高中数学一个代数不等式求助证明

一个不等式问题已知x,y,z为不

已知x,y,z为不全为零的非负实数,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥ [5(x^2+y^2+z^2)-2(yz+zx+xy)]/[2(x+y+z)] (1) 证明 由(1)的全对称性,不妨设x≥y≥z,则(z-x)^2≥(y-z)^2, (1)Σ(y+z)*Σx^2/(y+z)≥5Σx^2-2Σyz Σx^2+Σ[x^2/(y+z)](2x+y+z)≥5Σx^2-2Σyz 2Σx^2+2Σ[x^3/(y+z)]≥5Σx^2-2Σyz 2Σ[x^3/(y+z)]≥3Σx^2-Σx(y+z) Σx[2x^2-3x(y+z)+(y+z)^2]/(y+z)≥...全部

  已知x,y,z为不全为零的非负实数,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥ [5(x^2+y^2+z^2)-2(yz+zx+xy)]/[2(x+y+z)] (1) 证明 由(1)的全对称性,不妨设x≥y≥z,则(z-x)^2≥(y-z)^2, (1)Σ(y+z)*Σx^2/(y+z)≥5Σx^2-2Σyz Σx^2+Σ[x^2/(y+z)](2x+y+z)≥5Σx^2-2Σyz 2Σx^2+2Σ[x^3/(y+z)]≥5Σx^2-2Σyz 2Σ[x^3/(y+z)]≥3Σx^2-Σx(y+z) Σx[2x^2-3x(y+z)+(y+z)^2]/(y+z)≥0 Σx(x-y-z)(2x-y-z)/(y+z)≥0 Σ[x(x-y-z)/(y+z)][(x-y)-(z-x)]≥0 Σ[x(x-y-z)/(y+z)-y(y-z-x)/(z+x)](x-y)≥0 Σ{[x(z+x)(x-y)-zx(z+x)+y(y+z)(x-y)+yz(y+z)]/[(y+z)(z+x)]}(x-y)≥0 Σ{[x(z+x)(x-y)-(x-y)z^2-(x+y)(x-y)z+y(y+z)(x-y)]/[(y+z)(z+x)]}(x-y) ≥0 Σ(x+y)(x^2+y^2-z^2)(x-y)^2≥0 (2) 而 Σ(x+y)(x^2+y^2-z^2)(x-y)^2 =(x+y)(x^2+y^2-z^2)(x-y)^2+(y+z)(y^2+z^2-x^2)(y-z)^2 +(z+x)(z^2+x^2-y^2)(z-x)^2 ≥(y+z)(y^2+z^2-x^2)(y-z)^2+(z+x)(z^2+x^2-y^2)(z-x)^2 ≥(y+z)(y^2+z^2-x^2)(y-z)^2+(y+z)(z^2+x^2-y^2)(y-z)^2 =2(y+z)z^2*(y-z)^2 ≥0。
   (3) 所以不等式(2)成立,从而不等式(1)成立。 。收起