Triangle inequality for norms

Statement

Let $V$ be a vector space over $\R$ or $\C$ with a norm $\norm{\cdot}$. Then, for all $x, y \in V$, $$\norm{x + y} \le \norm{x} + \norm{y} \qquad \text{and} \qquad \bigl| \norm{x} - \norm{y} \bigr| \le \norm{x - y}.$$

Remarks

The first inequality is an axiom of a norm; the second, the reverse triangle inequality, follows from it. For the norm induced by an inner product, the first inequality follows from the Cauchy–Schwarz inequality. In $\R^n$, it says that one side of a triangle is at most the sum of the other two sides.

Explanations

L1 human

The straight way from $0$ to $x + y$ is never longer than the way through $x$. This is the triangle inequality. The reverse form says that two vectors with a small difference have almost equal lengths.