Answer: all functions of the form $f_{c, d}(x)=\left\{\begin{array}{ll}c x^{2} & \text { if } x>0 \\ d & \text { if } x=0\end{array}\right.$ with $c \geq 0$ and $d \leq 0$.

Substituting $x=0$ and $y=1$ yields $f(1) \geq 3 f(0) z^{2}$ for all $z \geq 0$. If $f(0)>0$, then the right side of this inequality is unbounded, a contradiction. Therefore, $f(0) \leq 0$.

Substituting $z=0$ yields $y f(y) \geq 0$, so $f(y) \geq 0$ for all $y>0$. In particular, $f(1) \geq 0$.
Substituting $x=y$ and $z=1$ yields

$$2 y^{3} f(1)+y f(y) \geq 3 y f(y)$$

so $2 y^{3} f(1) \geq 2 y f(y)$, so $f(y) \leq y^{2} f(1)$ for $y>0$.
Substituting $z=y$ and $x=1$ yields

$$2 y f(y)+y f(y) \geq 3 y^{3} f(1)$$

so $3 y f(y) \geq 3 y^{3} f(1)$, so $f(y) \geq y^{2} f(1)$ for $y>0$.
Together, this gives $f(y)=y^{2} f(1)$ for $y>0$.
Write $c=f(1)$ and $d=f(0)$. Then we now know that $f(x)=\left\{\begin{array}{ll}c x^{2} & \text { if } x>0 \\ d & \text { if } x=0\end{array}\right.$ with $c \geq 0$ and $d \leq 0$. We will now check these functions.

First note that $x f(x) \geq 0$ for all $x \in \mathbb{R}_{\geq 0}$: for $x=0$ this is trivial and for $x>0$ we have $x f(x)=c x^{3} \geq 0$ since $c \geq 0$.

For $x=0$ we have $y f(y) \geq 3 y z^{2} f(0)$, and this is true since $y f(y) \geq 0$ and $3 y z^{2} f(0)=$ $3 y z^{2} d \leq 0$ for $y, z \geq 0$.
For $y=0$ we have $2 x^{3} z f(z) \geq 0$ and this is true. For $z=0$ we have $y f(y) \geq 0$ and this is true.
Now assume that $x, y, z>0$. Then we have $f(x)=c x^{2}, f(y)=c y^{2}, f(z)=c z^{2}$ and we need to prove that

$$2 x^{3} z c z^{2}+y c y^{2} \geq 3 y z^{2} c x^{2}$$

Since $c \geq 0$ it is sufficient to show that $2 x^{3} z^{3}+y^{3} \geq 3 y z^{2} x^{2}$ for all $x, y, z>0$. This follows from the arithmetic mean-geometric mean inequality on the three terms $x^{3} z^{3}, x^{3} z^{3}$ and $y^{3}$. So all these functions satisfy the condition.