AUT_ABooklet_2023

We order the variables by size: For $a\ge b\ge c$, all three factors are positive and we have $(a-b)(b-c)(a-c)\ge 0$. For $b\ge c\ge a$ and $c\ge a\ge b$, two of the factors are negative and one factor is positive, so we have again $(a-b)(b-c)(a-c)\ge 0$. For all the other orderings of variables, we have either three negative factors or one negative and two positive factors. This implies $(a-b)(b-c)(a-c)\le 0$, so the inequality holds for these cases and there is no case of equality.

Let us now consider $a\ge b\ge c$. With the AM-GM-inequality, we get $$(a-b)(b-c)\le \frac{(a-b+b-c{)}^2}{4}=\frac{(a-c{)}^2}{4}.$$ So we obtain $$(a-b)(b-c)(a-c)\le \frac{(a-c{)}^2}{4}(a-c)=\frac{(a-c{)}^3}{4}\le \frac{2^3}{4}=2.$$ The two remaining cases of orderings can be treated analogously.

We see that equality holds for $a-c=2$ and $a-b=b-c$, which implies $a=2$, $b=1$ and $c=0$. Taking into account the analogous cases, we see that equality holds exactly for the triples $(2,1,0)$, $(1,0,2)$ and $(0,2,1)$.

(Karl Czakler) □