Question

已知a>b>0,求a+16/(b(a-b))的最小值.和a+16/(b(a-b))的最小值

Answer 1

a>b>0,则a>0,b>0且a-b>0.

故依基本不等式得

a+16/[b(a-b)]

≥a+16/[(b+a-b)/2]^2

＝a+64/a^2

＝(a/2)+(a/2)+(64/a^2)

≥3(a/2·a/2·64/a)^(1/3)

＝6×2^(1/3).

∴a/2＝64/a^2且b＝a-b,

即a＝4×2^(1/3),b＝2×2^(1/3)时,

所求最小值为：6×2^(1/3）