You have a canonical inner product in $\Bbb C^n$ given by $x\cdot y=\displaystyle \sum_{i=1}^n x_i\overline{y_i}$. Note $x\cdot x=\displaystyle \sum_{i=1}^n x_i\overline{x_i}=\displaystyle \sum_{i=1}^n |x_i|^2$
Claim Let $V$ be a $\Bbb C$-vector space, and $\langle \cdot,\cdot\rangle$ an inner product. If we define a norm $\lVert x\rVert :=\langle x,x\rangle^{1/2}$, then we always have $$|\langle x,y\rangle|^2\leqslant \lVert x\rVert ^2\lVert y\rVert^2$$
Proof If $x$ or $y$ are zero, the inequality is true, since $\langle x,0\rangle=0$ and $\lVert 0\rVert =0$. We may assume then that both norms are nonzero. Since $\langle z,z\rangle\geqslant 0$ for any choice of $z$, we may take $$z=x-\frac{\langle x,y\rangle y}{\lVert y\rVert ^2} $$
Using the inner product is (sesqui)linear we get
$$\begin{align}
\langle z,z\rangle
&= \langle {x,x} \rangle - \frac{1}{\lVert y\rVert^2}\langle {x,\langle x,y \rangle y} \rangle - \frac{1}{\lVert y\rVert^2}\langle {\langle x,y \rangle y,x} \rangle + \frac{{\langle x,y \rangle \overline {\langle x,y \rangle} }}{{\lVert y\rVert^2\lVert y\rVert^2}}\langle {y,y} \rangle \\
&= {\| x \|^2} - \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} - \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} + \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} \\
&= {\| x \|^2} - \frac{{| \langle x,y \rangle |^2}}{\| y \|^2} \geqslant 0\end{align} $$
as we wanted. We made repeated use of $$\langle a+b,c\rangle=\langle a,c\rangle+\langle b,c\rangle\\ \langle a,b+c\rangle=\langle a,b\rangle+\langle a,c\rangle\\\langle \alpha a,b\rangle=\alpha\langle a,b\rangle\\\langle a,\beta b\rangle=\overline \beta\langle a,b\rangle \\z\overline z=|z|^2$$
