If the product of n positive numbers is 1, then their sum is

Options

A. a positive integer
B. divisible by n
C. equal to `n+(1)/(n)`
D. greater than or equal to n.

Explanation

To solve the problem, we need to determine the sum of n positive numbers given that their product is equal to 1. We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find the relationship between the sum and the product of these numbers.

### Step-by-Step Solution:

1. **Define the Numbers**:
Let the n positive numbers be \( a_1, a_2, a_3, \ldots, a_n \).

2. **Given Condition**:
We know that the product of these numbers is:
\[ a_1 \times a_2 \times a_3 \times \ldots \times a_n = 1 \]

3. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for any set of positive numbers, the arithmetic mean is greater than or equal to the geometric mean. This can be expressed as:
\[ \frac{a_1 + a_2 + a_3 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 \times a_2 \times a_3 \times \ldots \times a_n} \]

4. **Substitute the Product**:
Since we know the product of the numbers is 1, we can substitute this into the inequality:
\[ \frac{a_1 + a_2 + a_3 + \ldots + a_n}{n} \geq \sqrt[n]{1} \]
Simplifying this gives:
\[ \frac{a_1 + a_2 + a_3 + \ldots + a_n}{n} \geq 1 \]

5. **Multiply by n**:
To isolate the sum, we multiply both sides of the inequality by n:
\[ a_1 + a_2 + a_3 + \ldots + a_n \geq n \]

6. **Conclusion**:
Therefore, the sum of the n positive numbers is greater than or equal to n:
\[ a_1 + a_2 + a_3 + \ldots + a_n \geq n \]

### Final Answer:
The sum of the n positive numbers is greater than or equal to n.

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