existence of a prime as close as possible to $\sqrt[3]{n}$, for $n$ large enough

Question

I'm studying the property of some extremal graphs in Graph Theory, and in an old paper (1966) I encountered a number theory theorem, that goes as follow :

For any $\varepsilon \in ]0,1[$, there exist an integer $N_\varepsilon$ such that for all $n>N_\varepsilon$, there exist a prime $p$ for which $$ n^{1/3} > p > (1-\varepsilon)^{1/5}n^{1/3}$$

However, there is an error in the referenced bibliography. I would be looking for a proof, or at least a sketch of the proof if too complicated. I could start from Bertrand's postulate (exists prime $p$ between $n$ and $2n$), but I don't know how to go from there.

Any idea, or references?

Thanks a lot.

Answer 1

I'm writing a small proof in order to close the topic.

As stated by others, the question is equivalent to the following:

For any $x\in ]0,1[$, there exist an integer $N_x$ such that for all $n>N_x$, there exist a prime $p$ for which $$ n > p > xn$$

Using the Prime Number Theorem, we know that : $$\pi(x) \approx \frac{x}{\log x}$$ and the result follow immediatly. Thanks all.