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[不等式] 一个轮换对称不等式

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力工 posted 2017-5-6 21:55 |Read mode
Last edited by 力工 2017-5-6 22:14已知$x,y,z>0$,且$x+y+z=1$,求证:$x^2+y^2+z^2+9xyz\leqslant  \dfrac{2}{3}$.

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kuing posted 2017-5-6 22:08
已知$x,y,z>0$,且$x+y+z=1$,求证:$x^2+y^2+z^2+9xyz\geqslant \dfrac{2}{3}$.
力工 发表于 2017-5-6 21:55
不等式不成立(try x->0, y=z->1/2),这次又抄错哪个数字呢?

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original poster 力工 posted 2017-5-6 22:12
回复 2# kuing
无颜啊。不等号方向错了,应该是最大值。

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kuing posted 2017-5-6 22:27
回复 3# 力工

反向也不成立啊,try x=y->0, z->1

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kuing posted 2017-5-6 22:41
事实上我上面给的两个反例就是原式的上下界

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original poster 力工 posted 2017-5-7 07:54
回复 5# kuing

是的,最后我也想到了,只是上不了 网了。

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