Turkey — Team Selection Test

Lemma: Let $X$, $Y$, $Z$, $T$ be points on a plane. Then
$$XY^2 + YZ^2 + YT^2 + TX^2 \ge XZ^2 + YT^2.$$
Proof: Let $x = \overrightarrow{XY}$, $y = \overrightarrow{YZ}$, $z = \overrightarrow{ZT}$. Note that $\overrightarrow{XZ} = x + y$, $\overrightarrow{YT} = y + z$ and $\overrightarrow{XT} = x + y + z$. Then $XY^2 + YZ^2 + YT^2 + TX^2 - XZ^2 - YT^2$ is equal to
$$\begin{aligned} & x \cdot x + y \cdot y + z \cdot z + (x + y + z) \cdot (x + y + z) - (x + y) \cdot (x + y) - (y + z) \cdot (y + z) \\ & = x \cdot x + z \cdot z + 2(x \cdot z) = (x + z) \cdot (x + z) \ge 0. \end{aligned}$$

Let the point $G'$ be the centroid of the triangle $A'B'C'$. Applying the lemma for $A$, $G'$, $A'$, $G$ gives $AG'^2 + G'A'^2 + A'G^2 + GA^2 \ge G'G^2 + AA'^2$. As $AA'^2 = A'G^2 + AA_1^2$ we have
$$AA_1^2 \le AG'^2 + G'A'^2 + GA^2 - G'G^2.$$

By similar inequalities for $B$ and $C$, we see that $AA_1^2 + BB_1^2 + CC_1^2$ is less than or equal to
$$AG'^2 + G'A'^2 + GA^2 + BG'^2 + G'B'^2 + GB^2 + CG'^2 + G'C'^2 + GC^2 - 3G'G^2. \quad (*)$$
By the Leibniz's theorem we obtain that $AG'^2 + BG'^2 + CG'^2 - 3G'G^2 = GA^2 + GB^2 + GC^2$. It is well known that $GA^2 + GB^2 + GC^2 = \frac{1}{3}(AB^2 + BC^2 + CA^2)$. As $ABC \cong A'B'C'$, we also have $G'A'^2 + G'B'^2 + G'C'^2 = \frac{1}{3}(AB^2 + BC^2 + CA^2)$. These three results conclude that $(*)$ is equal to $AB^2 + BC^2 + CA^2$ and we are done.