-2+9(xy+yz+xz)-27xyz)
So 9(xy+yz+xz)-27xyz=2+(1-3x)(1-3y)(1-3z)
(xy+yz+xz)-3xyz=2/9+(1-3x)(1-3y)(1-3z)/9
By AM>=GM
(1-3x)(1-3y)(1-3z)
<={[(1-3x)+(1-3y)+(1-3z)]/3}^3
=0
So
(xy+yz+xz)-3xyz
<=2/9
<1/4
if 1-3x=1-3y=1-3z then the equality holds
That is x=y=z