Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$. Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$.
I tried geometric solution or using triangle inequality but doesn't work  because I lose  equality case.

There exists a very beautiful graphic solution.

We have $ |z| \le 1$ the unit disk bounded by the green circle, with centre $C_1$.
Let's assume: $|z-1| = r$, drawn as the purple circle $(C_2,r)$.
Note that we must have $0 \le r \le 2 $.
Now our inequality is $|z^2-1||z-1|^2 \le 3\sqrt{3} $ which is equivalent to: $|z+1||z-1|^3 \le 3\sqrt{3}$.
Applying the above conditions we are left to prove $|z+1| \le \dfrac{3\sqrt{3}}{r^3} $.
Geometrically, $|z+1|$ is the distance between $z$ and the point $C_3=-1$. We must now show this distance will always be less than or equal to $\dfrac{3\sqrt{3}}{r^3} .$
We can now see that the allowed $z$ under our conditions lie on the arc $AB$. The points $A$ and $B$ both have maximum distance to $C_3$.
An important thing to note is $\angle C_2AC_3 = \frac{\pi}{2} \implies {AC_2}^2 + {AC_3}^2 = {C_3C_2}^2$.
As, $AC_2 = r$ and ${C_3C_2} = 2,$  we get: $ AC_3 =\vert z+1\vert = \sqrt{4-r^2}$.
Hence we should prove that:
$\sqrt{4-r^2} \le \dfrac{3\sqrt{3}}{r^3}$ which is equivalent to $\sqrt{4-r^2} \le 3\sqrt{3}$, for  $r > 0 $.
(The case $r=0$ is trivial.)
Using the derivative of the function $\displaystyle f(r) = r^3 \sqrt{4-r^2}$, we can see $f(r)$ attains a global maximum at
$$\displaystyle r =\sqrt{3} \implies f(r) \le 3\sqrt{3}.$$
