In the book Linear Algebra Done Right, the author asserts that the Triangle Inequality is an equality iff $\langle u,v\rangle $ $=$ $\|u\|\|v\|$.
My Attempt:
Claim: The Triangle Inequality is an equality iff $\langle u,v\rangle $ $=$ $\|u\|\|v\|$
($\Longrightarrow$) Suppose $\|u+v\|=\|u\|+\|v\|$. Then $\|u+v\|^2 = (\|u\|+\|v\|)^2$. Thus $\langle u+v,u+v\rangle = \langle u,u\rangle + 2\|u\|\|v\| +\langle v,v\rangle $. LHS is $\langle u,u\rangle + \langle v,u\rangle + \langle u,v\rangle + \langle v,v\rangle $. So $\langle u,v\rangle + \overline{\langle u,v\rangle} = 2\|u\|\|v\|$
($\Longrightarrow$) $2Re\langle u,v\rangle = 2\|u\|\|v\|$
($\Longrightarrow$) $Re\langle u,v\rangle = \|u\|\|v\|$
($\Longrightarrow$) $\frac{1}{2}(\langle u,v\rangle +\overline{\langle u,v\rangle}) = \|u\|\|v\|$
($\Longrightarrow$) $\frac{1}{2}(\langle u,v\rangle +\langle v,u\rangle ) = \|u\|\|v\|$
($\Longrightarrow$) $\frac{1}{2}(2\langle u,v\rangle ) = \|u\|\|v\|$
($\Longrightarrow$) $\langle u,v\rangle = \|u\|\|v\|$ as claimed.
($\Longleftarrow$) Suppose $\langle u,v\rangle = \|u\|\|v\|$. Then $\cos(\theta)=1$ since $\langle u,v\rangle = \|u\|\|v\| \cos(\theta)$. Because $\cos(\theta)=1$, $\theta=0$, and it follows that $u$ and $v$ point in the same direction. Moreover, one of $u$ and $v$ is a nonnegative scalar of the other since $\|u\|\|v\|\ge 0$ for all $u,v\in V$, where $V$ is an inner product space. Thus $\|u+v\| = \|u\|+ \|v\|$; that is, the length of the sum of $u$ and $v$ is equal to the sum of the lengths of $u$ and $v$ as claimed.
For part 2 of the proof, I know that the condition for equality in the Cauchy-Schwarz Inequality implies that one of $u$,$v$ is a scalar multiple of the other whereby that scalar is nonnegative. Is that because $\|u\|\|v\|\ge 0$? (i.e. lengths can't be negative obviously). Also, if $|\langle u,v\rangle | = \|u\|\|v\|$, then wouldn't the absolute value also force the scalar to be negative? The book says that the Cauchy-Schwarz Inequality is an equality iff one of $u$,$v$ is a scalar multiple of the other. My intuition is that because the cosine of the angle between the two vectors is $0$, they must lie on the same line and point in the same direction which means that one of the vectors is a nonnegative scalar multiple of the other.
I think that it is simpler to do it as follows\begin{align}\|u+v\|=\|u\|+\|v\|&\iff\|u+v\|^2=\bigl(\|u\|+\|v\|\bigr)^2\\&\iff\|u\|^2+\|v\|^2+2\langle u,v\rangle=\|u\|^2+\|v\|^2+2\|u\|\|v\|\\&\iff\langle u,v\rangle=\|u\|\|v\|.\end{align}