Question

若实数a,b满足ab-4a-b+1=0(a>1) ,则（a+1）（b+2）的最小值是

Answer 1

ab-4a-b+1=0
b(a-1)+(4-4a)-3=0
(b-4)(a-1)=3
由于a>1
所以b>4
ab+2a+b+2=ab-4a-b+1+6a+2b+1
=6a+2b+1=2√3ab+1
当6a=2b时有最小值
将b=3a代入方程
ab-4a-b+1=0
3a^2-7a+1-0
因为a>1
所以
a=(7+√35)/2
再代入6a+2b+1中
我的计算是这样的。呵呵。

Answer 2

ab-4a-b+1=0===>a(b-4)=b-1===>a=(b-1)/(b-4)=1+[3/(b-4)]===>a+1=2+[3/(b-4)]且由a>1知b-4>0===>(a+1)(b+2)={2+[3/(b-4)]}[(b-4)+6].令t=b-4,则t>0且（a+1)(b+2)=[2+3/t][t+6]=15+2t+(18/t)》15+12=27.====>(a+1)(b+2)》27.等号仅当t^2=9时取得。即当a=2,b=7时，(a+1)(b+2)min=27.

Answer 3

ab-4a-b+1=0
→b=(4a-1)/(a-1)=4+3/(a-1)
(a>1)
(a+1)(b+2)
=ab+b+2a+2
=a[4+3/(a-1)
]+4+3/(a-1)
+2a+2
=6a+6a/(a-1)+3
=6a+6/(a-1)+9
=6(a-1)+6/(a-1)+15
=6[（a-1）+1/（a-1）]
当
a-1=1/（a-1）时取最大值
这时
a=2
最大值为：6×2+15=27