[不等式] Inequality use chebyshev?

hbghlyj

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这一步的变换是怎么搞的呢

2021-4-8 16:21

hbghlyj

a,b,c>0,$a+b+c=a^{-1}+b^{-1}+c^{-1}$,求证\[\sum{\frac{(a-1)(a-2)}{a^2+8}}\ge0\]
注:$\frac{1-a}{a^2+8}+\frac{1-b}{b^2+8}+\frac{1-c}{c^2+8}+1\ge3 \left(\frac{1}{a^2+8}+\frac{1}{b^2+8}+\frac{1}{c^2+8}\right)\iff \sum{\frac{(a-1)(a-2)}{a^2+8}}\ge0$
证:若$a,b,c∈\left[\frac{2}{3},\frac{8}{3}\right],$由Chebyshev不等式,$\sum\frac{\left(a-1\right)\left(a-2\right)}{a^2+8}=\sum\left[\frac{a^2-1}{a}\cdot\frac{a\left(a-2\right)}{\left(a+1\right)\left(a^2+8\right)}\right]\geq\frac{1}{3}\sum\frac{a^2-1}{a}\cdot\sum\frac{a\left(a-2\right)}{\left(a+1\right)\left(a^2+8\right)}=0.$
下设$a\in\left(0,\frac{2}{3}\right)\cup\left(\frac{8}{3},+\infty\right)$,则$f\left(a\right)>\frac{1}{20}.$易知在(0,+∞)上$f\left(x\right)=\frac{\left(x-1\right)\left(x-2\right)}{x^2+8}>-\frac{1}{40},$故$f(a)+f(b)+f(c)≥f(a)-2\cdot\frac{1}{40}>0$
(by qianqiangzhen3)

hbghlyj

顶一下
n=3时,1#的不等式如何证明呢

hbghlyj

回复 5# 色k
把此题贴到MSE了,暂时还没有解答.
3#证明了$\frac{1-a}{a^2+8}+\frac{1-b}{b^2+8}+\frac{1-c}{c^2+8}+1\ge3 \left(\frac{1}{a^2+8}+\frac{1}{b^2+8}+\frac{1}{c^2+8}\right)$,但是对此题应该没啥帮助.....