Taiwan IMO Selection Camp, 2022

令點 $G, G'$ 分別為 $\triangle ABC$ 與 $\triangle A'B'C'$ 的重心。由重心公式知，對於平面上的任一點 $P$，均有
$$PA^2 + PB^2 + PC^2 = 3PG^2 + \frac{AB^2 + BC^2 + CA^2}{3}.$$
分別令 $P$ 為 $A', B', C'$ 三點，所得的三條式子加總，再使用上述公式在三角形 $\triangle A'B'C'$ 及 $P=G$ 上，得
$$\begin{aligned} A'A^2 + A'B^2 + A'C^2 + B'A^2 + B'B^2 + B'C^2 + C'A^2 + C'B^2 + C'C^2 \\ &= AB^2 + BC^2 + CA^2 + 3(A'G^2 + B'G^2 + C'G^2) \\ &= AB^2 + BC^2 + CA^2 + A'B'^2 + B'C'^2 + C'A'^2 + 9GG'^2 \\ &\geq 3a^2 + 3a'^2. \end{aligned}$$
因此，最前面的 9 項之中，存在一項不小於 $\frac{1}{9}(3a^2 + 3a'^2) = \frac{a^2 + a'^2}{3}$，證畢。

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Let points $G, G'$ be the centroids of $\triangle ABC$ and $\triangle A'B'C'$ respectively. By the centroid formula, for any point $P$ in the plane, we have
$$PA^2 + PB^2 + PC^2 = 3PG^2 + \frac{AB^2 + BC^2 + CA^2}{3}.$$
Letting $P$ be each of the three points $A', B', C'$ respectively, and summing the three resulting equations, then applying the above formula to triangle $\triangle A'B'C'$ with $P=G$, we get
$$\begin{aligned} A'A^2 + A'B^2 + A'C^2 + B'A^2 + B'B^2 + B'C^2 + C'A^2 + C'B^2 + C'C^2 \\ &= AB^2 + BC^2 + CA^2 + 3(A'G^2 + B'G^2 + C'G^2) \\ &= AB^2 + BC^2 + CA^2 + A'B'^2 + B'C'^2 + C'A'^2 + 9GG'^2 \\ &\geq 3a^2 + 3a'^2. \end{aligned}$$
Therefore, among the 9 terms at the beginning, there exists one term that is not less than $\frac{1}{9}(3a^2 + 3a'^2) = \frac{a^2 + a'^2}{3}$, which completes the proof.