<s>[INST] Given complex $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$ : when and what is $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$ 
If $z_{1},z_{2},z_{3}$ are three complex number Such that $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$
Then $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$

$\bf{My\; Try::}$ Let $z_{1}=2\left(\cos \alpha+i\sin \alpha\right)$ and $z_{2}=3\left(\cos \beta+i\sin \beta\right)$ and $z_{3}=4\left(\cos \gamma+i\sin \gamma\right)$
So $$f(\alpha,\beta,\gamma) = 58-\left[12\cos(\alpha-\beta)+24\cos(\beta-\gamma)+16\cos(\gamma-\alpha)\right]$$
Now How can I calculate $\max$ of $f(\alpha,\beta,\gamma)$
Help me
Thanks
 [/INST] $$\dfrac S4=3\cos(A-B)+6\cos(B-C)+4\cos(C-A)$$
$$=\cos A(3\cos B+4\cos C)+\sin A(3\sin B+4\sin C)+6\cos(B-C)$$
$$=\sqrt{25+24\cos(B-C)}\cos\left(A-\arccos\dfrac{3\cos B+4\cos C}{3\sin B+4\sin C}\right)+6\cos(B-C)$$
$$\le\sqrt{25+24\cos(B-C)}+6\cos(B-C)$$
If $\sqrt{25+24\cos(B-C)}=y, 1\le y\le7$ and $\cos(B-C)=\dfrac{y^2-25}{24}$
$$S\le4y+24\cdot\dfrac{y^2-25}{24}=4y+y^2-25=(y+2)^2-29\le-29$$
$$\implies f(\alpha+\beta+\gamma)\ge58+29$$
The equality occurs if $y=2$ and $A-\arccos\dfrac{3\cos B+4\cos C}{3\sin B+4\sin C}=2n\pi$ where $n$ is any integer
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