If a, b, c are three distinct positive real numbers such that `(b+c)/(a)+(c+a)/(b)+(a+b)/( c )gt` k, then the grealtest value of k, is

Options

A. 6
B. 3
C. 4
D. 9

Explanation

To solve the problem, we need to find the greatest value of \( k \) such that the following inequality holds for distinct positive real numbers \( a, b, c \):

\[ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} > k \]

### Step-by-step Solution:

1. **Apply the AM-GM Inequality**:
We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For any two distinct positive real numbers \( x \) and \( y \), we have:
\[ \frac{x+y}{2} \geq \sqrt{xy} \]
with equality when \( x = y \).

2. **Consider Each Pair**:
We will apply AM-GM to the pairs \( \frac{b+c}{a} \) and \( \frac{c+a}{b} \), and \( \frac{a+b}{c} \).

- For \( \frac{b+c}{a} \) and \( \frac{c+a}{b} \):
\[ \frac{b+c}{a} + \frac{c+a}{b} \geq 2 \sqrt{\frac{(b+c)(c+a)}{ab}} \]
- For \( \frac{c+a}{b} \) and \( \frac{a+b}{c} \):
\[ \frac{c+a}{b} + \frac{a+b}{c} \geq 2 \sqrt{\frac{(c+a)(a+b)}{bc}} \]
- For \( \frac{a+b}{c} \) and \( \frac{b+c}{a} \):
\[ \frac{a+b}{c} + \frac{b+c}{a} \geq 2 \sqrt{\frac{(a+b)(b+c)}{ca}} \]

3. **Combine the Inequalities**:
Adding these three inequalities together gives:
\[ \left( \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \right) \geq 6 \]
This implies:
\[ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \geq 6 \]

4. **Conclusion**:
Therefore, we can conclude that:
\[ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} > k \text{ holds for } k = 6. \]
Since the inequality is strict, the greatest value of \( k \) is:
\[ \boxed{6} \]