While playing around with some complex numbers in Geogebra, I came up with this conjecture:
Show that given $3$ distinct complex numbers $a,b,c$, you can choose two of them such that the absolute value of their sum is greater than or equal to the absolute value of the third.
I was not able to find a counterexample, so I tried proving this by assuming the opposite:that $|a+b| < |c|, |b+c| < |a|, |c+a| < |b|$. Summing, I expected to be able to prove that $$ |a+b| + |b+c| + |c+a| \geq |a| + |b| + |c|,$$ in order to obtain a contradiction, but it turns out that this inequality is not always true.
Any other ideas?
Let $p = a + b + c$, the conjecture is equivalent to $$|p-a| \ge |a|\quad\lor\quad |p-b| \ge |b|\quad\lor\quad |p-c| \ge |c| \tag{*1}$$ If $p = 0$, this is trivially true.
If $p \ne 0$, let $\ell$ be the perpendicular bisector of the line segment joining $0$ and $p$. This bisector $\ell$ separate $\mathbb{C}$ into two half-planes, one contains the $0$ and the other contains $p$. The condition $(*1)$ is equivalent to at least one of $a, b, c$ is lying on the closed half plane containing $0$.
If this isn't true, all three points $a, b, c$ lie on the open half plane containing $p$. All the projections of $a, b, c$ along direction $p$ will be greater than $\frac12 |p|$. This force the projection of $p$ onto itself is at least $\frac32|p|$. This is impossible. $a + b + c$ cannot sum to $p$ and this contradicts with the definition of $p$.
As a result, $(*1)$ is also true when $p \ne 0$.
Above proof is geometric in nature. We can also prove the conjecture in a more algebraic manner.
When $(*1)$ fails, we have
$$\begin{align} & |p-a|^2 + |p-b|^2 + |p-c|^2 < |a|^2 + |b|^2 + |c|^2\\ \iff & 3|p|^2 - p(\bar{a}+\bar{b}+\bar{c}) - \bar{p} (a+b+c) < 0\\ \iff & |p|^2 = 3|p|^2 - 2p\bar{p} < 0 \end{align}$$ The last assertion $|p|^2 < 0$ is absurd, so $(*1)$ is true.