false impression Note that $\frac{a^{2} +1}{bc} +\frac{b^{2} +1}{ca} +\frac{c^{2} +1}{ab} =\frac{a^{3} +b^{3} +c^{3} +a+b+c}{abc}$. Let $d=\gcd( a,b,c)$. We have $d^{3} \mid abc$ and $d^{3} \mid a^{3} +b^{3} +c^{3}$. So we must have $d^{3} \mid a+b+c$. It follows that $d^{3} \leq a+b+c$.

@kaede: what do you mean by false impression? As you've correctly put, $d^3\leqslant a+b+c$, that is, $d\leqslant \sqrt[3]{a+b+c}$. Since $d\in\mathbb{Z}^+$, $d\leqslant \lfloor \sqrt[3]{a+b+c}\rfloor$, as the OP correctly wrote.

@grupyorum No worries. This problem looks difficult but it's very easy indeed.