求证ay+bz+cx<1 方法越多越好
证法一:
(a+x)(b+y)(c+z)
=(ab+ay+bx+xy)(c+z)
=abc+acy+bcx+cxy+abz+ayz+bxz+xyz
=abc+(bxz+abz)+(bcx+cxy)+(acy+ayz)+xyz
=abc+(a+x)bz+(b+y)cx+(c+z)ay+xyz
=abc+bz+cx+ay+xyz
=1
abc>0
bz+cx+ay<1
证法二