Making use of given hint, $ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq {(a + b +c)(a^2 + b^2 + c^2)\over ab + bc +ac} = {(a + b +c)(a^2 + b^2 + c^2)\over (a+b+c)} = a^2 + b^2 +c^2$ The minimum value of $a^2+b^2+c^2 \geq a+b+c \geq 3 $ for given criteria.
Let ${a^2 + b^2 + c^2 = z}$ then $f(z, n) = z + {3n \over z}$ for $n \leq 3$ and $z \geq 3$ $f(z,n) = z + {3n \over z} \geq 6 \text{ for n=3}$ For $n < 3$ the minimum value occurs at $f'(z) = 0 \implies z = \sqrt{3n } < 3 \leq z$ so further more $f'(z \geq 3) > 0$ suggest that $f(z)$ is increasing. So, the min-value must occur at $z=3$ which gives the desired result.