, 2019

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