TITLE: The Lagrangian formulation of mechanics without going through variational principles.
QUESTION [5 upvotes]: In some texts on classical mechanics and not only, the Euler--Lagrange equations of motion are directly obtained as solution of variational problems.
On the other side, sometimes reading about hamiltonian mechanics, one find the expression that this latter formulation is preferred to the lagrangian one because of it does completely avoid the appeal to variational principles.
This observation suggested to myself the following question:
Is the variational approach to the Euler--Lagrange equations the only one viable?
If not, is there some reason that explain why the geometry of the Euler-Lagrange eqns is much more hidden than the geometry of the Hamilton eqns?

I was searching for suggestion of reading for best tackle this question.

As usual any feedback is welcome.

REPLY [3 votes]: I'm just going to make a basic point, so apologies if this is completely obvious to you (it's also contained in Igor Khavkine's much more thorough answer).
Let $Q$ be the configuration manifold of the system. The corresponding cotangent bundle $T^*Q$ has an intrinsic symplectic form $\omega=-d\Theta$, where $\Theta$ is the tautalogical one-form on $T^*Q$. For a Hamiltonian $H:T^*Q\rightarrow \mathbb{R}$, Hamilton's equations can be expressed in terms of the Hamiltonian vector field $X_H$ (defined by $i_{X_H}\omega=dH$). Note $\omega$ is intrinsic to the phase space $T^*Q$, and doesn't depend on the Hamiltonian $H$.
Now given a Lagrangian $L:TQ\rightarrow\mathbb{R}$, and corresponding Legendre transform $\mathbb{F}L:TQ\rightarrow T^*Q$, and assuming here for simplicity that $\mathbb{F}L$ is a diffeomorphism ("L is hyperregular"), one can use $\mathbb{F}L$ to pull everything back to $TQ$. $TQ$ becomes a symplectic manifold, with symplectic form $\omega_L=(\mathbb{F}L)^*\omega$, and the Euler-Lagrange equations are just the equations for the flow of the Hamiltonian vector field $X_E$ defined by $i_{X_E}\omega_L = dE$, where $E=(\mathbb{F}L)^*H = H\circ\mathbb{F}L$ is the energy function on $TQ$. I guess the main point though is that $TQ$ is not intrinsically a symplectic manifold. The symplectic form $\omega_L$ depends also on the choice of Lagrangian. This is one possible answer to why the geometric formulation is more common on the Hamiltonian side, and appears to be `hidden' on the Lagrangian side: the geometry of $TQ$ (as it pertains to the E-L eqns) is tied up with the particular Lagrangian $L$, whereas the geometry of $T^*Q$ is independent of the particular Hamiltonian $H$. 
I'd also recommend any of the books by Jerry Marsden mentioned in Spiro Karigiannis' answer as the best place to learn this (my notation is consisent with his).
Edit: I should clarify that by `Legendre transform' above I mean (in coordinates) the map $p_i(x, \dot{x}) = \frac{\partial L}{\partial \dot{x}^i}(x, \dot{x})$. This is standard in the literature, but it differs from the classical meaning $H(x, p) = p\dot{x}-L(x, \dot{x})$ (where $p_i = \frac{\partial L}{\partial \dot{x}^i}$).