The least value of $a \in \mathbb{R}$ for which $4ax^{2}+\frac{1}{x} \ge 1$, for all $x > 0$, is

The least value of a ∈ R for which 4ax²+1/x ≥ 1, for all x > 0, is

$4ax^{2}+\dfrac{1}{x} \ge 1 \quad \text{for all } x>0$

$\text{Find the least } a \text{ with } 4a \ge \dfrac{1-1/x}{x^{2}} \text{ for all } x>0$

$4ax^{2} \ge 1-\dfrac{1}{x} \Rightarrow 4a \ge \dfrac{1-1/x}{x^{2}}$

$4a \ge \dfrac{x-1}{x^{3}} = \phi(x)$

$\text{Need } 4a \ge \max_{x>0}\phi(x)$

$\phi(x) = x^{-2}-x^{-3}$

$\phi'(x) = -2x^{-3}+3x^{-4} = \dfrac{3-2x}{x^{4}}$

$\phi'(x)=0 \Rightarrow x=\dfrac{3}{2}$

$\phi'\ \text{changes } +\ \text{to}\ - \Rightarrow \text{max at } x=\dfrac{3}{2}$

$\phi\!\left(\dfrac{3}{2}\right)=\dfrac{1/2}{27/8}=\dfrac{4}{27}$

$4a \ge \dfrac{4}{27} \Rightarrow a \ge \dfrac{1}{27}$

$\boxed{a_{\min} = \dfrac{1}{27}\ \text{— option (3)}}$