第2个回答 2011-04-15
x>0,y>0,x≠y,且x+y=x²+y²+xy,求证1<x+y<4/3
证明:∵x>0,y>0,x≠y,∴x+y=x²+y²+xy<x+y=x²+y²+2xy=(x+y)²
即有(x+y)²-(x+y)=(x+y)(x+y-1)>0,∵x>0,y>0,x+y>0,故得x+y-1>0,即有x+y>1.
又∵x>0,y>0,x≠y,∴(x+y)²=x²+y²+2xy>4xy,∴xy<(x+y)²/4...........(1)
又已知x+y=x²+y²+xy=(x+y)²-xy>(x+y)²-(x+y)²/4=3(x+y)²/4
即有1>3(x+y)/4,∴x+y<4/3.
故1<x+y<4/3得证.