Why if $x,y\in\mathbb{R}^n-\{0\} $ and $\|x\| + \|y\| =\|x+y \|$, then exist $\lambda > 0$ s.t. $x=\lambda y$ Why if $x,y\in\mathbb{R}^n-\{0\} $ and $\|x\| + \|y\| =\|x+y \|$, then there exists $\lambda > 0$ such that $x=\lambda y$?
I don't know how to proceed.

 A: The equality $$\|x\|+ \|y\| = \|x+y\|$$ is equivalent to
$$(\|x\|+ \|y\|)^2 = \|x+y\|^2$$
or
$$\|x\| \cdot \|y\| = \langle x, y\rangle $$
Now calculate
$$\| y - \frac{\langle y, x\rangle }{\|x\|^2}\cdot x\|^2$$
and in a few steps we get
$$\|y\|^2 - \frac {\langle x, y \rangle^2}{\|x\|^2}$$
and hence $0$. We conclude that
$$y - \frac{\langle y, x\rangle }{\|x\|^2}\cdot x=0$$ so $y$  is proportional to $x$. Now it's easy to see that the proportionality constant is positive.
Note: the expression $\frac{\langle y, x\rangle }{\|x\|^2}\cdot x$ is the projection of $y$ on $x$, so it has a geometrical meaning.
A: Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n).$ Then $$\|x\|+\|y\|=\|x+y\|\implies$$ $$\implies 0=(\|x\|+\|y\|)^2-\|x+y\|^2= 2\|x\|\cdot \|y\|-2<x,y>\implies$$ $$\implies 0=\|x\|^2\cdot \|y\|^2-<x,y>^2=\sum\sum_{1\le i\le j\le n}(x_iy_j-x_jy_i)^2\implies$$ $$\implies \forall i,j\,(x_iy_j-x_jy_i=0).$$ Now $y\ne 0$ so take some $i$ such that $y_i\ne 0$ and let $\lambda=(x_i/y_i).$ Then  we have $$\forall j\,(x_j-\lambda y_j=\frac {x_jy_i-x_iy_j}{y_i}=0).$$ $$\text {So we have } x=\lambda y.$$ And $\lambda\ne 0$ because $\lambda y=x\ne 0.$ Now we have $$0\ne (|\lambda|+1)\cdot \|y\|=\|\lambda y\|+\|y\|=$$ $$=\|x\|+\|y\|=\|x+y\|=$$ $$=\|\lambda y+y\|=|\lambda +1|\cdot \|y\|.$$ So $|\lambda|+1=|\lambda +1|,$ which is not possible for $0\ne \lambda\in\Bbb R$ unless $\lambda>0.$
Remark: The identity $\|x\|^2\cdot \|y\|^2-<x,y>^2=\sum\sum_{1\le i\le j\le n}(x_iy_j-x_jy_i)^2$ arises in the usual proof of the $\triangle$  inequality for the Cartesian norm on $\Bbb R^n.$
