maximum value of $|a-b||b-c|+|b-c||c-a|+|c-a||a-b|$ is If $a,b,c$ are complex numbers and $|a|=|b| = |c| = 2$
finding maximum value of $|a-b||b-c|+|b-c||c-a|+|c-a||a-b|$ is
$|a-b|^2=|a|^2+|b|^2-2(a\cdot b) = 8-8\cos 2A = 16\sin^2 A$
$|b-c|^2=|b|^2+|c|^2-2(b\cdot c) = 8-8\cos 2B = 16\sin^2 B$
$|a-c|^2=|a|^2+|c|^2-2(a\cdot c) = 8-8\cos 2C = 16\sin^2 C$
maximum value of $16(\sin A\sin B+\sin B\sin C+\sin C\sin A)$
$2A,2B,2C$ are angle between vectors $a,b$ and $b,c$ and $c,a$
i want be able to proceed after that, could some help me
 A: $a,b,c$ are points on a circle of radius $2$, so $A=|b-c|,B=|c-a|,C=|a-b|$ are lengths of the sides of the triangle formed between them. Let the angles subtended at the origin by the chords $A,B,C$ be $\alpha,\beta,\gamma$ respectively.
By Cauchy–Schwarz,
$$AB+BC+CA\ \le\ \sqrt{A^2+B^2+C^2}\sqrt{B^2+C^2+A^2}\ =\ A^2+B^2+C^2$$
By the cosine rule,
$$A^2\ =\ 2^2+2^2-2\cdot2\cdot2\cos\alpha\ =\ 8-8\cos\alpha$$
Similarly $B^2=8-8\cos\beta$ and $C^2=8-8\cos\gamma$. Hence
$$AB+BC+CA\ \le\ A^2+B^2+C^2\ =\ 24-8(\cos\alpha+\cos\beta+\cos\gamma)$$
Now we have $\cos\alpha+\cos\beta+\cos\gamma\ge-\dfrac32$ (see below); hence
$$AB+BC+CA\ \le\ 24-8(\cos\alpha+\cos\beta+\cos\gamma)\ \le\ 24+8\cdot\frac32=36$$
So the maximum value is $36$, attained when, e.g.
$$a=2,\ b=-1+i\sqrt3,\ c=-1-i\sqrt3$$

To prove that $\cos\alpha+\cos\beta+\cos\gamma\ge-\dfrac32$, we have $\alpha+\beta+\gamma=2\pi$ $\implies$ $\cos\gamma=\cos(\alpha+\beta)=2\cos^2\dfrac{\alpha+\beta}2-1$. Also $\cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}2\cos\dfrac{\alpha-\beta}2$. So we want to prove
$$2\cos^2\frac{\alpha+\beta}2+2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2-1\ \ge\ -\frac32$$
$$\iff\quad4\cos^2\frac{\alpha+\beta}2+4\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2+1\ \ge\ 0$$
$$\iff\quad\left(2\cos\frac{\alpha+\beta}2+\cos\frac{\alpha-\beta}2\right)^2+1\ \ge\ \cos^2\frac{\alpha-\beta}2$$
The last inequality must be true otherwise $\cos^2\dfrac{\alpha-\beta}2 > \left(2\cos\dfrac{\alpha+\beta}2+\cos\dfrac{\alpha-\beta}2\right)^2+1 \ge 1$.
