I am stuck on the following proof and am really in need of some advice. I figured it would be similar to the cauchy-schwarz inequality proof but would like some help.
Prove that for any $\alpha > 0$, and for any complex numbers $a_1, > a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$, we have $$\left| \sum_{j=1}^{n}a_j\bar{b_j} \right| \leq \frac{1}{\alpha}\sum_{j=1}^n|a_j|^2 + \frac{\alpha}{4}\sum_{j=1}^n|b_j|^2$$
Any help or advice would greatly be appreciated.
Thank you
Hint:
Since $|a_j|$ and $|b_j|$ are real positive numbers then $$0\le\left(\frac{|a_j|}{\sqrt{\alpha}}-\frac{\sqrt{\alpha} |b_j|}{2}\right)^2$$ which implies $$|a_j\overline{b_j}|= |a_j||b_j| \le \frac{|a_j|^2}\alpha+\frac{\alpha |b_j|^2}4$$ Then $$\sum_{j=1}^n|a_j\overline{b_j}|\le\frac1{\alpha}\sum_{j=1}^n|a_j|^2+\frac{\alpha}{4}\sum_{j=1}^n|b_j|$$ Now use the triangular inequality.