Prove that the shortest path between two points in a Euclidean space is a straight line.

Question

I've tried using the following distance formula: $$d(A,B)=\int_{A_{x}}^{B_{x}}\sqrt{1+(f'(x))^2}\, dx.$$ But couldnt do much with it. Help would be appreciated.

Answer 1

$$d(A,B)=\int_{A}^{B}\sqrt{1+(f'(x))^2}\, dx \geq \int_{A}^{B}\sqrt{1+0}\, dx $$

Answer 2

Correct me if wrong.

1) Original $X,Y$ system.

2) Translated system $X'Y':$

$O'$ at $A$, $X',Y'$ parallel to $X,Y$ ; $B'$ at $B$.

3) Rotated $X'',Y''$ system,

$O'$ at $O''$, $B''$ at $B$, where $B''$ lies on $X''$ axis.

4) In $X", Y''$ :

$f_2(0'')=0$; $f_2(b'')=0$.

5) Shortest distance

$I''= \int_{0''}^{b''}\sqrt{1+(f_2'(x''))^2)}dx''$ .

$(f_2)'(x'')= 0''$, i.e. $f_2(x'')= c=0''$, does the trick(Why?),

assuming rotation and translation do not change the length of a curve.