Let $k\geq 2$. Consider the following norm of exponenetial sum:
$$
I(N,p,k)=\int\_0^1\int\_0^1 \left|\sum\_{n=0}^N e^{2\pi i (n x+n^k y)}\right|^p dxdy.
$$
Bourgain mentioned on Page 118 of 


<https://math.mit.edu/classes/18.158/bourgain-restriction.pdf>


that $I(N,6,2)\gtrsim N^3\log N$, where he referenced the following article:
<https://www.researchgate.net/publication/259308546_The_method_of_trigonometric_sums_in_number_theory>.


But I did not find an explicit result in the above article that leads directly to the lower bound above.


So my questions are:


1. What is the idea to prove the above lower bound? The famous Vinogradov's mean value theorem deals with upper bounds of $I(N,p,2)$, but not lower bounds.
2. What is a reasonably sharp lower bound for $I(N,p,3)$, or particularly, $I(N,6,3)$? Note that this may not be in the direct form of Vinogradov's mean value theorem, as the $n^2$ term is missing here.



Lower bound on exponential sums
There are a few things to clear up.


The first is that, on the page in the Bourgain paper you mention, he actually proves the lower bound $I(N,6,2)\gg N^3\log N$ from the fact that 


$$ \left\lvert\sum\_{n=0}^N e(nx+n^2y)\right\rvert \gg N/q^{1/2}$$


whenever $\lvert x-b/q\rvert \ll 1/N$ and $\lvert y-a/q\rvert \ll 1/N^2$ for some fixed $1\leq a< q\leq N^{1/2}$ with $(a,q)=1$ and $1\leq b\leq q$. (The proof is simply summing the contribution from all such $a$ and $b$). It is this estimate which he invokes a reference for, rather than the lower bound for $I(N,6,2)$.


Secondly, the reference he gives is not to the paper you link to (which is a 1986 paper by Karatsuba-Vinogradov) but instead to Vinogradov's 1954 book with a similar title, usually translated to 'The Method of Trigonometric Sums in the Theory of Numbers'. I don't have a copy of this to hand to check the reference, but a quick search turned up a short note by Tamahiro Oh (<https://www.maths.ed.ac.uk/~toh/Files/WeylSum.pdf>) proving exactly this Weyl sum lower bound.


Finally, for $I(N,6,3)$, the situation is quite different, and here in fact an asymptotic formula is known:
$$ I(N,6,3) = 6N^3 + O(N^2(\log N)^5).$$
This is a result of Vaughan and Wooley (On a certain nonary cubic form and related equations. Duke Mathematical Journal, 80(3), 669–735, 1995).


