Mongolian Mathematical Olympiad

It is obvious that $a<2<c$. Therefore $$3(a-1)(a-3)-(a-b)(a-c)=3(b-1)(b-3)-(b-1)(b-a)=$$ $$3(c-1)(c-3)-(c-a)(c-b)=0$$ $$\left(\begin{array}{l}0=9-ab-bc-ca=9+a^2-2a(b+c)-(a-b)(a-c)=\\ =9+a^2-2a(6-a)-(a-b)(a-c)=3(a-1)(a-3)-(a-b)(a-c)\end{array}\right)$$ 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.