Question

xy>0@,+求+3xy+2y分之一的+最小值

Answer 1

问：xy>0@,+求+3xy+2y分之一的+最小值

答：x>0,y>0且x+y+1＝3xy,则(1)x+y+1＝3xy ≤3[(x+y)/2]^2→3(x+y)^2-4(x+y)-4≥0→(x+y-2)[3(x+y)+2]≥0.显然，x>0,y>0时，3(x+y)+2>0,∴x+y-2≥0,故所求最小值为：2.(2)3xy＝x+y+1 ≥2√(xy)+1→3(√(xy))^2-2√(xy)-1≥0→[√(xy)-1][3√(xy)+1]≥0.显然，3√(xy)+1≥0，∴√(xy)-1≥0,故所求最小值为：1。