I think there is some sort of bug near the end of the first answer (joshphysics), possibly in the variational step which involves integration by parts. Unless some further assumptions were used. Otherwise the implied conclusion seems to be that the EL eqns for any functional with integrand $F(x(s),\dot{x}(s))$ are generally the same as for the functional with integrand $F^2$.
[A simple counterexample is $F = x(s)$, though I realise this is getting away from geodesics. 359
Then the EL eqn for $F^2$ would be $x=0$ whereas the EL eqn for $F$ itself would be $1=0$.]
Anyway, getting back to the original question, often the words "Cauchy-Schwarz" appear when relating $\int f^2$ to $\int f$.
Ignoring that, and looking just at EL equations, consider minimizing/maximizing the two functionals $\int F^2 ds$ and $\int F ds$ (with the same parameter and bcs), where $F = F(x(s),\dot{x}(s))$. When are these two problems equivalent (or at least, one implies the other)?
The EL eqns for the two cases are generally different (cf counterexample above), but one condition for equivalence is $dF/ds = 0$ (when evaluated on a solution to either of the two EL eqns).
We also have the Beltrami identities for each case, which are first integrals of the EL eqns (cf energy conservation): for $F$ we have $\dot{x}_i\frac{\partial F}{\partial \dot{x}_i} - F = c_1$ and for $F^2$ we have $\dot{x}_i\frac{\partial (F^2)}{\partial \dot{x}_i} - F^2 = c_2$ (*).
Restricting to the problem of geodesics on some surface, we have an integrand of the form $F = \sqrt{g_{ij}(x)\dot{x}^i\dot{x}^j}$ and this satisfies $\dot{x}_i \frac{\partial F}{\partial \dot{x}_i} = F$ (identically), and also satisfies $\dot{x}_i \frac{\partial (F^2)}{\partial \dot{x}_i} = 2F^2$ (+).
Let's suppose we have a solution to the EL eqn for $F^2$, and thus to the Beltrami identity $()$ for $F^2$. Then () and (+) together give $F^2 = c_2$ i.e. $F$ is constant when evaluated on our solution.
[this is essentially point 9 of <PERSON>'s answer] Hence the equivalence condition mentioned above ($dF/ds = 0$) is satisfied, and so the EL eqn for $F$ is also satisfied.
The converse doesn't hold in general since the <PERSON> identity for such an $F$ is satisfied identically (with $c_1 = 0$), so does not give anything "extra". 374