For any n positive numbers `a_(1),a_(2),…,a_(n)` such that
`sum_(i=1)^(n) a_(i)=alpha`, the least value of `sum_(i=1)^(n) a_(i)^(-1)`, is

Options

A. `2n-alpha`
B. `(3n)/(alpha)`
C. `(n(n-1))/(alpha)`
D. `(n^(2))/(alpha)`

Explanation

To find the least value of the sum \( S = \sum_{i=1}^{n} a_i^{-1} \) given that \( \sum_{i=1}^{n} a_i = \alpha \), we can use the Cauchy-Schwarz inequality.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to minimize the expression \( S = \sum_{i=1}^{n} \frac{1}{a_i} \) under the constraint \( \sum_{i=1}^{n} a_i = \alpha \).

2. **Applying Cauchy-Schwarz Inequality**: According to the Cauchy-Schwarz inequality, we have:
\[ \left( \sum_{i=1}^{n} a_i \right) \left( \sum_{i=1}^{n} \frac{1}{a_i} \right) \geq n^2 \]
This means:
\[ \alpha \cdot S \geq n^2 \]

3. **Rearranging the Inequality**: From the inequality above, we can rearrange it to find \( S \):
\[ S \geq \frac{n^2}{\alpha} \]

4. **Finding the Least Value**: The least value of \( S \) occurs when the equality condition of the Cauchy-Schwarz inequality holds. This happens when all \( a_i \) are equal. If we set \( a_1 = a_2 = \ldots = a_n = \frac{\alpha}{n} \), we can substitute this back into our expression for \( S \):
\[ S = \sum_{i=1}^{n} \frac{1}{a_i} = \sum_{i=1}^{n} \frac{n}{\alpha} = \frac{n^2}{\alpha} \]

5. **Conclusion**: Therefore, the least value of \( \sum_{i=1}^{n} a_i^{-1} \) is:
\[ \boxed{\frac{n^2}{\alpha}} \]