If $P$ is a polynomial with $|P(1)| = \max\limits_{|z| =1} |P(z)|$, then its root on the unit circle is separated away from 0

EDIT. Mulpiplying on a constant, we may assume wlog that $P(1)=1$. Consider the function $$f(t)=\frac{1}{2}\left( P(e^{it})+\overline{ P(e^{it})}\right)=\frac{1}{2}\left( P(e^{it})+P^*(e^{-it})\right),$$ where $P^*$ is the polynomial with complex conjugate coefficients.

This $f$ is an entire function of exponential type $n$, real on the real line, and and $|f(t)|\leq1$ on the real line, $f(0)=1$.

For such functions we have a generalization of Bernstein's inequality due to Szego and Boas: $$f'^2+n^2f^2\leq n^2,$$ which we rewrite as $$|f'|\leq n\sqrt{1-|f|^2}.$$ Now define $g(t)=\arcsin f(t)$.

Suppose that $f(a)=0$ and $f(b)=1$, and $0<f(t)<1$ for $t\in(a,b)$. Then $g$ is well defined (with the principal branch of arcsin) and analytic on $(a,b)$, and $g(a)=0,\; g(b)=\pi$. Thus in view of our inequality above $$\pi\leq\int_a^b|g'(t)|dt\leq\int_a^b \frac{|f'(t)|}{\sqrt{1-f^2(t)}}dt\leq n|b-a|.$$ So $|b-a|\geq\pi/n,$ as requested. The statement on the case of equality follows from the case of equality in Szego's inequality.

Refs.

Szegö, Schriften der Königsberger gelehrten Gesellschaft, Naturwissenschaftliche Klasse, Fünftes Jahr 4 (1928), p. 69.

R. Duffin and A. Shaeffer, Some inequalities for functions of exponential type, Bull AMS, 43 (1937), 554. (Freely available online.)

EDIT. Further research shows that the result is not new: it was proved in

MR0018268
Turán, P. On rational polynomials. Acta Univ. Szeged. Sect. Sci. Math. 11, (1946). 106–113.