Question

已知x>0,y>0,且x+y+3=xy,求x+y的最小值

Answer 1

x>0,y>0,则x+y>0.
依基本不等式得
x+y+3＝xy≤[(x+y)/2]²
→(x+y)²-4(x+y)-12≥0
→(x+y+2)(x+y-6)≥0.
而x+y>0→x+y+2>0,
∴x+y-6≥0,
∴(x+y)|min＝6.