设x,y,z为正实数k>=1，求证:

设x,y,z为正实数k>=1，求证:
x^2/(x^2+y^2+kxy)+y^2/(y^2+z^2+kyz)+z^2/(z^2+x^2+kzx)≥3/(k+2) (1)
证明 当k≥1，首先给出三个局部不等式:
x^2/(x^2+y^2+kxy)≥[3/(k+2)]*[x^2/(x^2+y^2+xy)] (2-1)
y^2/(y^2+z^2+kyz)≥[3/(k+2)]*[y^2/(y^2+z^2+yz)] (2-2)
z^2/(z^2+x^2+kzx)≥[3/(k+2)]*[z^2/(z^2+x^2+zx)] (2-3)
(2-1)<==> (k-1)(x-y)^2≥0，当k≥1，显然成立。同样可证另外两式.
再证下列不等式:
x^2/(x^2+y^2+xy)+y^2/(y^2+z^2+yz)+z^2/(z^2+x^2+zx)≥1 (3)
(3)式去分母得:
x^2*(y^2+z^2+yz)*(z^2+x^2+zx)+y^2*(z^2+x^2+zx)*(x^2+y^2+xy)+z^2*(x^2+y^2+xy)(y^2+z^2+yz)≥(y^2+z^2+yz)*(z^2+x^2+zx)*(x^2+y^2+xy)
展开化简为:
x^4*y^2+y^4*z^2+z^4*x^2≥xyz(zx^2+xy^2+yz^2)
<==> y^2*(x^2-yz)^2+x^2*(z^2-xy)^2+z^2*(y^2-zx)^2≥0
最后一式显然成立，故不等式(3)成立。
据此可证
x^2/(x^2+y^2+kxy)+y^2/(y^2+z^2+kyz)+z^2/(z^2+x^2+kzx)≥
[3/(k+2)]*[x^2/(x^2+y^2+xy)+y^2/(y^2+z^2+yz)+z^2/(z^2+x^2+zx)]≥3/(k+2).证毕。

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