TITLE: Lower bound on clique number of graph through vertex degrees?
QUESTION [5 upvotes]: Suppose $\Gamma$ is a finite simple graph with $n$ vertices, such as the $i$-th vertex has degree $d_i$. Suppose $c$ is the clique number of $\Gamma$. Is it always true, that $c \geq \sum_{i=1}^n \frac{1}{n - d_i}$?
This inequality definitely holds for small graphs ($n \leq 5$) and for some "nice" classes of graphs (complete graphs and complete bipartite graphs). Also it holds for all graphs without vertices of degree exceeding $\frac{n}{2}$ (because in that case $\sum_{i=1}^n \frac{1}{n - d_i} \leq 2$). However, I do not know, whether the inequality is correct in general.

REPLY [5 votes]: Yes, it is true. Dually, you can show that any graph has an independence number of at least $$\sum_{i=1}^n \frac1{1+\deg v_i},$$ then apply that result to the complement of $\Gamma$. This bound is was first shown by Caro, and Wei, and is proven as follows. Choose a random permutation of the $n$ vertices, and build an independent set $A$ consisting of all vertices $v$ such that none of $v$'s neighbors appear earlier in the permutation (i.e. a greedy algorithm on a random order). The probability a particular vertex $v$ is included in $A$ is the probability that $v$ occurs before all of its neighbors, yielding the expected size of $A$ as the expression above.