If $a_1,...,a_n$ and $b_1,...,b_n$ are complex numbers then
$$|\sum_{j=1}^na_j\overline{b_j}|^2 \leq \sum_{j=1}^n|a_j|^2\sum_{j=1}^n|b_j|^2$$
Proof: Put $A=\sum_{j=1}^n|a_j|^2$, $B=\sum_{j=1}^n|b_j|^2$, $C=\sum_{j=1}^na_j\overline{b_j}$. If $B=0$, then $b_1=...=b_n = 0$ and the conclusion is trivial. Assume therefore that $B \gt 0$. Then we have
$$\sum_{j=1}^n|Ba_j-Cb_j|^2 = \sum_{j=1}^n(Ba_j-Cb_j)(B\overline{a_j}-\overline{Cb_j}) $$
$$= B^2\sum_{j=1}^n|a_j|^2-B\overline{C}\sum_{j=1}^na_j\overline{b_j}-BC\sum_{j=1}^n\overline{a_j}b_j+|C|^2\sum_{j=1}^n|b_j|^2$$
$$= B^2A-B|C|^2$$
$$(...)$$
How did $-B\overline{C}\sum_{j=1}^na_j\overline{b_j}-BC\sum_{j=1}^n\overline{a_j}b_j$ disappear? Why is it always $0$?
\begin{align} &B^2\sum_{j=1}^n|a_j|^2-B\overline{C}\sum_{j=1}^na_j\overline{b_j}-BC\sum_{j=1}^n\overline{a_j}b_j+|C|^2\sum_{j=1}^n|b_j|^2 &\\ &=B^2A-B\overline{C}C-BC\overline{C}+|C|^2B\\ &=B^2A-2B|C|^2+|C|^2B \\&= B^2A-B|C|^2\end{align}