是否存在常數(shù)c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)對任意正數(shù)x,y恒成立
[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)]
=[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)]
=[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)]
=(2xy-x^2-y^2)/[(x+2y)(2x+y)]
=-(x-y)^2/[(x+2y)(2x+y)]≤0{對任意正數(shù)x,y}
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其中,x=y時,等號成立
x=y時,x/(2x+y) +y/(x+2y)=x/3x+y/3y=1/3+1/3=2/3
所以,c=2/3時,
不等式x/(2x+y) +y/(x+2y)