This inequality is known as Schur's inequality. By symmetry, we may assume
$0 \le x \le y \le z$ and introduce $t = y - x$ and $s = z - y$, which satisfy $t, s \ge 0$.
Using $y = t + x$ and $z = y + s = t + s + x$ the desired inequality can be
restated as follows.
$$\begin{align*} 0 &\le \sum_{\text{cyc}} (x^3 - x^2(y+z) + xyz) \\ &= x(x-y)(x-z) + y(y-z)(y-x) + z(z-x)(z-y) \\ &= xt(t+s) - (t+x)ts + (t+s+x)(t+s)s \\ &= t(x(t+s) - (t+x)s) + (t+s+x)(t+s)s \\ &= t^2(x-s) + (t^2 + t(2s+x) + s(s+x))s \\ &= t^2x + ts(2s+x) + s^2(s+x), \end{align*}$$
which holds, because $x, t, s$ are non-negative. Equality holds iff $s = 0$ and
$xt = 0$. i.e. $y = z$ and either $x = 0$ or $y = x$. Thus, there is equality iff either
(i) $x = 0$ and $y = z$ or
(ii) $y = 0$ and $z = x$ or
(iii) $z = 0$ and $x = y$ or
(iv) $x = y = z$.