Prove $||u+v||=||u||+||v||\iff u=\alpha v, \alpha>0$ 
Let $V$ be an inner product space $u,v\in V$ and $u,v\neq 0$
Prove: $$||u+v||=||u||+||v||\iff u=\alpha v, \alpha>0$$

$(\Leftarrow):$
$$||\alpha v+v||^2=\langle \alpha v+v,\alpha v+v\rangle=\\=\langle \alpha v,\alpha v\rangle+\langle \alpha v, v\rangle+\langle v,\alpha v\rangle+\langle v,v\rangle=\\=|a|^2||v||^2+2Re(\langle \alpha v,v\rangle)+||v||^2=\\=|a|^2||v||^2+2 \alpha \cdot Re(\langle v,v\rangle)+||v||^2=||av||^2+2 \alpha \cdot ||v||^2+||v||^2=(||\alpha v||+||v||)^2$$
So we got  $||\alpha v+v||=|| \alpha v||+||v||$
$(\Rightarrow):$
$$||u+v||=||u||+||v||\iff (||u+v||)^2=(||u||+||v||)^2\iff\\\langle u+v,u+v\rangle=||u||^2+2||u||\cdot||v||+||v||^2\iff \\\langle u,u\rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle=\langle u,u\rangle +2\sqrt{\langle u,u\rangle\langle v,v\rangle}+\langle v,v\rangle\iff\\ 2Re(\langle u,v\rangle)=2\sqrt{\langle u,u\rangle\langle v,v\rangle}\iff\\Re(\langle u,v\rangle)=\sqrt{\langle u,u\rangle\langle v,v\rangle}\iff \\ Re(\langle u,v\rangle)^2={\langle u,u\rangle\langle v,v\rangle}$$
How should I continue?
 A: You have
$$
\mathrm{Re}\langle u,v\rangle \le |\langle u,v\rangle| \le \sqrt{\|u\| \|v\|}
$$
for all $u,v$. The second Cauchy–Schwarz inequality is an equality if and only if $v=\lambda u$ for some $\lambda \in \mathbf{C}$. The first inequality is a equality iff $\mathrm{Re}\langle u,v\rangle \ge 0$ and $\mathrm{Im}\langle u,v\rangle=0$. Setting $v=\lambda u$, you need
$$
\mathrm{Re}\langle u,\lambda u\rangle = \|u\|^2\mathrm{Re}\overline{\lambda}= \|u\|^2\mathrm{Re}\lambda\ge 0\,\,\text{ and }\,\,\mathrm{Im}\langle u,\lambda u\rangle=\|u\|^2\mathrm{Im}\overline{\lambda}=-\|u\|^2\mathrm{Im}\lambda=0.
$$
If $\lambda=0$ then $v=0$, which is against our assumptions $u,v \neq 0$. Otherwise $\lambda \neq 0$, hence by force $\mathrm{Re}\lambda\ge 0$ and $\mathrm{Im}\lambda=0$, i.e., $\lambda$ is a positive real number.
A: Hint. 
Let $\lambda \in \mathbb {C}$ such that $\lambda =\langle u,v\rangle /\|v\|^{2}$, then it is possible to prove that
$$\|u-\lambda \cdot v\|^{2}= \dots=\|u\|^{2}-{\frac {|\langle u,v\rangle |^{2}}{\|v\|^{2}}}\, . \ \ \ \ \ \ (*)$$
Putting the RHS equal to zero means
$$|\langle u,v\rangle |^{2}=\|u\|^2 \|v\|^2\, ,$$
a step can be derivable from your:
$$\langle u,u\rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle=\langle u,u\rangle +2\sqrt{\langle u,u\rangle\langle v,v\rangle}+\langle v,v\rangle\, .$$
But, if the RHS of (*) is equal to zero, this means that  $ \|u-\lambda \cdot v\|=0$, and so $u-\lambda \cdot v=0$.
