If `a,b,c` are three positive real numbers then the minimum value of the expression `(b+c)/a+(c+a)/b+(a+b)/c` is

Options

A. 1
B. 2
C. 3
D. none of these

Explanation

To find the minimum value of the expression \(\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}\) for positive real numbers \(a\), \(b\), and \(c\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step Solution:

1. **Rewrite the Expression**:
We start with the expression:
\[ E = \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \]

2. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for any positive real numbers \(x_1, x_2, \ldots, x_n\):
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \]
We can apply this to the three fractions in \(E\):
\[ \frac{\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}}{3} \geq \sqrt[3]{\frac{(b+c)(c+a)(a+b)}{abc}} \]

3. **Simplify the Right-Hand Side**:
We need to simplify \((b+c)(c+a)(a+b)\):
\[ (b+c)(c+a)(a+b) = (b+c)(ca + ab + ac + bc) = bca + bcb + bca + cba + cbc + cab + cba + abc \]
This expression is complex, but we can use the AM-GM inequality on each pair:
\[ b+c \geq 2\sqrt{bc}, \quad c+a \geq 2\sqrt{ca}, \quad a+b \geq 2\sqrt{ab} \]

4. **Combine the Inequalities**:
Multiplying these inequalities gives:
\[ (b+c)(c+a)(a+b) \geq 2\sqrt{bc} \cdot 2\sqrt{ca} \cdot 2\sqrt{ab} = 8\sqrt{(abc)^2} = 8abc \]

5. **Substituting Back**:
Now substituting back into the AM-GM inequality:
\[ \frac{\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}}{3} \geq \sqrt[3]{\frac{8abc}{abc}} = \sqrt[3]{8} = 2 \]
Therefore,
\[ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \geq 6 \]

6. **Conclusion**:
The minimum value of the expression \(\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c}\) is \(6\).

### Final Answer:
Thus, the minimum value of the expression is:
\[ \boxed{6} \]