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Algebra Difficulty 8.2 Shortlist Find the answer

Given real numbers b0,b1,…,b2019b_0, b_1, \dots, b_{2019} with b2019≠0b_{2019} \neq 0, let z1,z2,…,z2019z_1,z_2,\dots,z_{2019} be the roots in the complex plane of the polynomial P(z)=∑k=02019bkzk. P(z) = \sum_{k=0}^{2019} b_k z^k. Let μ=(∣z1∣+⋯+∣z2019∣)/2019\mu = (|z_1| + \cdots + |z_{2019}|)/2019 be the average of the distances from z1,z2,…,z2019z_1,z_2,\dots,z_{2019} to the origin. Determine the largest constant MM such that μ≥M\mu \geq M for all choices of b0,b1,…,b2019b_0,b_1,\dots, b_{2019} that satisfy 1≤b0<b1<b2<⋯<b2019≤2019. 1 \leq b_0 < b_1 < b_2 < \cdots < b_{2019} \leq 2019.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

The answer is M=2019−1/2019M = 2019^{-1/2019}. For any choices of b0,…,b2019b_0,\ldots,b_{2019} as specified, AM-GM gives μ≥∣z1⋯z2019∣1/2019=∣b0/b2019∣1/2019≥2019−1/2019. \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 b0,…,b2019b_0,\ldots,b_{2019} given by bk=2019k/2019b_k = 2019^{k/2019} for all kk. Then P(z/20191/2019)=∑k=02019zk=z2020−1z−1 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)P(z) have modulus 2019−1/20192019^{-1/2019}, and so μ=2019−1/2019\mu = 2019^{-1/2019} in this case.

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