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

4αx² + 1/x ≥ 1 ⇒ y = 4αx² + 1/x ⇒ y = dy/dx = 8αx - 1/x² = 0 ⇒ x = (1/8α)^(1/3) ⇒ f(x) = 4αx³ + 1/x = 1/2 + 1/(8α)^(1/3) ⇒ 3√(8α) ≥ 1 ⇒ α^(1/3) ≥ 1/2 ⇒ α ≥ 127