Let $x, y$ and $z$ be real numbers such that $\frac{4}{x}+\frac{2}{y}+\frac{1}{z}=1$. Find the minimum of $x+8 y+4 z$.
According to the Cauchy-Schwarz inequality,

$$
\left(\frac{4}{x}+\frac{2}{y}+\frac{1}{z}\right)(x+8 y+4 z) \geqslant\left(\sqrt{\frac{4}{x}} \cdot \sqrt{x}+\sqrt{\frac{2}{y}} \cdot \sqrt{8 y}+\sqrt{\frac{1}{z}} \cdot \sqrt{4 z}\right)
$$

Since $\frac{4}{x}+\frac{2}{y}+\frac{1}{z}=1$, we obtain $x+8 y+4 z \geqslant(2+4+2)^{2}=64$.

To show that this minimum is achieved, let's examine the equality case of the Cauchy-Schwarz inequality: we must have $\frac{\sqrt{\frac{4}{x}}}{\sqrt{x}}=\frac{\sqrt{\frac{2}{y}}}{\sqrt{8 y}}=\frac{\sqrt{\frac{1}{z}}}{\sqrt{4 z}}$, which means $\frac{2}{x}=\frac{1}{2 y}=\frac{1}{2 z}$. The constraint $\frac{4}{x}+\frac{2}{y}+\frac{1}{z}=1$ then allows us to find $x=16, y=4$ and $z=4$ as a suitable triplet.