己知x,y为正实数,且xy=4x+y+12,求xy最小值己知x,y为正实数,且xy=4x+y+12,求xy的最小值
问题描述:
己知x,y为正实数,且xy=4x+y+12,求xy最小值
己知x,y为正实数,且xy=4x+y+12,求xy的最小值
答
己知x,y为正实数,且xy=4x+y+12,
有xy=4x+y+12>=2√(4xy)+12=4√(xy)+12,
令t=√(xy)>0,有t^2-4t-12>=0,得t=6,
所以xy的最小值为36