简单的一个不等式

陈洪葛 posted @ May 05, 2014 05:54:43 PM in 不等式 , 783 阅读

设$x_{1},x_{2},\cdots,x_{n}$是非负实数,求证:
\[ \prod_{k=1}^{n}(1+x_{k})\leq 1+\sum_{k=1}^{n}\frac{1}{k!}\left(1-\frac{k}{2n}\right)^{k-1}\left(\sum_{k=1}^{n}x_{k} \right)^{k}\]
证明:为了方便,我们记
\[ S=\sum_{k=1}^{n}x_{k}\]
由AM-GM不等式,得到
\[ \prod_{k=1}^{n}(1+x_{k})\leq \left(1+\frac{S}{n}\right)^{n}\]
这时,只要证明
\[ \left(1+\frac{S}{n}\right)^{n}\leq 1+\sum_{k=1}^{n}\frac{1}{k!}\left(1-\frac{k}{2n}\right)^{k-1}\left(\sum_{k=1}^{n}x_{k} \right)^{k}\]
注意到
\[ \left(1+\frac{S}{n}\right)^{n}=1+C_{n}^{1}\cdot\frac{S}{n}+\cdots+C_{n}^{k}\left(\frac{S}{n}\right)^{k}+\cdots+C_{n}^{n}\left(\frac{S}{n}\right)^{n}\]
兹证明
\[ C_{n}^{k}\left(\frac{S}{n}\right)^{k}\leq \frac{1}{k!}\left(1-\frac{k}{2n}\right)^{k-1}\left(\sum_{k=1}^{n}x_{k} \right)^{k}\]
稍微化简下,就是
\[ [2n-2(k-1)][2n-2(k-2)]\cdots (2n-2)\leq (2n-k)^{k-1} \]
注意到首尾项的细节,就有AM-GM
\[ [2n-2(k-1)](2n-2)\leq \left(\frac{2n-2(k-1)+2n-2}{2}\right)^{2} \]
把它们配对,马上得到不等式。
 


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