Geodesics are extremals with constant speed proof

Let $$L:(t,q,\dot q)\mapsto L(t,q,\dot q)$$ a Lagrangian defined on $$U\times \mathbb{R}^n$$, with $$U\subset \mathbb{R}^n$$.

A Lagrangian variational problem consists in, given two points $$x$$ and $$y$$ of $$U$$, finding the curves $$c:[a,b]\to U$$ such that $$c(a)=x, c(b)=y$$ and the quantity $$\Phi(c(.)) = \int_a^b L(c(t),\dot c(t))dt \tag{1}$$ is minimised.

An extremal of a Lagrangian variational problem is a curve which cancels the differential of $$\Phi$$ (defined on all curves with fixed extremeties).

Consider a Riemannian metric $$g$$ on $$\mathbb{R}^n$$ and $$L(q,\dot q)=\sqrt{g_q(\dot q)}$$. A geodesics is an extremal of the variational problem associated to the Lagrangian $$L^2(q,\dot q)=g_q(\dot q)$$.

Question : Show that the geodesics are extremals of $$L$$ with constant speed.

Proof : The Euler Lagrange equations associated to $$L^2$$ can be written as $$2L\frac{\partial L}{\partial q} -\frac{d}{dt}\left(2\dot L \frac{\partial L}{\partial \dot q}\right)=0$$ that is $$\frac{\partial L}{\partial q} -\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right) -2\dot L \frac{\partial L}{\partial \dot q}=0. \qquad (*)$$

Now if $$q(.)$$ is a gedesics, we can show first that $$\frac{d}{dt}(2L^2(q,\dot q))=\frac{d}{dt}(L^2(q,\dot q))$$ using only the Euler Lagrange equations for $$L^2$$. Thsi proves that $$q(.)$$ is with constant speed i.e. $$\dot L=0$$. Then from (*) we get $$\frac{\partial L}{\partial q} -\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right)=0$$ and $$q(.)$$ is an extremal of $$L$$.

Conversely, if $$q(.)$$ is an extremal of $$L$$ with constant speed then $$\frac{\partial L^2}{\partial q} -\frac{d}{dt}\left(\frac{\partial L^2}{\partial \dot q}\right)=-2\dot L \frac{\partial L}{\partial \dot q}=0$$ since $$\dot L = \frac{\partial g_q}{\partial \dot q}\ddot q=0$$.

Is it correct ?

2. On the other hand, one may show that the solutions to the Euler-Lagrange (EL) equation associated with the squared Lagrangian $$L^2$$ are the affinely parametrized geodesics, and hence with constant speed. Briefly this is because $$L^2$$ has no explicit $$t$$-dependence, so that the corresponding energy (which happens to be $$L^2$$) is constant, cf. e.g. this Math.SE post and this Phys.SE post.