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I have the following inequality: $\alpha x<1+\beta x^4$ and this equality should hold for all $x \geq 0$ and some $\alpha,\beta \geq 0$ to be determined ($\alpha,x,\beta$ should all be real). I am considering the pairs $(\alpha,\beta)$ for which this holds given that it must be true for all $x \geq 0$.

For $\alpha=0$ I find $\beta \geq 0$ but for $\alpha>0$ I got stuck. Wolfram Alpha tells me that $\beta>27\alpha^4/(256)$ should hold. I guess this is related to the discriminant of the above polynomial $\beta x^4 +1 - \alpha x$ which is also precisely giving the above condition. But I don't see why this condition suffices?

Mathphys meister
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    Find $\alpha$, such as $\alpha x$ is a tangent to $1 + \beta x^4$. – user58697 Jun 04 '21 at 17:58
  • @user58697 . The derivative of $\alpha x$ is constant, it is just $\alpha$, but the derivative of $1+\beta x^4$ is $4x^3 \beta$. But these cannot be equal for all $x \geq 0$ right? – Mathphys meister Jun 04 '21 at 18:10
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    To be a tangent, it must have $x$ such that $\alpha x = 1 + \beta x^4$ (as well as $\alpha = 4\beta x^3$). Now exclude $x$, and obtain a relation between $\alpha$ and $\beta$. – user58697 Jun 04 '21 at 18:15

1 Answers1

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For $x \gt 0$ the inequality is equivalent to $\beta x^3 + \frac{1}{x} \ge \alpha$. By AM-GM, the LHS:

$$ \require{cancel} \beta x^3 + \frac{1}{x} = \beta x^3 + \frac{1}{3x} + \frac{1}{3x} + \frac{1}{3x} \ge 4 \sqrt[4]{\beta \cancel{x^3} \frac{1}{3^3 \cancel{x^3}}} $$

The minimum is attained when $\beta x^3 = \frac{1}{3x}$, so the condition for the inequality to hold for all $x \gt 0$:

$$ \alpha \lt 4 \sqrt[4]{\beta \cdot \frac{1}{3^3}} \;\;\;\;\iff\;\;\;\; \alpha^4 \lt 256 \cdot \frac{\beta}{27} $$

dxiv
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