Find maximum value of$ |z_1 -z_2 |^2 + |z_2 -z_3 |^2 + |z_3 -z_1 |^2 $ if $|z_1 | = 2, |z_2 | = 3, |z_3 | = 4 $ [duplicate]
i know $ |z_1 -z_2 |^2 = |z_1 |^2 + |z_2 |^2 - 2|z_1 | |z_2 | \cos \alpha $ where $ \alpha $ is angle between $z_1$ and $z_2 $. Similarly i get $ |z_1 -z_2 |^2 + |z_2 -z_3 |^2 + |z_3 -z_1 |^2 = 2(4+9+16) - 2(6 \cos \alpha+8 \cos \beta+12\cos \gamma)$ What will be minimum value of $ 6 \cos \alpha+8 \cos \beta+12\cos \gamma $ if $ z_1 , z_2, z_3 $ lie on the 3 given circles
$$\sum_{cyc}\|z_1-z_2\|^2 = 3\left(\|z_1\|^2+\|z_2\|^2+\|z_3\|^2\right)-\|z_1+z_2+z_3\|^3 \leq \color{red}{87}$$ hence we just need to prove that there is some triangle $ABC$ with centroid $G$ such that $AG=2,BG=3,CG=4$ to prove that the previous inequality is tight.
But $AG^2=4,BG^2=9,CG^2=16$ give the squared lengths of the medians, then the squared lengths of the sides of $ABC$, so here it is our maximal triangle:
