Least value of $\alpha \in \mathbb R$ ($\forall x >0)$ for which $4 \alpha x^2 + \frac{1}{x} \geq 1$ 
What is the least value of $\alpha \in \mathbb R$ ($\forall x >0)$  for which 
  $$4 \alpha x^2 + \frac{1}{x} \geq 1?$$ 

I've tried applying the A.M $\geq$ G.M inequality-
$$\dfrac{4\alpha x^2 +\frac{1}{x}}{2} \geq \sqrt{4\alpha x^2 \times \frac{1}{x}}$$
$$\implies \dfrac{4\alpha x^2 +\frac{1}{x}}{2} \geq \sqrt{4\alpha x }$$
I'm not sure how I can simplify this further to obtain a condition for $\alpha$. 
How do I proceed with this question? 
 A: An alternative approach using calculus: when $\alpha<0,4\alpha x^2+\frac1x\to-\infty<1$ as $x\to\infty.\alpha=0$ may be rejected similarly. Thus, $\alpha>0$.
$4\alpha x^2+\frac1x$ is differentiable for $x>0$ and attains global minimum of $3\alpha^{1/3}$ at $x=\frac1{2\alpha^{1/3}}$. Thus,$$3\alpha^{1/3}\ge1\\\implies\alpha\ge\frac1{27}$$
A: Clearly you can see, x does not cancel on RHS
You need to use the weighted AM-GM inequality. 
$$\frac {(4\alpha x^2)+2(\frac 1 {2x})} {3} \ge \sqrt[3]{ (4\alpha x^2)\Big(\frac 1 {2x}\Big)^2}$$
See if you can proceed now.
A: $$4\alpha x^2+\frac{1}{2x}+\frac{1}{2x}\geq 3\sqrt[3]{4\alpha x^2\cdot \frac{1}{2x}\cdot \frac{1}{2x}}$$
A: The least value is $1/27$. Here is why.
Suppose that $\alpha\in\mathbb{R}$ satisfies 
$$
(\forall x>0)\quad
4\alpha x^2+\frac{1}{x}\geq 1.
$$
We therefore must have $(\forall x>0)\;\alpha\geq(x-1)/(4x^2)$. A straightforward calculation shows that the maximum value of the function $x\mapsto(x-1)/(4x^2)$ on $\left]0,+\infty\right[$ is $1/27$. Thus $\alpha$ must be at least $1/27$. Let us show that the value $1/27$ satisfies the requirement. Indeed, by Cauchy--Schwarz,
$$
(\forall x>0)\quad
\frac{4}{27}x^2+\frac{1}{x}
=\frac{4}{27}x^2+\frac{1}{2x}+\frac{1}{2x}
\geq
3\sqrt[3]{\frac{4}{27}x^2\times\frac{1}{2x}\times\frac{1}{2x}}=1,
$$
as claimed. Show the least value of $\alpha$ is $1/27$.
