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[不等式] 两个最值问题

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Rachmani Posted 2023-8-26 21:07 |Read mode
Last edited by Rachmani 2023-8-26 23:07$1.$已知$a^2+b^2=1$,求$\displaystyle a+\frac{1}{b+2}$的取值范围。由于没有相关计算软件,不能确定此问题是否为高次方程不可解问题。
$2.$已知正实数$a,b,c$满足$2a^4+b^4+c^4-2b^2c^2-2a^2b^2-2a^2c^2=0$,求$a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)$的最大值。
感谢各位。

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第一题:https://www.zhihu.com/question/615869439  Posted 2023-8-26 23:54

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isee Posted 2023-8-26 23:56
第一题,化一元求求导啊,不然怎么知道会不会遇到高次方程呢
isee=freeMaths@知乎

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 Author| Rachmani Posted 2023-8-26 23:59
isee 发表于 2023-8-26 23:56
第一题,化一元求求导啊,不然怎么知道会不会遇到高次方程呢
求导后分子后出现了七次,吓得直接扔了

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kuing Posted 2023-8-27 00:43
2. 记 `x=a^2`, `y=b^2`, `z=c^2`,条件可化为
\begin{align*}
4yz&=2x^2+(y+z)^2-2x(y+z)\\
&\geqslant\bigl(2\sqrt2-2\bigr)x(y+z),
\end{align*}
所以
\[x\left(\frac1y+\frac1z\right)\leqslant\frac2{\sqrt2-1}=2\sqrt2+2,\]
取等略。

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 Author| Rachmani Posted 2023-8-27 10:22 From mobile phone
感谢(≧∇≦)

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