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.
 A: We have $|z^2 -1|\cdot |z-1|^2 = \lvert p(z) \rvert$ with $p(z) = (z^2-1)(z-1)^2$. Since $p(z)$ is holomorphic, $ \lvert p(z) \rvert$ attains its maximum on the closed disk $D^2 = \{ z \in \mathbb C \mid \lvert z \rvert \le 1\}$ on the boundary $\partial D^2 = \{ z \in \mathbb C \mid \lvert z \rvert = 1\}$. This follows from the maximum modulus principle.
Each $z$ with $\lvert z \rvert = 1$ has the form
$$z = x \pm i \sqrt{1-x^2} .$$
We get
$$z^2 -1 = x^2 -(1-x^2) \pm 2i x  \sqrt{1-x^2} -1 = 2(x^2 -1 \pm ix \sqrt{1-x^2}), $$
$$\lvert z^2 -1 \rvert = 2\sqrt{(x^2-1)^2 + x^2(1-x^2)} = 2\sqrt{1-x^2},$$
$$z -1 = x -1 \pm i \sqrt{1-x^2} ,$$
$$|z-1|^2 = (x-1)^2 + 1- x^2 = 2(1-x) .$$
Thus
$$\lvert p(z) \rvert  = 4\sqrt{1-x^2}(1-x) .$$
It therefore suffices to find the maximum of $q(x) = \sqrt{1-x^2}(1-x)$ for $x \in[-1,1]$. Since $q(x)$ is differentiable in $(-1,1)$, the maximum is either attained at one of the boundary points $x = \pm 1$ of the interval $[-1,1]$ or at a point $x \in (-1,1)$ with $q'(x) = 0$. Since $q(-1)= q(1) = 0$ and $q(0) = 1$, we have to find $x$ with $q'(x) = 0$. We get
$$q'(x) = \frac 1 2 (1-x^2)^{-1/2}(-2x)(1-x) + \sqrt{1-x^2}(-1) = \frac{2x^2 -x -1}{\sqrt{1-x^2}} .$$
We have $2x^2 -x -1 = 0$ for $x = \frac 1 4 \pm \frac 3  4$, i.e. for $x = 1$ and for $x = -\frac 1 2$. But $x =1$ is not contained in $(-1,1)$ (in fact $q'(x)$ is not defined for $x = 1$), thus the maximum is attained at $x = -\frac 1 2$.
We therefore get
$$\lvert p(z) \rvert \le 4q(-\frac 1 2) = 3 \sqrt 3 .$$
A: 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
$$\displaystyle r =\sqrt{3} \implies f(r) \le 3\sqrt{3}.$$
