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When working on this MSE question, I was led to conjecture the following :

If $z_1,z_2,z_3,z_4,z_5$ are five complex numbers with modulus $1$, such that $z_1+z_2+z_3+z_4+z_5=0$, then

$$ \bigg|\sum_{cyc} z_kz_{k+1}\bigg|^2 \geq \bigg(\frac{7-3\sqrt{5}}{4}\bigg)\Bigg(\sum_{cyc} \big|z_k^2-z_{k-1}z_{k+1}\big|^2\Bigg) $$

Or in other words,

$$ \begin{array}{l} \bigg|z_1z_2+z_2z_3+z_3z_4+z_4z_5+z_5z_1\bigg|^2 \geq \big(\frac{7-3\sqrt{5}}{4}\big) \times \\ \Bigg( \big|z_1^2-z_5z_2\big|^2+\big|z_2^2-z_1z_3\big|^2+ \big|z_3^2-z_2z_4\big|^2+\big|z_4^2-z_3z_5\big|^2+ \big|z_5^2-z_4z_1\big|^2\Bigg) \end{array} $$

Does anyone have an idea on how to prove (or disprove) this ?

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  • $\begingroup$ How precisely is your conjecture related to the MSE question? If I knew that, I would perhaps see my way ahead. $\endgroup$ – chizhek Aug 28 '14 at 10:22
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What follows is not a definitive answer. I have only managed to convince myself that one should try to prove the inequality, not to disprove it, because it is (almost) certainly true. So far I have only a visual 'empirical evidence' for this belief of mine, which, however, is very compelling. $\newcommand{\conjug}{\overline} $$\newcommand{\RR}{\mathbb{R}} $$\newcommand{\defeq}{:=} $$\newcommand{\abs}[1]{{\left|#1\right|}} $$\newcommand{\union}{\cup} $

Let $S$ be the unit circle in the plane of complex numbers. Write the difference LHS $-$ RHS of the conjectured inequality as $f(z_1,z_2,z_3,z_4,z_5)$, with $z_1,z_2,z_3,z_4,z_5\in S$. The function $f$ is of course invariant under cyclic permutations of its arguments; it is also invariant under rotations of the complex plane $z_i\mapsto uz_i$ (where $u\in S$), and under the complex conjugation $z_i\mapsto\conjug{z}_i$, $1\leq i\leq 5$. With a suitable rotation we can make $x\defeq z_1+z_2$ a nonnegative real number, and if need be we apply complex conjugation to get $z_1=\exp(i\varphi)$ and $z_2=\exp(-i\varphi)$ for some $0\leq\varphi\leq\pi/2$, so that $x=2\cos\varphi$. Write $a\defeq z_1+z_2+z_3$. Now if we know $z_1$, $z_2$, and $z_3=\exp(i\theta)$, then we almost know also $z_4$ and $z_5$, because $z_4+z_5=-a$. If $0<\abs{a}<2$, then there are two possibilities: if $z_4=u$, $z_5=v$ is one, then $z_4=v$, $z_5=u$ is the other one. When $\abs{a}=2$, then $z_4=z_5=-a/2$. Finally, if $a=0$, then $z_4=z_5=u$, where $u\in S$ is arbitrary, so in this case we have a singularity with an infinity of possibilities and with infinitely many possible values of $f$; it occurs when $z_1=\exp(i\pi/3)$, $z_2=\exp(-i\pi/3)$, and $z_3=-1$. To eliminate the need to separately observe two 'branches' of $f$, we allow $\varphi$ in the range $-\pi/2\leq\varphi\leq\pi/2$, while we require, for $a\neq0$, that $a+z_4$ lies on the left side of the line (perhaps on the line itself) through the origin $0$ and the point $a$, with the line directed from $0$ to $a$; this time we use the complex conjugation to get the point $a+z_4$ to the left side of the line.

We have parameterized the representative $5$-tuples $(z_1,z_2,z_3,z_4,z_5)$ by $\varphi\in[-\pi/2,\pi/2]$ and by $\theta$. If $x=2\cos\varphi\leq 1$, then $\theta$ can range through the whole interval $[0,2\pi]$. If $x>1$, then $\theta$ is restricted to the interval $[\alpha,2\pi\!-\!\alpha]$, $\alpha\defeq\arccos\bigl((3-x^2)/2x\bigr)$; this restriction comes from the condition $\abs{z_1+z_2+z_3}\leq 2$. Here is the diagram of $f$ as a function of the parameters $\varphi$ and $\theta$:

$\qquad$diagram of f

Note the discontinuities at the points $(\varphi,\theta)=(-\pi/3,\pi)$ and $(\varphi,\theta)=(\pi/3,\pi)$. At each of these two singularities the function $f$ can have any value between $-11-7\sqrt{3}+6\sqrt{5}+3\sqrt{15}\doteq1.911$ and $5(19-3\sqrt{5})/8\doteq6.432$ -- every one of these values is actually reachable as the limit of $f(\varphi,\theta)$ as $(\varphi,\theta)$ approaches the singularity from appropriate direction.

This is the view of the diagram from a viewpoint at the level $0$ and some distance away from the diagram to its left:

$\qquad$diagram of f, side view

This view of the diagram sits on the line that marks the level $0$. The function $f$ attains the zero value at two points, the point $(-\pi/5,3\pi/5)$ and the point $(-2\pi/5,6\pi/5)$. Let's look at the vectors $z_1$, $z_2$, $z_3$, $z_4$, $z_5$ which, added in this order, yield the result $0$. The vectors corresponding to the first point are shown on the left, and the vectors corresponding to the second point on the right:

$\qquad\qquad$ zi's for the first point $\qquad\qquad\quad$ zi's for the second point

So this is my non-definitive answer. Perhaps it already contains a germ of a bona fide proof of the inequality.

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