Mongolian Mathematical Olympiad

It is obvious that $a < 2 < c$. Therefore
$$3(a-1)(a-3) - (a-b)(a-c) = 3(b-1)(b-3) - (b-1)(b-a) =$$
$$3(c-1)(c-3) - (c-a)(c-b) = 0$$
$$\left( \begin{array}{l} 0 = 9 - ab - bc - ca = 9 + a^2 - 2a(b + c) - (a - b)(a - c) = \\ \qquad = 9 + a^2 - 2a(6 - a) - (a - b)(a - c) = 3(a - 1)(a - 3) - (a - b)(a - c) \end{array} \right)$$
Since $a < b < c$ we get $(a-b)(b-c) > 0$, $(b-a)(b-c) < 0$, $(c-a)(c-b) > 0$ and $(a-1)(a-3) > 0$, $(b-1)(b-3) > 0$, $(c-1)(c-3) > 0$. Therefore $a < 1 < b < 3 < c$. It follows that $(1-a)(1-b) < 0$.
Since $0 = 9 - ab - bc - ca = (c-1)(c-4) - (1-a)(1-b)$ we get $(c-1)(c-4) < 0$ and $c < 4$.
Similarly, $(1-a)(1-b) < 0$ and $0 = 9 - ab - bc - ca = a(a-3) - (3-b)(3-c)$ we get $a(a-3) < 0$ and $a > 0$. It implies $0 < a < 1 < b < 3 < c < 4$ and $a(1-a) + (b-1)(3-b) + (c-3)(4-c) > 0$ from which follows required inequality. There is no value to attain equality.