The least value of `a in RR` for which `4ax^2+1/x >= 1`, for all `x > 0`, is

Options

A. `(1)/(64)`
B. `(1)/(32)`
C. `(1)/(27)`
D. `(1)/(25)`

Explanation

To find the least value of \( a \) in \( \mathbb{R} \) for which the inequality \( 4ax^2 + \frac{1}{x} \geq 1 \) holds for all \( x > 0 \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Here’s the step-by-step solution:

### Step 1: Rewrite the expression
We start with the inequality:
\[ 4ax^2 + \frac{1}{x} \geq 1 \]
To apply the AM-GM inequality, we can break down \( 4ax^2 \) and \( \frac{1}{x} \) into two parts. We can express \( 4ax^2 \) as:
\[ 4ax^2 = 2ax^2 + 2ax^2 \]
Now, we can rewrite the inequality as:
\[ 2ax^2 + 2ax^2 + \frac{1}{x} \geq 1 \]

### Step 2: Apply the AM-GM inequality
According to the AM-GM inequality:
\[ \frac{2ax^2 + 2ax^2 + \frac{1}{x}}{3} \geq \sqrt[3]{(2ax^2)(2ax^2)\left(\frac{1}{x}\right)} \]
This simplifies to:
\[ \frac{4ax^2 + \frac{1}{x}}{3} \geq \sqrt[3]{4a^2x^4 \cdot \frac{1}{x}} = \sqrt[3]{4a^2x^3} \]

### Step 3: Simplify the inequality
From the AM-GM result, we have:
\[ \frac{4ax^2 + \frac{1}{x}}{3} \geq \sqrt[3]{4a^2x^3} \]
Thus:
\[ 4ax^2 + \frac{1}{x} \geq 3\sqrt[3]{4a^2x^3} \]
For the inequality \( 4ax^2 + \frac{1}{x} \geq 1 \) to hold for all \( x > 0 \), we need:
\[ 3\sqrt[3]{4a^2x^3} \geq 1 \]

### Step 4: Analyze the inequality
To ensure this holds for all \( x > 0 \), we can set \( x = 1 \):
\[ 3\sqrt[3]{4a^2} \geq 1 \]
This simplifies to:
\[ \sqrt[3]{4a^2} \geq \frac{1}{3} \]

### Step 5: Cube both sides
Cubing both sides gives:
\[ 4a^2 \geq \left(\frac{1}{3}\right)^3 = \frac{1}{27} \]
Thus:
\[ a^2 \geq \frac{1}{108} \]

### Step 6: Solve for \( a \)
Taking the square root of both sides, we find:
\[ a \geq \frac{1}{\sqrt{108}} = \frac{1}{6\sqrt{3}} = \frac{\sqrt{3}}{18} \]

### Step 7: Find the least value of \( a \)
The least value of \( a \) that satisfies this inequality is:
\[ a = \frac{1}{27} \]

### Final Answer
Thus, the least value of \( a \) in \( \mathbb{R} \) for which \( 4ax^2 + \frac{1}{x} \geq 1 \) for all \( x > 0 \) is:
\[ \boxed{\frac{1}{27}} \]