I wrote this proof of the following identity and I want to verify that a certain step is correct.
$\newcommand{\conj}[1]{\overline{\vphantom{b}#1}}$ $\newcommand{\on}[1]{\operatorname{#1}}$ $$\left(\sum_{k=1}^n|a_k|^2\right)\left(\sum_{k=1}^n|b_k|^2\right) - \left|\sum_{k=1}^na_kb_k\right|^2 = \sum_{i=1}^{n-1}\sum_{j=i+1}^n|a_i\conj{b_j}-a_j\conj{b_i}|^2$$
Expand and re-write the identity to find that:
$$\sum_{i=1}^{n}\sum_{j=1}^{n}|a_ib_j|^2 - \sum_{i=1}^{n}|a_ib_i|^2 - \sum_{i=1}^{n}\sum_{\substack{j=1\\j\neq i}}^{n}|a_ib_i\conj{a_jb_j}| = \sum_{i=1}^{n-1}\sum_{j=i+1}^n|a_i\conj{b_j}-a_j\conj{b_i}|^2$$
The first and second terms simplify. Additionally, one may use symmetry to combine terms in the third sum and then rewrite it using $\on{Re}(z)$. Now we have $$\sum_{i=1}^{n-1}\sum_{\substack{j=1\\j\neq i}}^{n}|a_ib_j|^2 - 2\on{Re}\left(\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}a_i\conj{a_j}b_i\conj{b_j}\right) = \sum_{i=1}^{n-1}\sum_{j=i+1}^n|a_i\conj{b_j}-a_j\conj{b_i}|^2$$
My question involves the use of $\on{Re}(z)$ as I cannot justify this step in the proof from the second line to third line.
The step in the proof is justified as follows:
$$ \sum_{i=1}^{n}\sum_{\substack{j=1\\j\neq i}}^{n}|a_ib_i\overline{a_jb_j}| = \sum_{i < j}a_ib_i \overline{a_j b_j} + \sum_{i < j}\overline{a_i b_i}a_j b_j = \sum_{i < j}a_ib_i \overline{a_j b_j} + \overline{\sum_{i < j}a_ib_i \overline{a_j b_j}} = \sum_{i < j}2 \operatorname{Re}(a_ib_i \overline{a_j b_j})$$
Note that for better visibility I have replace your notation $ \sum_{i=1}^{n}\sum_{\substack{j=1\\j\neq i}}^{n}$ with $\sum_{i<j}$.