Question

设a,b,c都是正实数,且a+b+c=1,则(1/a-1)(1/b-1)1/c-1)的取值范围是?

麻烦写详细过程

Answer 1

解：
由题意可知0<a<1,0<b<1,0<c<1
令y=(1/a-1)(1/b-1)(1/c-1)
=1/abc-1/ac-1/bc+1/c-1/ab+1/a+1/b-1
=1/abc+(1/a+1/b+1/c)-(1/ab+1/bc+1/ac)-1
=1/abc-(a+b+c)/abc+(1/a+1/b+1/c)-1
代入：a+b+c=1
则y=1/abc-1/abc+(1/a+1/b+1/c)-1
=1/a+1/b+1/c-1
当a=b=c=1/3时，y取最小值：y=1/(1/3)+1/(1/3)+1/(1/3)-1=8
当a,b,c中任何一个取趋向于0的小数，y取值为无穷大
所以y的取值范围为：y>8
即：(1/a-1)(1/b-1)(1/c-1)>8

Answer 2

显然有0<a<1,0<b<1,0<c<1故1/a-1>0,1/b-1>0,1/c-1>0 (1/a-1)(1/b-1)(1/c-1)≤(1/a+1/b+1/c-3)^3/27