Let $\gamma_i$ be such that for even $i$ $\gamma_i=1$ and for odd $i$ $\gamma_i$ shall have absolute value $1$ and the product of all of the odd ones is also on the complex unit circle but not 1 or -1.
I ask:
Is there a lower bound for the real part
$$\Re \sum_{i=0}^n \gamma_i x_i \overline{x_{i+1}}$$ such that $x_{n+1}:=x_0$ which is better than the trivial one $-\left\lVert x \right\rVert^2$ ?
Let $\alpha \in \mathbb{C}$ be a point on the unit circle that is close to $1$, but $\alpha \neq 1$. Take the vector $$x = (1,-1,1,-1,\ldots)$$ of length $n$.Then $$-\Vert x \Vert^2 = -n.$$ Set $\gamma_1 = \alpha$ and $\gamma_i = 1$ for all other odd $i$. Then the product of the $\gamma_i$ is $\pm \alpha$ which is not $\pm 1$ and we get $$\sum_{i=1}^n \gamma_i x_i\overline{x_{i+1}} = -\alpha -(n-1).$$ As we chose $\alpha$ to be close to $1$, the real part of this expression should be close to $-n$.
Formalizing this idea properly with $\epsilon \to 0$ you should be able to show that there is no better bound in general.
