Let $x, y$ and $z$ be positive real numbers satisfying $x^{3} y^{2} z=1$. What is the minimum value of $x+2 y+3 z$?
It is about minimizing the sum $x+2 y+3 z$ by exploiting the condition $x^{3} y^{2} z=1$ using the AM-GM inequality. Unfortunately, there are no terms in $x^{3}$ or $y^{2}$ in the expression $x+2 y+3 z$. Therefore, we need to split the term $x$ into three terms so that the application of the AM-GM inequality will introduce a term in $x^{3}$ which should simplify due to the assumption $x^{3} y^{2} z=1$. Similarly, we need to split the term $2 y$ into two to get an exponent of two and leave the term $3 z$ as is because the exponent 1 already matches. Thus, we write

$$
\begin{aligned}
x+2 y+3 z & =\frac{1}{3} x+\frac{1}{3} x+\frac{1}{3} x+y+y+3 z \\
& \geqslant 6 \sqrt[6]{\frac{1}{3} x \times \frac{1}{3} x \times \frac{1}{3} x \times y \times y \times 3 z} \text { by the AM-GM inequality } \\
& =6 \sqrt[6]{\frac{1}{9} x^{3} y^{2} z}=6 \sqrt[6]{\frac{1}{9}} \text { since } x^{3} y^{2} z=1
\end{aligned}
$$

Be careful, we still need to verify that the value $6 \sqrt[6]{\frac{1}{9}}$ can be achieved to conclude that it is the minimum! Indeed, since $x, y$, and $z$ are positive, we could easily say without much risk that $x+2 y+3 z \geqslant 0$, but 0 is not the minimum. Here, by recalling the equality case of the AM-GM inequality, it is easily verified that taking $\frac{1}{3} x=y=3 z=\sqrt[6]{\frac{1}{9}}$ works.