$a_{1}, a_{2}, \cdots, a_{n}$ 为实数，如果它们中任意两数之和非负，那么对于满足 $x_{1}+x_{2}+\cdots+x_{n}=1$ 的任意非负实数 $x_{1}, x_{2}, \cdots, x_{n}$，有不等式 $a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \geqslant a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2}$ 成立．请证明上述命题及其逆命题．
【难度】
【出处】
1986第1届CMO试题
【标注】
【答案】
略
【解析】
先来证明逆命题．对于 $1,2, \cdots, n$ 中的任何不同的两个数 $i,j$，取 $x_{i}=x_{j}=\frac{1}{2}$，其余的 $x_{k}=0, k \neq i, j$．由于这一组数能使不等式成立，得到 $\frac{1}{2}\left(a_{i}+a_{j}\right) \geqslant \frac{1}{4}\left(a_{i}+a_{j}\right)$ 即 $a_{i}+a_{j} \geqslant 0$，这就证完了逆命题．至于命题本身，有许许多多的证法，限于篇幅，这里只列举三种．
证法一
由于 $x_{1}+x_{2}+\cdots+x_{n}=1$，所以
$ a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=\left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right) \left(x_{1}+x_{2}+\cdots+x_{n}\right)\\= a_{1} x_{1}^{2}+a_{1} x_{1} x_{2}+\cdots+a_{1} x_{1} x_{n}+ a_{2} x_{1} x_{2}+a_{2} x_{2}^{2}+\cdots+a_{2} x_{2} x_{n}+\cdots+ a_{n} x_{1} x_{n}+a_{n} x_{2} x_{n}+\cdots+a_{n} x_{n}^{2} $
由此可知
$a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}-\left(a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2}\right)\\=\left(a_{1}+a_{2}\right) x_{1} x_{2}+\left(a_{1}+a_{3}\right) x_{1} x_{3}+\cdots+\left(a_{1}+a_{n}\right) x_{1} x_{n}+\left(a_{2}+a_{3}\right) x_{2} x_{3}+\cdots+\left(a_{2}+a_{n}\right) x_{2} x_{n}+\cdots+\left(a_{n-1}+a_{n}\right) x_{n-1} x_{n}$
由题设可知，最后的这个和式是非负的．证毕．
证法二
当时0 $\leqslant x_{i} \leqslant 1$，总有 $x_{i}^{2} \leqslant x_{i}$；因此如果 $a_{6} \geqslant 0$ 对 $i=1,2, \cdots, n$ 均成立时，不等式显然是成立的．设在 $a_{1}, a_{2}, \cdots, a_{n}$ 中有一个负数，无妨设 $a_1<0$，由条件 $a_1+a_{j} \geqslant 0(j=2,3, \cdots, n)$，可知 $a_{j} \geqslant-a_{1}>0$．因此
$a_{1} x_{1}\left(1-x_{1}\right)+a_{2} x_{2}\left(1-x_{2}\right)+\cdots+a_{n} x_{n}\left(1-x_{n}\right) \geqslant$
$a_{1} x_{1}\left(1-x_{1}\right)-a_{1} x_{2}\left(1-x_{2}\right)-\cdots-a_{1} x_{n}\left(1-x_{n}\right)=$
$-a_{1}\left(x_{2}\left(1-x_{2}\right)+\cdots+x_{n}\left(1-x_{n}\right)-x_{1}\left(1-x_{1}\right)\right)=$
$-a_{1}\left(x_{2}\left(1-x_{2}\right)+\cdots+x_{n}\left(1-x_{n}\right)-x_{1}\left(x_{2}+\cdots+x_{n}\right)\right)=$
$-a_{1}\left(x_{2}\left(1-x_{1}-x_{2}\right)+\cdots+x_{n}\left(1-x_{1}-x_{n}\right)\right) \geqslant$
$-a_{1}\left(x_{2} \cdot 0+\cdots+x_{n} \cdot 0\right)=0$
证毕．
证法三
有许多同学试图用数学归纳法来证明本题，遗憾的是，只有个别同学是正确的．本题是可以通过数学归纳法来解的，其中的一种证法如下．$n=2$ 时是很容易证明的，这时 $x_1=1-x_2,x_2=1-x_1$，故有 $ a_{1} x_{1}\left(1-x_{1}\right)+a_{2} x_{2}\left(1-x_{2}\right)= a_{1} x_{1} x_{2}+a_{2} x_{1} x_{2}=\left(a_{1}+a_{2}\right) x_{1} x_{2} $ 它显然不是负数．设 $n=k$ 时命题成立．当 $x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}=1$ 时，若 $x_{k+1}=1$，那么只能是 $x_{1}=x_{2}=\cdots=x_{k}=0$，不等式成为 $a_{k+1} \geqslant a_{k+1}$，这是合理的，故设 $0 \leqslant x_{k+1}<1$，这时 $\frac{x_{1}}{1-x_{k+1}}+\frac{x_{2}}{1-x_{k+1}}+\cdots+\frac{x_{k}}{1-x_{k+1}}=1$ 依归纳假设有 $\displaystyle \sum\limits_{i=1}^{k} a_{i} \frac{x_{i}}{1-x_{k+1}} \geqslant \sum_{i=1}^{k} a_{i}\left(\frac{x_{i}}{1-x_{k+1}}\right)^{2}$（这里及以后，均设想读者掌握了求和记号 $\displaystyle \sum$ 的意义和性质）．由上式可得 $\displaystyle \left(1-x_{k+1}\right) \sum\limits_{i=1}^{k} a_{i} x_{i} \geqslant \sum_{i=1}^{k} a_{i} x_{i}^{2}$ 即 $\displaystyle \sum\limits_{i=1}^{k} a_{i} x_{i} \geqslant \sum_{i=1}^{k} a_{i} x_{i}^{2}+x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}$ 将 $a_{k+1} x_{k+1}$ 同时加到上式两边得 $\displaystyle \sum\limits_{i=1}^{k+1} a_{i} x_{i} \geqslant \sum_{i=1}^{k+1} a_{i} x_{i}^{2}+x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}-a_{k+1} x_{k+1}^{2}$ 让我们看
$x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}-a_{k+1} x_{k+1}^{2}=$
$x_{k+1} \sum_{i \neq 1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}\left(1-x_{k+1}\right)=$
$x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}\left(x_{1}+x_{2}+\cdots+x_{k}\right)=$
$x_{k+1} \sum_{i=1}^{k}\left(a_{i}+a_{k+1}\right) x_{i} \geqslant 0$
因此就得证了 $\displaystyle \sum\limits_{i=1}^{k+1} a_{i} x_{i} \geqslant \sum_{i=1}^{k+1} a_{i} x_{i}^{2}$ 这样就完成了归纳法证明．
证法一
由于 $x_{1}+x_{2}+\cdots+x_{n}=1$，所以
$ a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=\left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right) \left(x_{1}+x_{2}+\cdots+x_{n}\right)\\= a_{1} x_{1}^{2}+a_{1} x_{1} x_{2}+\cdots+a_{1} x_{1} x_{n}+ a_{2} x_{1} x_{2}+a_{2} x_{2}^{2}+\cdots+a_{2} x_{2} x_{n}+\cdots+ a_{n} x_{1} x_{n}+a_{n} x_{2} x_{n}+\cdots+a_{n} x_{n}^{2} $
由此可知
$a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}-\left(a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{n} x_{n}^{2}\right)\\=\left(a_{1}+a_{2}\right) x_{1} x_{2}+\left(a_{1}+a_{3}\right) x_{1} x_{3}+\cdots+\left(a_{1}+a_{n}\right) x_{1} x_{n}+\left(a_{2}+a_{3}\right) x_{2} x_{3}+\cdots+\left(a_{2}+a_{n}\right) x_{2} x_{n}+\cdots+\left(a_{n-1}+a_{n}\right) x_{n-1} x_{n}$
由题设可知，最后的这个和式是非负的．证毕．
证法二
当时0 $\leqslant x_{i} \leqslant 1$，总有 $x_{i}^{2} \leqslant x_{i}$；因此如果 $a_{6} \geqslant 0$ 对 $i=1,2, \cdots, n$ 均成立时，不等式显然是成立的．设在 $a_{1}, a_{2}, \cdots, a_{n}$ 中有一个负数，无妨设 $a_1<0$，由条件 $a_1+a_{j} \geqslant 0(j=2,3, \cdots, n)$，可知 $a_{j} \geqslant-a_{1}>0$．因此
$a_{1} x_{1}\left(1-x_{1}\right)+a_{2} x_{2}\left(1-x_{2}\right)+\cdots+a_{n} x_{n}\left(1-x_{n}\right) \geqslant$
$a_{1} x_{1}\left(1-x_{1}\right)-a_{1} x_{2}\left(1-x_{2}\right)-\cdots-a_{1} x_{n}\left(1-x_{n}\right)=$
$-a_{1}\left(x_{2}\left(1-x_{2}\right)+\cdots+x_{n}\left(1-x_{n}\right)-x_{1}\left(1-x_{1}\right)\right)=$
$-a_{1}\left(x_{2}\left(1-x_{2}\right)+\cdots+x_{n}\left(1-x_{n}\right)-x_{1}\left(x_{2}+\cdots+x_{n}\right)\right)=$
$-a_{1}\left(x_{2}\left(1-x_{1}-x_{2}\right)+\cdots+x_{n}\left(1-x_{1}-x_{n}\right)\right) \geqslant$
$-a_{1}\left(x_{2} \cdot 0+\cdots+x_{n} \cdot 0\right)=0$
证毕．
证法三
有许多同学试图用数学归纳法来证明本题，遗憾的是，只有个别同学是正确的．本题是可以通过数学归纳法来解的，其中的一种证法如下．$n=2$ 时是很容易证明的，这时 $x_1=1-x_2,x_2=1-x_1$，故有 $ a_{1} x_{1}\left(1-x_{1}\right)+a_{2} x_{2}\left(1-x_{2}\right)= a_{1} x_{1} x_{2}+a_{2} x_{1} x_{2}=\left(a_{1}+a_{2}\right) x_{1} x_{2} $ 它显然不是负数．设 $n=k$ 时命题成立．当 $x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}=1$ 时，若 $x_{k+1}=1$，那么只能是 $x_{1}=x_{2}=\cdots=x_{k}=0$，不等式成为 $a_{k+1} \geqslant a_{k+1}$，这是合理的，故设 $0 \leqslant x_{k+1}<1$，这时 $\frac{x_{1}}{1-x_{k+1}}+\frac{x_{2}}{1-x_{k+1}}+\cdots+\frac{x_{k}}{1-x_{k+1}}=1$ 依归纳假设有 $\displaystyle \sum\limits_{i=1}^{k} a_{i} \frac{x_{i}}{1-x_{k+1}} \geqslant \sum_{i=1}^{k} a_{i}\left(\frac{x_{i}}{1-x_{k+1}}\right)^{2}$（这里及以后，均设想读者掌握了求和记号 $\displaystyle \sum$ 的意义和性质）．由上式可得 $\displaystyle \left(1-x_{k+1}\right) \sum\limits_{i=1}^{k} a_{i} x_{i} \geqslant \sum_{i=1}^{k} a_{i} x_{i}^{2}$ 即 $\displaystyle \sum\limits_{i=1}^{k} a_{i} x_{i} \geqslant \sum_{i=1}^{k} a_{i} x_{i}^{2}+x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}$ 将 $a_{k+1} x_{k+1}$ 同时加到上式两边得 $\displaystyle \sum\limits_{i=1}^{k+1} a_{i} x_{i} \geqslant \sum_{i=1}^{k+1} a_{i} x_{i}^{2}+x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}-a_{k+1} x_{k+1}^{2}$ 让我们看
$x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}-a_{k+1} x_{k+1}^{2}=$
$x_{k+1} \sum_{i \neq 1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}\left(1-x_{k+1}\right)=$
$x_{k+1} \sum_{i=1}^{k} a_{i} x_{i}+a_{k+1} x_{k+1}\left(x_{1}+x_{2}+\cdots+x_{k}\right)=$
$x_{k+1} \sum_{i=1}^{k}\left(a_{i}+a_{k+1}\right) x_{i} \geqslant 0$
因此就得证了 $\displaystyle \sum\limits_{i=1}^{k+1} a_{i} x_{i} \geqslant \sum_{i=1}^{k+1} a_{i} x_{i}^{2}$ 这样就完成了归纳法证明．
答案
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