For some $s, k$, let $J\_{s, k}(X; \mathbf{n})$ be the number of solutions to the system $$\sum\_{i\le s} (x\_i^j - y\_i^j) = n\_j$$ for $j\le k$ with $x\_1, \dots, x\_s, y\_1, \dots, y\_s\in [1, X]\cap\mathbb{Z}$. It is not hard to show that $$J\_{s, k}(X; \mathbf{n}) = \int\_{[0, 1)^k} |f(\mathbf{\alpha}; X)|^{2s}e(-n\_1\alpha\_1-\dots-n\_k\alpha\_k)d\mathbf{\alpha}$$ where 
$$f(\alpha; X) = \sum\_{n\le X} e(\alpha\_1 n^1 + \dots + \alpha\_k n^k)$$ from which it easily follows that for all $\mathbf{n}\in\mathbb{Z}^k$, $J\_{s, k}(X;\mathbf{n})\le J\_{s, k}(X; \mathbf{0})$. Is there any way to show that this is the case only by considering the underlying Diophantine equations which doesn't use the fact that $J\_{s, k}(X;\mathbf{n})$ is equal to the above integral?



Counting solutions of a certain diophantine equation
Let $N\_{\bf n}$ be the number of solutions to $\sum x\_i^s=n\_i$. Them you only need to show $\sum N\_{\bf m}N\_{\bf m+n}\le\sum N\_{\bf n}^2$, which is just Cauchy-Schwarz.


