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[不等式] 不等式

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lrh2006 posted 2017-7-30 21:47 |Read mode
已知a,b,c都是正实数,ab+bc+ca=3,求证a^3/(b+2c)+b^3/(c+2a)+c^3/(a+2b)>=1
请教各位,谢谢咯

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kuing posted 2017-7-30 21:55
提示:由CS
\[\LHS=\sum\frac{(a^2)^2}{a(b+2c)}\ge\cdots\ge\cdots\]

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original poster lrh2006 posted 2017-7-30 22:04
回复 2# kuing


    没看懂。。。kk能不能说详细点啊?觉得不等式好难。。。身边没电脑,打字很不方便。。。

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original poster lrh2006 posted 2017-7-30 22:05
回复 3# lrh2006


    cs是什么

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kuing posted 2017-7-30 22:07
回复 4# lrh2006

forum.php?mod=viewthread&tid=4096 附件中的CS部分

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original poster lrh2006 posted 2017-7-30 22:27
回复 5# kuing


    嗯嗯,我看下先,估计不简单。。。

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色k posted 2017-7-30 22:44
回复 6# lrh2006

不复杂。
爪机看PDF没问题吧?

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original poster lrh2006 posted 2017-7-30 22:53
回复 7# 色k


    嗯嗯,还好,能打开
kk你又骗人啦,哪里是什么常用不等式。。。不用柯西,就高中数学的知识能做吗?
不过你教的方法我已经做出来了,哈哈,还是kk最好,关键时刻总能帮我,谢谢谢谢

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kuing posted 2017-7-31 00:01
回复 8# lrh2006

easy,由均值有
\[\frac{a^3}{b+2c}+\frac{a(b+2c)}9\ge\frac23a^2,\]
下略……

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original poster lrh2006 posted 2017-7-31 07:28
回复 9# kuing


    k神啊,有没有你做不出来的题目?你easy,我可不easy。按照你的提示我做出来了,可是自己是想不到这样做的,你怎么就想到了呢,有没有兴趣说一说呀?

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kuing posted 2017-7-31 12:14
说不出来的,因为一下就想到了

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original poster lrh2006 posted 2017-7-31 12:45
回复 11# kuing


    好吧,我能理解,还是谢谢你噢

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