For fixed $z_i$s inside the unit disc, can we always choose $a_i$s such that $\left|\sum_{i=1}^n a_iz_i\right|<\sqrt3$?

Let $z_1,z_2,\ldots,z_n$ be complex number such that $|z_i|<1$ for all $i=1,2,\ldots,n$. Show that we can choose $a_i \in\{-1,1\}$, $i=1,2,\ldots,n$ such that $$\left|\sum_{i=1}^n a_iz_i\right|<\sqrt3.$$

I was not able to think it through properly, but here's a sketch:

Use induction as suggested by Berci, but with a little twist. The main idea is that for two numbers $z_i$ and $z_j$ such that $|z_i| < 1$ and $|z_j| < 1$ we can obtain $|z_i\pm z_j| < 1$ as long as some angle (out of four) between them (the difference of arguments) is smaller or equal than $\frac{\pi}{3}$. However, as long as we have 3 or more numbers, we will be able to find such a pair.

Quick illustration of the lemma: $z_i$ is somewhere on the blue line, red cross is the $z_j$ and the violet is their sum. The point is that as long as the red cross belongs to darker green, the violet line will stay in light green region.

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I don't know if I will find enough time to work out all the details, so should this idea suit you, feel free to use it.

Cheers!

Claim: If $z_1, z_2, z_3, z_4$ are four numbers inside the open unit disk, then there is a pair of them $z_k, z_j$ with $z_k \pm z_j$ also in the unit disk, for the correct choice of sign.

Proof: If $z_1 = 0$, then $z_3 = z_3 + z_1$ and we're done. Otherwise, rotate the disk so that without loss of generality we can consider $z_1$ to be a positive real number. Let $b_i = \pm 1$ so that $b_2 z_2, b_3 z_3,$ and $b_4 z_4$ have non-negative imaginary part. Let $\theta_i = \arg(b_i z_i)$, with $0 \le \theta_i \le \pi$. Re-order the $z_i$ in terms of increasing argument so that $0 = \theta_1 \le \theta_2 \le \theta_3 \le \theta_4$. $\theta_4 = (\theta_2 - \theta_1) + (\theta_3 - \theta_2) + (\theta_4 - \theta_3) \le \pi $. There must be an index $j$ with $\theta_{j+1} - \theta_j$ no more than $\displaystyle \frac{\pi}{3}$. Let $w_1 = b_{j+1} z_{j+1}, w_2 = b_{j} z_{j}$. Then $e^{-i\theta_{j}}w_2$ is a positive real, and $0 \le \arg(e^{-i\theta_{j}}w_1) \le \displaystyle \frac{\pi}{3}$. It's easy to show that $$ |b_{j} z_{j} - b_{j+1} z_{j+1}| = |w_2- w_1| = |e^{-i\theta_{j}} w_2 - e^{-i\theta_{j}} w_1 | = |1 - e^{-i\theta_{j}}w_1| \lt 1 $$

But $|z_{j} \pm z_{j+1}| = |b_{j} z_{j} - b_{j+1} z_{j+1}|$ for one choice of sign, so we get the claim.

Now that we have the claim the rest is easy. Starting with any collection $z_1, z_2, \cdots, z_n$ with $n \ge 3$, repeatedly apply the claim so that we are left with three numbers $w_1, w_2, w_3$ inside the disk. One of these, say $w_3$, is of the form $a_1 z_1 a_2 z_2 + \cdots + a_{n-2} z_{n-2}$. Rotating the disk does not change the modulus of the sum of points in the disk, so again WLOG we can take $w_3$ to be a non-negative real.

We now need to show that we can find $a_1, a_2 = \pm 1$ so that $|w_3 + a_1 w_1 + a_2 w_2|^2 \lt 3$. Let $w_k = x_k + i y_k$. Expand out $|w_3 + a_1 w_1 + a_2 w_2|^2$ to get

$$ |w_3 + a_1 w_1 + a_2 w_2|^2 = \Big\{ x_3 ^2 + x_1 ^2 + x_2 ^2 + y_1 ^2 + y_2 ^2 \Big\} + 2 f(a_1, a_2) $$

where $f(a_1, a_2) = a_1 a_2 (x_1x_2 + y_1 y_2) + a_1 x_1 x_3 + a_2 x_2 x_3$. It is easy to show $f(a_1, a_2) \le 0$ for the right choices of $\pm1$ for the $a_i$.

In this case then $$ |w_3 + a_1 w_1 + a_2 w_2|^2 = \Big\{ x_3 ^2 + x_1 ^2 + x_2 ^2 + y_1 ^2 + y_2 ^2 \Big\} + 2 f(a_1, a_2) \le x_3 ^2 + x_1 ^2 + x_2 ^2 + y_1 ^2 + y_2 ^2 \le 3 $$