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

网友回答

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======以下答案可供参考======
供参考答案1:
左边=Σx^2/(2x^2+xy+xz)由于(Σx)^2=Σx^2+2Σxy所以Σx^2>=(Σx)^2/3
所以左边=(Σx)^2/(Σx^2+2Σxy+Σx^2)=(Σx)^2/((Σx)^2+Σx^2)