Is the following true (and if yes, where the best proof is written?)? 


For any $c>0$ for large enough positive integers $N$ we have $\sum\_{k=0}^{N-1} \cos(k^2t)\geqslant -cN$ for all real $t$?


Hm, if true, it should be hard: it allows to get signs of certain Gauss type sums, for example.



uniform one-sided van der Corput inequality
The answer is negative. 


As you said if you take $\frac{t}{2\pi}$ to be a rational number, say $\frac{p}{q}$ then it converges to the Gaus-sums:
$$
\lim\_{N \to \infty} \frac{1}{N} \sum\_{k=0}^{N-1} \cos(k^{2} t) = \Re \frac{1}{q} \sum\_{k=0}^{q-1} \exp\left( 2 \pi i \cdot \frac{p}{q} k^{2}\right)
$$
Now if you take $t=\frac{4 \pi}{5}$ (i.e., $p=2, q=5$) I think it should no be difficult to show that the latter sum is a negative number $< - 0.4$ (even without taking the real part). 


