Find all triples $(x, y, z)$ of real positive numbers, which satisfy the system

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=3 \\
x+y+z \leq 12
\end{array}\right.
$$
If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have

$$
\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right) \leq 22
$$

From AM-GM we have

$$
\frac{4 x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9 x}{z} \geq 6, \quad \frac{4 z}{y}+\frac{9 y}{z} \geq 12
$$

Therefore

$$
22 \leq\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right)
$$

Now from (1) and (3) we get

$$
\left(\frac{4 x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9 x}{z}\right)+\left(\frac{4 z}{y}+\frac{9 y}{z}\right)=22
$$

which means that in (2), everywhere equality holds i.e. we have equality between means, also $x+y+z=12$.

Therefore $\frac{4 x}{y}=\frac{y}{x}, \frac{z}{x}=\frac{9 x}{z}$ and, as $x>0, y>0, z>0$, we get $y=2 x, z=3 x$. Finally if we substitute for $y$ and $z$, in $x+y+z=12$, we get $x=2$, therefore $y=2 \cdot 2=4$ and $z=3 \cdot 2=6$.

Thus the unique solution is $(x, y, z)=(2,4,6)$.