Find the minimum value of $\frac {2n^5+n\sqrt [5]{2}+1}{n^2}$ where $n\in\Bbb R^{+}$

Question

Find the minimum value of $$\frac {2n^5+n\sqrt [5]{2}+1}{n^2}$$ where $n\in\Bbb R^{+}$.

I tried to write this expression as $\dfrac {2n^5+n\sqrt [5]{2}+1}{n^2}=a$. From here we have,

$$2n^5-an^2+n\sqrt [5]{2}+1=0$$

This is a quintic equation. Then the question becomes, find the minimum value of $a$, such that the quintic has at least one positive real root. Here I am stuck.

Answer 1

Rewrite your expression as $$\frac {2n^5+n\sqrt [5]{2}+1}{n^2}= 2n^3 + \frac{\sqrt [5]{2}}{n} + \frac 1{n^2} \stackrel{AM-GM}{\geq}3\sqrt[3]{2n^3\cdot\frac{\sqrt [5]{2}}{n}\cdot \frac 1{n^2}} = 3\sqrt[5]{4}$$ Equality is achieved if $2n^3=\frac{\sqrt [5]{2}}{n}=\frac 1{n^2}$ which happens for $n=\frac 1{\sqrt[5]{2}}$.

Answer 2

Hint Prove that $$\frac {2n^5+n\sqrt [5]{2}+1}{n^2} \geq 2^{-\frac{3}{5}} $$ for all $n\in\Bbb R^{+}$.