If a, b, c are the distinct positive numbers than `(a+b+c)(ab+bc+ac)` is :

Options

A. greater than 9abc
B. less than 8abc
C. equal to 10abc
D. greater than 25abc.

Answer

To solve the problem, we need to evaluate the expression \((a+b+c)(ab+bc+ac)\) where \(a\), \(b\), and \(c\) are distinct positive numbers.

Let's break it down step by step:

### Step 1: Understand the components of the expression
We have two parts in the expression:
1. \(a + b + c\) - This is the sum of the three distinct positive numbers.
2. \(ab + bc + ac\) - This is the sum of the products of the numbers taken two at a time.

### Step 2: Apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM)
For three distinct positive numbers \(a\), \(b\), and \(c\), the Arithmetic Mean (AM) is always greater than or equal to the Geometric Mean (GM). Thus, we have:
\[ \frac{a+b+c}{3} \geq \sqrt[3]{abc} \]
This implies:
\[ a + b + c \geq 3\sqrt[3]{abc} \]

### Step 3: Apply the AM-GM Inequality to the products
Similarly, we can apply the AM-GM inequality to the products:
\[ \frac{ab + ac + bc}{3} \geq \sqrt[3]{(abc)^2} \]
This implies:
\[ ab + ac + bc \geq 3\sqrt[3]{(abc)^2} \]

### Step 4: Combine the inequalities
Now we can combine the inequalities we derived:
\[ (a + b + c)(ab + ac + bc) \geq (3\sqrt[3]{abc})(3\sqrt[3]{(abc)^2}) = 9abc \]

### Step 5: Conclusion
Thus, we conclude that:
\[ (a + b + c)(ab + ac + bc) \geq 9abc \]
Since \(a\), \(b\), and \(c\) are distinct positive numbers, the minimum value of \((a + b + c)(ab + ac + bc)\) is \(9abc\).

### Final Answer
The expression \((a+b+c)(ab+bc+ac)\) is greater than or equal to \(9abc\).

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