Complex number question on proving an inequality. If $|z_1|=1,|z_2|=1$, how can one prove $|1+z_1|+|1+z_2|+|1+z_1z_2|\ge2$
 A: $$ \mid 1+z_1 \mid + \mid 1+ z_2 \mid + \mid 1+ z_1z_2 \mid \  \ge \mid 1+z_1 \mid + \mid 1+z_1z_2-1-z_2 \mid$$ [ Using Triangle inequality]
$$  \mid 1+z_1 \mid + \mid z_1z_2-z_2 \mid =  \mid 1+z_1 \mid + \mid z_1 -1 \mid  \ \ge |1+z_1+z_1-1| = 2  $$ [Again using triangle inequality]
So we are done :)
A: You have that $|1+w| \geq |\Re(1+w)| = |1 + \Re(w)|$. Since you only consider $|w| = 1$, you have that $|1 + Re(w)| = 1 + Re(w)$. Thus, it is enough to show that
$$1 + \cos(t_1) + 1 + \cos(t_2) + 1 + \cos(t_1+t_2) \geq 2,$$
where $z_1 = e^{it_1}$ and $z_2 = e^{it_2}$.
If $t_1$ is fixed, then the LFH obtains its extremal values when
$$-\sin(t_2) -\sin(t_1+t_2) = 0,$$
that is,
$$\sin(t_2) = - \sin(t_1+t_2),$$
which implies that either
$$t_1 = -t_1 - t_2+2\pi k \Leftrightarrow t_2 = -2 t_1 + 2\pi k$$
or
$$t_1 = \pi + t_1 + t_2 + 2\pi k\Leftrightarrow t_2 = \pi + 2\pi k.$$
Unless I'm mistaken, the first case corresponds to local maxima. The second case, which corresponds to local minima, gives
$$LHS = 2 + \cos(t_1) + \cos(t_1 + \pi) = 2.$$
That is, for each fixed $t_1$, the LFH is greater than $2$ as a function of $t_2$. Which of course implies that it is greater than $2$ for all $(t_1, t_2)$.
