Conjugate in Cauchy-Schwarz Inequality for Complex Numbers Simple question: do we really need the conjugate in the inequality?

$$ |\sum_{j=1}^n a_j \overline{b_j}|^2 \leq \sum_{j=1}^n |a_j|^2 \sum_{j=1}^n |b_j|^2 $$
The Cauchy-Schwarz inequality says that $|\langle a,b \rangle| \leq \| a \| \|b\|$.  In $\mathbb C^n$, the inner product is
\begin{equation}
\langle a, b \rangle = \sum_{j=1}^n a_j \bar{b_j}.
\end{equation}

That's why the Cauchy-Schwarz inequality in $\mathbb C^n$ has conjugates in it.

While it is true that you could omit the conjugates in your inequality and still have a true statement, that would only take us further away from the nice statement that $|\langle a,b \rangle| \leq \| a \| \|b\|$.
Simple answer, yes. Look back to the definition of the inner product (for complex numbers) and note the complex conjugate symmetry (as opposed to pure symmetry)