23 Cauchy–Schwarz inequality

All $z,w\in V$ satisfy

$$(i)\mathit{\hspace{1em}\hspace{1em}}|⟨z,w⟩|\le \parallel z\parallel \parallel w\parallel ,$$

$$(ii)\mathit{\hspace{1em}\hspace{1em}}\parallel z+w\parallel \le \parallel z\parallel +\parallel w\parallel .$$

Proof. (i) There exists $u\in 𝐂$ such that $u\overline{u}=1$ and $u⟨z,w⟩=|⟨z,w⟩|$. Now we have a quadratic in $t$ which is non negative

$$0\le {\parallel tw+uz\parallel }^2=⟨tw+uz,tw+uz⟩$$

$$=t^2⟨w,w⟩+t⟨w,uz⟩+⟨uz,tw⟩+⟨uz,uz⟩$$

$$=t^2{\parallel w\parallel }^2+2t|⟨z,w⟩|+{\parallel z\parallel }^2.$$

So this quadratic has discriminant $b^2-4ac\le 0$, so

$$4{|⟨z,w⟩|}^2\le 4{\parallel z\parallel }^2{\parallel w\parallel }^2.$$