Let `a_(1),a_(2)…,a_(n)` be a non-negative real numbers such that `a_(1)+a_(2)+…+a_(n)=m` and let `S=sum_(iltj) a_(i)a_(j)`, then

Options

A. `Sle(m^(2))/(2)`
B. `Sgt(m^(2))/(2)`
C. `Slt(m)/(2)`
D. `Sgt(m^(2))/(2)`

Explanation

To solve the problem, we need to find the relationship between the sum of the non-negative real numbers \( a_1, a_2, \ldots, a_n \) and the sum \( S = \sum_{i < j} a_i a_j \). Given that \( a_1 + a_2 + \ldots + a_n = m \), we can proceed as follows:

### Step-by-Step Solution:

1. **Understanding the Sum of Products**:
The expression \( S = \sum_{i < j} a_i a_j \) represents the sum of the products of all pairs of the numbers \( a_i \) and \( a_j \) where \( i < j \).

2. **Using the Identity**:
We can express \( S \) in terms of the square of the sum of the \( a_i \)'s:
\[ S = \frac{1}{2} \left( \left( \sum_{i=1}^{n} a_i \right)^2 - \sum_{i=1}^{n} a_i^2 \right) \]
This is derived from the expansion of \( \left( \sum_{i=1}^{n} a_i \right)^2 \), which includes all \( a_i^2 \) terms and the cross terms \( 2a_i a_j \).

3. **Substituting the Given Condition**:
Since we know that \( a_1 + a_2 + \ldots + a_n = m \), we can substitute this into our equation:
\[ S = \frac{1}{2} \left( m^2 - \sum_{i=1}^{n} a_i^2 \right) \]

4. **Applying Cauchy-Schwarz Inequality**:
By the Cauchy-Schwarz inequality, we have:
\[ n \sum_{i=1}^{n} a_i^2 \geq \left( \sum_{i=1}^{n} a_i \right)^2 = m^2 \]
This implies:
\[ \sum_{i=1}^{n} a_i^2 \geq \frac{m^2}{n} \]

5. **Finding an Upper Bound for S**:
Substituting this back into our expression for \( S \):
\[ S \leq \frac{1}{2} \left( m^2 - \frac{m^2}{n} \right) = \frac{1}{2} m^2 \left( 1 - \frac{1}{n} \right) = \frac{m^2 (n-1)}{2n} \]

### Conclusion:
Thus, we have shown that:
\[ S \leq \frac{m^2 (n-1)}{2n} \]