<s>[INST] 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.
 [/INST] 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.
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