Question

If the product of n positive real numbers is unity, then their sum is necessarily

• A. a multiple of \( n \)
• B. equal to \( n + 1/n \)
• C. never less than \( n \)
• D. a positive integer

Hint:

Compare the arithmetic mean of the numbers with their geometric mean. The geometric mean is easy to find when the product is known.

Answer 1

The Correct Option is C

Step 1: Understand the condition.
We have n positive real numbers \( a_1, a_2, \dots, a_n \) with \( a_1 a_2 \cdots a_n = 1 \). We want a statement about their sum that is always true.

Step 2: Use the AM-GM inequality.
For positive numbers, the arithmetic mean is never less than the geometric mean:
\[ \frac{a_1 + a_2 + \dots + a_n}{n} \ge (a_1 a_2 \cdots a_n)^{1/n} \]

Step 3: Put in the product.
The product is 1, so the right side is \( 1^{1/n} = 1 \). Therefore
\[ a_1 + a_2 + \dots + a_n \ge n \]
The sum is never less than n. Equality holds when all the numbers are 1.

Step 4: Check statement (1).
Take \( n = 2 \) with the numbers 2 and \( \frac{1}{2} \). Their product is 1 and their sum is 2.5, which is not a multiple of 2. So (1) is not necessarily true.

Step 5: Check statements (2) and (4).
For (2), the claim is that the sum always equals \( n + \frac{1}{n} \). With \( n = 2 \) that value is 2.5. The numbers 1 and 1 have product 1 but sum 2, so (2) fails.
For (4), the numbers 2 and \( \frac{1}{2} \) have sum 2.5, which is not an integer. So (4) fails.

Step 6: Check statement (3).
We proved that the sum is at least n for every such set of numbers, so (3) always holds.

Final Answer:
The sum is never less than n, which is option 3. \[ \boxed{3} \]