Question

已知x,y,z是实数,且x+Y+z=m.x+y+z=m/2 .m大于等于0.求证 0≤x≤2m/3 0≤y≤2m/3 0

Answer 1

已知x+y+z=m,所以,x=(m-y-z).将x=(m-y-z)代入x+y+z=m/2,我们有(m-y-z）+y+z=m/2.化简后可得,2y+(2z-2m)y+m/2-2mz+2z=0.将上式看作关于y的二次方程,我们可以写出判别式:Δ=(2z-2m）-8(m/2-2mz+2z)=4(2m-3z)z.因为y为实数,我们要求Δ≥0.由此可得,4(2m-3z)z≥0,化简得到0≤z≤2m/3.同理,我们可以将y=m-x-z,z=m-x-y分别代入x+y+z=m/2,然后根据判别式Δ≥0,分别可得0≤x≤2m/3,0≤y≤2m/3.