stylistic:tone
C: n -dimensional Complex space

If u, v ∈ C n {\displaystyle \mathbf {u},\mathbf {v} \in \mathbb {C} ^{n}} with u = (u 1, …, u n) {\displaystyle \mathbf {u} =\left(u_{1},\ldots,u_{n}\right)} and v = (v 1, …, v n) {\displaystyle \mathbf {v} =\left(v_{1},\ldots,v_{n}\right)} (where u 1, …, u n ∈ C {\displaystyle u_{1},\ldots,u_{n}\in \mathbb {C} } and v 1, …, v n ∈ C {\displaystyle v_{1},\ldots,v_{n}\in \mathbb {C} }) and if the inner product on the vector space C n {\displaystyle \mathbb {C} ^{n}} is the canonical complex inner product (defined by ⟨ u, v ⟩:= u 1 v 1 ¯ + ⋯ + u n v n ¯, {\displaystyle \langle \mathbf {u},\mathbf {v} \rangle :=u_{1}{\overline {v_{1}}}+\cdots +u_{n}{\overline {v_{n}}},} where the bar notation is used for complex conjugation), then the inequality may be restated more explicitly as follows: | ⟨ u, v ⟩ | 2 = | ∑ k = 1 n u k v ¯ k | 2 ≤ ⟨ u, u ⟩ ⟨ v, v ⟩ = (∑ k = 1 n u k u ¯ k) (∑ k = 1 n v k v ¯ k) = ∑ j = 1 n | u j | 2 ∑ k = 1 n | v k | 2. {\displaystyle |\langle \mathbf {u},\mathbf {v} \rangle |^{2}=\left|\sum _{k=1}^{n}u_{k}{\bar {v}}_{k}\right|^{2}\leq \langle \mathbf {u},\mathbf {u} \rangle \langle \mathbf {v},\mathbf {v} \rangle =\left(\sum _{k=1}^{n}u_{k}{\bar {u}}_{k}\right)\left(\sum _{k=1}^{n}v_{k}{\bar {v}}_{k}\right)=\sum _{j=1}^{n}\left|u_{j}\right|^{2}\sum _{k=1}^{n}\left|v_{k}\right|^{2}.}

That is, | u 1 v ¯ 1 + ⋯ + u n v ¯ n | 2 ≤ (| u 1 | 2 + ⋯ + | u n | 2) (| v 1 | 2 + ⋯ + | v n | 2). {\displaystyle \left|u_{1}{\bar {v}}_{1}+\cdots +u_{n}{\bar {v}}_{n}\right|^{2}\leq \left(\left|u_{1}\right|^{2}+\cdots +\left|u_{n}\right|^{2}\right)\left(\left|v_{1}\right|^{2}+\cdots +\left|v_{n}\right|^{2}\right).}