Bound on location of critical points of a rational function 
Let $f$ be a rational function with $j$ zeros and $k$ poles, all of which reside in the closed unit disk (excepting of course the zeros or pole at $\infty$ when $j\neq k$).  What is the smallest number $R>0$ such that all the critical points of $f$ must lie in the closed disk centered at the origin with radius $R$?

When $j\neq k$, a lower bound for the answer is $\dfrac{j+k}{|j-k|}$, as can be seen by inspecting the example $f(z)=\dfrac{(z-1)^j}{(z+1)^k}$. I suspect that $R=\dfrac{j+k}{|j-k|}$ is the answer in general (again assuming $j\neq k$).
 A: $\def\Re{\mathop{\mathrm{Re}}}\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\peq{\mathrel{\phantom{=}}{}}$Define $D = \{z \in \mathbb{C} \mid |z| \leqslant 1\}$.

Lemma 1: For any $z_0 \in D$ and $z = r \e^{\i θ} \in \mathbb{C} \setminus D$,$$
\Re\left( \frac{\e^{\i θ} + z_0}{r \e^{\i θ} - z_0} \right) \geqslant 0.
$$
Proof: It suffices to prove that $\Re((\e^{\i θ} + z_0)(r \e^{-\i θ} - \overline{z_0})) \geqslant 0$. Suppose $z_0 = r_0 \e^{\i θ_0}$, then\begin{align*}
&\peq \Re((\e^{\i θ} + z_0)(r \e^{-\i θ} - \overline{z_0})) = \Re(r- \overline{z_0} \e^{\i θ} + rz_0 \e^{-\i θ} - |z_0|^2)\\
&= \Re(r- r_0 \e^{\i (θ - θ_0)} + rr_0 \e^{\i (θ_0 - θ)} - r_0^2)\\
&= r - r_0 \cos(θ - θ_0) + rr_0 \cos(θ_0 - θ) - r_0^2\\
&= r + r_0 (r - 1) \cos(θ_0 - θ) - r_0^2\\
&\geqslant r - r_0 (r - 1) - r_0^2 = (r + r_0)(1 - r_0) \geqslant 0.
\end{align*}

Lemma 2: For any $z_1, \cdots, z_n \in D$ and $z \in \mathbb{C} \setminus D$,$$
\frac{n}{|z| + 1} \leqslant \left| \sum_{k = 1}^n \frac{1}{z - z_k} \right| \leqslant \frac{n}{|z| - 1}.
$$
Proof: For the RHS,$$
\left| \sum_{k = 1}^n \frac{1}{z - z_k} \right| \leqslant \sum_{k = 1}^n \frac{1}{|z - z_k|} \leqslant \sum_{k = 1}^n \frac{1}{|z| - |z_k|} \leqslant \frac{n}{|z| - 1}.
$$
For the LHS, suppose $z = r \e^{\i θ}$, then\begin{align*}
&\peq \left| \sum_{k = 1}^n \frac{1}{z - z_k} \right| = \frac{1}{|r \e^{\i θ} + \e^{\i θ}|}·\left| \sum_{k = 1}^n \frac{r \e^{\i θ} + \e^{\i θ}}{r \e^{\i θ} - z_k} \right| = \frac{1}{r + 1} \left| n + \sum_{k = 1}^n \frac{\e^{\i θ} + z_k}{r \e^{\i θ} - z_k} \right|\\
&\geqslant \frac{1}{r + 1} \Re\left( n + \sum_{k = 1}^n \frac{\e^{\i θ} + z_k}{r \e^{\i θ} - z_k} \right) = \frac{1}{r + 1} \left( n + \sum_{k = 1}^n \Re\left( \frac{\e^{\i θ} + z_k}{r \e^{\i θ} - z_k} \right) \right)\\
&\geqslant \frac{n}{r + 1} = \frac{n}{|z| + 1},
\end{align*}
where the last inequality is from Lemma 1.

Now, suppose$$
f(z) = \frac{\prod\limits_{m = 1}^j (z - a_m)}{\prod\limits_{m = 1}^k (z - b_m)},
$$
where $a_1, \cdots, a_j \in D$ are zeros of $f$ and $b_1, \cdots, b_k \in D$ are poles, and there are no $1 \leqslant m_1 \leqslant j$ and $1 \leqslant m_2 \leqslant k$ such that $a_{m_1} = b_{m_2}$.
For any critical point $z_0$ of $f$, i.e. $f'(z_0) = 0$, if $z_0 \in D$, then $|z_0| \leqslant 1 \leqslant \dfrac{j + k}{|j - k|}$. Otherwise, $|z_0| > 1$ implies $z_0 \not\in \{a_1, \cdots, a_j\} \cup \{b_1, \cdots, b_k\}$. Note that$$
0 = f'(z_0) = f(z_0) \left( \sum_{m = 1}^j \frac{1}{z_0 - a_m} - \sum_{m = 1}^k \frac{1}{z_0 - b_m} \right),
$$
thus$$
\sum_{m = 1}^j \frac{1}{z_0 - a_m} = \sum_{m = 1}^k \frac{1}{z_0 - b_m}.
$$
From Lemma 2,$$
\frac{j}{|z_0| + 1} \leqslant \left| \sum_{m = 1}^j \frac{1}{z_0 - a_m} \right| = \left| \sum_{m = 1}^k \frac{1}{z_0 - b_m} \right| \leqslant \frac{k}{|z_0| - 1},
$$
then $\dfrac{|z_0| + 1}{|z_0| - 1} \geqslant \dfrac{j}{k}$. Analogously, $\dfrac{|z_0| + 1}{|z_0| - 1} \geqslant \dfrac{k}{j}$. Without loss of generality, assume $j > k$, then$$
\frac{|z_0| + 1}{|z_0| - 1} \geqslant \frac{j}{k} \Longrightarrow k|z_0| + k \geqslant j|z_0| - j\\
\Longrightarrow |j - k|·|z_0| = (j - k)|z_0| \leqslant j + k \Longrightarrow |z_0| \leqslant \frac{j + k}{|j - k|}.
$$
