We are given that $x, y, z$ are natural numbers such that $1 < x < y < z$ and $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$. We are asked to find all possible triples $(x, y, z)$ satisfying these conditions.

1. Problem Description

We are given that $x, y, z$ are natural numbers such that $1 < x < y < z$ and $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$. We are asked to find all possible triples $(x, y, z)$ satisfying these conditions.

2. Solution Steps

Since $1 < x < y < z$, we have $x \ge 2, y \ge 3, z \ge 4$.

Also, $1 + \frac{1}{x} > 1$, $1 + \frac{1}{y} > 1$, $1 + \frac{1}{z} > 1$.

We also know that $1 + \frac{1}{x} > 1 + \frac{1}{y} > 1 + \frac{1}{z} > 1$.

Since $x \ge 2$, we have $1 + \frac{1}{x} \le 1 + \frac{1}{2} = \frac{3}{2}$.

Since $y \ge 3$, we have $1 + \frac{1}{y} \le 1 + \frac{1}{3} = \frac{4}{3}$.

Since $z \ge 4$, we have $1 + \frac{1}{z} \le 1 + \frac{1}{4} = \frac{5}{4}$.

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

If $x = 2$, then $(1 + \frac{1}{2})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{3}{2}(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$.

So $(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5} \cdot \frac{2}{3} = \frac{8}{5}$.

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

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

If $y = 3$, then $(1 + \frac{1}{3})(1 + \frac{1}{z}) = \frac{4}{3}(1 + \frac{1}{z}) = \frac{8}{5}$.

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 $z = 5$.

Therefore, $x = 2, y = 3, z = 5$.

3. Final Answer

The only solution is $(2, 3, 5)$.