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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.

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

1 Answer 1

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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*}

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  • $\begingroup$ So are you fixing $z_1$ and considering cases for the modulus of $z_2$? $\endgroup$
    – anonymous
    Jul 21, 2022 at 0:27
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    $\begingroup$ @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. $\endgroup$
    – Carl Schildkraut
    Jul 21, 2022 at 0:28
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    $\begingroup$ Makes sense. As the same would hold if you fixed $z_2$ and put those condition on modulus of $z_1$ $\endgroup$
    – anonymous
    Jul 21, 2022 at 0:36
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    $\begingroup$ Is there a solution which doesn't split the two cases of $|z_2|$? I had a go but couldn't find one. $\endgroup$
    – Suzu Hirose
    Jul 21, 2022 at 0:58

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