Least value of real parameter alpha for algebraic inequality

To find the least value of $a \in \mathbb{R}$ for which the inequality $4 a x^2 + \frac{1}{x} \ge 1$ holds for all $x > 0$, we can approach the problem using two methods: calculus and the AM-GM inequality.

Method 1: Using Calculus

Since $x > 0$, we can rearrange the given inequality to isolate $a$: $4a x^2 \ge 1 - \frac{1}{x}$ $4a x^2 \ge \frac{x - 1}{x}$ $a \ge \frac{x - 1}{4x^3}$

For the inequality to hold for all $x > 0$, the parameter $a$ must be greater than or equal to the maximum value of the function $f(x)$ on the interval $(0, \infty)$, where $f(x) = \frac{x - 1}{4x^3} = \frac{1}{4} \left( \frac{1}{x^2} - \frac{1}{x^3} \right)$

To find the maximum of $f(x)$, we differentiate $f(x)$ with respect to $x$: $f'(x) = \frac{1}{4} \left( -\frac{2}{x^3} + \frac{3}{x^4} \right) = \frac{3 - 2x}{4x^4}$

Setting $f'(x) = 0$ gives: $3 - 2x = 0 \implies x = \frac{3}{2}$

For $x \in \left(0, \frac{3}{2}\right)$, $f'(x) > 0$, meaning $f(x)$ is strictly increasing. For $x \in \left(\frac{3}{2}, \infty\right)$, $f'(x) < 0$, meaning $f(x)$ is strictly decreasing.

Thus, $f(x)$ attains its absolute maximum at $x = \frac{3}{2}$: $f_{\text{max}} = f\left(\frac{3}{2}\right) = \frac{\frac{3}{2} - 1}{4 \left(\frac{3}{2}\right)^3} = \frac{\frac{1}{2}}{4 \times \frac{27}{8}} = \frac{\frac{1}{2}}{\frac{27}{2}} = \frac{1}{27}$

Therefore, $a \ge \frac{1}{27}$, and the least value of $a$ is $\frac{1}{27}$.

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Method 2: Using the AM-GM Inequality

For any positive real number $x > 0$ and $a > 0$, split the term $\frac{1}{x}$ into two equal parts: $4ax^2 + \frac{1}{x} = 4ax^2 + \frac{1}{2x} + \frac{1}{2x}$

By the Arithmetic Mean - Geometric Mean (AM-GM) inequality: $\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{4ax^2 \cdot \frac{1}{2x} \cdot \frac{1}{2x}}$ $4ax^2 + \frac{1}{x} \ge 3 \sqrt[3]{\frac{4ax^2}{4x^2}} = 3 \sqrt[3]{a}$

The minimum value of $4ax^2 + \frac{1}{x}$ over $x > 0$ is $3 \sqrt[3]{a}$ (achieved when $4ax^2 = \frac{1}{2x} \implies x = \frac{1}{2a^{1/3}}$).

For $4ax^2 + \frac{1}{x} \ge 1$ to hold for all $x > 0$, the minimum value must satisfy: $3 \sqrt[3]{a} \ge 1$ $\sqrt[3]{a} \ge \frac{1}{3}$ $a \ge \frac{1}{27}$

Hence, the least value of $a$ is $\frac{1}{27}$.

Correct Option: (C) $\frac{1}{27}$