设 $n\geqslant 2$，$x_1, x_2,\cdots,x_n$ 均为实数，且 $\displaystyle \sum\limits_{i=1}^nx^2_i+\sum_{i=1}^{n-1}x_ix_{i+1}=1$．对于每一个固定的 $k(k\in N,1\leqslant k\leqslant n)$，求 $|x_k|$ 的最大值．
【难度】
【出处】
1998第13届CMO试题
【标注】
【答案】
略
【解析】
由已知条件得 $\displaystyle 2\sum\limits_{i=1}^nx_i^2+2\sum_{i=1}^{n-1}x_ix_{i+1}=2$ 即 $x_1^2+(x_1+x_2)^2+(x_2+x_3)^2+\cdots+(x_{n-2}+x_{n-1})^2+(x_{n-1}+x_n)^2+x_n^2=2$ ① 由均值不等式得
$\sqrt{\dfrac{x_{1}^{2}+{{\left({{x}_{1}}+{{x}_{2}} \right)}^{2}}+\cdots +{{\left( {{x}_{k-1}}+{{x}_{k}}\right)}^{2}}}{k}}\geqslant \dfrac{\left| {{x}_{1}} \right|+\left|{{x}_{1}}+{{x}_{2}} \right|+\cdots +\left| {{x}_{k-1}}+{{x}_{k}} \right|}{k}\\\geqslant\dfrac{\left| {{x}_{1}}-\left( {{x}_{1}}+{{x}_{2}} \right)+\cdots +{{\left( -1\right)}^{k-1}}\left( {{x}_{k-1}}+{{x}_{k}} \right) \right|}{k}=\dfrac{\left|{{x}_{k}} \right|}{k}$
即 $x_{1}^{2}+{{\left({{x}_{1}}+{{x}_{2}} \right)}^{2}}+\cdots +{{\left( {{x}_{k-1}}+{{x}_{k}}\right)}^{2}}\geqslant \dfrac{x_{k}^{2}}{k}$ ② 同理
$\sqrt{\dfrac{{{\left( {{x}_{k}}+{{x}_{k+1}}\right)}^{2}}+\cdots +{{\left( {{x}_{n-1}}+{{x}_{n}}\right)}^{2}}+x_{n}^{2}}{n-k+1}}\geqslant \dfrac{\left| {{x}_{k}}+{{x}_{k+1}}\right|+\cdots +\left| {{x}_{n-1}}+{{x}_{n}} \right|+\left| {{x}_{n}}\right|}{n-k+1}\\\geqslant \dfrac{\left| {{x}_{n}}-\left( {{x}_{n-1}}+{{x}_{n}} \right)+\cdots+{{\left( -1 \right)}^{n-k}}\left( {{x}_{k}}+{{x}_{k+1}} \right)\right|}{n-k+1}=\dfrac{\left| {{x}_{k}} \right|}{n-k+1}$
即 ${{\left({{x}_{k}}+{{x}_{k+1}} \right)}^{2}}+\cdots +{{\left( {{x}_{n-1}}+{{x}_{n}}\right)}^{2}}+x_{n}^{2}\geqslant \dfrac{x_{k}^{2}}{n-k+1}$ ③
由 ② + ③ 及 ① 得 $2\geqslant (\dfrac{1}{k}+\dfrac{1}{n-k+1})x_k^2$ 即
$\left|{{x}_{k}} \right|\leqslant \sqrt{\dfrac{2k\left( n+1-k \right)}{n+1}},k=1,2,\cdots,n$ ④
由上可知，当且仅当 ${{x}_{1}}=-\left({{x}_{1}}+{{x}_{2}} \right)={{x}_{2}}+{{x}_{3}}=\cdots ={{\left( -1\right)}^{k-1}}\left( {{x}_{k-1}}+{{x}_{k}} \right)$ 及 ${{x}_{k}}+{{x}_{k+1}}=-\left({{x}_{k+1}}+{{x}_{k+2}} \right)=\cdots ={{\left( -1 \right)}^{n-k}}{{x}_{n}}$ 时，④ 中的等号成立，即当且仅当 ${{x}_{i}}={{x}_{k}}{{\left(-1 \right)}^{i-k}}\dfrac{i}{k},i=1,2,\cdots ,k-1$ 且 ${{x}_{j}}={{x}_{k}}{{\left( -1\right)}^{j-k}}\dfrac{n+1-j}{n-k+1},j=k+1,k+2,\cdots ,n$ 时 $|x_k|=\sqrt{\dfrac{2k(n+1-k)}{n+1}}$ 综上所述，可得 $|x_k|_\max=\sqrt{\dfrac{2k(n+1-k)}{n+1}}$
$\sqrt{\dfrac{x_{1}^{2}+{{\left({{x}_{1}}+{{x}_{2}} \right)}^{2}}+\cdots +{{\left( {{x}_{k-1}}+{{x}_{k}}\right)}^{2}}}{k}}\geqslant \dfrac{\left| {{x}_{1}} \right|+\left|{{x}_{1}}+{{x}_{2}} \right|+\cdots +\left| {{x}_{k-1}}+{{x}_{k}} \right|}{k}\\\geqslant\dfrac{\left| {{x}_{1}}-\left( {{x}_{1}}+{{x}_{2}} \right)+\cdots +{{\left( -1\right)}^{k-1}}\left( {{x}_{k-1}}+{{x}_{k}} \right) \right|}{k}=\dfrac{\left|{{x}_{k}} \right|}{k}$
即 $x_{1}^{2}+{{\left({{x}_{1}}+{{x}_{2}} \right)}^{2}}+\cdots +{{\left( {{x}_{k-1}}+{{x}_{k}}\right)}^{2}}\geqslant \dfrac{x_{k}^{2}}{k}$ ② 同理
$\sqrt{\dfrac{{{\left( {{x}_{k}}+{{x}_{k+1}}\right)}^{2}}+\cdots +{{\left( {{x}_{n-1}}+{{x}_{n}}\right)}^{2}}+x_{n}^{2}}{n-k+1}}\geqslant \dfrac{\left| {{x}_{k}}+{{x}_{k+1}}\right|+\cdots +\left| {{x}_{n-1}}+{{x}_{n}} \right|+\left| {{x}_{n}}\right|}{n-k+1}\\\geqslant \dfrac{\left| {{x}_{n}}-\left( {{x}_{n-1}}+{{x}_{n}} \right)+\cdots+{{\left( -1 \right)}^{n-k}}\left( {{x}_{k}}+{{x}_{k+1}} \right)\right|}{n-k+1}=\dfrac{\left| {{x}_{k}} \right|}{n-k+1}$
即 ${{\left({{x}_{k}}+{{x}_{k+1}} \right)}^{2}}+\cdots +{{\left( {{x}_{n-1}}+{{x}_{n}}\right)}^{2}}+x_{n}^{2}\geqslant \dfrac{x_{k}^{2}}{n-k+1}$ ③
由 ② + ③ 及 ① 得 $2\geqslant (\dfrac{1}{k}+\dfrac{1}{n-k+1})x_k^2$ 即
$\left|{{x}_{k}} \right|\leqslant \sqrt{\dfrac{2k\left( n+1-k \right)}{n+1}},k=1,2,\cdots,n$ ④
由上可知，当且仅当 ${{x}_{1}}=-\left({{x}_{1}}+{{x}_{2}} \right)={{x}_{2}}+{{x}_{3}}=\cdots ={{\left( -1\right)}^{k-1}}\left( {{x}_{k-1}}+{{x}_{k}} \right)$ 及 ${{x}_{k}}+{{x}_{k+1}}=-\left({{x}_{k+1}}+{{x}_{k+2}} \right)=\cdots ={{\left( -1 \right)}^{n-k}}{{x}_{n}}$ 时，④ 中的等号成立，即当且仅当 ${{x}_{i}}={{x}_{k}}{{\left(-1 \right)}^{i-k}}\dfrac{i}{k},i=1,2,\cdots ,k-1$ 且 ${{x}_{j}}={{x}_{k}}{{\left( -1\right)}^{j-k}}\dfrac{n+1-j}{n-k+1},j=k+1,k+2,\cdots ,n$ 时 $|x_k|=\sqrt{\dfrac{2k(n+1-k)}{n+1}}$ 综上所述，可得 $|x_k|_\max=\sqrt{\dfrac{2k(n+1-k)}{n+1}}$
答案
解析
备注
