If `a_(1), a_(2),….,a_(n)` are `n(gt1)` real numbers, then

Options

A. `n underset(i=1)overset(n)suma_(i)""^(2)le(underset(i=1)overset(n)suma_(i))^(2)`
B. ` underset(i=1)overset(n)suma_(i)""^(2)ge(underset(i=1)overset(n)suma_(i))^(2)`
C. ` underset(i=1)overset(n)suma_(i)""^(2)le(underset(i=1)overset(n)suma_(i))^(2)`
D. none of these

Explanation

To solve the problem step by step, we will apply the concept of the Cauchy-Schwarz inequality, which is a fundamental result in inequalities.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have \( n \) real numbers \( a_1, a_2, \ldots, a_n \) where \( n > 1 \). We need to analyze the relationship between these numbers using an inequality.

2. **Applying the Cauchy-Schwarz Inequality**:
The Cauchy-Schwarz inequality states that for any real numbers \( x_1, x_2, \ldots, x_n \) and \( y_1, y_2, \ldots, y_n \):
\[ (x_1^2 + x_2^2 + \ldots + x_n^2)(y_1^2 + y_2^2 + \ldots + y_n^2) \geq (x_1y_1 + x_2y_2 + \ldots + x_ny_n)^2 \]
We can choose \( x_i = a_i \) and \( y_i = 1 \) for all \( i \).

3. **Setting Up the Inequality**:
Substituting into the Cauchy-Schwarz inequality gives us:
\[ (a_1^2 + a_2^2 + \ldots + a_n^2)(1^2 + 1^2 + \ldots + 1^2) \geq (a_1 \cdot 1 + a_2 \cdot 1 + \ldots + a_n \cdot 1)^2 \]
The second term simplifies to \( n \) since there are \( n \) terms of 1.

4. **Simplifying the Expression**:
This leads to:
\[ (a_1^2 + a_2^2 + \ldots + a_n^2) \cdot n \geq (a_1 + a_2 + \ldots + a_n)^2 \]

5. **Rearranging the Inequality**:
We can rearrange this to express it in a more standard form:
\[ n(a_1^2 + a_2^2 + \ldots + a_n^2) \geq (a_1 + a_2 + \ldots + a_n)^2 \]

6. **Conclusion**:
This inequality shows that the arithmetic mean of the squares of the numbers is greater than or equal to the square of the arithmetic mean of the numbers. Thus, we conclude that:
\[ n(a_1^2 + a_2^2 + \ldots + a_n^2) \geq (a_1 + a_2 + \ldots + a_n)^2 \]

### Final Result:
The result of the inequality is:
\[ n \sum_{i=1}^{n} a_i^2 \geq \left( \sum_{i=1}^{n} a_i \right)^2 \]