Can we generate the Proca action (spin-1) from the Dirac action (spin-1/2)? The Lagrangian for a spin-$\frac{1}{2}$ particle is the Dirac Lagrangian, while for a spin-$1$ particle is the Proca Lagrangian.
But $1$ should just be $\frac{1}{2} \bigoplus \frac{1}{2}$, so is it possible to infer the Proca action from combining Dirac ones?
 A: The closest thing to what you're looking for consists of decomposing the field $A_\mu$ using two component Weyl spinors; exhibiting the isomorphism $SO(1,3) = SL(2,\mathbb{C})$, the vector representation of the Lorentz group goes into $X \to P X P^\dagger$ with $X$ the complex $2x2$ matrix associated to the vector $X^\mu$ and $P \in SL(2, \mathbb{C})$. Exhibiting the $SL(2, \mathbb{C})$ indices explicitly, the field becomes $A_{a\dot{a}}$ and the field equations can be written in a 2-component spinor notation. Warren Siegel discusses these issues at length in his field theory text. 
A: Remember that the Dirac equation describes fermions with half-integer spin, like electrons or quarks. A proca Lagrangian describes a massive boson with spin 1. The statistics followed by these two types of particles are different: Bosons obey Bose-Einstein statistics while fermions obey Fermi-Dirac statistics. 

But 1 should just be $1/2⨁1/2$, so is it possible to infer the Proca action from combining Dirac ones?

The answer is no, because the proca action (or Lagrangian) and the Dirac action describe two different type of particles (and two very different mathematical objects). The proca Lagrangian describes a real vector field $A^{\mu}$ while the Dirac Lagrangian describes a complex valued 4-component spinor field $\Psi$. Summing two spinors (two 1/2 spin particles) would result in a spinor, not a vector field. 
Even more so, the Lagrangians don't even have the same symmetries! For example, the Proca Lagrangian is not invariant under global $U(1)$ transformations, while the Dirac Lagrangian is.
