Question

基本不等式已知a>b>c,求(a-c)-1\(b^2-ab+c(a-b))的最小值

Answer 1

^2-ab+c(a-b)=b(b-a)+c(a-b)=-(a-b)(b-c)

(a-c)-1/(b^2-ab+c(a-b))=(a-b)+(b-c)+1/[(a-b)(b-c)]≥3{(a-b)(b-c)*1/[(a-b)(b-c)]}^3=3