The following was proved here:
Let $p(z) = az^3+bz^2+cz+d$, where $a, b, c, d $ are complex numbers with $|a| = |b| = |c| = |d| = 1.$ Then $|p(z)| \ge \sqrt{6}$ for at least one complex number $z$ satisfying $|z| = 1$.
I noticed that this can be generalized as follows:
Let $p(z) = a_n z^n + a_{n-1}z^{n-1} + \ldots + a_1 z + a_0$ be a polynomial of degree $n \ge 1$ with $|a_k|=1$ for all its coefficients. Then $|P(z)| \ge \sqrt{n+3}$ for at least one complex number $z$ with $|z|=1$.
Proof (sketch): For $|z| = 1$ we have $\overline z = 1/z$, so that expanding $|p(z)|^2 = p(z)\overline{p(z)}$ gives $$ |p(z)|^2 = n+1 + 2 \operatorname{Re} \sum_{0 \le j < k \le n} \overline{a_j} a_k z^{k-j} \, . $$ Now let $\omega = e^{2 \pi i /n}$ be a $n^{\text{th}}$ root of unity. Then $$ \sum_{l=0}^{n-1} |p(\omega^l z)|^2 = (n+1)n + 2n \operatorname{Re}(\overline{a_0} a_n z^n) $$ since all other terms cancel. We can choose $z_0$ such that $\overline{a_0} a_n z_0^n = 1$. Then $ \sum_{l=0}^{n-1} |p(\omega^l z_0)|^2 = (n+3)n $ and the desired conclusion follows. $\Box$
Now (out of pure curiosity) I wonder if that bound is sharp. In other words:
Is there for any $n \ge 1$ a polynomial $p_n$ of degree $n$ with coefficients of absolute value one such that $|p_n(z)| \le \sqrt{n+3}$ for all $z$ with $|z|=1$?
Here are the partial results that I have so far:
The case $n=1$ is trivial: $|p(z)| = |a_1 z + a_0| \le 2 = \sqrt{1+3}$ for $|a_0| = |a_1| = 1$ and $|z|=1$.
For $n=2$ we can choose $p(z) = z^2+z-1$. Then $$ |p(z)|^2 = 3 + 2 \operatorname{Re}(z-z^2-z) = 3 - 2 \operatorname{Re}(z^2) \le 5 \, . $$
For $n=3$ it already gets difficult. Playing around with Geogebra I found the following polynomial which almost satisfies the desired limit: $$ p(z) = \left(\frac{\sqrt 7}{4} - \frac 3 4 i\right)z^3 + \left( -\frac 1 4 + \frac{\sqrt{15}}{4} i\right)z^2 + z +1 \, . $$ The following plot shows $|p(e^{2 \pi it})|$ and, for comparison, the constant value $\sqrt 6$ (created with wxMaxima):
This polynomial satisfies $|p(z)| < 2.5$ on the unit circle, which is not too far from $\sqrt 6 \approx 2.449$.
This is a complicated problem and the one result that I know is from Kahane (the paper shifts the degree by one taking $P(0)=0$ so one sees $\sqrt n$ there) who proved that there is a sequence of numbers $\epsilon_n \to 0, \epsilon_n=O(n^{-1/17}\sqrt {\log n})$ and a sequence of unimodular polynomials $P_n(z)=\sum a_{kn}z^k, |a_{nk}|=1, k=0,..n$ in degree $n$ for which on the unit circle (so for all $|z|=1$) we have:
$(1-\epsilon_n)\sqrt {n+1} < |P_n(z)| < (1+ \epsilon_n)\sqrt {n+1}$
So in general the above problem is definitely hard but in degree $3$ one can show that $\sqrt 6$ is not sharp and the only polynomials $P(z)=z^3+az+bz+1, |a|=|b|=1$ for which $|P(1)|=|P(\omega)|=|P(\omega^2)|=\sqrt 6$ are the normalization of the one found above and its various symmetries as below:
$P(z)=z^3+az^2+az+1, a=\frac{-1 +i \sqrt 15}{4}$ and for that it is not hard to show directly that $\omega, \omega^2$ are not maxima of the modulus, so $||P||_{\infty} > \sqrt 6$
(normalized - we always can both take a unimodular constant in front and normalize the free term to $1$ and then shift the variable by a unimodular constant to make another coefficient $1$ and the condition for that sum to be $18$ makes it useful to normalize the leading coefficient to $1$ which gives $u^3=1, u=1, \omega, \omega^2$ for the three numbers in the solution - also one can shift coefficients $a,b$ by $\omega$ and $\omega^2$ say)
Sketch of the proof: (after normalization as above): let $a=e^{i \theta}, b=e^{i \phi}$ and then the condition $|P(1)|=|P(\omega)|=|P(\omega^2)|=\sqrt 6$ translates to
$2\cos \theta+2\cos \phi + \cos (\theta-\phi)=0$ and
$2\cos (\theta+4\pi/3)+2\cos (\phi+2\pi/3) + \cos (\theta-\phi+2\pi/3)=0$
By a little manipulation, we get (from second relation using first):
$-2\sin \theta+2\sin \phi + \sin (\theta-\phi)=0$
and then $\cos (\theta+\phi)=-7/8$
$(\sin 3(\theta-\phi)/2)( \sin (\theta+\phi)/2)=0$ and the relation above precludes $\sin (\theta+\phi)/2=0$ so $|\theta-\phi| =0, 2\pi/3, 4\pi/3$ and keeping in mind the symmetries mentioned above we can assume $\theta=\phi$ which immediately gives the required $a=b$ above (or its conjugate etc)
But now using the fundamental fact that for a local maximum modulus of an analytic function on a circle one must have $wP'(w)/P(w)=q \ge 0$ we can substitute for $w=1$ (works and since $\Re zP''/P'+1 >0$ there it is indeed a local maximum) but $\omega P'(\omega)/P(\omega)=q_1 \ge 0, \omega^2 P'(\omega^2)/P(\omega^2)=q_2 \ge 0$ leads to a contradiction by subtraction