Dimostrazione.
1.
Segue dalla definizione di prodotto scalare e dal fatto di essere definito positivo.
2.
$\sqrt{\alpha \mathbf{v} \cdot \alpha \mathbf{v}}= \sqrt{\alpha^{2} (\mathbf{v} \cdot \mathbf{v})}= \vert\alpha\vert \, \vert\vert\mathbf{v}\vert\vert$
3.
$\vert\vert\mathbf{v}+\mathbf{w}\vert\vert^{2}= (\mathbf{v}+\mathbf{w})(\mathbf{...
...\vert^{2}+\vert\vert\mathbf{w}\vert\vert^{2}+2 \mathbf{v} \cdot \mathbf{w} \leq$
$\vert\vert\mathbf{v}\vert\vert^{2}+\vert\vert\mathbf{w}\vert\vert^{2}+2\vert\ve...
...vert\vert = (\vert\vert\mathbf{v}\vert\vert+\vert\vert\mathbf{w}\vert\vert)^{2}$
4.
$\vert\vert\mathbf{u}+\mathbf{v}\vert\vert^2 +\vert\vert\mathbf{u}-\mathbf{v}\ve...
... (\mathbf{u}+\mathbf{v})+(\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v})=$
$= \vert\vert\mathbf{u}\vert\vert^2+\vert\vert\mathbf{v}\vert\vert^2+2\mathbf{u}...
...\mathbf{v}= 2\vert\vert\mathbf{u}\vert\vert^2+2\vert\vert\mathbf{v}\vert\vert^2$
c.v.d.