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[不等式] 来自粉丝群的简单二元不等式幻

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kuing posted 2013-9-29 20:00 |Read mode
Tenshi幻(6024*****)  19:52:50
正数x, y
x^3+y^3+xy^2+x^2y-2x^2-2y^2-2xy+2>=0
配方有
\[2(x^3+y^3+xy^2+x^2y-2x^2-2y^2-2xy+2)=(1+x+y)(x+y-2)^2+(x-y)^2(x+y-1),\]
若 $x+y\geqslant1$ 则显然成立,若 $x+y<1$,则易见 $(1+x+y)(x+y-2)^2>1$ 且 $(x-y)^2(x+y-1)>-1$,故亦成立。

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szl6208 posted 2013-9-30 19:17
Last edited by szl6208 2013-9-30 19:27配平方:
1/4*(x-1)^2*(3*x+y)+1/4*(y-1)^2*(3*y+x)+1/4*(x+y+2)*(x-2+y)^2.
初来咋到,环境不熟,多多关照,凑个热闹,谢谢!

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original poster kuing posted 2013-9-30 19:49
配平方:
1/4*(x-1)^2*(3*x+y)+1/4*(y-1)^2*(3*y+x)+1/4*(x+y+2)*(x-2+y)^2.
初来咋到,环境不熟,多多关照 ...
szl6208 发表于 2013-9-30 19:17
nice!
PS、公式输入可以看看置顶。
比如你的公式可以输入成 \$\frac{1}{4}(x-1)^2(3x+y)+\frac{1}{4}(y-1)^2(3y+x)+\frac{1}{4}(x+y+2)(x-2+y)^2\$
就显示 $\frac{1}{4}(x-1)^2(3x+y)+\frac{1}{4}(y-1)^2(3y+x)+\frac{1}{4}(x+y+2)(x-2+y)^2$
╰☆ヾo.海x posted 2013-10-1 06:51
回复 3# kuing

酷儿你怎么这么体贴。。。。

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睡神 posted 2013-10-1 11:03
回复 4# ╰☆ヾo.海x
感动天感动地,把海妹纸也给感动!
除了不懂,就是装懂

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