hint

Assume that $\overline{BC}$ is divided into $n$ equal segments by $n-1$ points $P_1,P_2,P_3 , ... , P_k.$ By Stewart theorem, we obtain the general expression $(AP_k)^2= \left ( \frac{k}{n} \right ) b^2+ \left (1- \frac{k}{n} \right ) c^2-\frac{k}{n}\left (1- \frac{k}{n} \right )a^2 \ , \ k=1,2,3,4,5 , ... , n$ $(AP_k)^2= c^2+ \left ( \frac{a \cdot k}{n} \right )^2 -\frac{k}{n} (a^2+c^2-b^2)$ Sum $S_A$ of the square of the $n-1$ cevians $AP_k$ is then $S_A = n c^2+\frac{a^2}{n^2} \cdot \sum_{k=1}^{n}k^2 - \left (\frac{a^2+c^2-b^2}{n} \right) \cdot \sum_{k=1}^{n}k - b^2$ Note that $k=n$ gives $AP_n \equiv AC,$ thus in the latter summation we have subtracted $b^2,$ since it does not count. Then, using sum formulas yields $S_A = n c^2+\frac{a^2}{n^2} \cdot \left [ \frac{n(n+1)(2n+1)}{6} \right ] - \left (\frac{a^2+c^2-b^2}{n} \right) \cdot \left [ \frac{n(n+1)}{2} \right ] - b^2$ $S_A=\frac{(n-1)(b^2+c^2)}{2}-\frac{(n^2-1)a^2}{6n}$ By cyclic permutation of elements, we obtain the expressions: $S_B=\frac{(n-1)(a^2+c^2)}{2}-\frac{(n^2-1) b^2}{6n}$ $S_C=\frac{(n-1)(b^2+a^2)}{2}-\frac{(n^2-1) c^2}{6n}$ $S=S_A+S_B+S_C= \left [ (n-1)-\frac{(n-1)(n+1)}{6n} \right ] \cdot (a^2+b^2+c^2)$ $\Longrightarrow \ \frac{S}{a^2+b^2+c^2}=\frac{(n-1)(5n-1)}{6n}.$

We use Barycentric coordinates Let $D_k$ be the point on $BC$ such that $BD_K=\frac knBC$. We define $E_k,F_k$ similarly. Here, $1\le k\le n-1$. We have $A=(1,0,0)$ and $D_k=\left(0,\frac{n-k}n,\frac kn\right)$. From the distance formula, it follows \[ AD_k^2=-a^2\frac{n-k}n\frac kn+b^2\frac kn+c^2\frac{n-k}n. \]Adding this up from $k=1$ to $n-1$, we get $\sum AD_k^2=\frac{-a^2+b^2+c^2}2(n-1)+a^2\frac{(n-1)(2n-1)}{6n}$. Adding the equivalent expressions for the other sides, \[ S = (a^2+b^2+c^2)\frac{n-1}2+(a^2+b^2+c^2)\frac{(n-1)(2n-1)}{6n} = (a^2+b^2+c^2)\left(\frac{n-1}2+\frac{(n-1)(2n-1)}{6n}\right). \]Thus, $S/(a^2+b^2+c^2)=\frac{(n-1)(5n-1)}{6n}\in\mathbb Q$, as desired. $\blacksquare$