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[不等式] 不等式求最小值要求用柯西

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依然饭特稀 Posted 2019-12-1 19:51 |Read mode
已知a,b∈R+,a+b=1,求srqt(1+2a^2)+2sqrt((5/12)^2+b^2)的最小值

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kuing Posted 2019-12-2 02:04
\(\require{cancel}\)与这帖 forum.php?mod=viewthread&tid=4447 同类——本质四次方程,凑好的数据才有简单解。
你只要用任意手段得知取等后,即可写出各种不同解法,如链接中我用了切线法,现在要柯西的话,可以这样:
\begin{align*}
\text{原式}&=\sqrt{(1+2a^2)\left( \frac8{17}+\frac9{17} \right)}+2\sqrt{\left( \left( \frac5{12} \right)^2+b^2 \right)\left( \frac{25}{34}+\frac9{34} \right)}\\
&\geqslant\sqrt{\frac8{17}}+\sqrt{\frac{18}{17}}a+\frac56\sqrt{\frac{25}{34}}+2\sqrt{\frac9{34}}b\\
&=\frac{49+36(a+b)}{6\sqrt{34}}\\
&\xcancel{=\frac5{12}\sqrt{17}}=\frac5{12}\sqrt{34}.
\end{align*}

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 Author| 依然饭特稀 Posted 2019-12-2 07:40
牛,牛,牛,谢谢

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 Author| 依然饭特稀 Posted 2019-12-2 22:36
回复 2# kuing

我算了一次,最后一行应该是sqrt34

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kuing Posted 2019-12-2 23:51
回复 4# 依然饭特稀

再算一次看看?

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 Author| 依然饭特稀 Posted 2019-12-3 09:21
我觉得没算错啊
8CLZ{KNT(T3Q9J9TWAU@U36.png

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kuing Posted 2019-12-3 13:19
回复 6# 依然饭特稀

哦,是我错了
……
我以为你说 sqrt34 是最终结果……

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