Graph Theory. Prove that $\sum_{v}^{} \frac{1}{1+d(v)} \ge \frac{n^2}{2e+n} $ Let e denote the number of edges and n the number of vertices. We can assume that the graph G is simple. Prove that
$\sum_{v}^{} \frac{1}{1+d(v)} \ge   \frac{n^2}{2e+n}   $
Any help/hints would be very much appreciated!

The inequality is wrong. If $G$ is a path of length 3, then the left hand side is $\frac{4}{3}$ and the right hand side is $\frac{9}{7}$.
