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[不等式] 求限制条件的三元分式的最值

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lemondian Posted at 2024-10-24 08:39:49 |Read mode
已知$a,b,c\in(0,\sqrt{3})$,且$a^4+b^4+c^4\geqslant 3$,求$\dfrac{a^3}{a^2-3}+\dfrac{b^3}{b^2-3}+\dfrac{c^3}{c^2-3}$的最大值。

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Aluminiumor Posted at 2024-10-24 11:53:26
$$x\in(0,\sqrt{3}),-\frac{x^4}{2}-\frac{x^3}{x^2-3}=\frac{x^3(x-1)^2(x+2)}{2(3-x^2)}\geq0$$
$$\sum\frac{a^3}{a^2-3}\leq\sum-\frac{a^4}{2}\leq-\frac32$$
取等略。

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谢谢!  Posted at 2024-10-24 15:19

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 Author| lemondian Posted at 2024-10-24 15:18:38
Aluminiumor 发表于 2024-10-24 11:53
$$x\in(0,\sqrt{3}),-\frac{x^4}{2}-\frac{x^3}{x^2-3}=\frac{x^3(x-1)^2(x+2)}{2(3-x^2)}\geq0$$
$$\sum\f ...
是不是可以推广一下子?
已知$a,b,c,M,t>0$,且$x_1^a+x_2^a+\cdots+x_n^a\geqslant M
$,求$\frac{x_1^b}{x_1^c-t}+\frac{x_2^b}{x_2^c-t}+\cdots+\frac{x_n^b}{x_n^c-t}$的最大值。($a,b,c,M,t$有什么限制条件?)

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kuing Posted at 2024-10-24 16:34:30
上次 kuing.cjhb.site/forum.php?mod=viewthread&tid=12696 这帖写的,楼主是完全没学会吗?

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 Author| lemondian Posted at 2024-10-24 17:03:02
我也是从那帖来学的,主要是这里的$t$的范围如何确定呢?

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