$|z_1|=r_1,|z_2|=r_2,|z_3|=r_3$, when $|z_1-z_2|+|z_1-z_3|+|z_3-z_2|$has max and what is it?
the question is from this question but no square. I had a geometric solution which is very long.(will post later)
I try to find a simple and short solution which similar to $|z_1+z_2+z_3|=0$
this question is the points on 3 co-center circle make a a triangle and find their max perimeter.
first to show that if a $\triangle ABC, BC$ is reference line ,then the co-center should at same side of $A$.
if $A$ and the co-center at two side of $BC$, then we can flip $BC$ to $B'C'$ and it is trivial $AB'>AB,AC'>AC,B'C'=BC$ (see picture above)
2nd : we show that there is only one special point of $A$
the $A$ point become $Q$ in the following picture.
and there is only one such point $Q_m$ BTW, if $L$ is on $JK$, the result is same.
3rd :we show that there is a circle which tangent the circle that $A$ lies on.
now let $Z_1(c,0)=C,Z_2(x,y)=A,Z_3(-c,0)=B, $ the co-center of three circles is$O_2$
WLOG, we can say $O_2(p,q)$ has $p>0,q>0$, we say the special point $Q_m(x_1,y_1)$ has $x_1>0,y_1>0$ ( but we can't figure out $(x_1,y_1)$ with given $p,q,r$)
connect $Z_2(x_1,y_1) O_2(p,q)$, and extend $Z_2O_2$ cross $Z_1Z_3$ at $E(m,0)$, since $Z_2O_2$ is the bisector of $\angle Z_1Z_2Z_3 \implies 0<m<c$
if we make a circle at $E$ with radius $EZ_2=r$, then two circles are tangent at $Z_2$ and $O_2$ is inside of circle $E$.
4th: for circle $E:(x-m)^2+y^2=r^2$, it has a ellipse $\dfrac{x^2}{b^2+c^2}+\dfrac{y^2}{b^2}=1$ which tangent circle $E$
eliminate $y^2$ and let $\Delta=0, \to b^2=\dfrac{c^2r^2}{c^2-m^2}$ so there is only one tangent point $Z_2'$ in 1st quart on the circle $E$. and the ellipse tell us that $Z_2'E$ is the bisector of $\angle Z_1Z_2'Z_3 \implies Z_2'=z_2$.
otherwise, circle $E$ has two points like $Q_m$ which conflict with step 2.
so we conclude that for circle $O_2$, when $Z_2Z_1+Z_2Z_3$ get max, $Z_2O_2$ will be the the bisector of $\angle Z_1Z_2Z_3$.
5th: with above results, we can say when and only when the co-center is the inner center of $\triangle Z_1Z_2Z_3,|z_1-z_2|+|z_1-z_3|+|z_3-z_2|$ has max,
Otherwise, we can always find a new triangle with bigger perimeter in step 4.
the value is $2(\sqrt{r_1^2-t^2}+\sqrt{r_2^2-t^2}+\sqrt{r_3^2-t^2}),t^2=\dfrac{\sqrt{r_1^2-t^2}*\sqrt{r_2^2-t^2}*\sqrt{r_3^2-t^2}}{\sqrt{r_1^2-t^2}+\sqrt{r_2^2-t^2}+\sqrt{r_3^2-t^2}}$.
I don't know if there is method with complex expression to show the proof and result.
