题目

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设实数 ${{x}_{1}}\text{,}{{x}_{2}}\text{,}\cdots \text{,}{{x}_{n}}$ 满足 $\displaystyle \sum\limits_{i\text{=}1}^{n}{x_{i}^{2}\text{=}1\left( n\geqslant 2 \right)}$ 。求证：$\displaystyle \sum\limits_{k\text{=}1}^{n}{{{\left( 1-\frac{k}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}} \right)}^{2}}}\frac{x_{k}^{2}}{k}\leqslant {{\left( \frac{n-1}{n+1} \right)}^{2}}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}$，并确定等号成立的条件。

【难度】

【出处】

2010第9届CGMO试题

【标注】

知识点
>
二试代数部分

【答案】

略

【解析】

注意到 $\displaystyle \sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}{{\left(1-\frac{k}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}}\right)}^{2}}}\text{=}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}-\sum\limits_{k\text{=}1}^{n}{\frac{2x_{k}^{2}}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}}+}\sum\limits_{k\text{=}1}^{n}{\frac{kx_{k}^{2}}{{{\left(\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}\right)}^{2}}}}\text{=}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}-\frac{2\sum\limits_{k\text{=}1}^{n}{x_{k}^{2}}}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}}+\sum\limits_{k\text{=}1}^{n}{\frac{kx_{k}^{2}}{{{\left(\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}} \right)}^{2}}}}\text{=}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}-\frac{1}{\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}}$ 。于是，要证原不等式只需证 $\displaystyle 1-{{\left( \frac{n-1}{n+1} \right)}^{2}}\leqslant\frac{1}{\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\cdot\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}}\Leftrightarrow \sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\cdot\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}\leqslant \frac{{{\left( n+1\right)}^{2}}}{4n}$ 。事实上，$\displaystyle \sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\cdot\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}\text{=}\frac{1}{n}\left[ \left(\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}} \right)\left(n\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}} \right) \right]\leqslant\frac{1}{n}\cdot \frac{1}{4}{{\left(\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}+n\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}\right)}^{2}}=\frac{1}{4n}{{\left[ \sum\limits_{k\text{=}1}^{n}{\left(k+\frac{n}{k} \right)x_{k}^{2}} \right]}^{2}}\leqslant \frac{1}{4n}{{\left[\sum\limits_{k=1}^{n}{\left( n+1 \right)}x_{k}^{2} \right]}^{2}}=\frac{{{\left(n+1 \right)}^{2}}}{4n}$ 。若要等号成立，首先必须有 $\displaystyle \sum\limits_{k=1}^{n}{\left( k+\frac{n}{k}\right)}x_{k}^{2}\text{=}\sum\limits_{k=1}^{n}{\left( n+1 \right)}x_{k}^{2}$ 。而当 $k=2,3,\cdots,n-1$ 时，有 $k+\frac{n}{k}$ < $n+1$ 。因此，${{x}_{2}}={{x}_{3}}=\cdots={{x}_{n-1}}=0$ 。其次，要 $\displaystyle \sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\text{=}n\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}$，即 $x_{1}^{2}+nx_{n}^{2}\text{=}nx_{1}^{2}+x_{n}^{2}$ 。故 $x_{1}^{2}\text{=}x_{n}^{2}$ 。由 $\displaystyle \sum\limits_{i\text{=}1}^{n}{x_{i}^{2}}\text{=}1$ 和 $x_{2}^{2}\text{=}x_{3}^{2}\text{=}\cdots\text{=}x_{n-1}^{2}\text{=}0$，得 $x_{1}^{2}\text{=}x_{n}^{2}\text{=}\frac{1}{2}$，而这也是等号成立的充分条件。因此，等号成立的充要条件为 ${{x}_{1}}\text{=}\pm\frac{\sqrt{2}}{2}$，${{x}_{2}}={{x}_{3}}=\cdots ={{x}_{n-1}}=0$，${{x}_{n}}\text{=}\pm\frac{\sqrt{2}}{2}$ 。

答案解析

注意到 $\displaystyle \sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}{{\left(1-\frac{k}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}}\right)}^{2}}}\text{=}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}-\sum\limits_{k\text{=}1}^{n}{\frac{2x_{k}^{2}}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}}+}\sum\limits_{k\text{=}1}^{n}{\frac{kx_{k}^{2}}{{{\left(\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}\right)}^{2}}}}\text{=}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}-\frac{2\sum\limits_{k\text{=}1}^{n}{x_{k}^{2}}}{\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}}}+\sum\limits_{k\text{=}1}^{n}{\frac{kx_{k}^{2}}{{{\left(\sum\limits_{i\text{=}1}^{n}{ix_{i}^{2}} \right)}^{2}}}}\text{=}\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}-\frac{1}{\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}}$ 。于是，要证原不等式只需证 $\displaystyle 1-{{\left( \frac{n-1}{n+1} \right)}^{2}}\leqslant\frac{1}{\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\cdot\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}}\Leftrightarrow \sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\cdot\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}\leqslant \frac{{{\left( n+1\right)}^{2}}}{4n}$ 。事实上，$\displaystyle \sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\cdot\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}\text{=}\frac{1}{n}\left[ \left(\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}} \right)\left(n\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}} \right) \right]\leqslant\frac{1}{n}\cdot \frac{1}{4}{{\left(\sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}+n\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}\right)}^{2}}=\frac{1}{4n}{{\left[ \sum\limits_{k\text{=}1}^{n}{\left(k+\frac{n}{k} \right)x_{k}^{2}} \right]}^{2}}\leqslant \frac{1}{4n}{{\left[\sum\limits_{k=1}^{n}{\left( n+1 \right)}x_{k}^{2} \right]}^{2}}=\frac{{{\left(n+1 \right)}^{2}}}{4n}$ 。 若要等号成立，首先必须有 $\displaystyle \sum\limits_{k=1}^{n}{\left( k+\frac{n}{k}\right)}x_{k}^{2}\text{=}\sum\limits_{k=1}^{n}{\left( n+1 \right)}x_{k}^{2}$ 。而当 $k=2,3,\cdots,n-1$ 时，有 $k+\frac{n}{k}$ < $n+1$ 。因此，${{x}_{2}}={{x}_{3}}=\cdots={{x}_{n-1}}=0$ 。其次，要 $\displaystyle \sum\limits_{k\text{=}1}^{n}{kx_{k}^{2}}\text{=}n\sum\limits_{k\text{=}1}^{n}{\frac{x_{k}^{2}}{k}}$，即 $x_{1}^{2}+nx_{n}^{2}\text{=}nx_{1}^{2}+x_{n}^{2}$ 。故 $x_{1}^{2}\text{=}x_{n}^{2}$ 。由 $\displaystyle \sum\limits_{i\text{=}1}^{n}{x_{i}^{2}}\text{=}1$ 和 $x_{2}^{2}\text{=}x_{3}^{2}\text{=}\cdots\text{=}x_{n-1}^{2}\text{=}0$，得 $x_{1}^{2}\text{=}x_{n}^{2}\text{=}\frac{1}{2}$，而这也是等号成立的充分条件。因此，等号成立的充要条件为 ${{x}_{1}}\text{=}\pm\frac{\sqrt{2}}{2}$，${{x}_{2}}={{x}_{3}}=\cdots ={{x}_{n-1}}=0$，${{x}_{n}}\text{=}\pm\frac{\sqrt{2}}{2}$ 。