When does equality hold for $|x+y| \leq |x|+|y| ?$ This is problem 1-2 in Calculus on Manifolds. Spivak. 
He gives two hints. 


*

*Examine the proof and 

*The answer is not "when y and y are linearly dependent."
This is the proof we are told to examine:
$|x+y|^2 = \sum^n_{i=1}(x^i+y^i)^2=\sum^n_{i=1}{x^i}^2+\sum^n_{i=1}{y^i}^2+2\sum^n_{i=1}{x^i}{y^i} \leq |x|^2+|y|^2+2|x||y|=(|x|+|y|)^2$
Then take the square root of both sides.
Based on this proof, I believe the answer is 
$|x+y| = |x|+|y|\;\longleftrightarrow\;\sum^n_{i=1}{x^i}{y^i}=<x,y>=|x||y|.$ 
The answer seems obvious, but I'm not sure because I came across this set of solutions (http://jianfeishen.weebly.com/uploads/4/7/2/6/4726705/calculus_on_manifolds.pdf) that has a much more complicated answer. My proof seems a bit too easy. I might be overlooking something.
 A: Your argument is good, but you can go further.

Suppose $\|x+y\|=\|x\|+\|y\|$ (sorry, but I can't use notation with single bars and upper indices) with $x\ne0$; this is equivalent to
$$
\langle x+y,x+y\rangle=\langle x,x\rangle+2\|x\|\,\|y\|+\langle y,y\rangle
$$
that is, expanding the left-hand side and simplifying,
$$
\langle x,y\rangle=\|x\|\,\|y\|\tag{*}
$$
This is clearly connected to the Cauchy-Schwarz inequality, so let's examine its proof.
Consider $\langle tx+y,tx+y\rangle\ge0$, for every scalar $t$. Then
$$
t^2\langle x,x\rangle+2t\langle x,y\rangle+\langle y,y\rangle\ge0
$$
for all $t$; therefore the (reduced) discriminant
$$
\langle x,y\rangle^2-\langle x,x\rangle\langle y,y\rangle\le0
$$
In case the equality (*) holds, the discriminant is zero, so, for $t=-\frac{\langle x,y\rangle}{\langle x,x\rangle}$, we have that
$$
\langle tx+y,tx+y\rangle=0
$$
that is, $tx+y=0$, so $y\in\operatorname{Span}\{x\}$.
Now, if $y=rx$, for some scalar $r$,
$$
\|x+y\|=\|(1+r)x\|=|1+r|\,\|x\|
$$
whereas $\|x\|+\|rx\|=(1+|r|)\|x\|$. So we need
$$
|1+r|=1+|r|
$$
Squaring we get $1+2r+r^2=1+2|r|+r^2$, so the condition is $r\ge0$.
Thus the solution is

either $y=rx$, for some $r\ge0$, or $x=ry$, for some $r\ge0$.

