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I have posted a similar question earlier, but in that question I have only posted a part of the problem...maybe it explains why I became stuck every single time... The problem goes as follows:

Given complex numbers $a,b,c$ such that $|a|=|b|=|c|=1$

And we need to show that

$$|a+2ab+b|^2+|b+2bc+c|^2+|c+2ac+a|^2 \le 8(3+Re(a+b+c))$$

What I have tried so far is writing all of the numbers in algebraic form. Here's what I mean:

$$|x_1+y_1i+2(x_1+y_1i)(x_2+y_2i)+x_2+y_2i|^2$$

This is the first term. Simplifying things we would get:

$$|x_1+y_1i+x_2+y_2i+2x_1x_2+2x_1y_2i+2x_2y_1i−2y_1y_2|^2$$

Gathering all the real and imaginary parts together, then squaring the magnitude left me with this expression:

$$6+2x_1x_2+2y_1y_2+4x_2+4x_1$$

And I became stuck. If the terms would be squared I could start something with them, but this way I can't. What I noticed though is that the minimum of the left hand side happens when $a=b=c=i$, and this value is 24, and the maximum when $a=b=c=1$, and this value is 48.

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  • $\begingroup$ are you conjecturing this or this is an actual problem from somewhere ? $\endgroup$
    – dezdichado
    Commented Jan 13 at 17:29
  • $\begingroup$ @dezdichado it is a problem from a math periodical. I'm currently learning complex numbers at school. I understand the basics, but perchance it requires some trick to solve it. $\endgroup$
    – fikooo
    Commented Jan 13 at 17:33
  • $\begingroup$ Please edit the post to credit the math periodical where you saw this, with a full citation. See math.stackexchange.com/help/referencing. Also, it would be helpful to explain what's the motivation, and why is this inequality relevant and useful to future users of this site? $\endgroup$
    – D.W.
    Commented Jan 16 at 5:10
  • $\begingroup$ @D.W. I could credit the periodical's website, however the problem is not available yet there. There is a delay of 4-5 months between the publishing of the physical version and the online version, and it is not written in English, which would make things harder to understand. $\endgroup$
    – fikooo
    Commented Jan 17 at 12:53
  • $\begingroup$ That's fine, you can still list the name of the periodical, the author, the volume/issue number / the date on the periodical, etc. We have been crediting (referencing) sources for long before the Internet was ever created. $\endgroup$
    – D.W.
    Commented Jan 17 at 19:36

2 Answers 2

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For unit vectors, you have $a= e^{ix}, b = e^{iy}, c = e^{iz}$ and the first squared norm on the LHS is: $$(e^{ix} + 2e^{i(x+y)}+e^{iy})((e^{-ix} + 2e^{-i(x+y)}+e^{-iy})) = 6+2e^{-iy}+2e^{-ix}+$$ $$+2e^{ix}+2e^{iy}+e^{i(x-y)}+e^{-i(x-y)}=6+4\cos x+4\cos y+2\cos(x-y).$$ So your inequality after cancellation is equivalent to: $$18+2(\cos(x-y)+\cos(y-z)+\cos(z-x))\leq 24$$ which is obviously true since $\cos\theta\leq 1$ for real arguments $\theta.$

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  • $\begingroup$ Thank you very much! One question though: $e^ik$ is the Euler form, right? And using this form...the conjugate of let's say $a$ is the same as multiplying the exponent by -1, right? $\endgroup$
    – fikooo
    Commented Jan 13 at 17:54
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    $\begingroup$ yes conjugate of $e^{ix}$ is $e^{-ix}.$ $\endgroup$
    – dezdichado
    Commented Jan 13 at 18:00
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$$ \begin{align} |a+2ab+b|^2 &= 6 + (a\overline{b} + \overline{a}b) + 2(a + \overline{a}) + 2(b + \overline{b}) \\ &= 6 + 2 \operatorname{Re}(\bar a b) + 4 \operatorname{Re}(a+b) \end{align} $$ has been shown in this answer. It follows that $$ |a+2ab+b|^2 \le 6 + 2|\bar a b|+ 4 \operatorname{Re}(a+b) = 8 + 4 \operatorname{Re}(a+b) $$ Applying this estimate to each of the three expressions on the left-hand side gives $$|a+2ab+b|^2+|b+2bc+c|^2+|c+2ac+a|^2 \le 24 + 8 \operatorname{Re}(a+b+c) \, . $$

Equality holds if (and only if) $\bar a b = \bar b c = \bar c a = 1$, that is if $a=b=c=1$.

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  • $\begingroup$ Yep...it seems I still need to learn a lot about complex numbers. Thank you very much! It helped a lot! $\endgroup$
    – fikooo
    Commented Jan 13 at 18:07

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