# inequality involving $a,b,c \in \mathbb{C}$

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.

• are you conjecturing this or this is an actual problem from somewhere ? Commented Jan 13 at 17:29
• @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. Commented Jan 13 at 17:33
• 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?
– D.W.
Commented Jan 16 at 5:10
• @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. Commented Jan 17 at 12:53
• 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.
– D.W.
Commented Jan 17 at 19:36

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 $$\theta.$$
• 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? Commented Jan 13 at 17:54
• yes conjugate of $e^{ix}$ is $e^{-ix}.$ Commented Jan 13 at 18:00
\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 $$a=b=c=1$$.