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 $$J_{s, k}(X; \mathbf{n}) = \int_{[0, 1)^k} f(\mathbf{\alpha}; X)^{2s}e(n_1\alpha_1\dotsn_k\alpha_k)d\mathbf{\alpha}$$ 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?
1 Answer
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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 CauchySchwarz.

1$\begingroup$ $N_mN_{n+m}\leqslant \frac12(N_m^2+N_{n+m}^2)$ is not even CauchySchwarz $\endgroup$ Apr 5, 2017 at 5:52