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 $4ax^2 + \frac{1}{x}$
Can I do that for least value of $a$, the value of $3\sqrt[3]{a} $ must greater than minimum value of $4ax^2 + \frac{1}{x}$
I find this strange because "minimum value of $4ax^2 + \frac{1}{x}$" is $3\sqrt[3]{a}$. So I guess that you meant that "the value of $3\sqrt[3]{a} $ must greater than $1$". Then, it is correct. I'll write the details in the following.
As some comment, $a$ has to be positive. (setting $x=1$ gives $a\ge 0$, but for $a=0$, the inequality does not hold for $x=2$.)
You have $$4ax^2+\frac 1x\ge 3\sqrt[3]{a}$$ The equality holds when $x=\frac{1}{2\sqrt[3]{a}}\ (\gt 0)$.
If $3\sqrt[3]{a}\ge 1$, i.e. $a\ge\frac{1}{27}$, $$4ax^2+\frac 1x\ge 3\sqrt[3]{a}\ge 1$$ holds for all $x\gt 0$.
If $0\lt 3\sqrt[3]{a}\lt 1$, i.e. $0\lt a\lt \frac{1}{27}$, $$4ax^2+\frac 1x\ge 1$$ does not hold for $x=\frac{1}{2\sqrt[3]{a}}$.
Therefore, the least value of $a$ is $\frac{1}{27}$.
Another way :
We can have $$a\ge\frac{x-1}{4x^3}$$ Now consider the graph of $y=\frac{x-1}{4x^3}$ for $x\gt 0$.
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 $\frac{1}{27}$.
