This can't be right as stated: if $h$ takes a negative value at some point $p$ of $M$ then $\Delta+ch$ has a negative eigenvalue for sufficiently large $c$.

Proof: let $f: M \rightarrow {\bf R}$ be a nonnegative smooth function that's positive at $p$ and supported on a small enough neighborhood of $p$ that $f(q)=0$ whenever $h(q)>0$. Then $\langle f, hf \rangle < 0$. Therefore if $c$ is large enough then $\langle f, (\Delta+ch) f \rangle < 0$, which would be impossible if every eigenvalue of $\Delta + ch$ where nonnegative.

This can't be right as stated: if $h$ takes a negative value at some point $p$ of $M$ then $\Delta+ch$ has a negative eigenvalue for sufficiently large $c$.

Proof: let $f: M \rightarrow {\bf R}$ be a nonnegative smooth function that's positive at $p$ and supported on a small enough neighborhood of $p$ that $f(q)=0$ whenever $h(q)>0$. Then $\langle f, hf \rangle < 0$. Therefore if $c$ is large enough then $\langle f, (\Delta+ch) f \rangle < 0$, which would be impossible if every eigenvalue of $\Delta + ch$ were nonnegative.