Let $ABC$ be an acute-angled triangle and $M$ a point inside it. We denote by $C_1$, $C_2$, $C_3$ the centers of the Nine-Point circles corresponding to the triangles $BMC$, $CMA$, and $AMB$, respectively. Show that $$(R+r)^2 \leq C_1C_2^2 + C_2C_3^2 + C_3C_1^2 \leq 2R^2 + r^2,$$
where $R$ is the radius of the circumcircle of $ABC$ and $r$ is the radius of the incircle $ABC$ and $M \neq H$ where H is the orthocenter.
I got stuck at this problem. Can it be solved using complex numbers? I would appreciate any help with finding the correct solution.
If $0,z_1,z_2\in\Bbb C$ are vertices of a triangle then its circumcenter is $$z_O=\frac{z_1z_2(\bar z_1-\bar z_2)}{\bar z_1z_2- z_1\bar z_2}.\tag1$$ For example if we let now $B=0, A=z_1=r_1e^{i\alpha_1}$ and $M=r_2e^{i\alpha_2}$ then $$C_1=Z_O(\triangle BMA)=\frac{r_1e^{-i\alpha_1}-r_2e^{-i\alpha_2}}{e^{-i2\alpha_1}-e^{-2i\alpha_2}}.\tag2$$ From $(2)$, we see that when $M$ approaches side $BA$ so that $r_1\neq r_2$, then $\alpha_2\to\alpha_1$ and $C_1\to\infty$ whearas $C_2$ and $C_3$ does not go to infinity. Hence, $C_1C_2^2+C_2C_3^2+C_3C_1^2\to\infty$ and your claim is false. Of course from law of sines it is more easy to this.
Also, the orthocenter of triangle with vertices $0,z_1,z_2$ is $$z_H=\frac{(\bar z_1z_2+ z_1\bar z_2)(z_2-z_1)}{\bar z_1z_2- z_1\bar z_2}.$$ Now our task is to show that when $M=H,$ $z_O(\triangle BMA)=z_O(\triangle BMC)$ so that $C_1C_2^2+C_2C_3^2+C_3C_1^2=0$ follows. Can you do that? It looks complicated. I gave up.