`P=x^3-1/x^3, Q=x-1/x` `x in (1,oo)` then minimum value of `P/(sqrt(3)Q^2)`

Options

A. `2sqrt(2)`
B. `-2sqrt(3)`
C. non-existent
D. non of these

Explanation

To find the minimum value of \( \frac{P}{\sqrt{3} Q^2} \) where \( P = x^3 - \frac{1}{x^3} \) and \( Q = x - \frac{1}{x} \) for \( x \in (1, \infty) \), we will follow these steps:

### Step 1: Express \( P \) in terms of \( Q \)

We know that:
\[ P = x^3 - \frac{1}{x^3} \]
Using the identity for the difference of cubes, we can rewrite \( P \):
\[ P = (x - \frac{1}{x}) \left( (x - \frac{1}{x})^2 + 3 \right) \]
Let \( Q = x - \frac{1}{x} \). Then, we can express \( P \) as:
\[ P = Q(Q^2 + 3) \]

### Step 2: Substitute \( P \) in the expression \( \frac{P}{\sqrt{3} Q^2} \)

Now we substitute \( P \) into the expression we want to minimize:
\[ \frac{P}{\sqrt{3} Q^2} = \frac{Q(Q^2 + 3)}{\sqrt{3} Q^2} \]
This simplifies to:
\[ \frac{P}{\sqrt{3} Q^2} = \frac{Q^2 + 3}{\sqrt{3}} \]

### Step 3: Minimize \( \frac{Q^2 + 3}{\sqrt{3}} \)

To minimize \( \frac{Q^2 + 3}{\sqrt{3}} \), we need to minimize \( Q^2 \). Since \( Q = x - \frac{1}{x} \), we can find the minimum value of \( Q \) for \( x \in (1, \infty) \).

### Step 4: Find the minimum value of \( Q \)

The function \( Q = x - \frac{1}{x} \) is increasing for \( x > 1 \). Thus, we evaluate \( Q \) at \( x = 1 \):
\[ Q(1) = 1 - \frac{1}{1} = 0 \]
As \( x \to \infty \), \( Q \to \infty \). Therefore, the minimum value of \( Q \) in the interval \( (1, \infty) \) is \( 0 \).

### Step 5: Calculate the minimum value of \( \frac{P}{\sqrt{3} Q^2} \)

Since \( Q \) approaches \( 0 \) as \( x \) approaches \( 1 \), we can substitute \( Q = 0 \) into our expression:
\[ \frac{P}{\sqrt{3} Q^2} = \frac{0 + 3}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

### Conclusion

Thus, the minimum value of \( \frac{P}{\sqrt{3} Q^2} \) is:
\[ \sqrt{3} \]