Suppose you know that $\Delta V=o(|\nabla V|^2)$ as $|x|\to\infty$. This implies that for any $\theta\in(0,1)$ you have eventually the inequality $\Delta V\le \theta|\nabla V|^2$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by $$ |\nabla V|^2((1-\alpha\theta)\phi'-\alpha \phi''). $$ Then you can play wiyh the choice of $\phi$. For instance, if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.

Suppose you know that $\Delta V=o(|\nabla V|^2)$ as $|x|\to\infty$. This implies that for any $\theta\in(0,1)$ you have eventually the inequality $\Delta V\le \theta|\nabla V|^2$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by $$ |\nabla V|^2((1-\alpha\theta)\phi'-\alpha \phi''). $$ Then you can play with the choice of $\phi$. For instance, if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.

Suppose you know that $\Delta V\le \theta|\nabla V|^2$ for $|x|$ sufficiently large, with $\theta<1$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by $$ |\nabla V|^2((1-\theta)\phi'-\alpha \phi'') $$ and if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.

Suppose you know that $\Delta V=o(|\nabla V|^2)$ as $|x|\to\infty$. This implies that for any $\theta\in(0,1)$ you have eventually the inequality $\Delta V\le \theta|\nabla V|^2$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by $$ |\nabla V|^2((1-\alpha\theta)\phi'-\alpha \phi''). $$ Then you can play wiyh the choice of $\phi$. For instance, if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.