The Lichnerowicz vanishing theorem says that if on a compact 4-dimensional spin manifold there exists a metric whose scalar curvature $R>0$, then there are no harmonic spinors; $$D\psi=0 \implies \psi=0$$

This follows from the Bochner-type formula $$D^2=\nabla^*\nabla + \displaystyle\frac{R}{4}$$ by applying it to a spinor field $\psi$ and pairing with $\psi$ to obtain $$\int \, \rvert D\psi\rvert^2 \,dV = \int \, \vert \nabla\psi \rvert^2 \, dV + \displaystyle\frac{R}{4}\int \, \rvert \psi \rvert^2 \, dV$$ from which the statement follows directly.

My question is about whether there's a sort of converse to this reasoning. That is, suppose we have a compact Riemannian spin manifold and we know that there are no solutions to $D\psi = 0$ except the trivial one $\psi=0$. Does it follow that $R\ge 0$ (hopefully with $R>0$ at some point)?

My vague intuition for the plausibility of this is the following: Staying away from the zero section, we know that as you range over all spinor fields, the LHS of the integral formula above is strictly positive. Suppose $R<0$ at some point. Is there enough "flexibility" in our choice of spinor field to localize near this negative value and make the RHS negative (by making the second term dominate the first in absolute value) so as to generate a contradiction?