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[不等式] 一个不等式

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转化与化归 Posted at 2013-11-26 15:07:15 |Read mode
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kuing Posted at 2013-11-26 15:21:25
有两个系数相同应该很好办啊,和差化积放缩一下变成一元函数

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战巡 Posted at 2013-11-26 15:38:22
这不难吧...
\[2\sqrt{2}\sin(a)+2\sqrt{2}\sin(b)+\sin(c)=2\sqrt{2}\sin(a)+2\sqrt{2}\sin(b)+\sin(a+b)\]
\[=4\sqrt{2}\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})+\sin(a+b)\]
\[\le 4\sqrt{2}\sin(\frac{a+b}{2})+\sin(a+b)\le \sqrt{13+16\sqrt{2}}\]
后面那个过程我懒得写了,导数神马的很容易弄出来

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其妙 Posted at 2013-11-27 13:27:34
人教里有一般结论吧?

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kuing Posted at 2013-11-27 13:50:36

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 Author| 转化与化归 Posted at 2013-11-27 15:39:07
回复 5# kuing
一般方法有了就好办了!

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2025-4-22 01:51 GMT+8

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