TITLE: show that $\{ nq^{\frac{1}{3}} \} + \{ nq^{\frac{2}{3}} \} \geq Cn^{-\frac{1}{2}}$
QUESTION [9 upvotes]: Let $q$ be a positive integer which is not a perfect cube. Prove that there exists a positive constant $C$ such that for all natural numbers $n$, one has
$$\{ nq^{\frac{1}{3}} \} + \{ nq^{\frac{2}{3}} \} \geq Cn^{-\frac{1}{2}}$$
where $\{ x \}$ denotes the fractional part of $x$

REPLY [2 votes]: This in not an answer but is too big for a comment to Noam Elkies' beautiful answer. What follows is a simple way of finding the small values in the P.P.S.
Let $a=[1,1,1;2,1,1;2,2,1]$, $c_i=a^i$, $n_i=c_i[1,3]$ and $f(n)=n^{1/2}( \|n 2^{1/3}\| + \|n2^{2/3}\|)$.  
The first few values of $f(n_i)$ are:
1  0.67251999792667369002
2  0.79333763029795146209
3  0.58150218789189747742
4  0.43461235216989424483
5  0.47947613209641674474
Note $n_9=38781$. For $i<=100000$ the smallest value of $f(n_i)$ occurs at $i=86880$ where $f(n_{86880})$ is approximately 0.36372453.
Two references for scaled error geometry type problems are:
Hensley, D., 2005. “Simultaneous diophantine approximation.
Hinkel, D., 2014. Constructing Simultaneous Diophantine Approximations of Certain Cubic Numbers. arXiv preprint arXiv:1412.3936.
The latter seems particularly appropriate here.