I'm reading the following proof of the triangle inequality 
However i'm a little confused. It says we have equality in 6.10 if and only if one of $u$ or $v$ is a non negative multiple which they're saying is equivalent to the statement 1 $\langle u,v \rangle = ||u||||v||$
Now the statement $ \exists a, u = av \Leftrightarrow \langle u,v \rangle = ||u||||v||$ i'm fine with but its getting to the requirement that i'm not.
See $|\langle u,v \rangle | = ||u||||v||$ holding only proves the second inequality is an equality namely $||u||^2+||v||^2 + 2|\langle u,v \rangle | = ||u||^2+||v||^2 + 2||u||||v||$. It does not however prove that $\text{Re}\langle u,v \rangle = |\langle u,v \rangle|$ or $\text{Re}\langle u,v \rangle = ||u||||v||$ if and only if $|\langle u,v \rangle| = ||u||||v||$ which is what we need for the first inequality to be an equality, or am i missing something? Can we prove this result from the axioms of norms and inner product spaces?
$\textbf{EDIT}$:ATTEMPT AT ANSWERING MY OWN QUESTION AFTER IT WAS POSED: We require $2\text{Re}\langle u,v \rangle = 2|\langle u,v \rangle|$, so let $\langle u,v \rangle = a+bi$ for $a,b \in \mathbb{R}$. Then $2\text{Re}\langle u,v \rangle = 2a = 2\sqrt{a^2+b^2} = 2|\langle u,v \rangle|$ if and only if $b=0$ in which case we know $\langle u,v \rangle \in \mathbb{R}$. Hence $\text{Re}(\langle u,v \rangle) = \langle u,v \rangle$ and so since we require $\text{Re}(\langle u,v \rangle) = ||u||||v||$ that is equivalent to saying $\langle u,v \rangle = ||u||||v||$. Is this correct?
What the author means is that if $\|u+v\|=\|u\|+\|v\|$, equality must hold in both 6.11 and 6.12. Since equality holds in 6.11, $\langle u,v\rangle$ must be a nonnegative real number. Therefore, the equality in 6.12, namely $|\langle u,v\rangle|=\|u\|\|v\|$, can be rephrased as $\langle u,v\rangle=\|u\|\|v\|$.