Here is an alternative formulation of a question related to yours.
Consider the case when $M=\Omega\subset\mathbb{R}^{n}$, a (smooth)
open bounded domain. Suppose that $u>0$ and
$$
-\Delta u+\psi u=f \mbox{ in }\Omega.
$$
Integrate against $\frac{w^{2}}{u}$ and rearranging you obtain for
all $w\in H_{0}^{1}(\Omega),$
$$
\int\nabla w\cdot\nabla w-\int_{\Omega}\left|\frac{u\left|\nabla w\right|-w\left|\nabla u\right|}{u}\right|^{2}+\int_{\Omega}\psi w^{2}=\int f\frac{w^{2}}{u}
$$
therefore for all $w\in H_{0}^{1}(\Omega)$
$$
\int\nabla w\cdot\nabla w+\int_{\Omega}\psi w^{2}\geq\int f\frac{w^{2}}{u}
$$
therefore, if $\lambda_{\psi}$ is the first dirichlet eigenvalue
of $-\Delta+\psi$, you obtain
$$
\lambda_{\psi}\int w^{2}\geq\int f\frac{w^{2}}{u}.
$$
So if there exists a positive solution such that $-\Delta u+\psi u>0$,
then $\lambda_{\psi}>0$. The converse is obsiously true, as we can
choose the first eigenvector to satisfy $u_{\psi}>0$ in $\Omega$,
and therefore $-\Delta u+\psi u=\lambda_{\psi}u>0$ in $\Omega$.
So your question could be written in this case:
Is it true that
$$
\mbox{if }\quad \psi+\phi>0,
\mbox{ then }
\lambda_{\psi}>0 \mbox{ or } \lambda_{\phi}>0\quad?
$$
It is not obvious, because $\lambda_{f}$ is concave in $f$: the fact that $\lambda_{\frac{1}{2}\psi+\frac{1}{2}\phi}>0$ does not help..
