The product of n positive numbers is unity. Then their sum is:

Options

A. a positive integer
B. divisible by n
C. equal to `(n + 1/n)`
D. never less than n

Answer

To solve the problem, we need to analyze the situation where the product of \( n \) positive numbers equals unity (1). Let's denote these \( n \) positive numbers as \( x_1, x_2, \ldots, x_n \).

### Step-by-Step Solution:

1. **Understanding the Product**:
We know that:
\[ x_1 \times x_2 \times \ldots \times x_n = 1 \]
Since all \( x_i \) are positive numbers, this means that some of these numbers must be less than 1 and some must be greater than 1 to balance out the product to equal 1.

2. **Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality)**:
The AM-GM inequality states that for any set of positive numbers, the arithmetic mean is greater than or equal to the geometric mean. Therefore, we can write:
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 \times x_2 \times \ldots \times x_n} \]
Given that the product \( x_1 \times x_2 \times \ldots \times x_n = 1 \), we have:
\[ \sqrt[n]{x_1 \times x_2 \times \ldots \times x_n} = \sqrt[n]{1} = 1 \]
Thus, we can rewrite the inequality as:
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq 1 \]

3. **Multiplying Both Sides by \( n \)**:
By multiplying both sides of the inequality by \( n \) (which is positive), we get:
\[ x_1 + x_2 + \ldots + x_n \geq n \]
This indicates that the sum of the \( n \) positive numbers is at least \( n \).

4. **Conclusion**:
Therefore, we conclude that the sum of the \( n \) positive numbers, given that their product is unity, is never less than \( n \).

### Final Answer:
The sum of the \( n \) positive numbers is **never less than \( n \)**.