x>=y>=z>0,求证：x^2y/z+y^2z/x+z^2*x/y>=x^2+y^2+z^2

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x>=y>=z>0,求证：x^2*y/z+y^2*z/x+z^2*x/y>=x^2+y^2+z^2

首先注意如下关系：
(x^2y/z+y^2z/x+z^2x/y) - (xy^2/z+yz^2/x+zx^2/y)
=(xy/z)(x-y) + (yz/x)(y-z) + (zx/y)(z-x)
=(xy/z)(x-y) + (yz/x)(y-z) - (zx/y)(x-y) - (zx/y)(y-z)
=(xy/z - zx/y)(x-y) + (yz/x - zx/y)(y-z)
=(y-z)(x-y)x(y+z)/yz - (x-y)(y-z)z(x+y)/xy
=(x-y)(y-z)((x/y+x/z) - (z/x+z/y))
最后一个括号中,前两项都不小于1,而后两项都不大于1,因此
(x^2y/z+y^2z/x+z^2x/y) >= (xy^2/z+yz^2/x+zx^2/y)
于是
2(x^2y/z+y^2z/x+z^2x/y)
>= (x^2y/z+y^2z/x+z^2x/y) + (xy^2/z+yz^2/x+zx^2/y)
= (x^2y/z+x^2z/y + (y^2z/x+y^2x/z) + (z^2x/y+z^2y/x)
>= 2(x^2+y^2+z^2)
再问：能否运用排序不等式的有关知识给出证明？

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