$\lVert u +v \rVert =\lVert u \rVert +\lVert v\rVert$ only when $u$ and $v$ have the same direction How can I prove that $\lVert u +v \rVert =\lVert u \rVert +\lVert v\rVert$ only when $u$ and $v$ have the same direction?


I started by raising the two parts of the equation to the second power, but I got stuck at $u \cdot v = \lVert u \rVert \, \lVert v \rVert$.
Use the fact that $u \cdot v = \|u\| \|v\| \cos \theta$, where $\theta$ is the angle between $u$ and $v$. You can show that $$\|u+v\|^2  = \left( \|u\| + \|v\| \right)^2$$ if and only if $\cos \theta = 1$, that is, if and only if $\theta = 0^\circ$.
If this condition holds then a quick calculation verifies $u = \alpha v$ for $\alpha = \frac{u \cdot v}{v \cdot v} &gt; 0$ because in this case $\|u - \alpha v\|^2 = 0$.
As stated in the comments earlier, start with $(u+v)^2$:
\begin{align}
(u + v)^2 &amp;= u^2 + 2 u\cdot v + v^2  \iff \\
\lVert u + v \rVert^2 &amp;= \lVert u \rVert^2 + \lVert v \rVert^2 + 
2 \lVert u \rVert \lVert v \rVert \cos \angle(u, v) \\
&amp;= (\lVert u \rVert + \lVert v \rVert)^2 +
2 \lVert u \rVert \lVert v \rVert (\cos \angle(u, v) - 1) \iff \\
\lVert u + v \rVert&amp;=
\sqrt{(\lVert u \rVert + \lVert v \rVert)^2 + 2 \lVert u \rVert \lVert v \rVert (\cos \angle(u, v) - 1)} 
\end{align}
where I assumed $\cdot$ to be the standard scalar product and $\lVert \cdot \rVert$ to be the Euclidean $\lVert \cdot \rVert_2$ norm. 

The claimed result holds if 
$$
\cos \angle(u, v) - 1 = 0
$$
which is true if $u$ and $v$ have angle $0$ between them, i.e. show in the same direction.