Find value of $x^3+y^3+z^3$ if $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$ If $x,y$ and $z$ are positive real numbers such that $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$, then what is the value of $x^3+y^3+z^3$?
I first thought of making them all equal because in that case the product is maximum, but obviously that was wrong and that is only valid for integers. By trial-and-error, I found that $(x,y,z)=(3,4,5)$ does satisfy the conditions, but I can't get a solid proof and method for this, other than just trial-and-error.
Thank you.

Applying the AM-GM inequality for 3 times $\frac{x}{3}$, 4 times $\frac{y}{4}$ and 5 times $\frac{z}{5}$, we have
\begin{align}
12 &= x+y+z \\ & = 3\times \frac{x}{3} + 4 \times \frac{y}{4} +5 \times \frac{z}{5} \ge (3+4+5)\sqrt[12]{\left(\frac{x}{3}\right)^3\left(\frac{y}{4}\right)^4\left(\frac{z}{5}\right)^5} = 12
\end{align}
The equality occurs if and only if $\frac{x}{3} = \frac{y}{4}= \frac{z}{5}$ and $x+y+z= 12$ or
$$(x,y,z) = (3,4,5)$$
Q.E.D
