Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$ 
If $x, y, z$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$

I genuinely have no clue on how to proceed. Is it proved using repeated CS?

Partial answer:
If we want to find the minimum, I think we can create a system from the implicit derivatives. We can multiply through by $xyz$, then subtract the right side to get:
$$f(x,y,z) = 30x^2yz+3xy^3z+\textstyle{\frac29}xyz^4+36x+36y+36z-84xyz \geqslant 0$$
And we want to show that this function has a minimum of $0$. Then we have:
$$
\begin{align}
\frac{\partial f}{\partial x} &= 60xyz+3y^3z+ \textstyle{\frac{2}{9}}yz^4-84yz+36 \\
\frac{\partial f}{\partial y} &= 30x^2z+9xy^2z+ \textstyle{\frac{2}{9}}xz^4 -84xz+36\\
\frac{\partial f}{\partial z} &= 30x^2y+3xy^3+ \textstyle{\frac{8}{9}}xyz^3-84xy+36
\end{align}
$$
To find minima, we want all of these partial derivatives to be equal to $0$. This is a system of three equations in three variables:
$$
\left\{
\begin{aligned} 
60xyz+3y^3z+ \textstyle{\frac{2}{9}}yz^4-84yz &= -36 \\ 
30x^2z+9xy^2z+ \textstyle{\frac{2}{9}}xz^4 -84xz &= -36 \\ 
30x^2y+3xy^3+ \textstyle{\frac{8}{9}}xyz^3-84xy &= -36 
\end{aligned} 
\right. 
$$
That's a bit... ugly. Let's try something:
$$
\left\{
\begin{aligned} 
60x^2yz+3xy^3z+ \textstyle{\frac{2}{9}}xyz^4-84xyz +36x &= 0 \\ 
30x^2yz+9xy^3z+ \textstyle{\frac{2}{9}}xyz^4-84xyz +36y &= 0 \\ 
30x^2yz+3xy^3z+ \textstyle{\frac{8}{9}}xyz^4-84xyz +36z &= 0 
\end{aligned} 
\right. 
$$
Now set $xyz=a$ and we have:
$$
\left\{
\begin{aligned} 
60ax+3ay^2z+ \textstyle{\frac{2}{9}}az^3 +36x &= 84a \\ 
30ax+9ay^2z+ \textstyle{\frac{2}{9}}az^3 +36y &= 84a \\ 
30ax+3ay^2z+ \textstyle{\frac{8}{9}}az^3 +36z &= 84a 
\end{aligned} 
\right. 
$$
I do believe we'll need to solve the system by substitution. I'll leave this here for the moment and return.
