, 2008

Solution:
If we multiply the given equation and inequality $(x>0, y>0, z>0)$, we have
$$\left(\frac{4x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9x}{z}\right)+\left(\frac{4z}{y}+\frac{9y}{z}\right) \leq 22$$
From AM-GM we have
$$\frac{4x}{y}+\frac{y}{x} \geq 4, \quad \frac{z}{x}+\frac{9x}{z} \geq 6, \quad \frac{4z}{y}+\frac{9y}{z} \geq 12$$
Therefore
$$22 \leq \left(\frac{4x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9x}{z}\right)+\left(\frac{4z}{y}+\frac{9y}{z}\right)$$
Now from (1) and (3) we get
$$\left(\frac{4x}{y}+\frac{y}{x}\right)+\left(\frac{z}{x}+\frac{9x}{z}\right)+\left(\frac{4z}{y}+\frac{9y}{z}\right)=22$$
which means that in (2), everywhere equality holds i.e. we have equality between means, also $x+y+z=12$.
Therefore $\frac{4x}{y}=\frac{y}{x}$, $\frac{z}{x}=\frac{9x}{z}$ and, as $x>0, y>0, z>0$, we get $y=2x$, $z=3x$. 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 $(x, y, z)=(2,4,6)$.