The given inequality can be written in the following form:
$$\alpha(x^3 + y^3 + z^3 - x^2y - y^2z - z^2x) \geq x^2y + y^2z + z^2x - xy^2 - yz^2 - zx^2.$$
From the rearrangement inequality it follows that the left-hand side of the previous inequality is positive. Therefore, it suffices to prove the desired inequality for $\alpha = \frac{1}{2}$:
$$x^3 + y^3 + z^3 + 2(xy^2 + yz^2 + zx^2) \geq 3(x^2y + y^2z + z^2x). \quad (1)$$
Since the given inequality is cyclic, without loss of generality we may assume that $z \leq x$ and $z \leq y$. Let $a, b \geq 0$ be real numbers such that $x = z + a$, $y = z + b$. Therefore, the inequality (1) can be written as follows:
$$a^3 - 3a^2b + 2ab^2 + b^3 + 2z(a^2 - ab + b^2) \geq 0.$$
By AM-GM inequality, we have:
$$a^2 - ab + b^2 \geq ab \geq 0.$$
Thus, it suffices to show:
$$a^3 + 2ab^2 + b^3 - 3a^2b \geq 0.$$
However, this can be written as follows:
$$\begin{aligned} & a^3 - 4a^2b + 4ab^2 + a^2b - 2ab^2 + b^3 \geq 0, \\ & a(a^2 - 4ab + 4b^2) + b(a^2 - 2ab + b^2) \geq 0, \\ & a(a - 2b)^2 + b(a - b)^2 \geq 0, \end{aligned}$$
which obviously holds.