# 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?

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,$$3\alpha^{1/3}\ge1\\\implies\alpha\ge\frac1{27}$$

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.

• That did not occur to me, thank you! – ExtremeRaider May 19 at 4:28

$$4\alpha x^2+\frac{1}{2x}+\frac{1}{2x}\geq 3\sqrt[3]{4\alpha x^2\cdot \frac{1}{2x}\cdot \frac{1}{2x}}$$

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 $$1/27$$.