Inside the disk with center $A(2,0)$ and a radius of $1$, a set of $n \geq 1$ points is considered, each having the respective affixes $z_1, z_2, \ldots, z_n$. Show that
$
\left| \sum_{i=1}^{n} z_i \right| \cdot \left| \sum_{i=1}^{n} \frac{1}{z_i} \right| > \frac{3}{4} n^2.
$
source: RMG 2023
I have tried writing $z_k = a_k + b_k \cdot i$, where $ |z_k – 2|<1
$ and then using the formula for the modulus of a complex number and it got me here
$\sqrt{\left(\sum_{k=1}^{n} a_k\right)^2 + \left(\sum_{k=1}^{n} b_k\right)^2} \cdot \sqrt{\left(\sum_{k=1}^{n} \frac{a_k}{a_k^2 + b_k^2}\right)^2 + \left(\sum_{k=1}^{n} \frac{b_k}{a_k^2 + b_k^2}\right)^2} > \frac{3}{4}n^2
$
How could I continue from here or what would be a better proof for it?
Best Answer
First note that $$ \tag{1} \sum_{k=1}^n |z_k| \cdot \sum_{k=1}^n \frac{1}{|z_k|} \ge n^2 $$ by the inequality between harmonic and arithmetic mean.
What we need next is a lower bound of $\left| \sum_{k=1}^n z_k \right|$ in terms of $\sum_{k=1}^n |z_k|$, and similarly for the sum of the reciprocals. This is where the restriction $|z_k - 2|<1$ comes into play: All $z_k$ in that disk lie in the sector $$ S = \{ z = r e^{i \phi} \mid r > 0, -\frac\pi 6 < \phi < \frac\pi 6\} \,. $$
For $z = x+iy \in S$ is $$ |y| \le \tan(\frac \pi 6) \cdot x = \frac{x}{\sqrt 3} $$ and therefore $$ |z| = \sqrt{x^2+y^2} \le \sqrt{x^2+ \frac{x^2}{3}} = \sqrt{\frac 4 3} \cdot x \, . $$ It follows that $$ \tag{2} \left| \sum_{k=1}^n z_k \right| \ge \operatorname{Re} \left( \sum_{k=1}^n z_k \right) = \sum_{k=1}^n \operatorname{Re}(z_k) \ge \sqrt{\frac 3 4} \sum_{k=1}^n |z_k| \, . $$ Also $z \in S \implies 1/z \in S$, so that $$ \tag{3} \left| \sum_{k=1}^n \frac{1}{z_k} \right| \ge \sqrt{\frac 3 4} \sum_{k=1}^n \frac{1}{|z_k|} $$ holds as well.
Combining the estimates $(1)$, $(2)$, and $(3)$ we get the desired inequality.