Let $P(z) = az^3+bz^2+cz+d$ , where $a, b, c, d $ are complex numbers with $|a| = |b| = |c| = |d| = 1.$

The following is inspired by Bound on a complex polynomial on AoPS.

For $|z| = 1$ we have $\overline z = 1/z$, so that expanding $|P(z)|^2 = P(z)\overline{P(z)}$ gives $$ |P(z)|^2 = 4 + 2 \operatorname{Re} \left( a \overline b z + a \overline c z^2 + a \overline d z^3 + b \overline c z + b \overline d z^2 + c \overline d z \right) \, . $$ Now let $\omega = e^{2 \pi i /3}$ be a third root of unity, and note that $1 + \omega + \omega^2 = 0$. It follows that $$ |P(z)|^2 + |P(\omega z)|^2 + |P(\omega^2 z)|^2 = 12 + 6 \operatorname{Re}(a \overline d z^3) $$ because all the terms with $z$ and $z^2$ cancel in the sum.

We can choose $z_0$ on the unit circle such that $a \overline d z_0^3 = 1$. Then $$ |P(z_0)|^2 + |P(\omega z_0)|^2 + |P(\omega^2 z_0)|^2 = 18 $$ so that one term on the left must be at least $6$, and that implies the desired conclusion.