Is it possible to show simply using the Lagrangian that a body in free fall (& $v_i=0$) follows the most " efficient" path ( i.e. a vertical line)? In this video lecture by M. van Biezen ( Loyola Marymount Uni)  https://www.youtube.com/watch?v=uFnTRJ2be7I&list=PLX2gX-ftPVXWK0GOFDi7FcmIMMhY_7fU9&index=2
it is shown how to apply the equation
$$\frac{d}{dt}(\frac{\partial L}{\partial x'})- \frac{\partial L}{\partial x}=0.$$
to an object in free fall ( with initial velocity $= 0$)
with $L = KE - PE$ , $x=$ height of an object in free fall and $x'$ = the derivative of $x$ with respect to time.
The lecturer shows one can derive the equation $F = ma$ from the previous one.
He also says in another lecture that one can show using the equation involving the lagrangian that the object in free fall follows the most " efficient" trajectory, that is, the vertical straight line.
Is it possible to do this not using too complicated mathematical methods?
The equation is an " equals zero" one. Maybe it means that the derivative of something ( that I cannot identify) is zero and therefore that this something is constant.
What is this unidentified quantity and how does it relate to the path of the object?
 A: For each path, we can assign the following quantity known as action:
$$ S[\vec{r}]= \int_{t_1}^{t_2} dt( L)$$
Where,
$$ L = \text{ total kinetic energy} - \text{total potential energy}$$
The quantity 'S' is called a functional. A functional can be stated loosely as a function of functions. A concrete example of such objects being used to solve problems can be found in this post I made on MSE.
Anyhow, the great discovery was that the path which makes the action stationary (i.e: extremum of S) would be the path which the object follows. It turns out that the condition that the action is minimized/maxed is given as:
$$ \frac{d}{dt} \frac{ \partial L}{\partial q} = \frac{\partial L}{\partial \dot{q} }$$
This similar to how if we have a function then $ \frac{df}{dx}=0$ gives us the condition of $x$ such that it is maximized, the above is the condition that should be satisfied by $L$ for the action to be minimized/maxed. Turns out this condition is same as newton's second law.
The real rigorous details of all the above are quite complicated but I've heard Leonard Susskind's book gives a good take on this in a simple way. Btw the condition for action being optimized comes another mathematical result known as Euler Lagrange equations (there are too many videos deriving this one equation  onyoutube, so check there)
A: This specific formulation of the problem does not demonstrate that the motion is a straight line, rather it assumes that the motion is a straight line.
By formulating the potential as :
V = m g X
One is already assuming the problem is one-dimensional, i.e. it depends only on the only parameter X (as oppose to be a function of X, Y, Z).
Therefore, the problem is independent of the Y and Z coordinates.
Therefore, the problem is formulated as being symmetrical around the X axis.
Therefore, the motion can only be a straight line along the X axis (otherwise the aforementioned symmetry is broken).
Conclusion:
Basically, saying that this formula leads to the conclusion that the motion is a straight line is a circular reasoning. One is trying to demonstrate what one has assumed to be true in the premises.
