For any set of real numbers R = `{a,b,c} let sum of pairwise product S = ab+bc+ca, If a+b+c = 1 then

Options

A. `S le 1/3`
B. `Slt1/3`
C. `S le sqrt3`
D. `S lt 1/9`

Answer

To solve the problem, we need to find the sum of pairwise products \( S = ab + bc + ca \) given that \( a + b + c = 1 \).

### Step-by-step Solution:

1. **Understand the Given Condition**:
We have three real numbers \( a, b, c \) such that their sum equals 1:
\[ a + b + c = 1 \]

2. **Express the Sum of Pairwise Products**:
The sum of pairwise products is defined as:
\[ S = ab + bc + ca \]

3. **Assume Equal Values for Simplicity**:
To explore the maximum value of \( S \), we can assume \( a = b = c \). Since \( a + b + c = 1 \), we set:
\[ a = b = c = \frac{1}{3} \]

4. **Calculate the Pairwise Products**:
Substitute \( a, b, c \) into the expression for \( S \):
\[ S = ab + bc + ca = \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right) + \left(\frac{1}{3} \cdot \frac{1}{3}\right) \]
This simplifies to:
\[ S = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \]

5. **Evaluate the Options**:
Now we need to analyze the options given in the problem:
- \( S \leq \frac{1}{3} \)
- \( S < \frac{1}{3} \)
- \( S < \sqrt{3} \)
- \( S < \frac{1}{9} \)

Since we found \( S = \frac{1}{3} \), the only valid option is:
\[ S \leq \frac{1}{3} \]

6. **Conclusion**:
Therefore, the correct answer is:
\[ \text{Option A: } S \leq \frac{1}{3} \]