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[不等式] 最值问题 $1\le a,b,c\le2$ 证 $(3a+4b+5c)(3/a+4/b+5/c)\le162$

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力工 posted 2017-2-28 08:02 |Read mode
若实数$a,b,c$满足 $1\leq a,b,c\leq 2$. 求证:$(3a+4b+5c)\left(\frac{3}{a}+\frac{4}{b}+\frac{5}{c}\right)\leq 162$.
并求代数式$(a+2b+3c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right)$的最大值。

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色k posted 2017-2-28 08:15
关于每个变量均下凸,只需计算 $a,b,c\in\{1,2\}$ 的所有值即可

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original poster 力工 posted 2017-2-28 09:02
回复 2# 色k
色k好早!

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realnumber posted 2017-2-28 13:51
回复 2# 色k


    鼓掌.....

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kuing posted 2017-2-28 15:16
回复 3# 力工

偶然早上醒来,看下爪机,习惯性地打开论坛扫一眼,看到这题简单就用爪机回了下,然后继续睡……

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isee posted 2017-2-28 17:31
8点了,能算早么。。。。

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original poster 力工 posted 2017-2-28 20:48
回复 6# isee


色k不到12点不起床,不早?

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色k posted 2017-2-28 20:57

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