can anyone help me solve the following minimisation problem ( w.r.t the matrix $A\in \mathbb{R}^{n\times n}$) using the Euler Lagrange equations?
$$ \inf \Big\{ \int_0^1\sqrt{\langle A\dot{\sigma},\dot{\sigma} \rangle},~~ \sigma:[0,1]\to \mathbb{R}^n ~\text{absolutely continuous on $[0,1]$},\sigma_0=x,\sigma_1=y \Big\} $$
On
Here is an approach not directly using the Euler-Lagrange equations, but supporting Ben Grossman's answer.
We can assume $A=A^T$; otherwise, replace $A$ by $\frac12\left(A+A^T\right)$. That is, since $A\in\mathbb{R}^{n\times n}$, $$ \begin{align} \langle Au,u\rangle &=\left\langle u,A^Tu\right\rangle\tag{1a}\\ &=\left\langle A^Tu,u\right\rangle\tag{1b}\\ &=\left\langle\tfrac12\left(A+A^T\right)u,u\right\rangle\tag{1c} \end{align} $$ Minimizing yields $$ \begin{align} 0 &=\delta\int_0^1\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}\,\mathrm{d}t\tag{2a}\\ &=\int_0^1\frac{\delta\langle A\dot\sigma,\dot\sigma\rangle}{2\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}}\,\mathrm{d}t\tag{2b}\\ &=\int_0^1\frac{\langle A\dot\sigma,\delta\dot\sigma\rangle}{\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}}\,\mathrm{d}t\tag{2c}\\ &=-2\int_0^1\frac{\langle A\ddot\sigma,\delta\sigma\rangle\langle A\dot\sigma,\dot\sigma\rangle-\langle A\dot\sigma,\delta\sigma\rangle\langle A\dot\sigma,\ddot\sigma\rangle}{\sqrt{A\dot\sigma\cdot\dot\sigma}^3}\,\mathrm{d}t\tag{2d}\\ &=-2\int_0^1\langle A(\kappa_1\ddot\sigma-\kappa_2\dot\sigma),\delta\sigma\rangle\,\mathrm{d}t\tag{2e} \end{align} $$ where $\kappa_1=\frac1{\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}}$ and $\kappa_2=\frac{\langle A\dot\sigma,\ddot\sigma\rangle}{\sqrt{\langle A\dot\sigma,\dot\sigma\rangle}^3}$
Therefore, if $A$ has full rank and $\dot\sigma$ doesn't vanish, $\ddot\sigma\parallel\dot\sigma$, which means that $\sigma$ follows a straight line.
We have $L(t, \sigma, \dot \sigma) = \sqrt{\langle A \dot \sigma, \dot \sigma \rangle}$. We see that $\frac{\partial L}{\partial \sigma} = 0$. To find $\frac{\partial L}{\partial \dot\sigma}$, we need a formula for the derivative of the function $$ f(x) = \sqrt{\langle A x,x \rangle}. $$ To do this, write $f(x) = \sqrt{g(x)}$ with $g(x) = \langle Ax,x \rangle$ and apply the chain rule. We have $$ g(x+h) = \langle A (x+h),(x+h) \rangle = \langle Ax,x \rangle + \langle (A + A^T)x, h \rangle + \langle Ah,h \rangle, $$ from which it follows that $\frac{\partial g(x)}{\partial x} = (A + A^T)x$. With that, the chain rule gives us $$ \frac{\partial f}{\partial x} = \frac 1{2 \sqrt{g(x)}}\cdot (A + A^T)x. $$ With that, we find that the Euler-Lagrange equation becomes $$ \frac d{dt} \frac{\partial L}{\partial \dot \sigma} = 0 \implies \frac d{dt} \left[\frac{(A + A^T)\dot \sigma}{2 \sqrt{\langle A\dot \sigma,\dot \sigma\rangle}} \right] = 0. $$ If something has derivative zero, then it must be constant. That is, there is a vector $c$ for which $$ \frac{(A + A^T)\dot \sigma}{2 \sqrt{\langle A\dot \sigma,\dot \sigma\rangle}} = c \implies \frac{A + A^T}{2} \cdot \frac{\dot \sigma}{\sqrt{\langle A \dot \sigma, \dot\sigma \rangle}} = c. $$ Without further assumptions on the matrix $A$ (e.g. that $A$ is positive definite), I see no way to move forward from here.
If we are given that $A$ is indeed (symmetric and) positive definite, then we can quickly see that $\dot \sigma$ must be a multiple of some constant vector, which to say that $\sigma$ moves within a straight line.
We might also have made the entire problem more straightforward with the substitution $\tau := A^{1/2}\sigma$, which would have led to the equivalent minimization of $$ \int_0^1 \sqrt{\langle \dot \tau, \dot \tau\rangle}\,dt, $$ which is the integral defining arc length (with respect to the usual metric).