It is in fact possible to explain Hamilton's action within the context of classical mechanics.

On physics.stackexchange I have discussed that, in an answer posted in oktober 2021. That discussion is illustrated with animated GIF's. The animated GIF's are composed of successive screenshots of interactive diagrams that are on my own website.

https://physics.stackexchange.com/a/670705/

Stackexchange has mathjax support, and support for uploading images, that is why I refer to my post on physics.stackexchange

The following is to give you an idea of what I discuss.

We have that if F=ma is granted as axiom then the Work-Energy theorem follows as theorem.

(As we know: the derivation of the Work-Energy theorem is subject to the following condition: it is only applicable if it is possible to define an unambiguous expression for potential energy. In order to have a well-defined expression for potential energy the force that is involved needs to be a conservative force.)

The Work-Energy theorem implies the following: In the process of interconversion of potential energy and kinetic energy: the rate of change of kinetic energy always matches the rate of change of potential energy. (If the potential energy is decreasing then the kinetic energy is increasing at the same rate)

In terms of exploring a variation space of trial trajectories:

The true trajectory has the following properties: A property of derivative with respect to time:
- At every point along the trajectory the derivative of the kinetic energy with respect to time matches the derivative of the potential energy with respect to time.

A property of derivative with respect to _position_:
- At every point along the trajectory the derivative of the kinetic energy with respect to _position_ matches the derivative of the potential energy with respect to _position_.

I want to highlight this: In mathematical models that describe changes taking place we are accustomed to taking the derivative with respect to _time_.

But:
When we are representing the physics taking place in terms of _Energy_ it is powerful to take the derivative with respect to _position_.

In classical mechanics: When you insert the Lagrangian in the Euler-Lagrange equation then the operation that the Euler-Lagrange equation performs is that it takes the derivative of the Lagrangian with respect to position.

You are looking for the point where the _derivative_ of the Lagrangian with respect to _position_ is zero.

when that derivative is zero the derivative-of-the-kinetic-energy-with-respect-to-position _matches_ the derivitive-of-the-potential-energy-with-respect-to-position.

For the concept of stationary action minimum or maximum is immaterial. Stationary action is about identifying the point in variation space such that at every point along the trajectory the derivative-of-the-kinetic-energy-with-respect-to-position matches the derivitive-of-the-potential-energy-with-respect-to-position

Finally: There is a concept that I will refer to as Jacob's lemma. (This concept was introduced by Jacob Bernoulli in the course of presenting his solution to the Brachistochrone problem. The Brachistochrone problem had been presented by Johann Bernoulli, as a challenge.)

Jacob's Lemma was stated decades before Euler started development of calculus of variations. Jacob's Lemma is crucial to understanding calculus of variations.

Take the Brachistochrone curve. Divide it in subsections. Then each subsection is in and of itself an instance of the Brachistochrone problem. This process of subdivision can be repeated indefinitely. In the end you have a concatenation of infinitissimally short subsections, and we know that each of those subsections is an instance of the Brachistochrone problem.

This tells us that a differential equation must exist that solves the Brachistochrone problem. (It does not narrow down what that differential equation is, but at least you have logical proof that it _must_ exist.)

In fact, Jacob Bernoulli succeeded in solving the Brachistochrone problem using a differential calculus approach.

The Euler-Lagrange equation takes the variational formulation, and converts it to differential equation form.