y均为正数,且xy=2x+y-1,则x+y的最小值

y均为正数,且xy=2x+y-1,则x+y的最小值
y均为正数,且xy=2x+y-1,则x+y的最小值

y均为正数,且xy=2x+y-1,则x+y的最小值
解由xy=2x+y-1
得（x-1）y=2x-1
即y=（2x-1）/(x-1)
故x+y
=x+（2x-1）/(x-1)
=[（x^2-x）+（2x-1）]/(x-1)
=(x^2+x-1)/(x-1)
=[（x-1）^2+3（x-1）+1]/(x-1)
=(x-1)+1/(x-1)+3
≥2√(x-1)*1/(x-1)+3
=5
当且仅当x=2时,等号成立
故x+y的最小值为5.