Please, i need help with proof of this theorem.
Let G be a connected graph with m edges and degree sequence $d_1,d_2,\dots ,d_n$, then
$λ_1≥\frac{m}{R(G)}$.
where $R(G) = \sum_{uv\in E} \frac{1}{\sqrt{d(u)d(v)}}$ is Randić index and $λ_1$ is the largest eigenvalue (spectral radius).
Please help, i have tried to use fact, that $λ_1 \le max_{v_iv_j \in E(G)}\{\sqrt{d(i)d(j)}\}$, but it's not right, it finally shows that $R(G) \ge \frac{m}{max_{v_iv_j \in E(G)}\{\sqrt{d(i)d(j)}\}}$ but it is not greater than or equal to $\frac{m}{\lambda_1}$.
EDIT
I also tried that $\lambda_1 \ge \frac{1}{m}\sum_{v_iv_j \in E(G)} \sqrt{d(i)d(j)} $ but it finally goes to $\lambda_1 \ge \frac{R(G)}{m}$. . .
This is Corollary 2.11 in this paper.
If you assume the inequality you made in the edit (the second inequality in Proposition 2.8), you can apply the Cauchy-Schwarz inequality $(\sum_{ij\in E(G)} \sqrt{d_id_j})(\sum_{ij\in E(G)} \frac{1}{\sqrt{d_id_j}})\geq m^2$ to get $\lambda_1\geq\frac{\sum_{ij\in E(G)} \sqrt{d_id_j}}{m}\geq\frac{m^2}{m\sum_{ij\in E(G)} \sqrt{d_id_j}}=\frac{m}{R(G)}$.