Let $x, y, z$ be positive real numbers such that $x^{3} y^{2} z=1$. What is the minimum value that $x+2 y+3 z$ can take?
To find the minimum value of $x+2y+3z$, we want to use the AM-GM inequality. But to make $x^{3} y^{2} z$ appear, we need 3 factors with $x$, 2 with $y$, and one with $z$. For this, we will split the $x$ into three parts, and the $2y$ into two. We have:

$$
x+2y+3z=\frac{x}{3}+\frac{x}{3}+\frac{x}{3}+y+y+3z \geqslant 6 \sqrt[6]{\frac{x^{3} y^{2} z}{3^{2}}}=6 \sqrt[3]{\frac{1}{3}}
$$

It remains to show that this value is achieved. For this, if we have equality in the previous inequality, each of the terms is equal to the geometric mean. So $x=3 \sqrt[3]{\frac{1}{3}}, y=\sqrt[3]{\frac{1}{3}}$ and $z=\frac{1}{3} \sqrt[3]{\frac{1}{3}}$.

If we fix these values, then $x^{3} y^{2} z=\frac{3^{3}}{3}\left(\sqrt[3]{\frac{1}{3}}\right)^{6}=\frac{3^{2}}{3^{2}}=1$, and $x+2y+3z=(3+2+1) \sqrt[3]{\frac{1}{3}}$