Proof of equality for triangle inequality I am trying to prove the following statement here:

Let $(V,\left<\cdot,\cdot\right>)$ be an inner product space and let $\vec{u},\vec{v}$ be vectors in $V$. Under the assumption that $\vec{u}\neq0$ show that $||\vec{u}+\vec{v}||=||\vec{u}||+||\vec{v}||$ if an only if $\vec{v}=c\vec{u}$ for some non-negative scalar $c$.

Here's my attempt. I use the general proof of the triangle inequality here.
$||\vec{u}+\vec{v}||^2$
$=\left<\vec{u}+\vec{v},\vec{u}+\vec{v}\right>$
$=\left<\vec{u},\vec{u}\right>+2\left<\vec{u},\vec{v}\right>+\left<\vec{v},\vec{v}\right>\leq\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},\vec{v}\right>|+\left<\vec{v},\vec{v}\right>$ (By Cauchy-Scharwz)
So,
$||\vec{u}+\vec{v}||^2\leq\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},\vec{v}\right>|+\left<\vec{v},\vec{v}\right>$
$||\vec{u}+\vec{v}||^2\leq\left<\vec{u},\vec{u}\right>+2||\vec{u}||\hspace{1mm}||\vec{v}||+\left<\vec{v},\vec{v}\right>$
$||\vec{u}+\vec{v}||^2\leq(||\vec{u}||+||\vec{v}||)^2$
$||\vec{u}+\vec{v}||\leq(||\vec{u}||+||\vec{v}||)$
I have a few comments here though. From what I understand, if I can turn the inequality parts into an equality, I am essentially done. I'm not really sure how to do that though. If I let $\vec{v}=c\vec{u}$, then I get:
$=\left<\vec{u},\vec{u}\right>+2\left<\vec{u},c\vec{u}\right>+\left<c\vec{u},c\vec{u}\right>$
but I am stuck at this point.
I know that:
$\left<\vec{u},\vec{u}\right>+2\left<\vec{u},c\vec{u}\right>+\left<c\vec{u},c\vec{u}\right>\leq\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},c\vec{u}\right>|+\left<c\vec{u},c\vec{u}\right>$
becomes:
$\left<\vec{u},\vec{u}\right>+2\left<\vec{u},c\vec{u}\right>+\left<c\vec{u},c\vec{u}\right>=\left<\vec{u},\vec{u}\right>+2|\left<\vec{u},c\vec{u}\right>|+\left<c\vec{u},c\vec{u}\right>$ if and only if $c\vec{u}$ and $\vec{v}$ are non negative real numbers.
So at this point, I've taken care of the Cauchy-Schwarz inequality portion since one vector is a scalar multiple of the other.
How can I show that $\vec{u}$ and $c$ are non-negative though? If $\vec{u}$ and $c$ are non-negative, that immediately implies $\vec{v}$ is non-negative.
Can someone guide me with this part here? Thank you!
 A: Suppose the equality is true for $u$, $v$. then, squaring both sides of the equation, using bilinearity of the inner product  and cancelling out terms gives $\left<u,v\right>=|u||v|$. As $\left<x,y\right>=|x||y|cos\theta$ for all $x,y\in V$, we must have $\theta=0$, so there is $c\geq0$ such that $v=cu$. Conversely, if $v=cu$, with $c\geq 0$, we have $|u+cu|=(1+c)|u|=|u|+|cu|$ as we wanted.
Note that, in general, expressions such as "$cu$ and $v$  are non negative real numbers" make no sense, as $u$ and $v$ are vectors.

EDIT: if $\left<u,v\right>=|u||v|$, then $\left<\frac{u}{|u|}-\frac{v}{|v|},\frac{u}{|u|}-\frac{v}{|v|}\right>=1+1-2=0$, so $\frac{u}{|u|}-\frac{v}{|v|}=0$, thus $v=cu$, where $c=\frac{|v|}{|u|}$.
A: Let $x,y \in V$ with $x \neq 0$ and notice that
$$
P(t)=||tx +y ||^2=t^2 || x||^2 + 2t\left<x,y\right>+||y||^2 \ge 0
$$
is a polynomial of degree $2$. By a classic result, $\Delta=4<x,y>^2 -4||x||^2||y||^2 \le 0$ that is $|\left<x,y\right>| \le ||x|| ||y||$. Suppose we have equality, therefore $\Delta=0$ and $P(t)=0=||tx +y ||^2$ with $t=\frac{-\left<x,y\right>}{||x||^2}$, so $y=-tx$.
Now, we have
$$
|| x+y ||=||x|| + ||y|| \iff <x,y>=||x|| ||y||
$$
so we have the result by the above (in case of equality $<x,y>\ge0$ so $-t\ge0$).
