x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)

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x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)

x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)
x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)

x,y,z为正实数 x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)
x/(2x+y+z)=[3*(2x+y+z)-(x+2y+z)-(x+y+2z)-]/(4(2x+y+z))
y/(x+2y+z)=[3*(x+y+2z)-(2x+y+z)-(x+y+2z)]/(4(x+2y+z))
z/(x+y+2z)=[3*(x+y+2z)-(2x+y+z)-(x+2y+z)]/(4(x+2y+z))
所以
左边=(3/4)*3 -(1/4)((2x+y+z)/(x+2y+z)+(x+2y+z)/(2x+y+z)+(x+2y+z)/(x+y+2z)+(x+y+2z)/(x+2y+z)+(x+y+2z)/(2x+y+z)+(2x+y+z)(x+y+2z))
又因为((2x+y+z)/(x+2y+z)+(x+2y+z)/(2x+y+z)+(x+2y+z)/(x+y+2z)+(x+y+2z)/(x+2y+z)+(x+y+2z)/(2x+y+z)+(2x+y+z)(x+y+2z))>=6
所以左边

左边=Σx^2/(2x^2+xy+xz)<=(柯西不等式的推论)(Σx)^2/(Σ(2x^2+xy+xz))=(Σx)^2/(2Σx^2+2Σxy)
由于(Σx)^2=Σx^2+2Σxy<=Σx^2+2Σx^2=3Σx^2
所以Σx^2>=(Σx)^2/3
所以左边=(Σx)^2/(Σx^2+2Σxy+Σx^2)=(Σx)^2/((Σx)^2+Σx^2)<=(Σx)^2/((Σx)^2+(Σx)^2/3)=3/4