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

$\frac{1}{27}$

Let $f(x)=4 \alpha x^2+\frac{1}{x}, x>0$. Then,

$f'(x)=8 \alpha x-\frac{1}{x^2}=\frac{8 \alpha x^3-1}{x^2}$ and, $f''(x)=8 \alpha+\frac{2}{x^3}$

At points of local maximum or minimum, we must have

$f'(x)=0 \Rightarrow 8 \alpha x^3-1=0 \Rightarrow x=\left(\frac{1}{8 \alpha}\right)^{\frac{1}{3}}$

Clearly, $f''(x)=8 \alpha+\frac{2}{x^3}>0$ for all $x>0$

So, $f(x)$ attains its minimum value at $x=\left(\frac{1}{8 \alpha}\right)^{\frac{1}{3}}$

The minimum value of $f$ at $x=\left(\frac{1}{8 \alpha}\right)^{\frac{1}{3}}$ is

$f\left(\left(\frac{1}{8 \alpha}\right)^{1 / 3}\right)=4 \alpha\left(\frac{1}{8 \alpha}\right)^{2 / 3}+2(8 \alpha)^{1 / 3}$

It is given that $f(x) \geq 1$ for all $x>0$. So, the minimum value of $f(x)$ is 1

∴  $4 \alpha\left(\frac{1}{8 \alpha}\right)^{2 / 3}+(8 \alpha)^{1 / 3}=1 \Rightarrow 3 \alpha^{1 / 3}=1 \Rightarrow \alpha=\frac{1}{27}$