Triangle inequality ($d(x,z)\le d(x,y)+d(y,z)$)

A metric satisfies $d(x,z)\le d(x,y)+d(y,z)$: visiting an intermediate point cannot beat the direct endpoint distance. A norm satisfies the corresponding inequality $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$.

Take $O=(0,0)$, $Q=(3,0)$, and $P=(3,4)$. For Euclidean distance, the triangle inequality reads $5\le 3+4=7$. The direct line segment has metric path length $5$, while the route through $Q$ has metric path length $7$. For Manhattan distance, the same comparison becomes $7=3+4$: both routes are shortest. The line segment and the bent route both become metric geodesics when progress is measured by metric path length. For Chebyshev distance, the direct line segment has metric path length $4$, while the same bent route still costs $7$.

Repeated application of the triangle inequality bounds the endpoint distance by the sum along any finite chain of intermediate points. This explains why metric path length is always at least the distance between endpoints. Equality reveals a shortest route, but its shape and uniqueness depend on the chosen metric.

In a metric space, applying the triangle inequality in both directions gives $|d(x,z)-d(y,z)|\le d(x,y)$. In a normed vector space, its corresponding form is $|\Vert x\Vert -\Vert y\Vert |\le \Vert x-y\Vert$. In particular, distance to a fixed point is a Lipschitz continuous real function.

For four points $x_1,x_2,y_1,y_2$ in a metric space,Both steps follow from the triangle inequality. The first also holds with an absolute value around the difference, by the reverse triangle inequality. This transports a difference of distances measured from one point into a sum measured from another.

For an integrable function $f$ with values in a finite-dimensional normed vector space,For the Euclidean norm, pair the integral with a unit vector in its direction and apply the Cauchy-Schwarz inequality pointwise.

For nonzero complex numbers $x,y$, equality $|x+y|=|x|+|y|$ holds exactly when $x/y$ is a positive real number, equivalently when the two numbers have the same complex argument modulo $2\pi$.

For every integrable real or complex function,