# Prove that $||u||\cdot ||v|| \geq |\langle u,v \rangle|$ for defined inner product

Let $K$ be a totally complex number field such that $[K:\mathbb{Q}]=n$ and $\mathcal{O}_K$ be its associated ring of integers. Let $\{\sigma_1, \dots, \sigma_n\}$ be the embeddings of $K$. For $u,v \in \mathcal{O}_K^2$, define the inner product:

$$\langle u,v \rangle := \sum_{i=1}^{n} \sigma_i(u_1)\overline{\sigma_i(v_1)} + \sigma_i(u_2)\overline{\sigma_i(v_2)},$$

where $u = (u_1,u_2)^T, v=(v_1,v_2)^T$ and $\bar{x}$ denotes complex conjugation. Define the norm of an element $u \in \mathcal{O}_K^2$ in the normal manner $||u|| = \langle u,u \rangle^{1/2}$. Is it possible to prove the following inequality:

$$|\langle u,v \rangle| \leq ||u||\cdot ||v||,$$

for all $u,v \in \mathcal{O}_K^2$?

$$|\langle u,v\rangle|^2=\Big|\sum_i\sigma_i(u_1)\overline{\sigma_i(v_1)}+\sigma_i(u_2)\overline{\sigma_i(v_2)}\Big|^2\leqslant\Big( \sum_i|\sigma_i(u_1)||\overline{\sigma_i(v_1)}|+|\sigma_i(u_2)||\overline{\sigma_i(v_2)}|\Big)^2$$ From Cauchy-Schwartz we get $$\Big( \sum_i|\sigma_i(u_1)||\overline{\sigma_i(v_1)}|+|\sigma_i(u_2)||\overline{\sigma_i(v_2)}|\Big)^2\\\leqslant\Big( \sum_i|\sigma_i(u_1)|^2+|\sigma_i(u_2)|^2\Big)^2\Big( \sum_i|\sigma_i(v_1)|^2+|\sigma_i(v_2)|^2\Big)^2\\=\Big( \sum_i\sigma_i(u_1)\overline{\sigma_i(u_1)}+\sigma_i(u_2)\overline{\sigma_i(u_2)}\Big)^2\Big( \sum_i\sigma_i(v_1)\overline{\sigma_i(v_1)}+\sigma_i(v_2)\overline{\sigma_i(v_2)}\Big)^2\\=||u||^2||v||^2$$