The problem asks us to find all triples of natural numbers $(x, y, z)$ such that $1 < x < y < z$ and $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$.

1. Problem Description

The problem asks us to find all triples of natural numbers $(x, y, z)$ such that $1 < x < y < z$ and $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$.

2. Solution Steps

We are given that $1 < x < y < z$ and $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5} = 2.4$.

Since $x, y, z$ are natural numbers greater than 1, we have $x \ge 2, y \ge 3, z \ge 4$.

If $x \ge 3$, then $y \ge 4$ and $z \ge 5$. Thus,

$1 + \frac{1}{x} \le 1 + \frac{1}{3} = \frac{4}{3}$,

$1 + \frac{1}{y} \le 1 + \frac{1}{4} = \frac{5}{4}$,

$1 + \frac{1}{z} \le 1 + \frac{1}{5} = \frac{6}{5}$.

Then, $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) \le \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} = 2 < \frac{12}{5} = 2.4$. Thus, we must have $x = 2$.

Now we have $(1 + \frac{1}{2})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{3}{2} (1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$, so $(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5} \cdot \frac{2}{3} = \frac{8}{5} = 1.6$.

Since $2 = x < y < z$, we have $y \ge 3$ and $z \ge 4$.

If $y \ge 4$, then $z \ge 5$, and $1 + \frac{1}{y} \le 1 + \frac{1}{4} = \frac{5}{4}$, $1 + \frac{1}{z} \le 1 + \frac{1}{5} = \frac{6}{5}$.

$(1 + \frac{1}{y})(1 + \frac{1}{z}) \le \frac{5}{4} \cdot \frac{6}{5} = \frac{6}{4} = \frac{3}{2} = 1.5 < \frac{8}{5} = 1.6$. Thus, we must have $y = 3$.

Now we have $(1 + \frac{1}{3})(1 + \frac{1}{z}) = \frac{4}{3} (1 + \frac{1}{z}) = \frac{8}{5}$, so $1 + \frac{1}{z} = \frac{8}{5} \cdot \frac{3}{4} = \frac{6}{5}$.

Then $\frac{1}{z} = \frac{6}{5} - 1 = \frac{1}{5}$, so $z = 5$.

We have $x = 2, y = 3, z = 5$, and $1 < x < y < z$.

$(1 + \frac{1}{2})(1 + \frac{1}{3})(1 + \frac{1}{5}) = \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{6}{5} = \frac{12}{5}$.

3. Final Answer

(2, 3, 5)