2. The numbers $a, b$, and $c$ are such that $a>b$ and $(a-b)(b-c)(c-a)>0$. Which is greater: $a$ or $c$?
Answer. $a>c$. Solution. Obviously, among the three given numbers, there are no equal ones. Since $a>b, a-b>0$, from which $(b-c)(c-a)>0$. Suppose $c>a$. Then $c-a>0$ and $b-c>0$, that is, $b>c>a$ - contradiction.