We are given that $x$, $y$, and $z$ are natural numbers such that $1 < x < y < z$, and we need to find all possible triples $(x, y, z)$ that satisfy the equation $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$.

1. Problem Description

We are given that $x$, $y$, and $z$ are natural numbers such that $1 < x < y < z$, and we need to find all possible triples $(x, y, z)$ that satisfy the equation $(1 + \frac{1}{x})(1 + \frac{1}{y})(1 + \frac{1}{z}) = \frac{12}{5}$.

2. Solution Steps

First, rewrite the equation as

$$\frac{x+1}{x} \cdot \frac{y+1}{y} \cdot \frac{z+1}{z} = \frac{12}{5}.$$

Since $1 < x < y < z$, we have $x \ge 2$, $y \ge 3$, $z \ge 4$. If $x \ge 3$, then $1 + \frac{1}{x} \le \frac{4}{3}$, $1 + \frac{1}{y} \le \frac{5}{4}$, $1 + \frac{1}{z} \le \frac{6}{5}$, so

$$$$

(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 $x = 2$. Substituting $x=2$ into the equation yields

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

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

Since $2 < y < z$, we have $y \ge 3$ and $z \ge 4$. If $y \ge 4$, then $1 + \frac{1}{y} \le \frac{5}{4}$ and $1 + \frac{1}{z} \le \frac{6}{5}$, so

$$$$

(1 + \frac{1}{y})(1 + \frac{1}{z}) \le \frac{5}{4} \cdot \frac{6}{5} = \frac{3}{2} = 1.5 < \frac{8}{5} = 1.

6. $$

Therefore, we must have $y = 3$. Substituting $y=3$ into the equation yields

$$\frac{4}{3} \cdot \frac{z+1}{z} = \frac{8}{5}.$$

$$\frac{z+1}{z} = \frac{8}{5} \cdot \frac{3}{4} = \frac{6}{5}.$$

$5(z+1) = 6z$, so $5z+5 = 6z$, which means $z=5$. Thus, $(x, y, z) = (2, 3, 5)$.

3. Final Answer

(2, 3, 5)