Derivation of rest energy in Landau & Lifshitz In Landau & Lifshitz The Classical Theory Of Fields there's a statement:

$$\mathscr E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\tag{9.4}$$
  This very important formula shows, in particular, that in relativistic mechanics the energy of a free particle does not go to zero for $v=0$, but rather takes on a finite value
  $$\mathscr E=mc^2.\tag{9.5}$$
  This quantity is called the rest energy of the particle.

But this is the direct result of choosing Lagrangian as
$$L=-mc^2\sqrt{1-\frac{v^2}{c^2}}.$$
We could as easily subtract $-mc^2$ from the Lagrangian to have $L^\prime=L+mc^2$ and have nothing changed in equations of motion, but now having the energy
$$E^\prime=mc^2\left(\frac1{\sqrt{1-\frac{v^2}{c^2}}}-1\right),$$
which would give rest energy $E^\prime_0=0$.
So my question is what is the physical significance of using $(9.4)$ and corresponding Lagrangian to define rest energy? Why is such a seemingly special-case result makes the authors say "in relativistic mechanics the energy of a free particle does not go to zero" as if it's a general result?
 A: 
So my question is what is the physical significance of using (9.4) and corresponding Lagrangian to define rest energy?

If the energy is defined as in 9.4, then this energy together with the three components of the momentum transform by Lorentz transformation in the same way as spacetime displacement four-vector, which is nice and preferable. If the energy was defined only as the kinetic energy, this could not be regarded as a component of such four-vector.
There are other reasons why the rest energy $mc^2$ is included, for example Einstein wrote many papers about the fact that change of energy of system in its rest frame leads to corresponding change in the rest mass according to relation $\Delta \mathscr{E} = \Delta mc^2$. This also suggests the above definition of energy.
A: When deriving Lagrangian, L&L demanded that action must be isotropic.
If we select Lagrangian as 
$$L=-mc^2\left(\sqrt{1-\frac{v^2}{c^2}}-1\right),$$
we automatically lose isotropy of infinitesimal action $L\text{d}t$. To see this, let's make a Lorentz transform on this form:
$$L^\prime\text{d}t^\prime=-mc^2\left(\sqrt{1-\frac{\left(\mathscr L_x(\vec v)\right)^2}{c^2}}-1\right)\text{d}\mathscr L_t(t)=-mc^2\Bigg(\sqrt{1-\frac{{v^\prime}
^2}{c^2}}-\underbrace{\frac{1+\frac{\Delta v}{c^2}v_x}{\sqrt{1-\frac{\Delta v^2}{c^2}}}}_{anisotropic}\Bigg)\text{d}t^\prime.$$
So, this means that the Lagrangian given by L&L in $(8.2)$ is unique, and thus energy given by $(9.4)$ is also unique, so there's always a rest energy, which can't be simply subtracted compatibly with Einstein relativity principle.
