An elementary way, i.e., without the use of the Hamilton principle of least action and symmetry arguments, to find the correct force law for electric point charges in an electromagnetic field is to recall the covariant form of the equation of motion for a point particle (using Heaviside-Lorentz units with c=1 for simplicity)
\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu},
where p^{\mu} is the energy-momenum-four vector given by
p^{\mu}=m u^{\mu}=m\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.
Further \tau is the proper time of the particle, defined by
\mathrm{d} \tau=\sqrt{\mathrm{d} x^{\mu} \mathrm{d} x_{\mu}}.
From this it immediately follows that
p_{\mu} p^{\mu}=m^2=\text{const},
which implies
p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} m}=p_{\mu} K^{\mu}=0. \qquad (1)
Next we must recall that the force of a particle at rest is proportional to its charge and given by \vec{F}=q \vec{E},
where \vec{E} are the electric components of the electromagnetic field.
In relativistically covariant notation, the electromagnetic field is given by the antisymmetric tensor
F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu},
where A^{\mu} is the four-potential of the electromagnetic field. Thus, to find the relativistic generalization of the static case, we note that F_{\mu \nu} consists of the electric and magnetic field components and that we can form a vector from it, which is (a) proportional to the field strengths and (b) to the electric charge of the particle and (c) fulfills the on-shell constraint (1), we come to the ansatz
K^{\mu}=q {F^{\mu}}_{\nu} u^{\mu}.
This also has the right dimensions, because u^{\mu} is a velocity (i.e., dimensionless in our natural units).
That we have also chosen the correct sign can be seen by looking at the spatial components and write everything in terms of the three-dimensional notation, i.e., in terms of
\vec{E}=-\vec{\nabla} A^0 - \partial_t \vec{A}, \quad \vec{B}=\vec{\nabla} \times \vec{B}.
In index notation we have
E^j=\partial^j A^0-\partial^0 A^j=F^{j0}, \quad B^{j}=-\epsilon^{jkl} \partial^{k} A^{l}=-\frac{1}{2} \epsilon^{jkl} F^{kl}.
From the latter equation we find (by inversion of the 3D Hodge duality):
F^{jk} = -\epsilon^{jkl} B^l.
This implies that
K^{j}=q (F^{j0} u^0 - F^{jk} u^k)=q (u^0 E^j + \epsilon^{jkl}u^k B^l).
In vector notation this reads
\vec{K}=q (u^0 \vec{E}+\vec{u} \times \vec{B}).
The spatial part of the equation of motion thus reads
\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau}=q (u^0 \vec{E}+\vec{u} \times \vec{B}).
This becomes more famliar by using the time derivative instead of the proper-time derivative:
\frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} \frac{\mathrm{d} \tau}{\mathrm{d} t}=\frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} \frac{1}{u^0}=q(\vec{E}+\vec{v} \times \vec{B}),
where we have used
\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \frac{1}{u^0}=\frac{\vec{u}}{u^0}.
This shows that the Lorentz force law is the natural extension of the electromagnetic force of a particle at rest in an electromagnetic field, which is \vec{F}=q \vec{E}, which is correct for a particle at rest (by definition of the electric field as force per unit charge).
