试解

来源: wxcfan123 2015-04-20 13:19:14 [] [旧帖] [给我悄悄话] 本文已被阅读: 次 (413 bytes)
回答: 数学不等式 48魁北克人2015-04-18 18:42:49
x^2/(x+y^2) = x - xy^2/(x+y^2)

原不等式等价于:

xy^2/(x+y^2) + yz^2/(y+z^2) + zx^2/(z+x^2) <= 3/2.

因x+y^2 >= 2sqrt(xy^2), 归结于证明

xsqrt(z) + ysqrt(x) + zsqrt(y) <= 3.

由柯西不等式,

LHS^2 <= (x+y+z)(xz+yx+zy) = 3(xz+yx+zy) <= 9

因xz+yx+zy <= (x+y+z)^2/3.
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