I am trying to show that any graph with n vertices and m edges has an independent set of size at least: $$\frac{n^2}{2m+n}$$
I have already shown that any graph has an independent set of size at least: $$\sum_{i=0}^n \frac{1}{d(v_i) + 1}$$
where $d(v_i)$ is the degree of vertex $v_i$.
I am trying to deduce the first result from what I have already shown.
I know that $$\sum_{i=0}^n {(d(v_i) + 1)} = 2m + n$$ by the handshaking lemma and I reckon this will be helpful, however I'm struggling to combine it with $\sum_{i=0}^n \frac{1}{d(v_i) + 1}$ to prove the required result.
Any help much appreciated!
Hint
You would be done if you could show that $$ \sum_i \frac1{d(v_i)+1}\ge \frac{n^2}{2m+n} $$ Letting $f(x)=\frac1{x+1}$, the LHS is $\sum_i f(d(v_i))$. Use the face that the function $f(x)$ is convex.