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

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

ab-4a-b+1=0, b=(4a-1)/(a-1)
(a+1)(b+2))=(a+1)(6a-3)/(a-1)
=3(a+1)(2a-1)/(a-1)
=3[(a-1)+2][2(a-1)+1]/(a-1)
=3[2(a-1)^2+5(a-1)+2]/(a-1)
=3[2(a-1) +2/(a-1)+5]
=15+6[(a-1)+1/(a-1)]
>=15+6*2=27
最小值 =27

ab-4a-b+1=0 a=(b-1)/(b-4)=1+3/(b-4), a>1, 3/(b-4)>0 b>4. (a-1)(b-4)=3 (a>1,b>4) y=(a-1+2)(b-4+6) =3+6(a-1)+2(b-4)+12=15+2(3(a-1)+(b-4)) 3(a-1)+(b-4)>=2根号[3(a-1)(b-4)=6 y=(a-1+2)(b-4+6) =3+6(a-1)+2(b-4)+12=15+2(3(a-1)+(b-4))>=15+2*6=27 所以最小值是27

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