I am trying to understand the proof of a root bound for polynomials which is rougly sketched in https://www.jstor.org/stable/2313703?origin=crossref. The proof makes use of Gershgorins circle theorem, which states that the eigenvalues of a complex $n\times n$ matrix $A$ all lie in the union of circles given by $$ |z| \leq |a_{ii}| + R_i,\tag{1}$$ where $a_{ii}$ are the diagonal elements of $A$ and $R_i$ are the sums of the moduli of the off-diagonal entries in the $i$-th row (can also be applied to the columns).
The theorem I am failing to understand is the following: The zeros of a complex polynomial $f(z)=a_n z^n + \cdots + a_0$ are all contained in the circle $$ \tag{2} \left| z + \frac{a_{n-1}}{2a_n}\right| \leq \left|\frac{a_{n-1}}{2a_n}\right| + \left|\frac{a_{n-2}}{a_n}\right|^{\frac{1}{2}} + \left|\frac{a_{n-3}}{a_n}\right|^{\frac{1}{3}} + \cdots + \left|\frac{a_0}{a_n}\right|^{\frac{1}{n}}. $$ The first step in proving this is to form the companion matrix $C(f)$ (whose eigenvalues are the zeros of $f$) and a diagonal matrix $P=\text{diag}(\rho^{n-1},\rho^{n-2},\cdots,\rho,1)$ and construct the matrix $PCP^{-1}$. Since $$ B = PCP^{-1} = \begin{pmatrix} 0 & 0 & \cdots & 0 & -\frac{a_0}{a_n \rho^{n-1}} \\ \rho & 0 & \cdots & 0 & -\frac{a_1}{a_n \rho^{n-2}} \\ \vdots & \rho & \ddots & \vdots & \vdots \\ \vdots & \ddots & \ddots & 0 & -\frac{a_{n-2}}{a_n \rho} \\ 0 & 0 & \cdots & \rho & -\frac{a_{n-1}}{a_n}\\ \end{pmatrix} $$ is a similarity transformation, the matrix $B$ retains the same eigenvalues as matrix $C$, which are the zeros of $f$. We then chose $$\rho = \max\left(\left|\frac{a_{n-j}}{a_n}\right|^{\frac{1}{j}}\right), \quad j=2,\cdots,n,$$ and apply Gershgorins theorem, which apparently shows that all Gershgorin circles (1) are contained in the circle (2). I don't understand this conclusion and I can't seem to fill in the missing steps. Any help would be much appreciated!