TITLE: Least value of $a$ for which $4ax^2 + \frac{1}{x} \geq 1$
QUESTION [5 upvotes]: Find the least value of $a \in R$ for which $4ax^2 + \frac{1}{x} \geq 1$, for all $x>0$.
The equation will transform into (Using $x>0$)
$4ax^3-x+1\geq 0$
But I don't know how to deal with this cubic inequality. Could someone give me some hint as how to proceed?

REPLY [5 votes]: If we make a change of variable $a=1/A^3$ and $x=Au/2$, then the problem becomes one of finding the largest value of $A$ for which the inequality
$$u^2+{2\over u}\ge A$$
holds for all $u\gt0$.
If you know calculus, then it's easy to see that $f(u)=u^2+{2\over u}$ is minimized at $u=1$, so that $f(1)=3$ is the largest that $A$ can be.  If you don't know calculus, then some extra work is required:
In order for the inequality to hold for all $u\gt0$, we clearly need $A\le1^2+{2\over1}=3$.  But it's also easy to see that, for $u\gt0$, we have
$$\begin{align}
u^2+{2\over u}\ge3&\iff u^3-3u+2\ge0\\
&\iff(u-1)^2(u+2)\ge0
\end{align}$$
Since the final inequality here is obviously true (the product of a square and a positive number is necessarily non-negative), we can conclude that $A=3$ is the largest number for which the inequality $u^2+{2\over u}\ge A$ holds for all $u\gt0$, and thus $a=1/27$ is the smallest value for which the original inequality holds for all $x\gt0$.