Let $u$ be a function such that $$ \Delta u + \lambda u = 0 $$ for some $\lambda \in \mathbb R$, also let $w$ be a function such that $$ \Delta w + \beta w < 0. $$ for some $\beta \in \mathbb R$. Set $v = \frac{u}{w}$, show that $$ L[v^2] := \Delta(v^2) + \frac{2}{w} \sum_{i=1}^n w_{x_i} (v^2)_{x_i} \ge 0 $$ for $\lambda \le \beta$
Any hints? I be trying with rearranging and some vector identities, but did not get anything useful.
I'll work in index notation with summation over doubled indices. In that same vein, I'll take $v_i$ rather than $v_{x_i}$ to be the the $x_i$-derivative of $v$.
We are given that $\Delta$ has an eigenfunction $u$, i.e. $u_{ii}+\lambda u=0$ for some real $\lambda$. We also have some function $w$ such that $w_{ii}+\beta w<0$ for some real $\beta$.
We now consider $L[v^2]:=(v^2)_{ii}+2 w^{-1}w_i (v^2)_i$ for $v=u/w$. Observe that
\begin{align} (v^2)_i&=(u^2 w^{-2})_i\\&=2 (u_i u )w^{-2}-2 u^2 (w_i w^{-3}),\\\\ (v^2)_{ii}&=(u^2 w^{-2})_{ii}\\&=2(u_i^2+u_{ii} u) w^{-2}-8 (u_i u)(w_i w^{-3})-2u^2(w_{ii} w^{-3}-3 w_{i}^2 w^{-4}) \end{align} so \begin{align} L[v^2]&=\left[2(u_i^2+u_{ii} u) w^{-2}-8 (u_i u)(w_i w^{-3})-2u^2(w_{ii} w^{-3}-3 w_{i}^2 w^{-4})\right]\\ &\hspace{1cm}+2w_i w^{-1}\left[2(u_i u)w^{-2}-2u^2(w_i w^{-3})\right]\\ &=2w^{-4}(u_i w-u w_i)^2+2 (u_{ii} u )w^{-2}-2u^2w_{ii} w^{-3}\\ &=2w^{-4}(u_i w-u w_i)^2+2(\beta-\lambda) u^2 w^{-2}+2(u_{ii}+\lambda u) u w^{-2}-2u^2(w_{ii}+\beta w) w^{-3} \end{align} With the rearrangements in the last line, we first notice that the first two terms are always non-negative. The third term is also non-negative under the assumption that $\beta\geq\lambda$, and the fourth term vanishes by assumption. For the last term, we know from assumptions that $-2u^2 (w_{ii}+\beta w_i)\geq 0$ (equality only if $u=0$).
That leaves the $w^{-3}$ term. If $w<0$, then this term will be non-positive; hence for the argument to go through, we need $w\geq 0$. If this is reasonable, we may conclude that $L[v^2]\geq 0$.