MT【178】平移不变性

(2008年北大自招)
已知$a_1,a_2,a_3;b_1,b_2,b_3$满足
$a_1+a_2+a_3=b_1+b_2+b_3$
$a_1a_2+a_2a_3+a_3a_1=b_1b_2+b_2b_3+b_3b_1$
$\min\{a_1,a_2,a_3\}\le \min\{b_1,b_2,b_3\}$;
求证:$\max\{a_1,a_2,a_3\}\le \max\{b_1,b_2,b_3\}$;

提示:由对称性,不妨设$a_1\le a_2\le a_3;b_1\le b_2\le b_3$如果注意到$a_i,b_i$的平移不变性,不妨每个数都加$-a_1$则条件变为非负数$x_i=a_i-a_1,y_i=b_i-a_1$满足:
$x_2+x_3=y_1+y_2+y_3$
$x_2x_3=y_1y_2+y_2y_3+y_3y_1$
消去$x_2$得
\begin{align*}
0&=x_3^2-(y_1+y_2+y_3)x_3+y_1y_2+y_2y_3+y_3y_1 \\
& =x_3^3-(y_1+y_2+y_3)x_3^2+(y_1y_2+y_2y_3+y_3y_1)x_3 \\
& \ge x_3^3-(y_1+y_2+y_3)x_3^2+(y_1y_2+y_2y_3+y_3y_1)x_3-y_1y_2y_3\\
&=(x_3-y_1)(x_3-y_2)(x_3-y_3)
\end{align*}
故必有$x_3\le y_3 $即$a_3\le b_3$

注:当然直接构造三次也可以，提示:不妨设$a_1\le a_2\le a_3;b_1\le b_2\le b_3$则

$f(x)=(x-a_1)(x-a_2)(x-a_3);g(x)=(x-b_1)(x-b_2)(x-b_3)$;

从而$f(x)=g(x)+a_1a_2a_3-b_1b_2b_3$

故$f(a_1)=f(a_3)=0$得$g(a_1)=g(a_3)$由$a_1\le b_1$易知$a_3\le b_3$