Can we understand the founding principle of Lagrangian mechanics in the following way?
If a mass particle follows the path $x(t)$, then we have the two types of energy: kinetic energy $T(x(t))$ and potential energy $V(x(t))$. The usual total energy is the sum $E=T+V$.
For every $t'\in \mathbb{R}$ we can define a scalar field $U(x(t')) = \sqrt{ T^2(x(t'))+V^2(x(t'))}$ that represents the generalized energy magnitude of the conjugate energies $T$ and $V$. Because the Nature favors uniformity, we wish to minimize the curvature $U$ over the time period $[t_i,t_f]$ when the particle moves from $x(t_i)$ to $x(t_f)$. We can simplify the problem by noting that $U=\sqrt{T^2+(E-T)^2}$ and treat $T$ as an independent variable. Thus, by taking the derivative, we have the minimum at $T=E/2$. Hence, we have $V=E-T = E/2=T$. Thus, if we wish to minimize $U$, then we must minimize the difference $|T-V|$. The remaining question is to find the path $x_*(t)$ that minimizes the following integral
$$\int_{t_i}^{t_f}(T-V)dt.$$
This seems like a good justification for the use of the quantity $T-V$ as the Lagrangian. A more qualitative way I’d phrase it as well is to say that the optimal path a particle follows is the one where the kinetic energy is driven solely by the potential energy constraints imposed on the system. Therefore, to guarantee that the kinetic energy matches the potential we would like $T\approx V$, or equivalently $T-V$ to be minimized.