Question

Let $a_{1},a_{2},....,a_{n}$ be positive non-zero real numbers. If $a_{1},a_{2},....,a_{n}=k$, then the minimum value of $a_{1}+a_{2}+....+a_{n}$ is ________.

• A. $n(k)^{2/n}$
• B. $n(k)^{1/n}$
• C. $(k)^{1/n}$
• D. $(k)^{2/n}$
• E. $2n(k)^{2/n}$

Hint:

Sum is minimum when all numbers are equal ($AM=GM$).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem asks for the minimum value of a sum of positive numbers, given that their product is a constant. This is a classic application of the Arithmetic Mean-Geometric Mean (AM-GM) inequality.
Step 2: Key Formula or Approach
The AM-GM inequality states that for any set of \(n\) non-negative real numbers \(a_1, a_2, \dots, a_n\), the arithmetic mean is greater than or equal to the geometric mean.
\[ \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n} \] Equality (which gives the minimum value for the sum) occurs if and only if \(a_1 = a_2 = \dots = a_n\).
Step 3: Detailed Explanation
1. Apply the AM-GM inequality to the given numbers.
\[ \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n} \] 2. Substitute the given product into the inequality.
We are given that the product \(a_1 a_2 \dots a_n = k\).
\[ \frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{k} \] 3. Isolate the sum to find its minimum value.
Multiply both sides by \(n\):
\[ a_1 + a_2 + \dots + a_n \geq n \sqrt[n]{k} \] This inequality shows that the sum \(a_1 + a_2 + \dots + a_n\) is always greater than or equal to \(n \sqrt[n]{k}\). Therefore, the minimum value of the sum is \(n \sqrt[n]{k}\).
4. Express the result using fractional exponents.
The n-th root of k can be written as \(k^{1/n}\).
Minimum value = \(n k^{1/n}\).
Step 4: Final Answer
The minimum value of the sum is \(n(k)^{1/n}\), which corresponds to option (B). The question might have been cancelled in the exam due to a printing error or some other issue, but the mathematical solution is straightforward.