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problem:if $z_1z_2z_3=1$ and $|z_1|=|z_2|=|z_3|$ prove that $|z_1+1|+|z_2+1|+|z_3+1|\ge 2$ I want a clean solution for this problem. I prove it is equivalent to this problem if $|z_1|=|z_2|=1$ then $|z_1+1|+|z_2+1|+|z_1 z_2 +1|\ge 2$ but I cant prove it. I want a hint.

Iuli
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ali
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    I'm not sure your statement is correct, if I take $z_1=z_2=z_3=1$ I get $3\leq 2$... – Chi Cheuk Tsang Oct 19 '16 at 06:13
  • sorry it is true now – ali Oct 19 '16 at 06:44
  • Notice $|z_1|=|z_2|=|z_3|=1$, so we can let $z_1=e^{i\theta}, z_2=e^{i\phi}$, then $z_3=e^{-i(\theta+\phi)}$. Meanwhile $|e^{i\theta}+1|=2+2cos(\theta)$ so your statement is equivalent to $cos(\theta)+cos(\phi)+cos(-\theta-\phi)\geq -4$. At this point I would use calculus to check this, but perhaps there is a neater way to verify this using inequalities – Chi Cheuk Tsang Oct 19 '16 at 07:08
  • The condition is redundant: https://math.stackexchange.com/questions/4497017/prove-left1z-1-right-left1z-2-right-left1z-1z-2-right-geq-2#4497017 – Dan Jul 21 '22 at 01:03

2 Answers2

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\begin{align*} |z_1+1|+|z_2+1|+|z_1z_2+1| &\geq |z_1+1| + |(z_2+1) - (z_1z_2 + 1)|\\ &\geq |z_1+1| + |z_2 - z_1z_2| \\ &=|z_1+1| + |1 - z_1|\\ &\geq |z_1+1 + 1-z_1|\\ &= 2 \end{align*} We have used $|a\pm b| \leq |a|+|b|$ for any two complex numbers $a,b$.

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Suppose you have $z_1, z_2, z_3 \in \mathbb{C}$ with the properties $z_1z_2z_3 = 1$ and $|z_1|=|z_2| =|z_3| = 1$, then we could rename them as following \begin{align} z_1 = \zeta_1, \ \ z_2 =\bar\zeta_2, \ \ z_3= \bar\zeta_1\zeta_2 \end{align} which means the original inequality could be written as \begin{align} |\zeta_1+1|+|\bar \zeta_2 +1| + |\bar\zeta_1\zeta_2 +1| \geq2. \ \ (\ast) \end{align}

To prove $(\ast)$, observe we have \begin{align} (|\zeta_1+1|+|\bar \zeta_2 +1| + |\bar\zeta_1\zeta_2 +1|)^2 \geq&\ |\zeta_1+1|^2+|\bar \zeta_2 +1|^2 + |\bar\zeta_1\zeta_2 +1|^2 +2|\zeta_1+1||\bar\zeta_2+1|\\ =&\ 6+\zeta_1+\bar\zeta_1 + \zeta_2+\bar\zeta_2+\bar\zeta_1\zeta_2+\zeta_1\bar\zeta_2+2|\zeta_1+1||\bar\zeta_2+1|\\ =&\ 4+(\zeta_1+1)(\bar\zeta_2+1) + (\bar\zeta_1+1)(\zeta_2+1)+2|\zeta_1+1||\bar\zeta_2+1|\\ =&\ 4+2\operatorname{Re}[(\zeta_1+1)(\bar\zeta_2+1)]+2|\zeta_1+1||\bar\zeta_2+1|\\ \geq&\ 4. \end{align}

Hence we have our desired inequality \begin{align} |\zeta_1+1|+|\bar \zeta_2 +1| + |\bar\zeta_1\zeta_2 +1|\geq 2. \end{align}

Jacky Chong
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