It is possible to prove that for odd natural number $n > 1$ and for arbitrary real numbers $a$, $b$, $c$ which satisfy the condition $a + b + c = 0$, the numbers $abc$ and $a^n + b^n + c^n$ have the same sign. It is obvious that after setting $n = 5$, $a = x - y$, $b = y - z$, $c = z - x$ follows required statement. Let us use following property.
Property.
a) If $x$, $y > 0$ and $k > 1$ then $x^k + y^k < (x + y)^k$;
b) If $x$, $y < 0$ and $k > 1$, $k$ odd then $x^k + y^k > (x + y)^k$.

First consider the case that one of $a$, $b$, $c$ is equal to $0$. For example, if $a = 0$ then $abc = 0$ and $a + b + c = b + c = 0$. From this we get $b = -c$ and $a^n + b^n + c^n = b^n + c^n = (-c)^n + c^n = 0$. In this case nothing to prove. Now consider the case $abc \neq 0$.
Without losing generality it is possible to assume $a \ge b \ge c$. From $a + b + c = 0$ follows $a > 0$ and $c < 0$.
i) $b < 0 \Rightarrow abc < 0$ and $a^n + b^n + c^n < (a + b)^n + c^n = (-c)^n + c^n = 0$ by property a).
ii) $b < 0 \Rightarrow abc > 0$ and $a^n + b^n + c^n > a^n + (b + c)^n = a^n + (-a)^n = 0$ by property b).