Editing Cauchy–Schwarz inequality (section)

=== {{math|R<sup>2</sup>}} - The plane ===
[[File:Cauchy-Schwarz inequation in Euclidean plane.gif|thumb|300px|Cauchy–Schwarz inequality in a unit circle of the Euclidean plane]]
The real vector space <math>\R^2</math> denotes the 2-dimensional plane. It is also the 2-dimensional [[Euclidean space]] where the inner product is the [[dot product]]. 
If <math>\mathbf{u} = (u_1, u_2)</math> and <math>\mathbf{v} = (v_1, v_2)</math> then the Cauchy–Schwarz inequality becomes:
<math display=block>\langle \mathbf{u}, \mathbf{v} \rangle^2 = \bigl(\|\mathbf{u}\| \|\mathbf{v}\| \cos \theta\bigr)^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2,</math>
where <math>\theta</math> is the [[angle]] between <math>\mathbf{u}</math> and <math>\mathbf{v}</math>.

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 <math>u_1</math>, <math>u_2</math>, <math>v_1</math>, and <math>v_2</math> as
<math display=block>\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),</math>
where equality holds if and only if the vector <math>\left(u_1, u_2\right)</math> is in the same or opposite direction as the vector <math>\left(v_1, v_2\right)</math>, or if one of them is the zero vector.