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[不等式] 比较简单的不等式

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icesheep Posted 2013-10-14 23:13 |Read mode
$0 < x < 1,0 < y < \infty$ 证明:\[y{x^y}\left( {1 - x} \right) < \frac{1}{e}\]

应该是有 \[{x^y}\left( {1 - x} \right) \leqslant {\left( {\frac{y}{{y + 1}}} \right)^y}\frac{1}{{y + 1}}\]

然后由 ${\left( {1 - \frac{1}{x}} \right)^x}$ 单调性即可,不过上面那个不等式怎么来的 =。=

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 Author| icesheep Posted 2013-10-14 23:21
真是老了老了,这个均值都想半天,沉了吧

\[\frac{x}{{y + 1}} + \frac{{1 - x}}{{y + 1}} \geqslant {\left( {\frac{x}{y}} \right)^{\frac{y}{{y + 1}}}}{\left( {1 - x} \right)^{\frac{1}{{y + 1}}}}\]

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kuing Posted 2013-10-14 23:25
居然和一年前秋风问的题一样 kkkkuingggg.haotui.com/viewthread.php?tid=838

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