Putnam 2019, Problem A-3

Given real numbers \(b_0, b_1, \ldots, b_{2019}\) with \(b_{2019} \neq 0\), let \(z_1, z_2, \ldots, z_{2019}\) be the roots in the complex plane of the polynomial
\[P(z) = \sum_{k=0}^{2019} b_k z^k.\]
Let \(\mu = (|z_1| + \cdots + |z_{2019}|)/2019\) be the average of the distances from \(z_1, z_2, \ldots, z_{2019}\) to the origin. Determine the largest constant \(M\) such that \(\mu \geq M\) for all choices of \(b_0, b_1, \ldots, b_{2019}\) that satisfy
\[1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019.\]

Solution:
The answer is $M = 2019^{-1/2019}$. For any choices of $b_0, \ldots, b_{2019}$ as specified, AM-GM gives
$$\mu \geq |z_1 \cdots z_{2019}|^{1/2019} = |b_0/b_{2019}|^{1/2019} \geq 2019^{-1/2019}.$$

To see that this is best possible, consider $b_0, \ldots, b_{2019}$ given by $b_k = 2019^{k/2019}$ for all $k$. Then
$$P(z/2019^{1/2019}) = \sum_{k=0}^{2019} z^k = \frac{z^{2020} - 1}{z - 1}$$
has all of its roots on the unit circle. It follows that all of the roots of $P(z)$ have modulus $2019^{-1/2019}$, and so $\mu = 2019^{-1/2019}$ in this case.