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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?

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

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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.

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  • $\begingroup$ That did not occur to me, thank you! $\endgroup$ – ExtremeRaider May 19 at 4:28
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$$4\alpha x^2+\frac{1}{2x}+\frac{1}{2x}\geq 3\sqrt[3]{4\alpha x^2\cdot \frac{1}{2x}\cdot \frac{1}{2x}}$$

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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 $1/27$.

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