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[不等式] 三元条件最值

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v6mm131 posted 2017-8-3 13:16 |Read mode
$x,y,z>0$满足$xyz+xy+yz+zx+x+y+z=3$求$xyz(x+y+z)$的最大值

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kuing posted 2017-8-3 13:25
《撸题集》第 439 页题目 4.6.17.

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original poster v6mm131 posted 2017-8-3 14:05
回复 2# kuing


    100贴这么水

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wanhuihua posted 2017-8-3 16:35
Last edited by hbghlyj 2025-3-19 19:09设 $x, y, z$ 为非负数且 $\sum x+\sum x y+x y z=3$
求 $x y z \sum x$ 的最大值
记 $\sum x=p, \sum x y=q, x y z=r . t=p r$ ,则 $q^2 \geq 3 p r$ .
问题转化 $p+r+\sqrt{3 p r} \leq 3, p^3 \geq 27 r$ 时,求 $t=p r$ 的最大值。
令 $f(p)=p+r+\sqrt{3 p r}=p+\frac{t}{p}+\sqrt{3 t}$ ,
由 $p \leq 3$ 得 $p^2 \geq \frac{p^4}{9} \geq 3 t, f^{\prime}(p)=1-\frac{t}{p^2} \geq 0,3 \geq f(p) \geq f\left((27 t)^{\frac{1}{4}}\right)$
解得 $t \leq 9\left(2^{\frac{7}{3}}-5\right)$ ,当 $x=y=z$ 时取等。

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