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[不等式] 一个著名的三元不等式

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hbghlyj posted 2021-4-4 22:10 |Read mode
a,b,c>0,求证\[\frac{a}{a^2+2bc}+\frac{b}{b^2+2ca}+\frac{c}{c^2+2ab}\le\frac{a+b+c}{ab+bc+ca}\]

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kuing posted 2021-4-4 22:33
任一个变量为零都是取等,所以如果直接作差通分,会出现 abc ,剩下的次数就是四次应该不难……

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original poster hbghlyj posted 2021-4-4 22:43
回复 2# kuing
谢谢. 我来补全一下过程
通分,约去abc,等价于证$3 ab c\sum  a+4 \sum a^4\le3\sum a^2 b^2+2 \sum a^3(b+c)$.
由四次舒尔不等式,$\sum a^4+abc\sum a\ge \sum a^3(b+c)$①
对右端使用AM-GM得$\sum a^4+abc\sum a\ge2 \sum b^2c^2$②
$\sum a^4\ge\sum b^2c^2$③
2①+②+③即可

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original poster hbghlyj posted 2021-4-4 22:48
Last edited by hbghlyj 2021-4-4 23:18一个等价形式:
\[ (a b+bc+ca) \left(a^2+b^2+c^2\right)+ \sum_\text{cyc}a^2 (a-b) (a-c)\ge\frac{1}{9} (a+b+c)^4\]

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kuing posted 2021-4-26 22:29
不完全去分母也行:
因为
\[\frac a{ab+bc+ca}-\frac a{a^2+2bc}=\frac{a(a-b)(a-c)}{(ab+bc+ca)(a^2+2bc)},\]所以原不等式等价于
\[\frac{a(a-b)(a-c)}{a^2+2bc}+\frac{b(b-c)(b-a)}{b^2+2ca}+\frac{c(c-a)(c-b)}{c^2+2ab}\geqslant0,\]不妨设 `c=\min\{a,b,c\}`,则上式第三个分式非负,而前两个分式之和为
\[\frac{a(a-b)(a-c)}{a^2+2bc}+\frac{b(b-c)(b-a)}{b^2+2ca}=\frac{(a-b)^2c(2a^2+2b^2-2ac-2bc+3ab)}{(a^2+2bc)(b^2+2ca)},\]显然也非负,所以不等式成立。

这样计算量就小多了,不过仍然不够美……

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