若实数a.b满足ab-4a-b+1=o(a>1),求实数(a+1)*(b+2)的最小值

问题描述:

若实数a.b满足ab-4a-b+1=o(a>1),求实数(a+1)*(b+2)的最小值

由ab-4a-b+1=0.===>b=(4a-1)/(a-1).===>b+1=(5a-2)/(a-1).===>(a+1)(b+1)=13+5(a-1)+[6/(a-1)].由a>1及均值不等式可知,(a+1)(b+1)≥13+2√30.等号仅当a=(5+√30)/5,b=(24+4√30)/30时取得。故[(a+1)(b+1)]min=13+2√30.

解;由a>1可得a-1>0因为ab-4a-b+1=0(a>1)所以b=(4a-1)/(a-1)设f(a)=(a+1)(b+2)将b=(4a-1)/(a-1)代入可得f(a)=(a+1)(b+2)=(a+1)((4a-1)/(a-1)+2)=(6a-3)(a+1)/(a-1)=6(a-1)+[6/(a-1)]+15由于a-1>0所以...