Question

已知a大于b大于0,求a的平方+16/b(a-b)的最小值

Answer 1

令a=b+t,t>0,b>0,则有：

a^2+16/[b(a-b)]=(b+t)^2+16/(bt)>=(2√bt)^2+16/(bt)=4bt+16/(bt)>=2√[4bt *16/(bt)]=16

当b=t,及4bt=16/(bt),即b=t=√2,即b=√2,a=2√2时取最小值16.