If a, b, c are three unequal positive numbers, then
Statement-1: The product of their sum and the sum of their reciprocals exceeds 9.
Statement-2: AM of n positive numbers exceeds their HM

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze both statements given in the question.

### Step 1: Analyze Statement 2
The second statement says that the Arithmetic Mean (AM) of \( n \) positive numbers exceeds their Harmonic Mean (HM). This is a well-known inequality in mathematics.

The formula for the Arithmetic Mean of \( n \) positive numbers \( x_1, x_2, \ldots, x_n \) is:
\[ AM = \frac{x_1 + x_2 + \ldots + x_n}{n} \]

The formula for the Harmonic Mean of \( n \) positive numbers is:
\[ HM = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \ldots + \frac{1}{x_n}} \]

By the AM-HM inequality, we know that:
\[ AM \geq HM \]
Since \( a, b, c \) are positive numbers, we can conclude that Statement 2 is correct.

### Step 2: Analyze Statement 1
The first statement claims that the product of the sum of \( a, b, c \) and the sum of their reciprocals exceeds 9.

Let’s denote:
- \( S = a + b + c \)
- \( R = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)

We need to show that:
\[ S \cdot R > 9 \]

Using the AM-HM inequality for the three positive numbers \( a, b, c \):
\[ AM \geq HM \implies \frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \]

Rearranging gives:
\[ (a + b + c) \cdot \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \geq 9 \]

Since \( a, b, c \) are unequal positive numbers, the equality does not hold, which means:
\[ (a + b + c) \cdot \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) > 9 \]

Thus, Statement 1 is also correct.

### Conclusion
Both statements are true, and Statement 2 serves as a basis for proving Statement 1.