(Self-recruited by Peking University in 2008)
It is known that $a_1,a_2,a_3;b_1,b_2,b_3$ satisfy
$a_1+a_2+a_3=b_1+b_2+b_3$
$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\}$;
Verify:$\max\{a_1,a_2,a_3\}\le \max\{b_1,b_2, b_3\}$;
Hint: From the symmetry, you might as well set $a_1\le a_2\le a_3;b_1\le b_2\le b_3$ If you notice the translation invariance of $a_i, b_i$, you might as well add $-a_1$ to each number, then the condition Become a non-negative number $x_i=a_i-a_1, y_i=b_i-a_1$ satisfy:
$x_2+x_3=y_1+y_2+y_3$
$x_2x_3=y_1y_2+y_2y_3+y_3y_1$
Eliminate $x_2$ to get
\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 *}
So there must be $x_3\le y_3$, that is, $a_3\le b_3$