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.
 A: 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
A: Let's find the minimal value of $x+y+z$ when $x^3y^4z^5 = C$.
At point of tangency between $x^3y^4z^5 = C$ and $x+y+z=A$ the normal is proportional to:
$$
\left(\frac{3}{x},\frac{4}{y},\frac{5}z\right)
$$
On the other hand, it should be proportional to $(1,1,1)$. It is possible when $x/3 = y/4= z/5 = k$. From
$$
(3k)^3(4k)^4(5k^5) = (3\times4\times 5\times10)^3 \frac{4\times 5^2}{1000}k^{12} = 0.1\times 600^3
$$
we find that $k=1$. By sheer luck (actually not), this point also lies on $x+y+z=12$. Since, it's a tangency point by construction and $x^3y^4z^5$ is convex in positive octant, there is no other solution.
A: Added by edit: It is pointed out in a comment that the problem was posed for $x,y,z \in \mathbb R$, and the solution below assumes $x,y,z \in \mathbb Z$, so I add the following preface:
One way to investigate whether there are real solutions is to see whether there are any integer solutions, if that can be done efficiently. Moreover, OP identifies an integer solution by trial and error, and asks for a mathematical exegesis of why that solution is valid. End of added material.
The clear answer comes from factoring.
$0.1(600)^3=2^83^35^5$.
$(xyz)^3(yz)(z)=x^3y^4z^5$
Plainly, the factor of $5^5$ in the product must be accounted for by $z^5$. If not, there is no other way of distributing the factors of $5$ among $x$ and $y$ that allow them to be perfect cubes and fourth powers, respectively. There are enough factors of $2$ so that $z$ might be either $5$ or $10$. But if you take five factors of $2$ and put them in $z$, there are not enough factors left for $y$ to be a fourth power. So $z=5$.
Now, the only way we can achieve $y^4$ is by making it from $(2^2)^4$, so $y=4$.
That leaves $x^3=3^3$, or $x=3$.
Conveniently, $3+4+5=12$.
The question posed, what is $x^3+y^3+z^3$ is simply $27+64+125=216$, which has the coincidence of being $6^3$.
