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 analyze the given conditions and derive the required inequality step by step.

### Step-by-Step Solution:

1. **Given Information**:
We have non-negative real numbers \( a_1, a_2, \ldots, a_n \) such that:
\[ a_1 + a_2 + \ldots + a_n = m \]
and we need to find the relationship involving \( S = \sum_{i < j} a_i a_j \).

2. **Square the Sum**:
We start by squaring the sum of the \( a_i \):
\[ (a_1 + a_2 + \ldots + a_n)^2 = m^2 \]
Expanding the left-hand side using the formula for the square of a sum:
\[ a_1^2 + a_2^2 + \ldots + a_n^2 + 2\sum_{i < j} a_i a_j = m^2 \]
Here, \( \sum_{i < j} a_i a_j \) is exactly \( S \).

3. **Rearranging the Equation**:
We can rearrange the equation to isolate \( S \):
\[ a_1^2 + a_2^2 + \ldots + a_n^2 + 2S = m^2 \]
This implies:
\[ 2S = m^2 - (a_1^2 + a_2^2 + \ldots + a_n^2) \]

4. **Non-negativity of Squares**:
Since \( a_i \) are non-negative real numbers, we know:
\[ a_1^2 + a_2^2 + \ldots + a_n^2 \geq 0 \]
Therefore, we can conclude:
\[ m^2 - (a_1^2 + a_2^2 + \ldots + a_n^2) \geq 0 \]

5. **Final Inequality**:
From the above, we have:
\[ 2S \leq m^2 \]
Dividing both sides by 2 gives:
\[ S \leq \frac{m^2}{2} \]

### Conclusion:
Thus, we have derived that:
\[ S \leq \frac{m^2}{2} \]