If x,y,z are three real numbers of the same sign, then the value of `(x/y+y/z+z/x)` lies in the interval _____

Explanation

To solve the problem, we need to find the value of the expression \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \) given that \( x, y, z \) are three real numbers of the same sign.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to analyze the expression \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \). Since \( x, y, z \) are of the same sign, we can assume they are all positive (the analysis will be similar if they are all negative).

2. **Applying the AM-GM Inequality**:
We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality to the three terms:
\[ \frac{\frac{x}{y} + \frac{y}{z} + \frac{z}{x}}{3} \geq \sqrt[3]{\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x}} \]

3. **Simplifying the Geometric Mean**:
The product inside the cube root simplifies as follows:
\[ \frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x} = \frac{x \cdot y \cdot z}{y \cdot z \cdot x} = 1 \]
Therefore, we have:
\[ \sqrt[3]{\frac{x}{y} \cdot \frac{y}{z} \cdot \frac{z}{x}} = \sqrt[3]{1} = 1 \]

4. **Applying the Result of AM-GM**:
From the AM-GM inequality, we can now say:
\[ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geq 3 \cdot 1 = 3 \]

5. **Finding the Upper Bound**:
Now, we need to check if there is an upper bound for the expression. Since \( x, y, z \) are positive, as any of these values approach zero, the terms \( \frac{x}{y}, \frac{y}{z}, \frac{z}{x} \) can become arbitrarily large. Thus, there is no upper limit for the expression.

6. **Conclusion**:
Therefore, we conclude that:
\[ \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \in [3, \infty) \]

### Final Answer:
The value of \( \frac{x}{y} + \frac{y}{z} + \frac{z}{x} \) lies in the interval \([3, \infty)\).