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[不等式] in R

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成龙之龙 posted 2015-8-28 15:53 |Read mode
a,b,c in R,prove that:
$a^4(b-c)^2+b^4(c-a)^2+c^4(a-b)^2 \geq \frac{1}{2}(a-b)^2(b-c)^2(c-a)^2.$

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kuing posted 2015-8-28 18:07
似曾相识,出处是哪里?

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kuing posted 2015-8-28 18:19
比想象中简单,看来出处也没所谓了

由对称性,不妨设 $a\geqslant b\geqslant c$,则
\begin{align*}
a^4(b-c)^2+c^4(a-b)^2 &\geqslant \frac12\bigl(a^2(b-c)+c^2(a-b)\bigr)^2 \\
& =\frac12(a-c)^2(ab+bc-ca)^2 \\
& =\frac12(a-c)^2\bigl(b^2+(a-b)(b-c)\bigr)^2 \\
& \geqslant \frac12(a-c)^2(a-b)^2(b-c)^2,
\end{align*}
即得证,当 $a=b=c$ 或 $b=a+c=0$ 及其轮换时取等。

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original poster 成龙之龙 posted 2015-8-28 19:10

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其妙 posted 2015-8-28 23:17
居然少一个加数也可以

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kuing posted 2015-8-28 23:38
回复 5# 其妙

但是要先设顺序才能拿掉相应的那个

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isee posted 2015-8-29 16:53
好逆天的证明,更逆天的题,这TND一个比一个狠。。。。。。。。

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色k posted 2015-8-29 17:49
好逆天的证明,更逆天的题,这TND一个比一个狠。。。。。。。。
isee 发表于 2015-8-29 16:53
一个比一个狠?是指楼主发的几个题?

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