Cauchy–Schwarz inequality
R - The plane

The real vector space R 2 {\displaystyle \mathbb {R} ^{2}} denotes the 2-dimensional plane. It is also the 2-dimensional Euclidean space where the inner product is the dot product. If u = (u 1, u 2) {\displaystyle \mathbf {u} =\left(u_{1},u_{2}\right)} and v = (v 1, v 2) {\displaystyle \mathbf {v} =\left(v_{1},v_{2}\right)} then the Cauchy–Schwarz inequality becomes: ⟨ u, v ⟩ 2 = (‖ u ‖ ‖ v ‖ cos θ) 2 ≤ ‖ u ‖ 2 ‖ v ‖ 2, {\displaystyle \langle \mathbf {u},\mathbf {v} \rangle ^{2}=(\|\mathbf {u} \|\|\mathbf {v} \|\cos \theta)^{2}\leq \|\mathbf {u} \|^{2}\|\mathbf {v} \|^{2},} where θ {\displaystyle \theta } is the angle between u {\displaystyle \mathbf {u} } and v {\displaystyle \mathbf {v} }.

The form above is perhaps the easiest in which to understand the inequality, since the square of the cosine can be at most 1, which occurs when the vectors are in the same or opposite directions. It can also be restated in terms of the vector coordinates u 1 {\displaystyle u_{1}}, u 2 {\displaystyle u_{2}}, v 1 {\displaystyle v_{1}}, and v 2 {\displaystyle v_{2}} as (u 1 v 1 + u 2 v 2) 2 ≤ (u 1 2 + u 2 2) (v 1 2 + v 2 2), {\displaystyle \left(u_{1}v_{1}+u_{2}v_{2}\right)^{2}\leq \left(u_{1}^{2}+u_{2}^{2}\right)\left(v_{1}^{2}+v_{2}^{2}\right),} where equality holds if and only if the vector (u 1, u 2) {\displaystyle \left(u_{1},u_{2}\right)} is in the same or opposite direction as the vector (v 1, v 2) {\displaystyle \left(v_{1},v_{2}\right)}, or if one of them is the zero vector.
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