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USAMO

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琉璃幻 Posted 2013-12-12 06:34 |Read mode
設實數$a,b\in R^{+}$, $a^n=a+1$, $b^{2n}=b+3a$.比較$a,b$大小

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realnumber Posted 2013-12-12 10:40
n是正整数吗?

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其妙 Posted 2013-12-12 12:59

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kuing Posted 2013-12-12 15:07
由 $a^n=a+1>1$ 知 $a>1$,故 $b^{2n}=b+3a>1$ 知 $b>1$,因为
\[a^{2n}-b^{2n}=(a+1)^2-(b+3a)=(a-1)^2+a-b>a-b \riff a^{2n}-a>b^{2n}-b,\]
而 $f(x)=x^{2n}-x$ 显然在 $(1,+\infty)$ 上递增,故 $f(a)>f(b) \iff a>b$。

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isee Posted 2013-12-12 21:23
帖的标题是什么意思?

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kuing Posted 2013-12-12 21:50
回复 5# isee

大概是美国数学奥林匹克

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其妙 Posted 2013-12-13 12:19
帖的标题是什么意思?
isee 发表于 2013-12-12 21:23
isee,救命啊!aurora下载了,输入 公式显示小正方形及其阴影,怎么回事啊?
详见此贴:forum.php?mod=viewthread&tid=1481&extra=page=1

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