If `a+b+c=1`, the greatest value of
`(ab)/(a+b)+(bc)/(b+c)+(ca)/(c+a)`, is

Options

A. `(1)/(3)`
B. `(1)/(2sqrt(2))`
C. `(1)/(2)`
D. `(1)/(4)`

Explanation

To find the greatest value of the expression \(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}\) given that \(a + b + c = 1\), we can use the method of inequalities.

### Step-by-Step Solution:

1. **Understanding the Expression**: We need to maximize the expression \(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}\).

2. **Using the AM-GM Inequality**: We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For any two non-negative numbers \(x\) and \(y\), we have:
\[ \frac{x+y}{2} \geq \sqrt{xy} \]
This implies:
\[ a + b \geq 2\sqrt{ab} \quad \Rightarrow \quad \frac{ab}{a+b} \leq \frac{ab}{2\sqrt{ab}} = \frac{\sqrt{ab}}{2} \]

3. **Applying AM-GM to Each Term**:
- For \(\frac{ab}{a+b}\):
\[ \frac{ab}{a+b} \leq \frac{\sqrt{ab}}{2} \]
- For \(\frac{bc}{b+c}\):
\[ \frac{bc}{b+c} \leq \frac{\sqrt{bc}}{2} \]
- For \(\frac{ca}{c+a}\):
\[ \frac{ca}{c+a} \leq \frac{\sqrt{ca}}{2} \]

4. **Summing the Inequalities**:
\[ \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a} \leq \frac{\sqrt{ab}}{2} + \frac{\sqrt{bc}}{2} + \frac{\sqrt{ca}}{2} \]

5. **Using \(a + b + c = 1\)**: Let \(a + b + c = 1\). We can express \(c\) in terms of \(a\) and \(b\):
\[ c = 1 - a - b \]

6. **Finding the Maximum Value**: To maximize the expression, we can use symmetry and set \(a = b = c\). Since \(a + b + c = 1\), we have:
\[ a = b = c = \frac{1}{3} \]

7. **Calculating the Expression**:
\[ \frac{ab}{a+b} = \frac{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)}{\frac{1}{3} + \frac{1}{3}} = \frac{\frac{1}{9}}{\frac{2}{3}} = \frac{1}{6} \]
Similarly, for the other two terms:
\[ \frac{bc}{b+c} = \frac{1}{6}, \quad \frac{ca}{c+a} = \frac{1}{6} \]
Thus, the total is:
\[ \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2} \]

8. **Conclusion**: The greatest value of the expression \(\frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ca}{c+a}\) is \(\frac{1}{2}\).