Suranyi不等式
Au邪
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个人记录
若x_k>0,则(n-1)\sum\limits_{k=1}^n x_k^n+n\prod\limits_{k=1}^n x_k\ge (\sum\limits_{k=1}^n x_k)(\sum\limits_{k=1}^n x_k^{n-1}).
Proof:
当n=1,2时显然成立.
假设当n\le k时成立
则当n=k+1时.即证k\sum\limits_{i=1}^{k+1}x_i^{k+1}+(k+1)\prod\limits_{i=1}^{k+1}x_i\ge(\sum\limits_{i=1}^{k+1}x_i)(\sum\limits_{i=1}^{k+1}x_i^k)
由齐次性和轮换对称性不妨设x_1\ge x_2\ge x_3 \ge...\ge x_{k+1}.\sum\limits_{i=1}^k x_i=1
\therefore x_{k+1}\le \frac{1}{k}.
又由归纳假设可得k x_{k+1}\prod\limits_{i=1}^k x_i\ge x_{k+1}\sum\limits_{i=1}^k x_i^{k-1}-(k-1)x_{k+1}\sum\limits_{i=1}^k x_i^k
因此只需证
(k\sum\limits_{i=1}^k x_i^{k+1}-\sum\limits_{i=1}^k x_i^k)-x_{k+1}(k\sum\limits_{i=1}^k x_i^{k}-\sum\limits_{i=1}^k x_i^{k-1})+x_{k+1}(\prod\limits_{i=1}^k x_i+(k-1)x_{k+1}^k-x_{k+1}^{k-1})\ge0
又由切比雪夫不等式得 k\sum\limits_{i=1}^k x_i^k-\sum\limits_{i=1}^k x_i^{k-1}\ge 0
又\because k x_i^{k+1}+\frac{1}{k}x_i^{k-1}\ge 2x_i^k
则k\sum\limits_{i=1}^k x_i^k-\sum\limits_{i=1}^k x_i^{k-1}\ge \frac{1}{k}(k\sum\limits_{i=1}^k x_i^k-\sum\limits_{i=1}^k x_i^{k-1})\ge x_{k+1}(k\sum\limits_{i=1}^k x_i^k-\sum\limits_{i=1}^k x_i^{k-1}).
\therefore (k\sum\limits_{i=1}^k x_i^{k+1}-\sum\limits_{i=1}^k x_i^k)-x_{k+1}(k\sum\limits_{i=1}^k x_i^{k}-\sum\limits_{i=1}^k x_i^{k-1})+x_{k+1}(\prod\limits_{i=1}^k x_i+(k-1)x_{k+1}^k-x_{k+1}^{k-1})
\ge x_{k+1}(\prod\limits_{i=1}^k x_i+(k-1)x_{k+1}^k-x_{k+1}^{k-1})
\ge x_{k+1}^k+x_{k+1}^{k-1}\sum\limits_{i=1}^k(x_i-x_{k+1})+(k-1)x_{k+1}^k-x_{k+1}^{k-1}
=0
因此n=k+1时成立
则由数学归纳法得对\forall n\in \mathbb{N}_+均成立