crazyfehmy wrote: $A_1$ be an intersection point of the circle with diameter $AA'$ and the circle with center $A'$ and passing through $G.$ If $A'G>AA'$ we get the small circle is completely in the bigger, may we also say $AA_1^2=AA'^2-A'G^2$ by the Pythagorean theorem and angle on the diameter. To prove : $\sum AA'^2 \le \sum AB^2+ A'G^2.$ equality when $\triangle ABC,A'B'C'$ are equilateral having same $G$ such that $G \in AA',BB',CC'.$

SCP wrote: To prove : $\sum AA'^2 \le \sum AB^2+ A'G^2.$ equality when $\triangle ABC,A'B'C'$ are equilateral having same $G$ such that $G \in AA',BB',CC'.$ Do you mean that the equality holds if two triangles are equilateral?

The equality case is when $G$ is the midpoint of $[BB'],[CC'],[AA']$ in general. (G=G') because $\sum XX_1^2= 3(AG^2+BG^2+CG^2)=a^2+b^2+c^2$ and hence done. When $G \not = G'$ it isn't proven by me, did you do that?

Let me complete the rest Denote the centroid of the triangle $A'B'C'$ as $G'$. First, we have a crucial lemma $Lemma:$ $AG^2+GA'^2+A'G'^2+G'A^2\geq AA'^2+GG'^2$ Proof:From Euler's quadrilateral theorem on the quadrilateral (which may not be convex, this is not important) $AG'A'G$, we have the equality $AG^2+GA'^2+A'G'^2+G'A^2$$=$$AA'^2+GG'^2+4.M_AM_G^2$ where $M_A$ and $M_G$ are the midpoints of $AA'$ and $GG'$ respectively. The rest is obvious. Equality occurs iff $M_A=M_G$ (Note that this is true for any non-degenerate quadrilateral) From the lemma, we have the inequality $AA'^2\leq AG^2+GA'^2+A'G'^2+G'A^2-GG'^2$. Using this inequality for $A$,$B$ and $C$ and the equation $AB^2+AC^2+BC^2=3.(GA^2+GB^2+GC^2)$ (which can be proven by the side bisector theorem), our desired inequality turns into $G'A^2+G'B^2+G'C^2-3.G'G^2\leq AG^2+BG^2+CG^2$ Finally, note that after writing the Cartesian coordinates of $A$,$B$,$C$ and $G'$ (indeed for any arbitrary point on plane), the left side does not depend on $G'$ (as $\frac{A+B+C}{3}=G$).If we take $G'=G$, the left side is obviously equal to the right side, which completes the proof Equality case occurs if and only if $M_G=M_A=M_B=M_C$, which means that $A'B'C'$ is the reflection of $ABC$ from an arbitrary point $P$ with ratio $-1$