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[不等式] 请教:$\sqrt a +\sqrt b <\frac 1a +\frac 1b$

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isee Posted 2014-5-10 22:41 |Read mode
两不相等的正实数$ab=1$,则$\sqrt a +\sqrt b <\frac 1a +\frac 1b$。

这个较不简单的不等式,是偶从三元降到二元的,结果发现并不好处理,如加个1,或者全部化一元再分解因式。

请教,这个有没秒证法?

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kuing Posted 2014-5-10 22:44
(a,b)->(x^4,y^4) 即 x^2+y^2<1/x^4+1/y^4  即 xy(x^2+y^2)<x^4+y^4 显然成立

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 Author| isee Posted 2014-5-10 22:49
Last edited by isee 2014-5-10 22:56回复 2# kuing

如果对新手,这个:$xy(x^2+y^2)<x^4+y^4$是移项分解因式么?、


应该是,这样升成4次比升成2次好多了

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 Author| isee Posted 2014-5-10 22:57
比升成2次好多了

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其妙 Posted 2014-5-10 23:36
回复 2# kuing
连美元符号都懒得打了,

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爪机专用 Posted 2014-5-11 00:50
回复 5# 其妙
题目和公式都简单,就懒了而且刚才累的,画图真要命
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爪机专用 Posted 2014-5-11 00:54
回复 1# isee

话说原题三元是怎么样的??
I am majia of kuing

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 Author| isee Posted 2014-5-11 07:21
加个根号c与c分之1

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kuing Posted 2014-5-11 12:21
回复 8# isee

噢,方法类似,一样简单,可以代换,也可以不代换,直接均值,还更简单些

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 Author| isee Posted 2014-5-11 15:07
回复 9# kuing


    所以 说  没想到二元其实相对不好处理

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