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
问题描述：
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
答
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)+...