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)
请注意括号的正确使用,以免造成误解.

不失一般性,令x≧y≧z>0.则:
x^2≧y^2≧z^2、x+y≧x+z≧y+z,∴1/(y+z)≧1/(x+z)≧1/(x+y).
考查下列两组数:x^2≧y^2≧z^2、1/(y+z)≧1/(x+z)≧1/(x+y).
由排序不等式:同序和不小于乱序和. 得:
x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≧y^2/(y+z)+z^2/(x+z)+x^2/(x+y)、
x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≧z^2/(y+z)+x^2/(x+z)+y^2/(x+y).

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^2)/(x+y),
∴(2z^2-x^2-y^2)/(x+y)+(2x^2-y^2-z^2)/(y+z)≧(x^2+z^2-2y^2)/(x+z).