In the proof below, how is the inequality arrived at? I don't understand how $W$ is broken down in terms of: (1) the contribution from nodes further than node $i$ and (2) the rest. Can someone please explain?
This is Lemma 6.2 in paper --> https://arxiv.org/pdf/1109.1077.pdf
By defining $${A_1\triangleq f'(W)\left[1+\sum_i X_i(W_i-W)\right]\\ A_2\triangleq \sum_i X_i(W_i-W)f'(W)+\sum_i X_if(W)}$$we obtain$$f'(W)-Wf(W){=A_1-A_2}$$since $W=\sum_i X_i$. Hence the inequality reduces to $$|\Bbb E\{A_1-A_2\}|\le |\Bbb E\{A_1\}|+|\Bbb E\{A_2\}|$$or$$|\Bbb E\{A_1\}+\Bbb E\{A_2\}|\le |\Bbb E\{A_1\}|+|\Bbb E\{A_2\}|$$which is the celebrated triangle inequality: $$|u+v|\le |u|+|v|$$