How to prove $|\sum_{j=1}^{n} a_{j}\overline{b_{j}}| \le \frac{1}{\alpha}\sum_{j=1}^{n} |a_{j}|^2 + \frac{\alpha}{4}\sum_{j=1}^{n} |b_{j}|^2$? I am trying to prove inequality: $|\sum_{j=1}^{n} a_{j}\overline{b_{j}}| \le \frac{1}{\alpha}\sum_{j=1}^{n} |a_{j}|^2 + \frac{\alpha}{4}\sum_{j=1}^{n} |b_{j}|^2$.
$\alpha>0$, $a_{j} $ and $ b_{j} \in\mathbb{C}$. 
Attempt: It looks somewhat similar to Cauchy-Schwarz inequality and I think I somehow need to use it. 
Cauchy-Schwarz inequality:
$|\sum_{j=1}^{n} a_{j}\overline{b_{j}}|^2 \le \sum_{j=1}^{n} |a_{j}|^2  \sum_{j=1}^{n} |b_{j}|^2$.
So I tried to multiply my inequality by $|\sum_{j=1}^{n} a_{j}\overline{b_{j}}|$ to equate its LHS to the LHS of Cauchy-Schwarz inequality, but all I managed to get is:
$|\sum_{j=1}^{n} a_{j}\overline{b_{j}}|^2 \le \frac{1}{\alpha}\sum_{j=1}^{n} |a_{j}|^2 |\sum_{j=1}^{n} a_{j}\overline{b_{j}}| + \frac{\alpha}{4}\sum_{j=1}^{n} |b_{j}|^2 |\sum_{j=1}^{n} a_{j}\overline{b_{j}}|$. I am not sure whether it is right approach and do not know how to proceed.
I can use everything from chapter 1 of baby Rudin on complex numbers and Euclidean spaces.
 A: Hint: Use
\begin{align}
ab \leq\frac{1}{2}a^2+\frac{1}{2}b^2
\end{align}
Then choose appropriate $a$ and $b$. Cough... $a = \sqrt{\frac{2}{ \alpha}}||a||$ and $b = \sqrt{\frac{ \alpha}{2}} ||b||$.
A: We have that
\begin{align*}
\frac1{\alpha}\sum_{j=1}^n(|a_j|-\frac{\alpha}2|\overline{b_j}|)^2&=\frac1{\alpha}\sum_{j=1}^n(|a_j|^2-\alpha|a_j\overline{b_j}|+\frac{\alpha^2}4|\overline{b_j}|^2)\\
&=\frac1{\alpha}\sum_{j=1}^n|a_j|^2-\sum_{j=1}^n|a_j\overline{b_j}|+\frac{\alpha}4\sum_{j=1}^n|\overline{b_j}|^2.
\end{align*}
Using the facts that $\alpha>0$, $(x-y)^2\ge0$ for any $x,y\in\mathbb R$ and $|\sum_{j=1}^na_j\overline{b_j}|\le\sum_{j=1}^n|a_j\overline{b_j}|$, we obtain
$$
\frac1{\alpha}\sum_{j=1}^n|a_j|^2+\frac{\alpha}4\sum_{j=1}^n|\overline{b_j}|^2\ge\sum_{j=1}^n|a_j\overline{b_j}|\ge\biggl|\sum_{j=1}^na_j\overline{b_j}\biggr|.
$$
We conclude by observing that $|\overline{b_j}|=|b_j|$.
A: Hint. Consider the quadratic polynomial 
$$\frac{x^2}{\alpha}\pm xu+\frac{\alpha u^2}{4}=
\left(\frac{x}{\sqrt{\alpha}}\pm \frac{\sqrt{\alpha}u}{2}\right)^2\geq 0.$$
A: This is a straightforward application of the so-called Young's inequality (or sometimes: Cauchy-Schwartz with at epsilon).
See https://en.wikipedia.org/wiki/Young%27s_inequality#Elementary_case
