prove $2Re (z_1 \bar{z_2}) \leq 2 |z_1||z_2|$ prove that  $\forall z_1,z_2 \in \mathbb{C}$
$$2Re (z_1 \bar{z_2}) \leq 2 |z_1||z_2|$$

Attempt
$Re(z_1 \bar{z_2})=x_1x_2+y_1y_2$ and $|z_1||z_2|=\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}$
so 
$$\begin{aligned}
2Re (z_1 \bar{z_2}) & \leq 2 |z_1||z_2|  && \text{div by 2}
\\ x_1x_2+y_1y_2 &\leq  \sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2}
\\ (x_1x_2+y_1y_2)^2 &\leq  (\sqrt{x_1^2+y_1^2}\sqrt{x_2^2+y_2^2})^2 && \text{square rooting}
\\ x_1^2*x_2^2+2x_1 x_2 y_1 y_2 +y_1^2 y_2^2 
         &\leq x_1^2 x_2^2 +x_1 ^2 y_2^2 +y_1^2 x_2^2 +y_1^2 y_2 ^2 
 \\2x_1 x_2 y_1 y_2 &\leq x_1^2 y_2^2 +y_1^2 x_2^2& 
\\ 0&\leq x_1^2 y_2^2 +y_1^2 x_2^2 -2x_1x_2 y_1 y_2
\\   0 &\leq (\text{stuck})^2
\end{aligned}$$ 
Not sure where to go from there. I want to say i have done this in the past but have a lapse. Also i am studying with someone else and is saying $Re(z_1 \bar{z_2}) \leq |z_1z_2|=|z_1||z_2|$ which would be simpler.  
 A: Since you proved that the inequality is equivalent to $(\text{stuck})^2\ge0$, and the latter is true because $\text{stuck}=x_1y_2 - x_2y_1 \in \mathbb R$, you are done. But there's a small issue that you need to take care of: LHS in the second inequality may be negative, so work with absolute values from the beginning 
Another way is prove that for any $z$, $|\Re z| \le |z|$ (it's pretty straightforward), so $\Re(z_1 \bar z_2) \le | z_1\bar  z_2| = |z_1||z_2|$
A: Your inequality is equivalent to
$$
[\Re(z_1\bar z_2)]^2\le|z_1|^2|z_2|^2
$$
Observe now that $|z_2|=|\overline {z_2}|$, so the inequality becomes
$$
[\Re(z_1\bar z_2)]^2\le|z_1\overline {z_2}|^2
$$
but putting $w=a+ib$ an arbitrary complex number, you always have
$$
|w|^2=a^2+b^2\ge a^2
$$
which is exactly your inequality with $w=z_1\overline{z_2}$.
A: $$
\because|z|^{2}=\operatorname{Re}^{2}(z)+\operatorname{Im}^{2}(z) \Rightarrow|z|^{2} \geqslant \operatorname{Re}^{2}(z) \Rightarrow|z| \geqslant Re(z){} 
$$
$$
\therefore 2 \operatorname{Re}\left(z_{1} \bar{z}_{2}\right) \leqslant 2\left|z_1\bar{z_{2}}\right|=2|z_1| |\bar{z_{2}}| =2|z_1| |z_{2}|
$$
