prove that minor arc of a great circle is the shortest distance How to prove that the minor arc of a great circle is the shortest path connecting two points on the surface of a sphere? 
The key to solving the problems is that we must take all curves connecting the two points , not just great-circle or parametrized differentiable curve, into account!
A similar problem is the straight line distance(shortest distance between two points): we must take all curves connecting the two points into account!
 A: Here is a low-tech approach which does not use derivatives much. Given distinct points $a,b$ on the sphere, let $\gamma$ be the shorter arc of great circle between them. Let $V$ be the surface obtained by rotating $\gamma$ about the line $ab$. Note that  $V$ is convex (i.e., bounds a convex body) and lies inside the sphere. The nearest-point projection onto a convex set does not increase Euclidean distance and therefore does not increase length of curves (defined by means of sums of Euclidean distances over partitions). Therefore, it suffices to prove that a curve connecting $a$ to $b$ on surface $V$ cannot be shorter than $\gamma$. 
Let $\Gamma$ be any such curve. Pick partition points $p_k$ on $\Gamma$ to approximate its length. For each $k$, draw the plane through $p_k$ that is orthogonal to line $ab$. This plane crosses $\gamma$ at a point $q_k$. It is easy to see that $|p_k-p_{k+1}|\ge |q_k-q_{k+1}|$ for each $k$. (Just write the Euclidean distance in cylindrical coordinates with $ab$ as the axis.) Since $\sum_k |q_k-q_{k+1}|$ is greater than or equal to the length of $\gamma$, the conclusion follows.
A: You know that such a length minimizing curve must be a geodesic. Then since every geodesic of the sphere is a great circle, such a length minimizing curve must be an arc of a great circle.
A: Frankly, the differentiability demand doesn't pose a serious threat to the proof. 
Assume there existed a continues but non-differentiable path, which was non-differentiable only at a finite number of points. for each such point which has slope $\phi'(x_-)$ on one side, and slope $\phi'(x_+)$ on the other, we can round the corner with a circle of radius $\epsilon$ who's contribution to the total length goes to zero.
