The first constraint is $x_0\geq 1$ and furthermore we have $x_i\geq x_{i-1}$. Therefore,
$x_i\geq 1, \forall i\geq 0$. Knowing, that all $x_i$ have to be positive, we derive that
$\frac{x_i}{x_{i-1}}\geq 1$ for every $i>0$. But this tells us, that our sum becomes minimal if each term equals its minimum, which is $1$. So we have to choose $x_i=c, \forall i\geq 0$ with $c\geq1$. Doing so, our minimum is $m$. I do not see how the constraint $x_m\geq N$ should be used. Assuming $N<1$ we already satisfy this constraint because $x_i\geq1$. If $N>1$ we can simply choose $c=N$, which does not change the result.