Triangle inequality

Euclidean geometry

In Euclidean geometry, the triangle inequality states that the sum of the lengths of two sides of a triangle is always larger than the third side.

For instance, in $\triangle ABC$

$$\begin{gathered} a+b>c \\ a+c>b \\ b+c>a \end{gathered}$$

Generalisation to vector spaces

Let $V$ be an inner product space over $\mathbb{F}$.

$$\forall u,v\in V,\; \left\| u+v \right\| \le \left\| u \right\| + \left\| v \right\|.$$

Proof:

$$ \begin{aligned} \left\| u+v \right\|^2 &=\langle u+v,u+v\rangle =\langle u,u\rangle+\langle v,v\rangle+\langle u,v\rangle+\langle v,u\rangle\\ &=\langle u,u\rangle+\langle v,v\rangle+\langle u,v\rangle+\overline{\langle u,v\rangle} =\left\| u \right\|^2+\left\| v \right\|^2+2\operatorname{Re}\langle u,v\rangle,\\ \text{Note }\operatorname{Re}\langle u,v\rangle &\le|\langle u,v\rangle| \stackrel{\text{Cauchy-Schwarz}}{\le}\left\| u \right\|\left\| v \right\|,\\ \left\| u \right\|^2+\left\| v \right\|^2+2\operatorname{Re}\langle u,v\rangle &\le\left\| u \right\|^2+\left\| v \right\|^2+2\left\| u \right\|\left\| v \right\| =(\left\| u \right\|+\left\| v \right\|)^2,\\ \Rightarrow \left\| u+v \right\|^2 \leq (\left\| u \right\|+\left\| v \right\|)^2,\\ \because \left\| . \right\| \geq 0 \\ \therefore \left\| u+v \right\|&\le\left\| u \right\|+\left\| v \right\| \end{aligned} $$