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[不等式] 求一组分式的最大值

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lemondian posted 2024-11-21 16:40 |Read mode
Last edited by lemondian 2024-11-21 19:381.已知$a,b,c>0$,求$\dfrac{a^2b}{a^2+b^2}$的最大值。
2.已知$a,b,c>0$,求$\dfrac{a^2b+b^2c}{a^3+b^3+c^3}$的最大值。
3.已知$a_i>0(i=1,2,\cdots ,n)$,$\dfrac{a_1^2a_2+a_2^2a_3+\cdots +a_{n-1}^2a_n}{a_1^3+a_2^3+\cdots +a_n^3}$有最大值吗?若有,如何求得?

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kuing posted 2024-11-21 17:00
1、显然打错;

2、待定 `0<k<1` 由均值
\begin{align*}
a^3+b^3+c^3&=\frac12a^3+\frac12a^3+kb^3+\frac{1-k}2b^3+\frac{1-k}2b^3+c^3\\
&\geqslant3\sqrt[3]{\frac k4}a^2b+3\sqrt[3]{\frac{(1-k)^2}4}b^2c,
\end{align*}

\[k=(1-k)^2\riff k=\frac{3-\sqrt5}2,\]
代入即得
\[a^3+b^3+c^3\geqslant\frac32\sqrt[3]{3-\sqrt5}(a^2b+b^2c),\]
所以
\[\frac{a^2b+b^2c}{a^3+b^3+c^3}\leqslant\frac13\sqrt[3]{6+2\sqrt5},\]
显然可取等,具体数值略;

3、最大值存在且不可求。

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我是照抄的,那应该是原题就出错了  posted 2024-11-25 22:56

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