Question

设M=(1/a-1)(1/b-1)(1/c-1),且a+b+c=1(a,b,c属于正实数),则M的取值范围是[0,1/8

Answer 1

因为a+b+c=1

M=(1/a-1)(1/b-1)(1/c-1)

=(b+c)/a*(a+c)/b*(b+c)/a

=(a+b)(b+c)(c+a)/abc

a+b≥2√ab

b+c≥2√bc

a+c≥2√ac

上面三式相乘,得

(a+b)(b+c)(a+c)≥8abc

(b+c)(a+c)(b+a)/abc≥8

(1/a-1)(1/b-1)(1/c-1)≥8

选D