I am trying to prove that the shortest path between two points is a straight line. I know how to prove this in $\mathbb{R}^1$ case using the E-L equation. However, I am encountering some difficulties when extending this problem to $\mathbb{R}^n$.
So Let $f:[a,b] \rightarrow \mathbb{R}^n$, so $f(t) = (f_1(t), f_2(t), ..., f_n(t))$. I am stuck at solving this system of differential equations: \begin{align*} \frac{f_i'(t)}{\sqrt{1 + |f'(t)|^2}} = C_i \text{ where } C_i \text{ is a constant} \end{align*} In $\mathbb{R}^1$ case this is easy, the solution is $f(t) = at + b$. But in $\mathbb{R}^n$, I don't know how to deal with the term $|f'(t)|^2$.
Square your equations and take the sum. You'll get $$ \frac{|f'(t)|^2}{1 + |f'(t)|^2} = \sum_{k = 1}^n C_i^2 = const. $$ It is easy to conclude that $|f'(t)|^2$ is constant. Thus, due to equation $$ \frac{f'_i(t)}{\sqrt{1 + |f'(t)|^2}} = C_i $$ function $f'_i(t)$ is also constant. Therefore, $f(t) = tp + q$ for some vectors $p$ and $q$ (actually $p = f'(t)$ and $q = f(a) - ap$).
Howewer, I don't think that your statement alone is enough to prove that the shortest path between two points is a straight line. It proves only that there is no other extremal path. You also have to prove that the shortest path exists (so it has to be extremal and therefore it is a straight line).