# Problem 2.

Find all positive $x, y, z$ for which the expression

$$
\frac{x^{2} y z}{324}+\frac{144 y}{x z}+\frac{9}{4 x y^{2}}
$$

takes the smallest value and among all such $x, y, z$ the expression

$$
\frac{z}{16 y}+\frac{x}{9}
$$

is also minimal.
# Solution:

According to the Cauchy inequality

$$
\frac{x^{2} y z}{324}+\frac{144 y}{x z}+\frac{9}{4 x y^{2}} \geq 3 \sqrt[3]{\frac{x^{2} y z}{324} \cdot \frac{144 y}{x z} \cdot \frac{9}{4 x y^{2}}}=3
$$

Equality is possible only when

$$
\frac{x^{2} y z}{324}=\frac{144 y}{x z}=\frac{9}{4 x y^{2}}
$$

From this,

$$
x^{3} z^{2}=46656, \quad 64 y^{3}=z
$$

Therefore, $x^{3}\left(64 y^{3}\right)^{2}=46656 \Rightarrow 64 x^{3} y^{6}=729 \Rightarrow 4 x y^{2}=9$.

In the end: $x=\frac{9}{4 y^{2}}, z=64 y^{3}$. Substituting the obtained values into the expression

$$
\frac{z}{16 y}+\frac{x}{9}
$$

we arrive at the expression

$$
4 y^{2}+\frac{1}{4 y^{2}} \geq 2 \sqrt{4 y^{2} \cdot \frac{1}{4 y^{2}}}=2
$$

Equality is achieved when $4 y^{2}=1$, i.e., $y=\frac{1}{2}$. Therefore, $x=$ $9, z=16$.

Answer: $x=9, y=\frac{1}{2}, z=16$.