Proof for norm of sum of vectors is less/equal to sum of norm of vectors 
Prove $||v+w|| \le ||v|| + ||w||$

My proof so far is:
$$ ||v+w|| = \sqrt{\langle v+w,v+w\rangle}$$
$$= \sqrt{\langle v,v\rangle +2\langle v,w\rangle +\langle w,w\rangle }$$
So, $||v+w||^2$
$$=||v||^2 +2\langle v,w\rangle + ||w||^2$$
When I squared and expanded the right side, I got
$$(||v||+||w||)^2=||v||^2 +2||v||\ ||w||+||w||^2$$
Which means I need to prove that $\langle v,w \rangle \le ||v||\ ||w||$.
I'm not sure how to prove this. Is there a different answer to the question that sidesteps this? How can I go about proving that statement?
 A: $$\langle v, w \rangle \le \|v\|\|w\|$$
is known as the Cauchy-Schwarz inequality.
A: \begin{align*}
\left<v+tw,v+tw\right>&=\|v\|^{2}+t^{2}\|w\|^{2}+2t\left<v,w\right>\geq 0,
\end{align*}
so $(2\left<v,w\right>)^{2}-4\|w\|^{2}\|v\|^{2}\leq 0$, and hence $|\left<v,w\right>|\leq\|v\|\|w\|$.
A: Recall the definition of the scalar product:
$\langle u, v \rangle = \lVert u \rVert \lVert v\rVert cos(\theta)$ where $\theta$ is the angle between $u$ and $v$. But since $\lvert cos(\theta)\rvert\leq 1$ then we must have $\langle u, v \rangle \leq \lVert u \rVert \lVert v\rVert $.
A: Maths made difficult version: maximum of
$$
(u_1,\dots,u_n,v_1,\dots,v_n)\longmapsto (u_1 + v_1)^2 + \cdots + (u_n + v_n)^2
$$
with the restriction $\sqrt{u_1^2 + \cdots u_n^2} + \sqrt{v_1^2 + \cdots v_n^2} = \|u\| + \|v\|= k$.
Applying Lagrange multipliers:
$$2(u_i + v_i) = \lambda\frac{u_i}{\sqrt{u_1^2 + \cdots u_n^2}},$$
$$2(u_i + v_i) = \lambda\frac{v_i}{\sqrt{v_1^2 + \cdots v_n^2}}$$
for $i = 1,\cdots,n$. I.e.:
$$\|v\| u = \|u\| v,\hbox{ or } u + v = 0,$$
in both cases, $u,v$ linearly dependent: $v = \mu u$ (or $u = \mu v$) and
$$\|u + v\| = \|u + \mu u\| = |1 + \mu|\|u\|\le
\|u\| + |\mu|\|u\| = \|u\| + \|\mu u\| = \|u\| + \|v\| = k.$$
