设a>b>0 求a^2+1/(ab)+1/[a(a-b)]的最小值

Answer 1

∵1/(ab)+1/[a(a-b)]=1/(ab)+1/(a^2-ab)=a^2/[ab(a^2-ab)]≥a^2*[2/(ab+a^2-ab)]^2=4/a^2
当且仅当a=2b时,等号成立
∴a^2+1/(ab)+1/[a(a-b)]≥a^2+4/a^2≥4
当且仅当a=√2时,等号成立
∴a^2+1/(ab)+1/[a(a-b)]的最小值为4.