Cyclic sum inequality involving five numbers with modulus one and zero sum 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 ?
 A: 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 $\theta$:
$\qquad$
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:
$\qquad$
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:
$\qquad\qquad$ 
$\qquad\qquad\quad$ 
So this is my non-definitive answer.
Perhaps it already contains a germ of a bona fide proof of the inequality.
