Question

已知a,b为正数,ab-3a-2b=0,求2a+3b的最小值和ab的最小值，求(a-1)(b-1)的最小值

Answer 1

b=3a/(a-2),因a>0,b>0,故a-2>0,2a+3b=2a+9a/(a-2)=2a+18/(a-2)+9=2(a-2)+18/(a-2)+13>=2√2(a-2)*18/(a-2)+13=25,即
2a+3b的最小值为25.
ab=3a+2b=3a+6a/(a-2)=3a+12/(a-2)+6=3(a-2)+12/(a-2)+12>=2√3(a-2)*12/(a-2)+12=24,ab的最小值为24.
(a-1)(b-1)=ab-(a+b)+1=3a+2b-(a+b)+1=2a+b+1=2a+3a/(a-2)+1=2a+6/(a-2)+4=2(a-2)+6/(a-2)+8>=2√2(a-2)*6/(a-2)+8=4√3+8,即(a-1)(b-1)的最小值为4√3+8。

Answer 2

(1)a>0、b>0.
ab-3a-2b＝0→2/a+3/b＝1.
故依Cauchy不等式得
(2a+3b)(2/a+3/b)≥(2+3)²
→2a+3b≥25.
故所求最小值为：25.

(2)依基本不等式得
ab＝3a+2b≥2√(6ab)
→(ab)²-24ab≥0.
显然，ab>0,故
ab≥24,
即所求最小值为：24.

(3)(a-1)(b-1)
＝ab-a-b+1
＝(3a+2b)-a-b+1
＝2a+b+1
＝2[2b/(b-3)]+b+1
＝(b-3)+12/(b-3)+8
≥2√[(b-3)·12/(b-3)]+8
＝4√3+8
即所求最小值为：4√3+8.