# Prove $\left|1+z_1\right| +\left|1+z_2 \right| + \left|1+z_1z_2\right|\geq 2$

I am struggling with this problem

Let $$z_1,z_2\in\mathbb{C}$$ prove that $$\left|1+z_1\right| +\left|1+z_2 \right| + \left|1+z_1z_2\right|\geq 2$$

I know a similar question has been solved: if $|z_i|=1$ prove $|z_1+1|+|z_2+1|+|z_1z_2+1|\ge 2$ . However, I don't have the $$|z_1|=|z_2|=1$$ condition, so I am not sure if the question is wrong or the last condition is implicit. This is the work I have done:

\begin{align*} |1+z_1|+|1+z_2|+|1+z_1 z_2| &= |1+z_1| + |1+z_2|+|-(1+z_1z_2)|\\ &\geq |1+z_1| + \left|(1+z_2)-(1+z_1z_2)\right|\\ &= |1+z_1|+ |z_2-z_1z_2|\\ &\geq |(1+z_1)+(z_2-z_1z_2)| \end{align*}

I'll appreciate your help. I am not sure what I am missing here.

• Are you sure there’s no bounds on the modulus of your $z_j$’s? $j=1,2,3$ Jul 21, 2022 at 0:22
• Yes, it's all information I got. Jul 21, 2022 at 0:25
• Someone just answered , pretty clean solution too ! Jul 21, 2022 at 0:26
• This question is superior to the linked question.
– Dan
Jul 21, 2022 at 1:05

If $$|z_2|\geq 1$$, then \begin{align*} |1+z_1|+|1+z_2|+|1+z_1z_2| &\geq |1+z_1|+|(1+z_2)-(1+z_1z_2)|\\ &=|1+z_1|+|z_2||1-z_1|\\ &\geq |1+z_1|+|1-z_1|\geq 2, \end{align*} where the last step uses the triangle inequality. If $$|z_2|\leq 1$$, then \begin{align*} |1+z_1|+|1+z_2|+|1+z_1z_2| &\geq|z_2||1+z_1|+|1+z_2|+|1+z_1z_2|\\ &\geq|(1+z_2)-(z_2+z_1z_2)|+|1+z_1z_2|\\ &=|1-z_1z_2|+|1+z_1z_2|\geq 2. \end{align*}
• So are you fixing $z_1$ and considering cases for the modulus of $z_2$? Jul 21, 2022 at 0:27
• @HossienS'MyMathYourMath' Yes, although I wouldn't necessarily think about it as "fixing $z_1$." If $z_1$ and $z_2$ are given to you, either $|z_2|\geq 1$ and you can follow the first computation, or $|z_2|\leq 1$ and you can follow the second computation. Jul 21, 2022 at 0:28
• Makes sense. As the same would hold if you fixed $z_2$ and put those condition on modulus of $z_1$ Jul 21, 2022 at 0:36
• Is there a solution which doesn't split the two cases of $|z_2|$? I had a go but couldn't find one. Jul 21, 2022 at 0:58