Minimize $\int_0^1 \frac{\left( \sum_{i=1}^n \left(\dot\sigma^i(t)\right)^2 \right)^{\frac{1}{2}}}{1+\sum_{i=1}^n \left(\sigma^i(t)\right)^2}dt$. 
We define a functional on the set of $C^1$ curves joining two given points in $\mathbb{R}^n$ by
  $$
\mathcal S(\sigma)=\int_0^1 \frac{\left( \sum_{i=1}^n \left(\dot\sigma^i(t)\right)^2 \right)^{\frac{1}{2}}}{1+\sum_{i=1}^n \left(\sigma^i(t)\right)^2}dt
$$
  where $\sigma:[0,1]\to\mathbb R^n$ satisfies $\sigma(0)=(0,\cdots,0)$ and $\sigma(1)=(0,\cdots,0,l)$, $l>0$. Find all the curves $\sigma_0$ such that for any such $\sigma$ we have $\mathcal S(\sigma_0)\le \mathcal S(\sigma)$.

Background
I was trying to compute the distance on spheres.
Put $U=S^n\setminus \{\underbrace{0,\cdots,0}_n,-1\}$, and
   $$
   \varphi:U\to\mathbb{R}^n,\quad
   x\mapsto \left(\frac{x^1}{1+x^{n+1}},\cdots,\frac{x^n}{1+x^{n+1}}\right)
   $$
In the coordinate the Riemannian metric is 
   $$
    g_1=\frac{4}{\left(1+\sum_{i=1}^n (u^i)^2\right)^2} \sum_{i=1}^n du^i\otimes du^i.
   $$
and the length is 
   \begin{align*}
      \int_a^b \left\lVert \sigma(t)\right\rVert_{g_1} dt 
      &= \int_a^b \left(g_{ij}\circ\sigma(t) \dot\sigma^i(t) \dot\sigma^j(t) \right)^{\frac{1}{2}} dt \\
      &= \int_a^b \left( \frac{4}{\left(1+\sum_{i=1}^n \left(\sigma^i(t)\right)^2\right)^2}
         \sum_{i=1}^n \left(\dot\sigma^i(t)\right)^2 \right)^{\frac{1}{2}} dt \\
      &= 2\int_a^b \frac{\left( \sum_{i=1}^n \left(\dot\sigma^i(t)\right)^2 \right)^{\frac{1}{2}}}{1+\sum_{i=1}^n \left(\sigma^i(t)\right)^2}dt
   \end{align*}
I want to minimize the length functional to get the distance.
My attempt
Put
   $$
   L(t,\sigma(t),\dot\sigma(t))=
   \frac{\left( \sum_{i=1}^n \left(\dot\sigma^i(t)\right)^2 \right)^{\frac{1}{2}}}{1+\sum_{i=1}^n \left(\sigma^i(t)\right)^2}
   $$
   Via Euler–Lagrange equation, we have
   $$\frac{\partial }{\partial \sigma^i}L(t,\sigma(t),\dot\sigma(t))=
   \frac{d}{dt}\frac{\partial }{\partial \dot\sigma^i}L(t,\sigma(t),\dot\sigma(t))$$
   where the boundary conditions are
   $$\sigma(a)=(\underbrace{0,\cdots,0}_n)\qquad
   \sigma(b)=(\underbrace{0,\cdots,0}_{n-1},l)$$
But this is too complex to analyze for me. I tried to use arc length to parametrize the curve, and use the uniqueness of the solution, to get that the minimum curve is the line segment. But I am not sure whether this is practicable. Any hints? Thanks in advance!
 A: Too long to write in the comment section.
Taking $\sigma^{i}=x^{i}$.
I think the minimal path should be the straight line from $(0,0,....,0)$ to $(0,0,....,l)$. 
Because, we know that $\int_{0}^{1} v_{x^n} dt=l$. Now, the given time integral is $S(x)=\int_{0}^{1} \frac{(\sum {(\dot{x^i})²})^{\frac{1}{2}}}{1+\sum (x^i)²} dt= \int_{0}^{1} \frac{v(t)}{1+r²} dt$ where, $v(t), r(t)$ is the velocity and distance from origin in the $n$-dimensional space. 
Now, the particle has to go $l$ distance along the $n-th$ axis in this Cartesian coordinate system in $1$ second. The integral or $S(x)$ will be minimum if $v(t)$ is minimum. If the particle takes a path other than the straight line, it would require more average velocity, like the Bernoulli's effect. And that would probably result in increasing the integral. Therefore, its most probable that the only minimum is along the straight line. And the velocity will be increasing. 

If we compare the straight path($P_1$) with another curve path($P_2$) and make the velocity $v_1(t)=v_2(t)$ for all $r_1(t)=r_2(t)$ then surely the particle taking curve path can never reach its destination in 1 second. As there is nothing "negative" we can't cancel anything. So, I think the only minimizing path is the straight line along the $x^n$.
