Triangle Inequality

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Theorem

Geometry

Given a triangle $ABC$, the sum of the lengths of any two sides of the triangle is greater than the length of the third side.


In the words of Euclid:

In any triangle two sides taken together in any manner are greater than the remaining one.

(The Elements: Book $\text{I}$: Proposition $20$)


Real Numbers

Let $x, y \in \R$ be real numbers.

Let $\size x$ denote the absolute value of $x$.


Then:

$\size {x + y} \le \size x + \size y$


Complex Numbers

Let $z_1, z_2 \in \C$ be complex numbers.

Let $\cmod z$ denote the modulus of $z$.


Then:

$\cmod {z_1 + z_2} \le \cmod {z_1} + \cmod {z_2}$


Vectors in $\R^n$

Let $\mathbf x, \mathbf y$ be vectors in the real Euclidean space $\R^n$.

Let $\norm {\, \cdot \,}$ denote vector length.

Then:

$\norm {\mathbf x + \mathbf y} \le \norm {\mathbf x} + \norm {\mathbf y}$

If the two vectors are scalar multiples where said scalar is non-negative, an equality holds:

$\exists \lambda \in \R, \lambda \ge 0: \mathbf x = \lambda \mathbf y \iff \norm {\mathbf x + \mathbf y} = \norm {\mathbf x} + \norm {\mathbf y}$


Triangle Inequality for Integrals

Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f: X \to \overline \R$ be a $\mu$-integrable function.


Then:

$\ds \size {\int_X f \rd \mu} \le \int_X \size f \rd \mu$


Also see


Sources