I am looking at a proof and I found a strange inequality.
Let $n\in \mathbb{Z}^d$ then it is stated that $\sum_{j=1}^d{(2\pi)^{2k}n_j^{2k}}>>\parallel n\parallel_2^{2k}$ due to the inequality $\parallel n\parallel_2\leq\sqrt{d}\max_{j=1,...,d}{\mid n_j\mid}$. I really don't see it.
$$\sum_{j=1}^d |n_j|^{2k} \geq \max_{j=1,\ldots,d} |n_j|^{2k}$$ implies, using the mentioned inequality,
$$(2\pi)^{2k} \sum_{j=1}^d n_j^{2k} \geq (2\pi)^{2k} \max_{j=1,\ldots,d} |n_j|^{2k} \geq \left( \frac{2\pi}{\sqrt{d}} \right)^{2k} \|n\|_2^{2k}$$