MT【148】凸数列

(2018浙江省赛13题)

设实数$x_1,x_2,\cdots,x_{2018}$满足$x_{n+1}^2\le x_nx_{n+2},(n=1,2,\cdots,2016)$和$\prod\limits_{k=1}^{2018}x_k=1$
证明:$x_{1009}x_{1010}\le1.$

证明:事实上,由$x_{n+1}^2\le x_nx_{n+2}$易知道,下标为奇数的项同号,下标为偶数的项同号.我们不妨考虑$x_k>0,(k=1,2,\cdots，2018)$(若都为负数只需每一项都变为原来的相反数即可.一正一负的情况下,$x_{1009}x_{1010}<0\le1$,显然)
记$a_n=\ln x_n,(n=1,2\cdots,2018)$.两边取对数,条件变为$2a_{n+1}\le a_{n}+a_{n+2},\sum\limits_{k=1}^{2018}{a_k}=0$,只需证明:$a_{1009}+a_{1010}\le0.\textbf{由凸函数性质}:$
$$a_m+a_n\le a_s+a_t,(m+n=s+t,1\le s\le m,n\le t)$$
则$a_{1009}+a_{1010}\le a_1+a_{2018},a_{1009}+a_{1010}\le a_2+a_{2017},\cdots,a_{1009}+a_{1010}\le a_{1009}+a_{1010}$故$1009(a_{1009}+a_{1010})\le\sum\limits_{k=1}^{2018}{a_k}=0$, 得证.

评论:$\{a_{n+1}-a_n\}$单调不减,则$\{a_n\}$称为凸数列,它有以下性质:

$1.a_n+a_{n+2}\ge 2a_{n+1}$

$2.a_n-a_m\ge(n-m)(a_{m+1}-a_m)$
$3.\dfrac{a_n-a_m}{n-m}\ge\dfrac{a_m-a_k}{m-k}(1\le k<m<n)$
$4.a_m+a_n\le a_s+a_t,(m+n=s+t,1\le s\le m,n\le t)$