Question

设a>b>0,则a²+1÷ab+1÷a(a-b)的最小值为_____

Answer 1

a²+1÷ab+1÷a(a-b)

=a²+(a-b)/[ab(a-b)]+b/[ab(a-b)]

=a²+a/[ab(a-b)]

=a²+1/[b(a-b)]

由均值不等式得：2√[b(a-b)]≤b+(a-b)=a

所以,b(a-b)≤a²/4

所以,1/[b(a-b)]≥4/a²

则,a²+1/[b(a-b)]≥a²+4/a²≥2√4=4

等号当且仅当b=a-b、a²=4/a²成立,即：a=√2,b=√2/2时成立

又a>b>0,满足题意,所以a²+1÷ab+1÷a(a-b)的最小值为4.