Suppose, $$p(x,y,z)=18x^2y+3y^2z+4z^2x-18xyz$$
Then prove that, $$\min_{x,y,z\in \mathbb{R}_{\geq 0}} p(x,y,z) =0$$
Clearly when $xyz=0$, $p$ is non-negative for $x,y,z\geq0$.
We need to handle the case when $xyz=0$. I have tried to find stationary points of $p$. However, this is very complicated as the gradient is system of second-order polynomials.
Also, I tried to show directly that $p\geq0$ for all non-negative $x,y,z$. It is clear that when $x\geq z$, $p$ is non-negative. But, when $z>x$, I am unable to draw any conclusion.
Since $x,y,z\in\mathbb R_{\ge 0}$, your polynomial $p(x,y,z)$ looks to me like it was specially constructed so that the AM-GM inequality could be applied:
$$\begin{aligned}p(x,y,z)+18xyz&=18x^2y+3y^2z+4z^2x\\ &\ge 3\sqrt [3]{18\cdot 3\cdot 4x^3y^3z^3}\\ &=18xyz\end{aligned} $$
This implies,
$$p(x,y,z)\ge 0$$
Finally, note that the equality $p(x,y,z)=0$ can hold, so you prove that
$$\min_{x,y,z\in \mathbb{R}_{\ge 0}} p(x,y,z) =0.$$