Q. (Caro-Wei) Given a graph $G$, there exists an independent set (i.e. no two vertices are adjacent) of size at least $$\sum_{i=1}^{|V|} \frac{1}{\deg v + 1}$$

Proof. Greedily pick this set. Select the vertex with minimum degree $w$. Then each neighbor $w'\in N(w)$ satisfies $\deg w' \ge \deg w$, and hence $$\sum_{w'\in N(w)\cup \{w\}} \frac{1}{\deg w' + 1}\le 1$$

After removing $N(w)\cup w$ from $G$ (to get $G'$), the size of the independent set obtained by applying the induction hypothesis on $G'$ is $$1 + \sum_{v\in |V(G')|} \frac{1}{\deg_{G'} v + 1} \ge \sum_{w'\in N(w)\cup \{w\}} \frac{1}{\deg w' + 1} + \sum_{v\in |V(G')|} \frac{1}{\deg v + 1}$$