The line is the shortest path between two points

Let $p, q$ be two points of $\mathbb{R}^n$, $u$ be a unit vector and let $\gamma$ be a curve passing through those points. ($\gamma(a)=p, \; \gamma(b)=q$). Prove that the shortest path between these two points is the line.

Solution:



We are using the following facts:

$\color{gray} \bullet$ $\gamma'(t)\cdot u \leqslant \left \| \gamma'(t) \right \|   \qquad  \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \;  (1)$
$\color{gray} \bullet$ $ \displaystyle \overrightarrow{q}-\overrightarrow{p}=\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,{\rm d} t}$

Integrating the first equation we get that:

$$ \int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \leqslant \int_{a}^{b} {\| \overrightarrow{\gamma}\,'(t) \|\,dt}=L(\overrightarrow{\gamma}) \tag{2}$$

whereas $L(\overrightarrow{\gamma})$ is the segment's length of the curve $\gamma$ between the points $p, q$. Taking $\overrightarrow{ u}=\frac{\overrightarrow{q}-\overrightarrow{p}}{ \|\overrightarrow{q}-\overrightarrow{p} \|}$ we have that:

$$\begin{aligned} \|{\overrightarrow{q}-\overrightarrow{p}}\|&= \frac{\|{\overrightarrow{q}-\overrightarrow{p}}\|^2}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\ &=\bigl({\overrightarrow{q}-\overrightarrow{p}}\bigr)\cdot\frac{\overrightarrow{q}-\overrightarrow{p}}{\|{\overrightarrow{q}-\overrightarrow{p}}\|} \\ &= \bigl({\overrightarrow{\gamma}(b)-\overrightarrow{\gamma}(a)}\bigr)\cdot\overrightarrow{ u}\\ &\stackrel{(1)}{=}\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\,dt}\cdot\overrightarrow{u}\\ &=\int_{a}^{b}{\overrightarrow{\gamma}\,'(t)\cdot\overrightarrow{u}\,dt} \\ &\stackrel{(2)}{\leqslant} \int_{a}^{b} { \| \overrightarrow{\gamma}\,'(t) \|\,dt}\\ &\stackrel{(2)}{=}L(\overrightarrow{\gamma})\,. \end{aligned}$$

which proves our claim. 

Source: mathimatikoi.org    &  mathematica.gr

The above is true in an Euclidean plane!