Let a,b,c be positive real parameter and `ax^2+ b/x^2gec, AA xepsilonR` then (A) `4abgec^2` (B) `4cgeb^2` (C) `4bcgec^2` (D) `4acltb^2`

Options

A. `4ab ge c^2`
B. `4ac ge b^2`
C. `4bc ge a^2`
D. `4ac lt b^2`

Answer

To solve the problem, we need to show that if \( ax^2 + \frac{b}{x^2} \geq c \) for all \( x \in \mathbb{R} \), then we can derive the inequality \( 4ab \geq c^2 \).

### Step-by-Step Solution:

1. **Understanding the Given Expression**:
We have the expression \( ax^2 + \frac{b}{x^2} \) and we know it is always greater than or equal to \( c \) for all positive values of \( x \).

2. **Applying the AM-GM Inequality**:
By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we can state:
\[ \frac{ax^2 + \frac{b}{x^2}}{2} \geq \sqrt{ax^2 \cdot \frac{b}{x^2}} \]
This simplifies to:
\[ \frac{ax^2 + \frac{b}{x^2}}{2} \geq \sqrt{ab} \]
Thus, we can write:
\[ ax^2 + \frac{b}{x^2} \geq 2\sqrt{ab} \]

3. **Relating to the Given Condition**:
Since we know that \( ax^2 + \frac{b}{x^2} \geq c \), we can combine this with our previous result:
\[ 2\sqrt{ab} \geq c \]

4. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ (2\sqrt{ab})^2 \geq c^2 \]
This gives us:
\[ 4ab \geq c^2 \]

5. **Conclusion**:
From the derived inequality, we conclude that:
\[ 4ab \geq c^2 \]
Therefore, the correct answer is option (A) \( 4ab \geq c^2 \).