Prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$ Can anyone give me a proof or a reference for a proof that there exists a prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$, for $n$ sufficiently large? 
I am reading a lecture where the professor uses this fact but does not provide any reference.
Thank you!

I think this gets you in the ball park. 
Number of primes  $<x\approx \frac{x}{ln(x)}$
$\frac{\sqrt{x}}{ln(\sqrt{x})}=\frac{2}{\sqrt{x}}\frac{x}{ln(x)}$
$\frac{x^{1/3}}{ln(x^{1/3})}=\frac{3}{x^{2/3}}\frac{x}{ln(x)}$
So it suffices to prove $(\frac{2}{\sqrt{x}}-\frac{3}{x^{2/3}})\frac{x}{ln(x)}>1$
for sufficient large x.
Perhaps the ratios could help mitigate relative error issues. 
We want: $$\pi(\sqrt{x})-\pi(x^{\frac{1}{3}})>1$$
$$\frac{\pi(\sqrt{x})}{\pi(x^{\frac{1}{3}})}>1+\frac{1}{\pi(x^{\frac{1}{3}})}$$
$$\pi(\sqrt{x})>2\pi(x^{\frac{1}{3}})$$
Taking ratios of the approximations above, 
$\frac{\pi(\sqrt{x})}{\pi(x^{\frac{1}{3}})}\approx \frac{2}{3}x^{\frac{1}{6}}$
$\frac{2}{3}x^{\frac{1}{6}}\pi(x^{\frac{1}{3}})>\pi(x^{\frac{1}{3}})+1$
$\frac{2}{3}x^{\frac{1}{6}}>1+\frac{1}{\pi(x^{\frac{1}{3}})}$
$\frac{2}{3}x^{\frac{1}{6}}>2$
$x>729$
