Austria 2010

Since $$\begin{array}{rl}(x-y{)}^7-(x-y)(y-z)(z-x)(x-y{)}^4& =(x-y{)}^5((x-y{)}^2-(y-z)(z-x))\\ & =(x-y{)}^5(x^2+y^2+z^2-xy-yz-zx),\end{array}$$ we can write $$\sum _{cyclic}(x-y{)}^7-(x-y)(y-z)(z-x)\cdot \sum _{cyclic}(x-y{)}^4=\left(\sum _{cyclic}(x-y{)}^5\right)\cdot \left(\sum _{cyclic}{x}^2-\sum _{cyclic}xy\right)$$ It therefore follows that the left-hand side of the inequality can be written as $$\frac{\left({\sum }_{cyclic}(x-y{)}^5\right)\cdot \left({\sum }_{cyclic}{x}^2-{\sum }_{cyclic}xy\right)}{{\sum }_{cyclic}(x-y{)}^5}=\sum _{cyclic}{x}^2-\sum _{cyclic}xy.$$ We therefore need to consider the inequality $$x^2+y^2+z^2-xy-yz-zx\ge 3$$ for $x$, $y$, $z\in \mathbb{Z}$ and $x\ne y\ne z\ne x$. We note that $$x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}(x-y{)}^2+\frac{1}{2}(y-z{)}^2+\frac{1}{2}(z-x{)}^2.$$ Each pair of variables differs by at least one, and this is not possible for all three pairs at once. The smallest possible value is therefore obtained when two pairs differ by one, i.e. for three consecutive integers $m$, $m+1$ and $m+2$ in any order. Since $$\frac{1}{2}((m+2)-(m+1){)}^2+\frac{1}{2}((m+1)-m{)}^2+\frac{1}{2}(m-(m+2){)}^2=\frac{1}{2}+\frac{1}{2}+2=3$$ holds, we see that the given inequality is correct, and equality holds for $$(x,y,z)=(m,m+1,m+2)$$ or any permutation thereof.