Team Selection Test for IMO 2019

Define $a=x-2y$, $b=y-2z$, $c=z-2x$. Then $b,c>0$ and the problem statement is $b>2c$. Now since $$x=-\frac{a+2b+4c}{7},y=-\frac{b+2c+4a}{7},z=-\frac{c+2a+4b}{7}$$ we get $$\begin{array}{rl}S& =16(x^{2}y+y^{2}z+z^{2}x)+2xyz-2(x^3+y^3+z^3)\\ & -15(x{y}^2+y{z}^2+z{x}^2)=a{b}^2+b{c}^2+c{a}^2-2abc\\ & =c{\left(a-b+\frac{b^2}{2c}\right)}^2+b(b-c{)}^2+\frac{b^2(4c^2-b^2)}{4c}<0\end{array}$$ Since $b,c>0$, we get that $4c^2<b^2$ and hence $b>2c$.