Consider the following problem.
Fix $n \in \mathbb N$. Prove that for every set of complex numbers $\{z_i\}_{1\le i \le n}$, there is a subset $J\subset \{1,\dots , n\}$ such that
$$\left|\sum_{j\in J} z_j\right|\ge \frac{1}{4\sqrt 2} \sum_{k=1}^n |z_k|.$$
I believe I proven have a stronger statement. Is this proof correct, and if so, what is the optimal constant?
My proof. Consider all the $z_i$ with positive real part. Call the real part of the sum of these numbers $X^+$. In a similar way, form $X^-$, $Y^+$, and $Y^-$. Without loss of generality, let $X^+$ have the greatest magnitude of these.
Note that because $|\operatorname{Re}(z)|+|\operatorname{Im}(z)|\ge |z|$, we have
$$ \left(\sum_{k=1}^n |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| \right) \ge \sum_{k=1}^n |z_k|.$$
But note that $\sum \limits_{k=1}^n |\operatorname{Re}(z_k)|+|\operatorname{Im}(z_k)| = X^+ + |X^-|+ Y^+ +|Y^-|$, so we have
$$ 4X^+ \ge \sum_{k=1}^n |z_k|.$$ By choosing $J$ to be the set of complex number with positive real part, this proves a stronger statement, because the factor of $1/\sqrt 2$ isn't needed.
Best Answer
The constant $\frac{1}{4 \sqrt{2}}$ can be replaced by $\frac{1}{\pi}$, which is the best possible constant independent of $n$.
Let $R(z) = \max(0, \text{Re}(z))$. Choose $\theta \in [0,2\pi]$ to maximize $F(z_1,\ldots, z_n,\theta) = \sum_{j=1}^n R(e^{i\theta} z_j)$. Note that for any complex number $z$, $$\frac{1}{2\pi} \int_0^{2 \pi} R(e^{i \theta} z) \ d\theta = \frac{|z|}{2 \pi} \int_{0}^\pi \sin \theta \ d\theta = \frac{|z|}{\pi}$$ The maximal value of $F(z_1,\ldots,z_n,\theta)$ is at least the average value for $\theta \in [0,2\pi]$, namely $\frac{1}{\pi} \sum_{j=1}^n |z_j|$. Now note that if $J = \{j: R(e^{i\theta} z_j) > 0\}$, $$\left|\sum_{j \in J} z_j\right| \ge \text{Re} \sum_{j \in J} e^{i \theta} z_j = F(z_1,\ldots,z_n,\theta).$$
To see that this estimate is best possible, consider cases where $n$ is large and the $z_n$ are the $n$'th roots of unity.