If a, b, c all positive and not euqal, then the value of `((a+b+c)(ab+bc+ca))/(abc)` is :

Options

A. less than 9
B. greater than 9
C. less than or equal to 9
D. can't be determined

Answer

To solve the problem, we need to find the value of the expression \(\frac{(a+b+c)(ab+bc+ca)}{abc}\) given that \(a\), \(b\), and \(c\) are all positive and not equal.

### Step-by-Step Solution:

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

2. **Use Known Inequalities**: Since \(a\), \(b\), and \(c\) are positive and not equal, we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. According to AM-GM, for any positive numbers \(x_1, x_2, \ldots, x_n\),
\[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \]
For three numbers, this gives us:
\[ \frac{a+b+c}{3} \geq \sqrt[3]{abc} \]
Hence, \(a+b+c \geq 3\sqrt[3]{abc}\).

3. **Evaluate \(ab + bc + ca\)**: Similarly, we can apply the AM-GM inequality to the products \(ab\), \(bc\), and \(ca\):
\[ \frac{ab + bc + ca}{3} \geq \sqrt[3]{(abc)^2} \]
Therefore, \(ab + bc + ca \geq 3\sqrt[3]{(abc)^2}\).

4. **Combine the Inequalities**: Now, substituting these inequalities into our original expression:
\[ (a+b+c)(ab+bc+ca) \geq (3\sqrt[3]{abc})(3\sqrt[3]{(abc)^2}) = 9(abc) \]

5. **Final Evaluation**: Thus, we have:
\[ \frac{(a+b+c)(ab+bc+ca)}{abc} \geq \frac{9abc}{abc} = 9 \]

6. **Conclusion**: Since \(a\), \(b\), and \(c\) are not equal, the equality in AM-GM does not hold, which means the expression is strictly greater than 9. Therefore, the value of \(\frac{(a+b+c)(ab+bc+ca)}{abc}\) is greater than 9.

### Final Answer:
The value of \(\frac{(a+b+c)(ab+bc+ca)}{abc}\) is greater than 9.