Relationship between degrees and number of edges

Question

Why $(\sum_{v_iv_j \in E(G)} \sqrt{d(i)d(j)} )(\sum_{v_iv_j \in E(G)}\frac{1}{ \sqrt{d(i)d(j)}}) \ge m^2$ is true? I mean, how to proof it? $m$ is number of edges and $d_1,d_2,\dots ,d_n$ is degree sequence.

Answer 1

Use Cauchy–Schwarz inequality for $N$-dimensional space $$ (x_1^2 + \cdots + x_N^2)(y_1^2 + \cdots + y_N^2) \geq (y_1x_1 + \cdots + y_Nx_N)^2. $$ All degrees $d(i)$ and their square roots are positive, thus you can safely assume $\sqrt{d(i)d(j)} = x_k^2$ and $1/\sqrt{d(i)d(j)} = y_k^2$.

Note: For $n$ vertices there are $N = n(n-1)/2$ possible pairs of vertices. It is always the case $N \geq m$ with equality if and only if the graph is full.