The principle of stationary action from the principle of inertia? Can we infer the principle of stationary action from the principle of inertia that causes a mass particle to resist changes in its momentum? The following is my own speculation.
When a mass particle changes its position by $\Delta s$ then it tends to maintain its momentum $p$ with no increase, decrease or change in direction, ideally $\Delta p = 0$. Therefore, the inertia of motion causes a sum $\sum \overline{p} \cdot \Delta s$, where $\Delta s$ is an infinitesimal part of the overall path in 3D and $\overline{p}$ is the average momentum on each $\Delta s$, to obtain a minimum value. All this leads to the Euler's form of the stationary action principle, i.e.,
$$\delta \int p \cdot ds=0.$$
Moreover, in the presence of conservative force, the only possible source of momentum dependent kinetic energy $T$ is position dependent potential energy $V$, and vice versa. Now the inertia of motion resists the flow of energy between $T$ and $V$, ideally $\Delta T = 0$. Therefore, the inertia of motion causes a sum $\sum \overline{T} \Delta t$, where $\Delta t$ is an infinitesimal time step during the overall energy flow process and $\overline{T}$ is the average kinetic energy on each $\Delta t$, to obtain a minimum value. All this leads to the Maupertuis' form of the stationary action principle, i.e.,
$$\delta \int T dt=0.$$
On the other hand, the previous energy flow between $T$ and $V$ can be expressed as $\Delta (T-V)$, and this leads to the Lagrange's form of the stationary action principle, i.e.,
$$\delta \int (T-V) dt=0.$$

A pedagogical article Simple derivation of Newtonian mechanics from the principle of least action deals with this question and explains how the correct path, and also Newton's second law, arises from the infinitesimal parts $\Delta S$. However, the authors don't consider the principle of inertia as "starting point and the fundamental assumption of classical mechanics" (like the Encyclopædia Britannica states), but they explain the origins of the action principle as follows:

The principle of least action says that a particle moves on the path for which the action S is a stationary. After deriving Newton’s laws from the principle of least action, then, according to Feynman, some questions naturally arise: What is the origin of the principle of least action? How does the particle find the right path (or worldline)? Does it ‘‘smell’’ the neighboring paths to find out whether or not they have increased action? Newtonian mechanics cannot answer these questions. Indeed, the principle of least action has a deep explanation in quantum mechanics. There are three apparently different mathematical formulations of nonrelativistic quantum mechanics due to Schrödinger, Heisenberg, and Feynman. The last one provides a simple justification of a minimum principle that is very accessible to students.

 A: We have the following two concepts:
The relation between force, inertial mass, and acceleration:
$$ F=ma \tag{1.1} $$
The relation between mechanical potential energy and kinetic energy:
$$ \Delta E_k + \Delta E_p = 0 \tag{1.2} $$
You grant the concept of conservation of mechanical energy, but it appears you think of that as independent from $F=ma$. That is in fact not the case:


$ ds = v \ dt  \qquad \qquad (2.1) $
$ a \ dt = dv  \qquad \qquad (2.2) $

$ F = ma \qquad \qquad (2.3) $
$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \qquad \qquad (2.4) $
$ \int_{s_0}^s a \ ds = \int_{t_0}^t  a \ v  \ dt
= \int_{t_0}^t  v \ a \ dt = \int_{v_0}^v  v \ dv
= \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \qquad (2.5) $
$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \qquad \qquad (2.6) $
$ \Delta E_p = -\int_{s_0}^s F \ ds \qquad \qquad (2.7) $
$ \Delta E_k = -\Delta E_p \quad \Leftrightarrow \quad \Delta E_k + \Delta E_p = 0 \qquad \qquad (2.8) $

From (2.3) to (2.4):
To both sides of (2.3) the same operation is applied: integration with respect to the position coordinate.
In (2.5) the expression $ \int_{s_0}^s a \ ds $ is developed. The acceleration is unspecified, but we can develop nonetheless because position and acceleration are not independent of each other. In developing the differential is changed twice, according to (2.1) and (2.2), with corresponding change of limits.
(2.7) states the definition of potential energy.


Transformation
We have in general that in physics we can shift from one representation to another by application of a mathematical transformation. In the case of $F=ma$ and $\Delta E_k + \Delta E_p = 0$: those two representations are related by integration.
Conversely: to recover $F=ma$ from $\Delta E_k = -\Delta E_p$ you take the derivative with respect to the position coordinate on both sides.

My point here is: from a physics point of view: to grant $F=ma$ or to grant $\Delta E_k = -\Delta E_p$ is one and the same thing.


We have: in cases where $\Delta E_k + \Delta E_p = 0$ holds good: Hamilton's stationary action will hold good also (no additional assumption required).
Discussion of that is in an answer by me (oktober 2021) to a question about motivation for Lagrangian formalism
A: The basic flaw in your argument is that the sum is not zero. It tends to resist but it always doesn't, if you have some infinitesimal movement $\delta x$, you will have some associated change in momentum.
But your final equation is correct, then how would you go about deriving it? You see in the path of least action, you know the Lagrangian is zero for any path at the beginning and end points; so you can impose boundary conditions.
