设a=x^2,b=y^2+c=z^2,x,y,z为正实数.则x^2+y^2+z^2=1.所证式齐次式为:3Σx^2√[(x^2+y^2+z^2)^2-(yz)^2]>=2√2*(x^2+y^2+z^2)^2<===>9Σx^4*(Σx^2)^2-9(xyz)^2*Σx^2+18(yz)^2√[(Σx^2)^2-(zx)^2]*[(Σx^2)^2-(xy)^2]>=8(x^2+y^2+z^2)^4因为√[(Σx^2)^2-(zx)^2]*[(Σx^2)^2-(xy)^2]>=x^4+y^4+z^4+2(yz)^2+(zx)^2+(xy)^2+yzx^2===>9Σx^4*(Σx^2)^2-9(xyz)^2*Σx^2+18(yz)^2[Σx^4+(zx)^2+(xy)^2+2(yz)^2+yzx^2]>=8(x^2+y^2+z^2)^4<===>Σx^8+4Σ(y^2+z^2)x^6+6Σ(yz)^4-(xyz)^2*(33Σx^2-18Σyz)>=0<===>Σ[(y^4+10y^2z^2+z^4+15x^4)(y+z)^2-18(xyz)^2](y-z)^2>=0