Let $M$ be a closed Riemannian manifold with a spin$^\mathbb{C}$ bundle $S$. Now for a spin connection $A,$ and a spinor $\phi,$ it can be shown that $C\lvert\nabla_A\phi\rvert^2\geq \lvert D_A\phi\rvert^2$ for some $C>0$. My question is what's the best value of $C$ one can hope for? Ideally this should depend on the geometry and dimension of the manifold I would think. Then again for a flat manifold and with a trivial line bundle, the usual Weitzenböck formula \begin{align*} D_A^2=\nabla_A^*\nabla_A+\frac{s}{4}+\frac{1}{2}F_A \end{align*} would imply that $\int_M \lvert D_A\rvert^2=\int_M \lvert\nabla_A\rvert^2$. My hope is: can $C$ be $1$? Is there an easy counter example to see?

I am also interested in a similar question for any unitary connection $B$ on $S$, one can define $D_B:\nabla_B\xrightarrow{\text{Clifford mult.}}D_B$. Again we have $C\lvert\nabla_B\phi\rvert^2\geq \lvert D_B\phi\rvert^2$, does the answer really depend on what connection we use?