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?

Remark: Once we know the equality case $x = 1, y = 2, z = 3$, we apply AM-GM.
Using AM-GM, we have
\begin{align*}
 &30\cdot x + 12 \cdot (y/2)^2
 + 6\cdot (z/3)^3 + 18 \cdot \frac{1}{xy/2} +  6\cdot \frac{1}{yz/6} + 12 \cdot \frac{1}{zx/3}\\
 \ge\,& 84\sqrt[84]{x^{30}\cdot (y/2)^{24}\cdot (z/3)^{18} \cdot \left(\frac{1}{xy/2}\right)^{18}\left(\frac{1}{yz/6}\right)^6\left(\frac{1}{zx/3}\right)^{12} }\\
 =\,& 84.
\end{align*}
We are done.
