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[不等式] 一个四元不等 式证明

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力工 Posted 2019-2-25 20:39 |Read mode
Last edited by 力工 2019-2-25 21:15对于满足$abcd=1$的正数$a,b,c,d$,
证明:$\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}+\sqrt{d^2+1}\leqslant \sqrt{2}(a+b+c+d)$.
我想构造为:$\sqrt{a^2+1}-\sqrt{2}a\leqslant mlna$,我的老师说不行,如果不行,怎么做才好?
讨教于大神。

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 Author| 力工 Posted 2019-2-25 20:41
画图确实不可以,找不到这个$m$.

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kuing Posted 2019-2-25 20:56
1、条件写错了
2、编辑时请顺便选择主题分类

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 Author| 力工 Posted 2019-2-25 21:16
感谢k神提示!习惯性地按了加

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kuing Posted 2019-2-25 21:55
指数函数代换后画图看出是凸函数,所以 m 一定是存在的,算一下可知 `m=-\sqrt2/2`。

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kuing Posted 2019-2-25 21:57
其实既然凸函数,那根本就不必要去算切线了,直接琴生得了……
至此瞬间我就想起以前就写过:forum.php?mod=viewthread&tid=5289(该帖是三元,除证法二之外都可用到这里

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