Let $P(z) = az^3+bz^2+cz+d$ , where $a, b, c, d $ are complex numbers with $|a| = |b| = |c| = |d| = 1.$ 
Let $P(z) = az^3+bz^2+cz+d$
, where $a, b, c, d $ are complex numbers with $|a| = |b|
= |c| = |d| = 1.$ Show that $|P(z)| ≥ \sqrt{6}$ for at least one complex number z satisfying
$|z| = 1.$

Attempt
By triangle inequality $$|az^3+bz^2+cz+d|\ge ||az^3+bz^2|-|cz+d||\ge ||az+b|-|cz+d||$$
As you can see i am not utilising $|a|=|b|=|c|=|d|=1$
Then i tried using triangle inequality differently:$$|az^3+bz^2+cz+d|\ge ||az^3|-|bz^2+cz+d||$$
That's all i have tried please tell me how to start.
Thank you!
 A: 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.
A: An approach that shows a weaker result:
Calculate
$$\frac{1}{2\pi} \int_0^{2 \pi} |P(e^{i\theta})|^2 d \theta $$
Due to orthogonality of the functions $e^{i n \theta}$, the above integral equals
$$|a|^2+|b|^2 + |c|^2 + |d|^2 = 4$$
so for at least one $\theta$ we have
$$|P(e^{i\theta})|\ge \sqrt{4}$$
That is weaker than the inequality $|P(e^{i\theta})|\ge \sqrt{6}$
