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all Lorentz observers watching a particle move will compute the same value for the quantity

$$ds^2 = -(c \, dt)^2 + dx^2 + dy^2 + dz^2,$$ $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$ and ''ds/c'' is then an infinitesimal proper time. For a point particle not subject to external forces (''i.e.,'' one undergoing inertial motion), the relativistic action is:

$$S = -m_oc \int ds.$$

Remark: An edit to the comment. we can write a Lagrangian for a relativistic particle, which will be valid even if the particle is traveling close to the speed of light. To preserve Lorentz invariance, the action should only depend upon quantities that are the same for all (Lorentz) observers.

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The action

$$S= - E_0 ~ \Delta \tau $$

of a relativistic massive particle is minus the rest energy $E_0=m_0c^2$ times the change $\Delta \tau=\tau_f-\tau_i$ in proper time.

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You do not explain what do you mean by "better answer" but an alternative form is obtained by using $ds = \gamma^{-1}\cdot{c}\cdot{dt}$ with $\gamma$ the time-dilation factor. Then, $$S = -mc \int ds = \int - \frac{mc^2}{\gamma} dt$$ from where you can obtain the Lagrangian, recalling that $S = \int L dt$.

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