For vector x, y, when does |x+y| = |x|+|y|? In general $|x+y|\le|x|+|y|$. When does equality hold?
Spivak "Calculus on Manifolds" says 

the answer is not "when x and y are linearly dependent."

However, that is the answer I get.
 A: One way to note the conditions for equality is to run through a proof of the triangle inequality
$$\|\mathbf{x}+\mathbf{y}\| \le \|\mathbf{x}\| + \|\mathbf{y}\|$$
This holds if and only if
$$\|\mathbf{x}+\mathbf{y}\|^2 \le \left(\|\mathbf{x}\| + \|\mathbf{y}\|\right)^2 = \|\mathbf{x}\|^2 + 2\|\mathbf{x}\|\|\mathbf{y}\|+\|\mathbf{y}\|^2$$
Rewriting in terms of dot products, we get
$$(\mathbf{x}+\mathbf{y})\cdot (\mathbf{x}+\mathbf{y}) = \mathbf{x}\cdot\mathbf{x} + 2\mathbf{x}\cdot\mathbf{y} + \mathbf{y}\cdot\mathbf{y}\le \mathbf{x}\cdot\mathbf{x} + 2\|\mathbf{x}\|\|\mathbf{y}\|+\mathbf{y}\cdot\mathbf{y}$$
which reduces to
$$\mathbf{x}\cdot\mathbf{y} \le \|\mathbf{x}\|\|\mathbf{y}\|$$
Therefore the triangle inequality holds if and only if the above inequality holds. But the above inequality is immediately given by Cauchy-Schwarz since
$$\mathbf{x}\cdot\mathbf{y}\le|\mathbf{x}\cdot\mathbf{y}|\le\|\mathbf{x}\|\|\mathbf{y}\|$$
This proves the inequality. For the equality, we need both inequalities above to be equalities. The first equality holds if and only if $\mathbf{x}\cdot\mathbf{y} \ge 0$ and the second holds (via the equality case of Cauchy-Schwarz) if and only if $\mathbf{x} = c\mathbf{y}$ for some $c\in\mathbb{R}$. The first condition forces $c\ge 0$.
A: As long as the angle between $\bf{x}$ and $\bf{y}$ is 0 degree.
