Complex number question on proving an inequality.

Question

If $|z_1|=1,|z_2|=1$, how can one prove $|1+z_1|+|1+z_2|+|1+z_1z_2|\ge2$

Answer 1

$$ \mid 1+z_1 \mid + \mid 1+ z_2 \mid + \mid 1+ z_1z_2 \mid \ \ge \mid 1+z_1 \mid + \mid 1+z_1z_2-1-z_2 \mid$$ [ 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 :)

Answer 2

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 $(t_1, t_2)$.