To solve the problem, we need to show that if \( a, b, c \) are three non-zero numbers of the same sign, then the value of \( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \) lies in the interval \([3, \infty)\).
### Step-by-step Solution:
1. **Identify the Sign of the Numbers**:
Since \( a, b, c \) are non-zero numbers of the same sign, they can either all be positive or all be negative. This means that \( \frac{a}{b}, \frac{b}{c}, \frac{c}{a} \) will all be positive.
**Hint**: Remember that the ratio of two numbers of the same sign is always positive.
2. **Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality)**:
According to the AM-GM inequality, for any positive real numbers \( x_1, x_2, x_3 \):
\[
\frac{x_1 + x_2 + x_3}{3} \geq \sqrt[3]{x_1 x_2 x_3}
\]
In our case, let \( x_1 = \frac{a}{b}, x_2 = \frac{b}{c}, x_3 = \frac{c}{a} \).
**Hint**: The AM-GM inequality helps us relate the average of numbers to their geometric mean.
3. **Calculate the AM**:
We can write:
\[
\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}}
\]
4. **Simplify the Geometric Mean**:
The product \( \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} \) simplifies to:
\[
\frac{a \cdot b \cdot c}{b \cdot c \cdot a} = 1
\]
Therefore, we have:
\[
\sqrt[3]{1} = 1
\]
5. **Combine the Results**:
Plugging this back into the AM-GM inequality gives us:
\[
\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq 1
\]
Multiplying both sides by 3 results in:
\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3
\]
6. **Determine the Interval**:
Since \( \frac{a}{b}, \frac{b}{c}, \frac{c}{a} \) are positive, their sum can be greater than or equal to 3 but not bounded above. Thus, we conclude:
\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \in [3, \infty)
\]
### Final Answer:
The value of \( \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \) lies in the interval \([3, \infty)\).
