# 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?

$$\langle v, w \rangle \le \|v\|\|w\|$$

is known as the Cauchy-Schwarz inequality.

\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 $|\left<v,w\right>|\leq\|v\|\|w\|$.

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 $\langle u, v \rangle \leq \lVert u \rVert \lVert v\rVert$.

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 $$\|u + v\| = \|u + \mu u\| = |1 + \mu|\|u\|\le \|u\| + |\mu|\|u\| = \|u\| + \|\mu u\| = \|u\| + \|v\| = k.$$