x＞0,y＞0,x≠y,x+y=x²+y²+xy,求证1＜x+y＜4/3
∵ x>0,y>0,xy>0,x≠y
∴x+y=x²+xy+y²2xy+2xy=4xy
∴xy(x+y)²-(x+y)²/4=3(x+y)²/4
∴x+y