设x,y,z∈R+,求证 2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z)

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设x,y,z∈R+,求证2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z)设x,y,z∈R+,求证2z2-x2-y2/(x+y)+2x2-y2-z2/(y+

设x,y,z∈R+,求证 2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z)
设x,y,z∈R+,求证 2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z)

设x,y,z∈R+,求证 2z2-x2-y2/(x+y)+2x2-y2-z2/(y+z)≥x2+z2-2y2/(x+z)
分类:数学

上述两式相加,得: 2x^2/(y+z)+2y^2/(x+z)+2z^2/(x+y)≧(y^2+z^2)/(y+z)+(z^2+x^2)/(x+z)+(x^2+y^