existence of a set of discs on the complex plane Given $z_1,\dots,z_n\in\mathbb{C}$ and $T>0$, prove that there exists a set of discs such that $\sum r_i\le 2T$ (where $r_i$ is the radius of $i$-th disc) and $\prod\limits_{i=1}^{n}|z-z_i| > \left(\frac{T}{e}\right)^n$ for all $z$ outside the union of these discs.

It is somehow related with Cartan's lemma. I haven't studied complex analysis yet, can this problem be solved with not very advanced methods?
 A: I found this demonstration in Distribution of Zeros of Entire Functions by B. Ja. Levin, Theorem 10.
Let $z_1,z_2,\ldots,z_n \in \mathbb{C}$ and $T>0$. From among all circles with radius equal to $\lambda \frac{T}{n}$, where $\lambda$ is the number of points of the set $\{z_i\}$ (counted with multiplicity) inside the circle, choose one with the largest radius $\lambda_1 \frac{T}{n}$ and call it $C_1$. 
There is no circle with radius $\lambda\frac{T}{n}$, where $\lambda \geq \lambda_1$, such that it contains more than $\lambda$ point of the set $\{z_i\}$. Indeed if $C$ is such a circle with $\lambda' > \lambda$ points of the set $\{z_i\}$ in it, then consider the concentric circle of radius $\lambda'\frac{T}{n}$. It either cointains $\lambda'$ or $\lambda''>\lambda'$ points. In the second case consider the concentric circle with radius $\lambda'' \frac{T}{n}$ and so forth. Since the set $\{z_i\}$ is finite, we eventually come to a circle of radius $\lambda \frac{T}{n}$ with $\lambda > \lambda_1$ points in it. This is impossibile since $C_1$ is the largest circle having this property. The points inside $C_1$ will be said to be of rank $\lambda_1$. 
Now remove the points of rank $\lambda_1$ and take the largest circle $C_2$ with the same property of $C_1$ with the reamaining $n-\lambda_1$ points. Now $\lambda_2 \leq \lambda_1$, indeed the inequality $\lambda_2 > \lambda_1$ would contradict the maximality of $C_1$. The points of $C_2$ will be said to be of rank $\lambda_2$. Now remove these points too and take the largest circle $C_3$ with radius $\lambda_3\frac{T}{n}$ containing $\lambda_3$ points. Clearly $\lambda_3 \leq \lambda_2$. The points in $C_3$ will be said to be of rank $\lambda_3$, and so forth. 
We thus obtain a sequence of circles
    $$ C_1,C_2,\ldots,C_p $$
    with radii $r_i = \lambda_i \frac{T}{n}$ respectively, where $\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_p$ and
    $$ \lambda_1 + \lambda_2 + \ldots + \lambda_p = n \Rightarrow \sum_{i=1}^p r_i = T .$$
    Now construct the circles
    $$ \Gamma_1,\Gamma_2,\ldots,\Gamma_p $$
    concentric with the cirlces $C_1,C_2,\ldots,C_p$ but with radii twice as large. Let $z \in \mathbb{C}$ be an arbitrary point lying outside all of the new cirlces. Describe about $z$ the cirlce $C_z$ of radius $\lambda \frac{T}{n}$, where $\lambda$ is some natural number. This circle does not intersect any of the circles $C_i$ that have radius larger or equal than $\lambda \frac{T}{n}$. Thus this circle can only contain points whose rank is less than $\lambda$. By the definition of rank $C_z$ can contain at most $\lambda-1$ points. Enumerating the points $\{z_i\}$ in the order of increasing distance from $z$ we have
$$ |z-z_k| > k \frac{T}{n} \quad \forall k \in \{1,\ldots,n\} $$
and
$$ \prod_{i=1}^n |z-z_i| > \left( \frac{T}{n} \right)^n n! > \left( \frac{T}{e} \right)^n. $$
