For all positive values of x and y, the value of
`((1+x+x^(2))(1+y+y^(2)))/(xy)`,is

Options

A. `le9`
B. `lt9`
C. `ge9`
D. `gt9`

Explanation

To solve the problem, we need to evaluate the expression

\[ \frac{(1 + x + x^2)(1 + y + y^2)}{xy} \]

for all positive values of \(x\) and \(y\).

### Step 1: Simplify the expression

We can rewrite the expression as:

\[ \frac{(1 + x + x^2)}{x} \cdot \frac{(1 + y + y^2)}{y} \]

This gives us two separate functions:

\[ f(x) = \frac{1 + x + x^2}{x} \quad \text{and} \quad g(y) = \frac{1 + y + y^2}{y} \]

### Step 2: Analyze \(f(x)\)

Now, let's analyze \(f(x)\):

\[ f(x) = \frac{1}{x} + 1 + x \]

### Step 3: Apply the AM-GM Inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality on the terms \(x\) and \(\frac{1}{x}\):

\[ \frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1 \]

This implies:

\[ x + \frac{1}{x} \geq 2 \]

Adding 1 to both sides gives:

\[ f(x) = \frac{1}{x} + 1 + x \geq 2 + 1 = 3 \]

### Step 4: Analyze \(g(y)\)

Similarly, we can analyze \(g(y)\):

\[ g(y) = \frac{1 + y + y^2}{y} = \frac{1}{y} + 1 + y \]

Applying the same AM-GM inequality:

\[ \frac{y + \frac{1}{y}}{2} \geq 1 \implies y + \frac{1}{y} \geq 2 \]

Thus, we have:

\[ g(y) \geq 3 \]

### Step 5: Combine the results

Now, we can combine the results of \(f(x)\) and \(g(y)\):

\[ \frac{(1 + x + x^2)(1 + y + y^2)}{xy} = f(x) \cdot g(y) \geq 3 \cdot 3 = 9 \]

### Conclusion

Therefore, we conclude that:

\[ \frac{(1 + x + x^2)(1 + y + y^2)}{xy} \geq 9 \]

### Final Answer

The value of the expression for all positive values of \(x\) and \(y\) is:

\[ \boxed{9} \]