Editing Cauchy–Schwarz inequality (section)

===Proof via the Pythagorean theorem===

The special case of <math>\mathbf{v} = \mathbf{0}</math> was proven above so it is henceforth assumed that <math>\mathbf{v} \neq \mathbf{0}.</math> 
Let
<math display=block>\mathbf{z} := \mathbf{u} - \frac {\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}.</math>

It follows from the linearity of the inner product in its first argument that:
<math display=block>\langle \mathbf{z}, \mathbf{v} \rangle 
= \left\langle \mathbf{u} - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v}, \mathbf{v} \right\rangle 
= \langle \mathbf{u}, \mathbf{v} \rangle - \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{v}, \mathbf{v} \rangle 
= 0.</math>

Therefore, <math>\mathbf{z}</math> is a vector orthogonal to the vector <math>\mathbf{v}</math> (Indeed, <math>\mathbf{z}</math> is the [[vector projection|projection]] of <math>\mathbf{u}</math> onto the plane orthogonal to <math>\mathbf{v}.</math>) We can thus apply the [[Pythagorean theorem#Inner product spaces|Pythagorean theorem]] to
<math display=block>\mathbf{u}= \frac{\langle \mathbf{u}, \mathbf{v} \rangle} {\langle \mathbf{v}, \mathbf{v} \rangle} \mathbf{v} + \mathbf{z}</math>
which gives
<math display=block>\|\mathbf{u}\|^2 
= \left|\frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}\right|^2 \|\mathbf{v}\|^2 + \|\mathbf{z}\|^2 
= \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{(\|\mathbf{v}\|^2 )^2} \,\|\mathbf{v}\|^2 + \|\mathbf{z}\|^2 
= \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{\|\mathbf{v}\|^2} + \|\mathbf{z}\|^2 \geq \frac{|\langle \mathbf{u}, \mathbf{v} \rangle|^2}{\|\mathbf{v}\|^2}.</math>

The Cauchy–Schwarz inequality follows by multiplying by <math>\|\mathbf{v}\|^2</math> and then taking the square root. 
Moreover, if the relation <math>\geq</math> in the above expression is actually an equality, then <math>\|\mathbf{z}\|^2 = 0</math> and hence <math>\mathbf{z} = \mathbf{0};</math> the definition of <math>\mathbf{z}</math> then establishes a relation of linear dependence between <math>\mathbf{u}</math> and <math>\mathbf{v}.</math> The converse was proved at the beginning of this section, so the proof is complete. <math>\blacksquare</math>