Solve the following using AM-GM inequality The least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \ge 1 $for all $x \gt 0 $, is 
Using AM-GM inequality
 $$\frac{4ax^2 + \frac{1}{2x} + \frac{1}{2x}}{3} \ge \sqrt[3]{a}$$
$$4ax^2 + \frac{1}{x} \ge 3\sqrt[3]{a}$$
Now my question start from here .
Can I do  that for least value of $a$, the value of  $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$

No, I think. Because does not exist $a>0$, for which $3\sqrt[3]a>3\sqrt[3]a$. 
For the original problem we see that since for $4ax^2=\frac{1}{2x}$ we get an equality and we obtain $a\geq\frac{1}{27}$, so the answer is $\frac{1}{27}$.
